Chapter 15, Section 15.4, Exercise 1 Page 1,190

Step 1

A person has a chance of \(\frac{2}{3}\) shooting a ball.

P(shoot)= \(\frac{2}{3}\)

Chapter 15, Section 15.4, Exercise 2 Page 1,190

Step 1

The task is to find the probability that the basketball player made 1 out of 2 free throws if he is consistently making 2 out of 3 free throws. Also, in the exercise, we are given the solutions of Michael, Julie, and Erica and our task is to determine which is incorrect and which is correct.

Chapter 15, Section 15.4, Exercise 3 Page 1,192

Step 1

To get the probability of independent events, we multiply the probability of events A and B:

P(A and B)=P(A)⋅P(B)

A person shooting chance is \(\frac{2}{3}\) and not shooting the ball is \(\frac{1}{3}\)

P(shoot)=\(\frac{2}{3}\)

P(miss)=\(\frac{1}{3}\)

Chapter 15, Section 15.4, Exercise 4 Page 1,193

Step 1

The task is to find the probability that the basketball player will make 0,1,2, or 3 out of 3 free throws if he is consistently making 2 out of 3 free throws.

Chapter 15, Section 15.4, Exercise 5 Page 1,195

Step 1

The task is to calculate the probability that the player will make 0,1 and 2 out of 2 and 0,1 2, and out of free through using the combinations that determine in how many ways the required event can happen.

Chapter 15, Section 15.4, Exercise 6 Page 1,198

Step 1

The probability of independent events is equal to:

P(A and B)=P(A).P(B)

If there are multiply combinations of the events, we have:

⇒ \({ }_n C_r[P(A) \cdot P(B) \ldots]\)

A person has a shooting chance of \(\frac{2}{3}\) and missing chance of \(\frac{1}{3}\). Let us use the formulas above to solve the followin scenarios.

Chapter 15, Section 15.4, Exercise 7 Page 1,200

Step 1

The task is to complete the given table with a corresponding number of combinations that represent the number of ways a certain event can happen.

Chapter 15, Section 15.4, Exercise 8 Page 1,200

Step 1

Chapter 15, Section 15.4, Exercise 9 Page 1,201

Step 1

A person has a shooting chance of \(\frac{2}{3}\)
​, and the probability of missing is \(\frac{1}{3}\)

P(shoot)= \(\frac{2}{3}\)

P(miss)= \(\frac{2}{3}\)

Chapter 15, Section 15.4, Exercise 10 Page 1,202

Step 1

A person has a shooting of 2 out of 5 chance, and the probability of missing is \(\frac{3}{5}\)

P(shoot)= \(\frac{2}{5}\)

P(miss)= \(\frac{3}{5}\)

To get the probability of shooting r times in n shoots, we came up with the formula:

⇒ \({ }_n C_r \cdot[P(\text { shoot })]^r \cdot[P(\text { miss })]^{n-r}\)

Chapter 15, Section 15.4, Exercise 11 Page 1,202

Step 1

A person has a shooting of 9 out of 10 chance, and the probability of missing \(\frac{1}{10}\)

P(shoot)= \(\frac{9}{10}\)

P(miss)= \(\frac{1}{10}\)

To get the probability of shooting r times in n shoots, we came up with the formula:

⇒ \({ }_n C_r \cdot[P(\text { shoot })]^r \cdot[P(\text { miss })]^{n-r}\)

Chapter 15, Section 15.4, Exercise 1 Page 1,203

Step 1

Let

R be the event of getting a red face, and B be the event of getting a blue face. A fair cube has
4 red faces and 2 blue faces. We have the following probabilities:

P(R)= \(\frac{4}{6}=\frac{2}{3}\)

P(B)= \(\frac{2}{6}=\frac{1}{3}\)

When we roll the cube multiple times, we are doing multiple trials. To get the probabilities of multiple trials, we have:

⇒ \({ }_n C_r \cdot[P(A)]^r \cdot[P(B)]^{n-r}\)

Chapter 15, Section 15.4, Exercise 2 Page 1,204

Step 1

Let R be the event of getting a red face, and B be the event of getting a blue face. A fair cube has
4 red faces and 2 blue faces. We have the following probabilities:

P(R)= \(\frac{4}{6}=\frac{2}{3}\)

P(B)= \(\frac{2}{6}=\frac{1}{3}\)

When we roll the cube multiple times, we are doing multiple trials. To get the probabilities of multiple trials, we have:

⇒ \({ }_n C_r \cdot[P(A)]^r \cdot[P(B)]^{n-r}\)

Chapter 15, Section 15.4, Exercise 3 Page 1,205

Step 1

A tetrahedron has 3 blue faces and 1 red face. We have the following probabilities:

Chapter 15, Section 15.4, Exercise 4 Page 1,205

Step 1

Let R be the event of getting a red face, and B the event of getting a blue face.

A tetrahedron has 1 red face and 3 blue faces. We have the following probabilities:

P(R)= \(\frac{1}{4}\)

P(B)= \(\frac{3}{4}\)

When we roll the tetrahedron multiple times, we are doing multiple trials. To get the probabilities of multiple trials, we have:

⇒ \({ }_n C_r \cdot[P(A)]^r \cdot[P(B)]^{n-r}\)

The post Carnegie Learning Section 15.4: Understanding Trials in Geometry appeared first on Answer Key for Math.

Chapter 15, Section 15.5, Exercise 1 Page 1,208

Step 1

The task is to calculate which of the given dartboards will give the player the highest probability of hitting the shaded area.

Chapter 15, Section 15.5, Exercise 2 Page 1,208

Step 1

The task is to determine the possible and the desired outcomes when the player throws a dart at a dart board.

Chapter 15, Section 15.5, Exercise 3 Page 1,209

Step 1

The task is to determine whether the given reasoning is correct or incorrect.

Chapter 15, Section 15.5, Exercise 4 Page 1,209

Step 1

The task is to determine the probability of hitting the shaded region of each of three given dartboards.

Chapter 15, Section 15.5, Exercise 5 Page 1,211

Step 1

The task is to determine which of the given three dartboards will make the competition most difficult.

Chapter 15, Section 15.5, Exercise 6 Page 1,212

Step 1

We are to find the probability of hitting the shaded part of this board:

Chapter 15, Section 15.5, Exercise 1 Page 1,213

Step 1

In a spinning wheel, there are 8 equal sections. Three sections will give you a chance of winning $100. Two sections will give you a chance of winning $200 and one section each for $300,$400, and $500

Chapter 15, Section 15.5, Exercise 2 Page 1,213

Step 1

In a spinning wheel, there are 8 equal sections. Three sections will give you a chance of winning $100 Two sections will give you a chance of winning $200 and one section each for $300, $400, and $500

Chapter 15, Section 15.5, Exercise 3 Page 1,213

Step 1

The task is to determine how often is expected to win a certain amount when the wheel is spun.

Chapter 15, Section 15.5, Exercise 4 Page 1,214

Step 1

In a spinning wheel, there are 8 equal sections. Three sections will give you a chance of winning $100. Two sections will give you a chance of winning $200, and one section each for $300, $400, and $500.

Chapter 15, Section 15.5, Exercise 1 Page 1,215

Step 1

We are either to keep $300 or to spin a wheel.

In a spinning wheel, there are 8 equal sections. Two sections will give you a chance of winning $100. The other 6 sections will give you $200, $300, $400, $500,$600 and $700

Chapter 15, Section 15.5, Exercise 2 Page 1,215

Step 1

We are either to keep $300 or to spin a wheel.

Chapter 15, Section 15.5, Exercise 1 Page 1,216

Step 1

We have a dartboard that has dimensions on its parts. Let us try to get the probability of each part.

Chapter 15, Section 15.5, Exercise 2 Page 1,219

Step 1

The task is to calculate the expected value for a 1 dart throw for the given dart board.

The post Carnegie Learning Section 15.5: Advanced Geometry Concepts – To Spin or Not to Spin appeared first on Answer Key for Math.

Chapter 1 Whole Numbers Exercises 1.1 Solutions Page 7  Exercise 1.1  Problem 1

Given:   Figure

It is asked to write the number represented by the given image in figures.

Multiply the number of circles given in each column with its place value and then add all the obtained values form each places will give the required number.

Analyzing the given figure

There are no circles in the first column representing ten thousands.

⇒   10000 × 0 = 0

Similarly, there are no circles in the second column representing thousands.

⇒  1000 × 0 = 0

There are 4 circles in the third column representing hundreds.

⇒   100 × 4 = 400

There are 5 circles in the fourth column representing tens.

⇒  10 × 5 = 50

There are 3 circles in the fifth column representing ones.

⇒  1 × 3 = 3

Add the total values of each places to obtain the given number in figures. so, we will get

0 + 0 + 400 + 50 + 3 = 453

Therefore, the required number represented by the given image in the figures is 453.

Page 7  Exercise 1.1  Problem 2

Given: Figure

It is asked to write the number represented by the given image in figures.

Multiply the number of circles given in each column with its place value and then add all the obtained values form each places will give the required number.

Analyzing the given figure

There are 2 circles in the first column representing ten thousands.

⇒ 10000 × 2 = 20000

There are 3 circles in the second column representing thousands.

⇒  1000 × 3 = 3000

There are 4 circles in the third column representing hundreds.

⇒  100 × 4 = 400

There are zero circles in the fourth column representing tens.

⇒  10 × 0 = 0

There are 5 circles in the fifth column representing ones.

⇒  1 × 5 = 5

Add the total values of each places to obtain the given number in figures. so, we will get

20000 + 3000 + 400 + 0 + 5 = 23405

Therefore, the required number represented by the given image in the figures is 23405.

Page 7   Exercise 1.1  Problem 3

Given:

Mr. Royce sold his car for this amount of money represented by figure

It is asked to write the total amount of money Mr. Royce in figures.

By multiplying the number of each type notes with its value and adding all the amount obtained in each type of note will get the total amount received by Mr. Royce.

By analyzing the figure, we get that Mr. Royce received 3 notes of $10000.

So, the total amount received in $10000 note will be

⇒  $10000 × 3 = $30000

Now, Mr. Royce also received 2 of $1000 notes and 4 of $100 notes.

So, the total amount received in $1000 note will be

⇒  $1000  ×2 = $2000

And total amount received in $100 will be

⇒  $100 × 4 = $400

So, the total amount received by Mr. Royce for the car is given by  $30000 + $2000 + $400 = $32400

Therefore, the amount of money received by Mr. Royce for the car in figures is $32400.

Primary Mathematics 4A Chapter 1 Step-By-Step Solutions Page 7  Exercise 1.1 Problem 4

Given:  Mr. Royce sold his car for this amount of money represented by figure

It is asked to write the total amount of money Mr. Royce in words.

By multiplying the number of each type notes with its value and adding all the amount obtained in each type of note will get the total amount received by Mr. Royce.

By referring to Exercise 1, Problem 2

We get that amount of money received by Mr. Royce for the car in figures is $32400.

To represent this amount in words, write this by specifying number of each values such as ten thousands, thousands, hundreds, etc.

So, $32400 is represented in words as Thirty two thousand four hundred Dollars.

Therefore, amount of money received by Mr. Royce for the car represented in words is ‘Thirty two thousand four hundred Dollars’.

Page 8   Exercise 1.1 Problem 5

Given:  That there are 12 Thousands and 6 Hundreds.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number

Given values are 12 thousands and 6 hundreds.

So, total place value of thousands gives

12 × 1000 = 12000

Now, total place value of hundreds gives

6×100=900

Now, by adding the total values obtained from different place values together will give the required number.

⇒ 12000 + 900 = 12900

Therefore, the given number can be represented in figures as 12900.

Whole Numbers Exercises Primary Mathematics Workbook 4A Answers Page 8   Exercise 1.1  Problem 6

Given: There are 45 Thousands, 9 tens and 3 ones.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are 45 thousands, 9 tens and 3 ones.

So, total place value of thousands gives

45 × 1000 = 45000

Now, total place value of tens gives

9 × 10 = 90

And total place value of ones gives

3 × 1 = 3

Now, by adding the total values obtained from different place values together will give the required number.

⇒  45000 + 90 + 3 = 45093

Therefore, the given number can be represented in figures as 45093

Page 8   Exercise 1.1  Problem 7

Given: There are 73 Thousand, 8 Hundreds, and 2 ones.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are 73 thousand, 8 hundreds and 2 ones.

So, total place value of thousands gives

75 × 1000 = 75000

Now, total place value of hundreds gives

8 × 100 = 800 and, the total place value of ones gives

2 × 1=2

Now, by adding the total values obtained from different place values together will give the required number.

⇒ 75000 + 800 + 2 = 75802

Therefore, the given number can be represented in figures as 75802.

Common Core Primary Mathematics 4A Chapter 1 Solved Examples Page 8   Exercise 1.1 Problem 8

Given:  There are twelve Thousands, seven Hundreds and ninety three.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are twelve thousands and seven hundred ninety three.

So, total place value of thousands gives

12 × 1000 = 12000

And, another value given is seven hundred ninety three

⇒ 793

Now, by adding the total values obtained from different place values together will give the required number.

⇒ 12000 + 793= 12793

Therefore, the given number can be represented in figures as 12793.

Chapter 1 Whole Numbers Worked Solutions For Primary Mathematics 4A Page 8   Exercise 1.1 Problem 9

Given: There are ninety Thousand, five Hundred and eleven.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are ninety thousand and five hundred eleven.

So, total place value of thousands gives

90 × 1000 = 90000

And another value given is five hundred-eleven.

⇒ 511

Now, by adding the total values obtained from different place values together will give the required number.

⇒  90000 + 511 = 90511

Therefore, the given number can be represented in figures as 90511.

Primary Mathematics Workbook 4A Common Core Edition Practice Problems Page 8   Exercise 1.1 Problem 10

Given:   There are eighty-eight Thousands and eight.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are eighty-eight thousand and eight.

So, total place value of thousands gives

88 × 1000 = 88000

And another value given is eight

⇒  8

Now, by adding the total values obtained from different place values together will give the required number.

⇒  88000 + 8 = 88008

Therefore, the given number can be represented in figures as 88008.

Primary Mathematics Workbook 4A Chapter 1 Exercises Breakdown Page 8   Exercise 1.1 Problem 11

Given: There are 485 Thousand, 7 Hundred, and 2 ones.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are 485 thousand 7 hundreds and 2 ones.

So, total place value of thousands gives

485 × 1000 = 485000

Now, total place value of hundreds gives

7 × 100 = 700

And total place value of ones gives

2 × 1 = 2

Now, by adding the total values obtained from different place values together will give the required number.

⇒  485000 + 700 + 2 = 485702

Therefore, the given number can be represented in figures as 485702.

Step-By-Step Guide For Primary Mathematics Workbook 4a Chapter 1 Page 8   Exercise 1.1 Problem 12

Given: There are 600 Thousand and 3 Tens.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are 600 thousand and 3 tens.

So, total place value of thousands gives

600 × 1000 = 600000

Now, the total place value of tens gives

3 × 10 = 30

Now, by adding the total values obtained from different place values together will give the required number.

⇒  600000 + 30 = 600030

Therefore, the given number can be represented in figures as 600030.

Common Core 4A Chapter 1 Whole Numbers Answers Page 8   Exercise 1.1 Problem 13

Given: There are 999 Thousands, 9 tens and 9 ones.

Question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are 999 thousands, 9 tens and 9 ones.

So, total place value of thousands gives

999 × 1000 = 999000

Now, total place value of tens gives

9 × 10 = 90 and total place value of ones gives

9 × 1 = 9

Now, by adding the total values obtained from different place values together will give the required number.

⇒  999000  +  90  + 9 =  999099

Therefore, the given number can be represented in figures as 999099

Page 8   Exercise 1.1  Problem 14

Given:  That there are three hundred twelve Thousands and four hundred sixty.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are three hundred twelve thousands and four hundred sixty.

So, total place value of thousands gives

312  ×  1000 = 312000 and other value given is four hundred sixty

⇒  460

Now, by adding the total values obtained from different place values together will give the required number.

⇒  312000  +  460 =  312460

Therefore, the given number can be represented in figures as 312460.

Page 8   Exercise 1.1  Problem 15

Given:  That there are eight hundred two Thousands and three.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are 802 thousands and three.

So, total place value of thousands gives

802 × 1000 = 802000

And another value given is three.

⇒  3

Now, by adding the total values obtained from different place values together will give the required number.

⇒ 802000 + 3 = 802003

Therefore, the given number can be represented in figures as 802003.

Page 8   Exercise 1.1  Problem 16

Given:  That there are nine hundred Thousands and nine hundred nine.

The question is asked to write the given number in figures.

By multiplying the each number given with its place value and adding the total values of different place values together will give the required number.

Given values are 900 thousands and nine hundred nine.

So, total place value of thousands gives

900  ×  1000 = 900000

And another value given is nine hundred nine.

⇒  909

Now, by adding the total values obtained from different place values together will give the required number.

⇒  900000 + 909 = 900909

Therefore, the given number can be represented in figures as 900909.

Page 9  Exercise 1.1   Problem 17

Given:  Number is 2,080

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

2080 = 2000 + 80

We can find the multiples of each place values from this expression.

⇒  2080 = 2 × 1000 + 8 × 10

So, the number comprises of 2 thousands and 8 tens.

It can be expressed in words as ‘Two thousand eighty’.

Therefore, the given number can be represented in words as ‘Two thousand eighty’.

Page 9  Exercise 1.1   Problem 18

Given: The number is 9,215.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

9215 = 9000 + 200 + 10 + 5

We can find number of each place values from this expression.

9215 = 9 × 1000 + 2 × 100 + 1 × 10 + 5 × 1

So, the number comprises of 9 thousands, 2 hundreds, 1 ten and 5 ones.

It can be expressed in words as ‘Nine thousand two hundred fifteen’.

Therefore, the given number can be represented in words as ‘Nine thousand two hundred fifteen’.

Page 9  Exercise 1.1  Problem 19

Given: The number is 47,010.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

47010 = 40000 + 7000 + 10

We can find number of each place values from this expression.

47010 = 4 × 10000 + 7 × 1000 + 1 × 10

So, the number comprises of 4 ten thousands, 7 thousands and 1 ten.

It can be expressed in words as ‘Forty-seven thousand ten’.

Therefore, the given number can be represented in words as ‘Forty-seven thousand ten’.

Page 9  Exercise 1.1  Problem 20

Given:  The Number is 89,102.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

89102 = 80000 + 9000 + 100 + 2

We can find number of each place values from this expression.

89102 = 8 × 10000 + 9 × 1000 + 1 × 100 + 2 × 1

So, the number comprises of 8 ten thousands, 9 thousands, 1 hundred and 2 ones.

It can be expressed in words as ‘Eighty-nine thousand one hundred two’.

Therefore, the given number can be represented in words as ‘Eighty-nine thousand one hundred two’.

Page 9  Exercise 1.1  Problem 21

Given:   The Number is 40,900.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

40900 = 40000 + 900

We can find number of each place values from this expression.

40900 = 4 × 10000 + 9 × 100

So, the number comprises of 4 ten thousands and 9 hundreds.

It can be expressed in words as ‘Forty thousand nine hundred’.

Therefore, the given number can be represented in words as ‘Forty thousand nine hundred’.

Page 9  Exercise 1.1  Problem 22

Given: The Number is 78,999.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

78999 = 70000 +  8000 + 900 + 90 + 9

We can find number of each place values from this expression.

78999 = 7 × 10000 + 8 × 1000 + 9 × 100 + 9 × 10 + 9 × 1

So, the number comprises of 7 ten thousands, 8 thousands, 9 hundreds, 9 tens, and 9 ones.

It can be expressed in words as ‘Seventy-eight thousand nine hundred ninety-nine’.

Therefore, the given number can be represented in words as ‘Seventy-eight thousand nine hundred ninety-nine’.

Page 9  Exercise 1.1  Problem 23

Given: The number is 50,234.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

50234 = 50000 + 200 + 30 + 4

We can find number of each place values from this expression.

50234=5 × 10000 + 2 × 100 + 3 × 10 + 4 × 1

So, the number comprises of 5 ten thousands, 2 hundreds, 3 tens and 4 ones.

It can be expressed in words as ‘Fifty thousand two hundred thirty-four’.

Therefore, the given number can be represented in words as ‘Fifty thousand two hundred thirty-four’.

Page 9  Exercise 1.1  Problem 24

Given:  The Number is 26,008.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

26,008 = 20000 + 6000 + 8

We can find number of each place values from this expression.

26,008 = 2 × 10000 + 6 × 1000 + 8 × 1

So, the number comprises of 2 ten thousands, 6 thousands and 8 ones.

It can be expressed in words as ‘Twenty-six thousand eight’.

Therefore, the given number can be represented in words as ‘Twenty-six thousand eight’.

Page 9  Exercise 1.1  Problem 25

Given:  The Number is 73,506.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

73,506 = 70000 + 3000+ 500 + 6

We can find number of each place values from this expression.

73,506 = 7 × 10000 + 3 × 1000 + 5 × 100 + 6 × 1

So, the number comprises of 7 ten thousands, 3thousands, 5 hundreds and 6ones.

It can be expressed in words as ‘Seventy-three thousand five hundred six’.

Therefore, the given number can be represented in words as ‘Seventy-three thousand five hundred six’.

Page 9  Exercise 1.1  Problem 26

Given: The number is 367,450.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

367,450 = 300000 + 60000 + 7000 +  400 + 50

We can find number of each place values from this expression.

367,450 = 3 × 100000 + 6 × 10000 + 7 × 1000 + 4 ×100 + 5 × 10

So, the number comprises of 3 hundred thousands, 6 ten thousands, 7 thousands, 4 hundreds, and 5 tens.

It can be expressed in words as ‘Three hundred sixty-seven thousand four hundred fifty’.

Therefore, the given number can be represented in words as ‘Three hundred sixty-seven thousand four hundred fifty’.

Page 9  Exercise 1.1  Problem 27

Given:  The Number is 506,009.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

506,009 = 500000 + 6000 + 9

We can find number of each place values from this expression.

506,009 = 5 × 100000 + 6 × 1000 + 9 × 1

So, the number comprises of 5 hundred thousands, 6 thousands, and 9 ones.

It can be expressed in words as ‘Five hundred-six thousand six’.

Therefore, the given number can be represented in words as ‘Five hundred-six thousand six

Page 9  Exercise 1.1  Problem 28

Given:  Number is 430,016.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

430,016 = 400000 + 30000 + 10 + 6

We can find number of each place values from this expression.

430,016 = 4 × 100000 + 3 × 10000 + 1 × 10 + 6 × 1

So, the number comprises of 4 hundred thousands, 3 ten thousands, 1 ten and 6ones.

It can be expressed in words as ‘Four hundred thirty thousand sixteen’.

Therefore, the given number can be represented in words as ‘Four hundred thirty thousand sixteen’.

Page 9  Exercise 1.1  Problem 29

Given:  Number is 800,550.

Question is to express the given number in words.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

800,550 = 800000 + 500 + 50

We can find number of each place values from this expression.

800,550 = 8 × 100000 + 5 × 100 + 5 × 10

So, the number comprises of 8 hundred thousands, 5 hundreds, and 5 tens.

It can be expressed in words as ‘Eight hundred thousand five hundred fifty’.

Therefore, the given number can be represented in words as ‘Eight hundred thousand five hundred fifty’.

Page 10   Exercise 1.2   Problem  1

Given: The number is 23,529.

Question is to write the values of the digits.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

23,529= 2 × 10000 + 3 × 1000 + 5 × 100 + 2 × 10 + 9 × 1  or

We can write it as

23,529 = 20000 + 3000 + 500 + 20 + 9

From the above expression, the values of the digits can be written in the given form as

Therefore, the values of the digits in the given number 23,529 can be written as

Page 10   Exercise 1.2   Problem  2

Given:  The Number is 40618.

Question is to write the values of the digits.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

40,618 = 4 × 10000 + 6 ×100+1 × 10 + 8 × 1   or

We can write it as

40,618 = 40000 + 600 + 10 + 8

From the above expression, the values of the digits can be written in the given form as

Therefore, the values of the digits in the given number 40,618 can be

Page 10   Exercise 1.2   Problem  3

Given:  The Number is 45023.

Question is to write the values of the digits.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

45,023 = 4 × 10000 + 5 × 1000 + 2 × 10 + 3 × 1 or

We can write it as

45,023 = 40000 + 5000 + 20 + 3

From the above expression, the values of the digits can be written in the given form as

Therefore, the values of the digits in the given number 45,023 can be written as

Page 10   Exercise 1.2   Problem  4

Given: The number is 88,888.

Question is to write the values of the digits.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

88,888 = 8 × 10000 + 8 × 1000 + 8 × 100 + 8 × 10 + 8 × 1 or

We can write it as

88,888 = 80000 + 8000 + 800 + 80 + 8

From the above expression, the values of the digits can be written in the given form as

Therefore, the values of the digits in the given number 88,888 can be written as

Page 10   Exercise 1.2   Problem  5

Given: The number is 104,682.

Question is to write the values of the digits.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

104682 = 100000 + 4000 + 600 + 80 + 2

We have

From the above expression, the values of the digits can be written in the given form as 2,80,600,4000,100000

Therefore, the values of the digits in the given number 104,682 can be written as 2,80,600,4000,100000.

Page 10   Exercise 1.2   Problem  6

Given: The number is 989,219.

Question is to write the values of the digits.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as the sum of multiples of each place values. i.e.

989219 = 9000000 + 800000 + 9000 + 200 + 10 + 9

We have

From the above expression, the values of the digits can be written in the given form as 9,10,200,9000,80000,900000

Therefore, the values of the digits in the given number 989,219 can be written as 9,10,200,9000,80000,900000

Page 11   Exercise 1.2   Problem 7 

Given: The number is 78,243.

Question is to write the find the digit 7 stands for.

Split the given number according to place values and write it as an expression.

Analyze the table and find out the answer.

The given number can be written as the sum of multiples of each place values. i.e.

78,243 = 70000 + 8000 + 200 + 40 + 3

So, it is clear that digit 7 stands for ten thousands place in 700000.

In 78,243, the digit 7 stands for ten thousands place in 700000.

Page 11   Exercise 1.2   Problem  8

Given: The number is 78,243.

Question is to find the digit in hundreds place and its value.

Split the given number according to place values and write it as an expression.

Analyze the table and find out the answer.

The given number can be written as sum of multiples of each place values. i.e.

78,243 = 70000 + 8000 + 200 + 40 + 3

So, it is clear that the digit in hundreds place is 2.

The value of hundreds place is 200.

In 78,243, the digit 2 in hundreds place. The value of the digit is 200.

Page 11   Exercise 1.2   Problem 9

Given: The number is 78,243.

Question is to find the digit in tens place and thousands place.

Split the given number according to place values and write it as an expression.

Analyze the table and find out the answer.

From the table, it is clear that the digit in tens place is 4 and digit in thousands place 8.

Therefore, In 78,243, the tens digit is 4and the thousands digit is 8.

Page 11   Exercise 1.2   Problem 10

Given:  Numbers are 78,243 and 8.243.

Question is to find the difference between given numbers..

Subtracting given two numbers gives the required answer.

⇒  78,243 − 8,243 = 70,000

So, the difference is 70,000

Therefore, 78,243 is 70,000 more than 8,243.

Page 11   Exercise 1.2   Problem  11

Given: The number is 24,568.

Question is to find the place of the digit 4.

Split the given number according to place values and write it as an expression.

The given number can be written as sum of multiples of each place values. i.e.

24,568 = 20000 + 4000 + 500 + 60 + 8  or

We can write it as

24,568 = 2 × 10000 + 4 × 1000 + 5 × 100 + 6 × 10 + 8 × 1

From the above expression

At ten thousands place is the digit 2.

At thousands place is the digit 4.

At hundreds place is the digit 5.

At tens place is the digit 6.

At ones place is the digit 8.

So, the place of the digit 4is found to be Thousand.

Therefore, in 24,568 the digit 4 stands for thousands place.

Page 11   Exercise 1.2   Problem  12

Given: The number is 43,251.

Question is to find the digit in ten thousands place.

Split the given number according to place values and write it as an expression.

The given number can be written as sum of multiples of each place values. i.e.

43,251 = 40000 + 3000 + 200 + 50 + 1 or

We can write it as

43,251 = 4 × 10000 + 3 × 1000 + 2 × 100 + 5 × 10 + 1 × 1

From the above expression, the digit in the ten thousands place is found to be 4.

Therefore, in 43,251 the digit 4 is in the ten thousands place.

Page 11   Exercise 1.2   Problem  13

Given:  4,000 + 300 + 7 =_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

4,000 + 300 + 7 = 4,307

So, the required number is

Therefore, the solution of the given expression 4,000 + 300 + 7 = _____is 4,307.

Page 11   Exercise 1.2   Problem  14

Given:  50,000 + 6,000 + 400=_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

50,000+ 6,000 + 400 = 56,400

So, the required number is 56,400.

Therefore, the solution of the given expression 50,000 + 6,000 + 400 =_____ is 56,400

Page 11   Exercise 1.2   Problem  15

Given:  30,000 + 700 + 60 + 8 =_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

30,000 + 700 + 60 + 8 = 30,760

So, the required number is 30,760.

Therefore, the solution of the given expression 30,000 + 700 + 60 + 8 =_____is 30,760.

Page 11   Exercise 1.2   Problem  16

Given:  10,000 + 1,000 + 400 =_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

10,000 + 1,000 + 400 = 11,400

So, the required number is 11,400.

Therefore, the solution of the given expression 10,000 + 1,000 + 400 =_____is 11,400.

Page 11   Exercise 1.2   Problem  17

Given:   The number 90,000 + 90 =_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

90,000 + 90 = 90,090

So, the required number is 90,090.

Therefore, the solution of the given expression 90,000 + 90 = _____is  90,090.

Page 12   Exercise 1.2   Problem 18

Given: The number is 127,685.

Question is to find the place of the digit 1.

Split the given number according to place values and write it as an expression

The given number can be written as sum of multiples of each place values. i.e.

127,685 = 100,000 + 20,000 + 7,000 + 600 + 80 + 5  or

We can write it as

127,685 = 1 × 100,000 + 2 × 10,000 + 7 × 1 ,000 + 6 × 100 + 8 × 10 + 5 × 1

From the above expression

At hundred thousands place is the digit 1.

At ten thousands place is the digit 2.

At thousands place is the digit 7.

At hundreds place is the digit 6.

At tens place is the digit 8.

At ones place is the digit 5.

So, the place of the digit 1 is hundred thousands.

Therefore, in 127,685, the digit 1 stands for hundred thousands place or 100,000.

Page 12   Exercise 1.2   Problem 19

Given: The number is 561,260.

Question is to find the digit in ten thousands place and to find the value of digit 1.

Split the given number according to place values and write it as an expression.

The given number can be written as sum of multiples of each place values. i.e.

561,260 = 500,000 + 60,000+1,000 + 200 + 60 + 0  or

We can write it as

561,260 = 5 × 100,000 + 6 × 10,000 + 1 × 1,000 + 2 × 100 + 6 × 10 + 0

From the above expression, the digit in the thousands place is found to be 1.

So, the value of the digit 1 is 1000.

Therefore, in 561,260, the digit 1 is in the thousands place, and the value of the digit 1 is 1,000

Page 12   Exercise 1.2   Problem 20

Given: The number is 432,091.

Question is to find the place of the digit 0.

Split the given number according to place values and write it as an expression.

The given number can be written as sum of multiples of each place values. i.e.

432,091 = 400,000 + 30,000 + 2,000 + 0 + 90 + 1 or

We can write it as

432,091 = 4 × 100,000 + 3 × 10,000 + 2 × 1,000 + 0 × 100 + 9 × 10 + 1 × 1

From the above expression

At hundred thousands place is the digit 4.

At ten thousands place is the digit 3.

At thousands place is the digit 2.

At hundreds place is the digit 0.

At tens place is the digit 9.

At ones place is the digit 1.

So, the place of the digit 0 is hundred.

Therefore, in 432,091, the digit 0 is in the hundreds place.

Page 12   Exercise 1.2   Problem 21

Given:  The Number is 368,540.

Question is to value of the place of the digit 4.

Split the given number according to place values and write it as an expression.

The given number can be written as sum of multiples of each place values. i.e.

368,540 = 300,000 + 60,000 + 8,000 + 500 + 40  or

we can write it as

368,540 = 3 × 100,000 + 6 × 10,000 + 8 × 1,000 + 5 × 100 + 4 × 10

From the above expression, the values of the digits can be written as,.

The value at hundred thousands place is the digit is 300,000.

The value at ten thousands place is 60,000.

The value at thousands place is the digit is 8,000.

The value at hundreds place is the digit is 500.

The value at tens place is the digit 40.

So, the value of the digit 4 is 40.

Therefore, in 368,540 the digit 4 stands for tens place and its value is 40.

Page 12   Exercise 1.2   Problem 22

Given: The number is 760,835.

Question is to find how much more is 760,835 than 700,000

Finding the difference between the two numbers will give the required answer.

⇒ 760,835 − 700,000 = 60,835

So, the required answer is 60,835.

Therefore, 760,835 is 60,835 more than 700,000.

Page 12   Exercise 1.2   Problem 23

Given:   The number  is 40,000+2,000+100+8=_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

40,000 + 2,000 + 100 + 8 = 42,108

So, the required number is 42,108.

Therefore, the solution of the given expression 40,000 + 2,000 + 100 + 8= _____ is 42,108

Page 12   Exercise 1.2   Problem 24

Given:  The number  is  562,000 + 32 = _____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

562,000 + 32 = 562,032

So, the required number is 562,032.

Therefore, the solution of the given expression 562,000 + 32=_____is  562,032.

Page 12   Exercise 1.2   Problem 25

Given:   The number  is 700,000 + 70,000 + 70 + 7 =_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

700,000 + 70,000 + 70 + 7 = 770,077

So, the required number is 770,077.

Therefore, the solution of the given expression 700,000 + 70,000 + 70 + 7 =_____is 770,077.

Page 12   Exercise 1.2   Problem 26

Given: The number  is 900,000 + 214 =_____

Question is to find the missing number and fill the blank.

Add the value of the given place values to obtain required number.

900,000 + 214 = 900,214

So, the required number is 900,214.

Therefore, the solution of the given expression 900,000 + 214 =_____ is 900,214.

Page 12   Exercise 1.2   Problem 27

Given:   The number  is 25,830 = 25,000 + _____ + 30

Question is to find the missing number and fill the blank.

Subtract the sum of the given place values from the original number to obtain required number.

We have

25,830 = 25,000 + _____ + 30

Add the place values present at the right hand side.

25,830 = (25,000 + 30) +_____

From the above expression, bringing the number at right hand side to the left hand side and their difference will give the required number.

⇒  25,830 − (25,000 + 30) =_____

⇒  25,830 − 25,030 = 800

Therefore, the number missing in the given expression 25,830 = 25,000 +  _____+ 30 is 800.

Page 12   Exercise 1.2   Problem 28

Given:   The number  is  370,049 =_____+ 70,000 + 40 + 9

Question is to find the missing number and fill the blank.

Subtract the sum of the given place values from the original number to obtain required number.

We have

370,049 =_____+ 70,000 + 40 + 9

Add the place values present at the right-hand side.

370,049 =_____+ (70,000 + 40 + 9)

From the above expression, bringing the number at right-hand side to the left-hand side and their difference will give the required number.

⇒ 370,049 − (70,000 + 40 + 9) =_____

⇒ 370,049 − 70,049 = 300,000

Therefore, the number missing in the given expression 370,049 =_____+ 70,000 + 40 + 9 is 300,000

Page 12   Exercise 1.2   Problem 29

Given:   The number  is  603,804 = 600,000+_____+ 800 + 4

Question is to find the missing number and fill the blank.

Subtract the sum of the given place values from the original number to obtain required number.

We have

603,804 = 600,000 +_____+ 800 + 4

Add the place values present at the right-hand side.

603,804 = (600,000 + 800 + 4) ​+ _____

From the above expression, bringing the number at right-hand side to the left-hand side and their difference will give the required number.

⇒  603,804−(600,000  +800 + 4) =_____

⇒  603,804−600,804 = 3,000

Therefore, the number missing in the given expression 603,804 = 600,000 +_____+ 800 + 4 is 3,000.

Page 12   Exercise 1.2   Problem 30

Given:    The number  is  416,008 = 416,000 +_____

Question is to find the missing number and fill the blank.

Subtract the sum of the given place values from the original number to obtain required number.

⇒  416,008−416,000 = 8

So, the required number is 8.

Therefore, the number missing in the given expression 416,008 = 416,000+_____ is 8.

Page 12   Exercise 1.2   Problem 31

Given:    The number is   123,456.

Question is to find the number of ones.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.,

123,456 = 100,000 + 20,000 + 3,000 + 400 + 50 + 6

We can find number of each place values from this expression.

123,456 = 1 × 100,000 + 2 × 10,000 + 3 × 1,000 + 4 × 100 + 5 × 10 + 6 × 1

Analyzing the above expression

It can be concluded that there are 6 ones.

Therefore, the given number 123,456 has 6 ones.

Page 12   Exercise 1.2   Problem 32

Given:  The Number is 123,456

Question is to find the number of tens.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

123,456 = 100,000 + 20,000 + 3,000 + 400 + 50 + 6

We can find number of each place values from this expression.

123,456 = 1 × 100,000 + 2 × 10,000 + 3 × 1,000 + 4 × 100 + 5 × 10 + 6 × 1

Analyzing the above expression

It can be concluded that there are 5 tens.

Therefore, the given number 123,456 has 5 tens.

Page 12   Exercise 1.2   Problem 33

Given:  The  Number is 123,456

Question is to find the number of hundreds.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

123,456 = 100,000 + 20,000 + 3,000 + 400 + 50 + 6

We can find number of each place values from this expression.

123,456 = 1 × 100,000 + 2 × 10,000 + 3 × 1,000 + 4 × 100 + 5 × 10 + 6 × 1

Analyzing the above expression

It can be concluded that there are 4 hundreds.

Therefore, the given number 123,456 has 4 hundreds.

Page 12   Exercise 1.2   Problem 34

Given: The number is 123,456.

Question is to find the number of thousands.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

123,456 = 100,000 + 20,000 + 3,000 + 400 + 50 + 6

We can find number of each place values from this expression.

123,456 = 1 × 100,000 + 2 × 10,000 + 3 × 1,000 + 4 × 100 + 5 × 10 + 6 × 1

Analyzing the above expression

It can be concluded that there are 3 thousands.

Therefore, the given number 123,456 has 3 thousands.

Page 12   Exercise 1.2   Problem 35

Given:  The number is 123,456.

Question is to find the number of ten thousands.

Split the given number according to place values and write it as an expression.

Analyze the expression and find the multiples of each place values.

The given number can be written as sum of multiples of each place values. i.e.

123,456 = 100,000 + 20,000 + 3,000 + 400 + 50 + 6

We can find number of each place values from this expression.

123,456 =1 × 100,000 + 2 × 10,000 + 3 × 1,000 + 4 × 100 + 5 × 10 + 6 × 1

Analyzing the above expression

It can be concluded that there are 2 ten thousands.

Therefore, the given number 123,456 has 2 ten thousands.

Page 12   Exercise 1.2   Problem 36

Given:  The number 123,456

To calculate the number of hundred thousand.

Here, digit 1 is in the hundred thousand’s place so there is one hundred thousand in the given number.

Therefore, the number 123,456 has one hundred thousand.

Page 12   Exercise 1.2   Problem 37

Given:  50 ten thousand + 40 hundreds + 20 tens.

To write in standard form.

We have to write 20 at tens, 40 at hundreds and 50 at ten thousand’s place.

We will write the terms in numerical form. 50 ten thousand

=50×10000

40 hundreds = 40 × 100

20 tens = 20 × 10

Adding all the three terms we get:

50 × 10000 + 40 × 100 + 20 × 10 = 500,000 + 4,000 + 200

=  504,200

Hence, we have the standard form as 504,200.

Therefore, 50 ten thousand + 40 hundreds + 20 tens can be written in standard form as 504,200.

Page 12   Exercise 1.2   Problem 38

Given:  83 ten thousand + 4 tens + 7ones.

To write in standard form.

We have to write 7 at ones, 4 at tens and 83 at ten thousand’s place.

We will write the terms in numerical form. 83 ten thousand =83×10000

4tens = 4 × 10

7ones =7 × 1

Adding all the three terms we get:

83 × 10000 + 4 × 10 + 7 × 1= 830,000 + 40 + 7

=  8,30,047

Hence, we have the standard form as 8,30,047.

Therefore,83 ten thousand +4 tens + 7 ones can be written in standard form as 8,30,047.

Page 13   Exercise 1.3   Problem 1

Given:  The number line.

To find –  The number that each letter represents.

We will find the least count between the numbers and accordingly find the numbers.

First, we will find the number of divisions between 35,000 and 40,000.

There are 5 divisions which can be seen from the figure. Now, the difference between 35,000 and 40,000 can be calculated as 40,000−35,000=5,000.

Now dividing the difference by the number of divisions obtain the least count. Hence, the least count is given by

\(\frac{5,000}{5}\) = 1,000.

Now, the number that each letter represent can be found as follows

A is located two divisions before 35,000 so it can be calculated by subtracting 1,000 two times from 35,000.

Hence, A=35,000−1,000−1,000=33,000.

Now, B is located two divisions after 35,000 so it can be calculated by adding 1,000 two times to 35,000.

This implies, B = 35,000 + 1,000 + 1,000 = 37,000.

Solving further, no we find C.

C is located just after 40,000 so it can be calculated by adding 1,000 to 40,000.

Hence, A = 40,000 + 1,000 = 41,000.

Further, calculating D which is located before 45,000, we need to subtract 1,000 from 45,000.

So, D = 45,000 −1,000 = 44,000.

Therefore, the numbers that each letter represent are calculated as follows:

A=33,000

B=37,000

C=41,000

D=44,000

Page 13   Exercise 1.3   Problem 2

Given: The numbers 13,268 and 31,862.

To tell which number is greater.

We have to compare the digits on the extreme left.

We know that if two numbers have an equal number of digits, the number having a greater valued digit on the extreme left is greater.

Here, the digit on the extreme left is at ten thousand-place. So, we will compare the places.

From the given numbers, 13,268 and 31,862, 31,862 has a greater value digit at the ten thousand place as 3>1.

Hence, 31,862 is the greater number.

Therefore, 31,862 is greater as compared to 13,268

Page 13   Exercise 1.3   Problem 3

Given:  The numbers 49,650 and 42,650.

To tell which number is smaller.

We have to compare the digits on the extreme left.

We know that if two numbers have an equal number of digits, the number having a smaller valued digit on the extreme left is smaller.

If the numbers’ digits on the extreme left are equal, the digits to the right of the extreme left digits are compared, and so on.

Here, the digit on the extreme left is 4 which is equal for both numbers so, we will take the next digit that is at the thousand place.

From the given numbers 49,650 and 42,650, 42,650 has a smaller value digit at the thousand place as 2<9.

Hence, 42,650 is a smaller number.

Therefore, 42,650 is smaller as compared to 49,650

Page 13   Exercise 1.3   Problem 4

Given:   The numbers 33,856, 33,786, and 33,796.

To tell which number is greatest.

We have to compare the digits on the extreme left.

We know that if two numbers have an equal number of digits, the number having a greater valued digit on the extreme left is greater.

If the numbers’ digits on the extreme left are equal, the digits to the right of the extreme left digits are compared, and so on.

Here, the digits on the extreme left are equal. So, we will compare the places on the thousand-place.

Here too, the places are equal.

Moving on further, we will now compare hundreds place.

From the given numbers 33,856, 33,786, and 33,796,33,856 has a greater value digit at the hundreds place as 8>7.

Hence, 33,856 is the greatest number out of the three numbers.

Therefore, 33,856 is the greatest as compared to 33,786 & 33,796.

Page 13   Exercise 1.3   Problem 5

Given:  The numbers 65,730, 65,703, and 66,730.

To tell which number is smallest.

We have to compare the digits on the extreme left.

We know that if two numbers have an equal number of digits, the number having a smaller valued digit on the extreme left is smaller.

If the numbers’ digits on the extreme left are equal, the digits to the right of the extreme left digits are compared, and so on.

Here, the digits on the extreme left are equal.

So, we will compare the places on the thousand-place.

Here, numbers 65,730 and 65,703 are smaller than 66,730 as 5<6.

Moving on further, we will now compare hundreds place which are equal for both 65,730 and 65,703.

Now, we compare the tens place.

From the numbers 65,730 and 65,703, 65,703 has a smaller value digit at the tens place as 0<3.

Hence, 65,703 is the smallest number out of the three numbers.

Therefore, 65,703 is the smallest as compared to 65,730 and 66,730.

Page 13   Exercise 1. 3  Problem 6

Given:  The numbers 3,695, 3,956, 35,096, 30,965.

To arrange the numbers in increasing order.

We have to compare the numbers and arrange accordingly.

First, we count the number of digits in each number.

The number with the least number of digits is the smallest.

From the given numbers, 3,695 and 3,956 have fewer digits than 35,096 and 30,965.

Now we will compare the two smaller numbers.

We know that if the numbers’ digits on the extreme left are equal, the digits to the right of the extreme left digits are compared, and so on.

Here, the digits on the extreme left are equal. So, we will compare the places on the hundreds place.

From the given numbers, 3,695 and 3,956, 3,695 is smaller than 3,956 because it has a smaller value digit as 6<9.

Now, we compare the remaining two numbers which are 35,096 and 30,965.

Here, the digits on the extreme left are equal. So, we will compare the places on the thousands place.

From the numbers 35,096 and 30,965, 30,965 has a smaller value digit as 0<5.

Hence, we see that the numbers can be arranged from smallest to greatest as follows: 3,695<3,956<30,965<35,096.

Therefore, the numbers can be arranged in increasing order as follows:  3,695<3,956<30,965<35,096.

Page 13   Exercise 1. 3  Problem 7

Given:  The numbers 435,760, 296,870, 503,140, 462,540.

To arrange the numbers in increasing order.

We have to compare the numbers and arrange accordingly.

First, we count the number of digits in each number.

For the numbers having the same number of digits, start with comparing the numbers from the leftmost digit.

Write the number with the smallest digit.

From the given numbers, 296,870 has the smallest leftmost digit that is 2 so it is the smallest of all four numbers.

Also, the number with the greatest leftmost digit is 503,140 so it is the greatest number.

Now, we compare the remaining two numbers which are 435,760 and 462,540.

We know that if the numbers’ digits on the extreme left are equal, the digits to the right of the extreme left digits are compared, and so on.

Here, the digits on the extreme left are equal. So, we will compare the ten thousand-place.

From the numbers 435,760 and 462,540, 435,760 has a smaller value digit at ten thousand-place as 3<6.

Hence, we see that the numbers can be arranged from smallest to greatest as follows:

296,870<435,760<462,540<503,140.

Therefore, the numbers can be arranged in increasing order as follows: 296,870<435,760<462,540<503,140.

Page 13   Exercise 1. 3  Problem 8

Given:  The numbers 9,870 and 18,970.

To compare and write >,<, or =

The number having more digits is a greater number.

First, we need to count the number of digits in each number.

We see that 9,870 has four digits and 18,970 has five digits.

So, clearly, 18,970 has a greater number of digits and hence, it is greater.

Now, we will write the sign as follows: 9,870<18,970.

Therefore, we obtain the result as 9,870<18,970

Page 13   Exercise 1. 3  Problem 9

Given:  The numbers 50,972 and 49,827…

To compare and write >,<, or =.

We have to compare the digits on the extreme left.

First, we need to count the number of digits in each number.

We see that the number of digits are equal in each number.

We know that if two numbers have an equal number of digits, the number having a greater valued digit on the extreme left is greater.

Here, the digit at extreme left is at ten thousand-place so we will compare the places.

From the given numbers, 50,972 has a greater valued digit at the extreme left as 5>4.

So, clearly, 50,972 is greater than 49,827.

Now, we will write the sign as follows: 50,972>49,827.

Therefore, we obtain the result as 50,972>49,827.

Page 13   Exercise 1. 3  Problem 10

Given:  The numbers 326,548 and 326,593.

To compare and write >,<, or =.

We have to compare the digits on the extreme left.

First, we need to count the number of digits in each number.

We see that the number of digits are equal in each number.

We know that if two numbers have an equal number of digits, the number having a greater valued digit on the extreme left is greater.

If the numbers’ digits on the extreme left are equal, the digits to the right of the extreme left digits are compared, and so on.

Here, the digits at hundred thousand, ten thousand, thousand and hundreds place are equal so we will compare the places at tens place.

From the given numbers, 326,548 has a smaller valued digit at the tens place as 4<9.

So, clearly, 326,548 is smaller than 326,593.

Now, we will write the sign as follows: 326,548<326,593.

Therefore, we obtain the result as 326,548<326,593.

Page 14   Exercise 1.4   Problem 1

Given:  The number 42,628.

To find which number is 1,000 more than 42,628.

We will obtain the new number by adding 1,000 to 42,628.

We can write it as − 1000 + 42628 = 43,628.

Therefore, the number which is 1000 more than 42,628 is 43,628.

Page 14   Exercise 1.4   Problem 2

Given:  The number 2,63,240.

To find 2,63,240 is 10,000 more than that number.

We will obtain the new number by subtracting 10,00 from 2,63,240.

We can write it as − 2,63,240−10,000 = 2,53,240.

Therefore, 2,63,240 is 10,000 more than 2,53,240.

Page 14   Exercise 1.4   Problem 3

Given:  The number 90,000.

To find which number is 100 less than 90,000.

We will obtain the new number by subtracting 100 from 90,000.

We can write it as− 90,000 − 100 = 89,900

Therefore, ​ 89,990 is 100 less than 90,000.

Page 14   Exercise 1.4   Problem 4

Given:  The number 86,000,000.

To find 86,000,000 is 100,000 less than which number.

We will obtain the new number by adding 100,000 to 86,000,000.

We can write it as 86,000,000 + 100,000 = 86,100,000.

Therefore, 86,000,000 is 100,000 less than 86,100,000.

Page 14   Exercise 1.4   Problem 5

Given:  45,500 is ____ more than 45,600.

Question is to fill the blank

We will obtain the number by subtracting 45,500 from 45,600.

We can write it as −  45,600 − 45,500 = 100.

Therefore, 45,500 is 100 more than 45,600.

Page 14   Exercise 1.4   Problem 6

Given:  The numbers 384,000 and 39,400.

To find – 384,000 is a number less than 394,000.

We will obtain the number by subtracting 384,000 from 394,000.

We can write it as − 394,000 − 384,000 = 10,000.

Therefore, 384,000 is 10,000 less than 394,000.

Page 14   Exercise 1.4   Problem 7

Given:  The numbers 29,409 and 39,409.

To find – The number which when added to 29,409 gives 39,409.

We will obtain the number by subtracting 29,409 from 39,409.

We can write it as − 39,409 − 29,409 = 10,000.

Therefore, the missing number which when added to 29,409 gives 39,409 is 10,000.

Page 14   Exercise 1.4   Problem 8

Given:  The numbers 2,483,000 and 2,482,000.

To find the number which when subtracted from 2,483,000 gives 2,482,000.

We will obtain the number by subtracting 2,482,000 from 2,483,000.

We can write it as − 2,483,000 − 2,482,000 = 1,000.

Therefore, the missing number which when subtracted from 2,483,000 gives 2,482,000 is 1,000.

Page 14   Exercise  1.4   Problem 9

Given:  The sequence 35,552;____;____;38,552;39,552.

To complete the given regular pattern.

We identify the certain sequence which is followed in the pattern and accordingly, fill the blanks.

We notice that the numbers 38,552 and 39,552 follow a certain sequence.

It can be seen that the difference between the numbers is 39,552−38,552=1,000.

Hence, the pattern follows a certain sequence which is a difference of 1,000 between numbers.

The first number is given as 35,552.

The second number will be 1000 added to the first number 35,552+1,000=36,552.

The third number will similarly be 36,552+1,000=37,552.

Therefore, the number pattern is completed as follows: 35,552;  36,552; 37,552; 38,552; 39,552

Page 14   Exercise 1.4   Problem  10

Given:  The sequence 71,680;71,780;____;71,980;____.

To complete the given regular pattern.

We identify the certain sequence which is followed in the pattern and accordingly, fill the blanks.

We notice that the numbers 71,680 and 71,780 follow a certain sequence.

It can be seen that the difference between the numbers is 71,780−71,680=100.

Hence, the pattern follows a certain sequence which is a difference of 100 between numbers.

The first number is given as 71,680 and the second number is given as 71,780.

The third number will be 100added to the first number 71,780 + 100 = 71,880.

The fourth number is given as 71,980.

The fifth number will similarly be 100 added to the fourth number 71,980 + 100 = 72,080.

Therefore, the number pattern is completed as follows: 71,680; 71,780;  71,880; 71,980; 72,080.

Page 14   Exercise 1.4   Problem 11

Given:  The sequence 283,610;293,610;____;____;323,610.

To complete the given regular pattern.

We identify the certain sequence which is followed in the pattern and accordingly, fill the blanks.

We notice that the numbers 283,610 and 293,610 follow a certain sequence.

It can be seen that the difference between the numbers is  293,610 − 283,610 = 10,000.

Hence, the pattern follows a certain sequence which is a difference of 10,000 between numbers.

The first number is given as 283,610 and the second number is 293,610.

The third number will be 10,000 added to the second number which is 293,610 + 10,000 = 303,610 and the fourth number will be 10,000 subtracted from the fifth number.

Hence, the fourth number is  323,610−10,000  =  313,610.

Therefore, the number pattern is completed as follows: 283,610; 293,610; 303,610; 313,610; 323,610.

Page 14   Exercise  1. 4   Problem 12

Given: A regular number pattern starts with 493,070 and increases each number by 10,000.

To create the regular number pattern.

We will create the pattern by adding 10,000 to the first number and so on.

We are given that the pattern starts with 493,070.

The next number will be 493,070+10,000 = 503,070.

The third number will be 503,070+10,000 = 513,070.

Continuing the sequence

The fourth number will be 513,070+10,000=523,070.

The next number will be 523,070+10,000=533,070.

And, the last number will be 533,070+10,000=543,070.

Therefore, the regular number pattern that starts with 493,070 and in which each number is increased by 10,000 is 493,070; 503,070;513,070; 523,070; 533,070; 543,070.

Page 14   Exercise 1.4   Problem 13

Given: A regular number pattern starts with 493,070 and decreases each number by 1,000.

To create the regular number pattern.

We will create the pattern by subtracting 1,000 from the first number and so on.

We are given that the pattern starts with 493,070.

The next number will be 493,070−1,000 = 492,070.

The third number will be 492,070−1,000 = 491,070.

Continuing the sequence

The fourth number will be 491,070−1,000 = 490,070.

The next number will be 490,070−1,000 = 489,070.

And, the last number will be 489,070−1,000 = 488,070.

Therefore, the regular number pattern that starts with 493,070 and in which each number is decreased by 1,000 is 493,070;492,070;491,070;490,070;489,070;488,070.

Page 15 Exercise  1. 4 Problem 14

Given the figures:

To draw figure 6.

We see that these figures follow a ceratain pattern.

Accordingly, we can draw figure 6 as follows:

Therefore, figure 6 is drawn below:

Page 15 Exercise  1. 4 Problem 15

Given the table.

To complete the given table.

We identify the sequence of the pattern and accordingly solve it.

We can see that the difference between the numbers  4−1 =3 and  7−4 =3 is 3.

Moving on, the number of squares for figure 4 will be 7 + 3 = 10.

Accordingly, for figure 5 it will be 10 + 3 = 13.

Furthermore, the number of squares for figure 6 will be 13 + 3 = 16.

The number of squares for figure 7 and figure 8 will be 16 + 3 = 19 and 19 + 3 = 22 , respectively.

Therefore, the completed table is given below:

Page 15 Exercise  1. 4 Problem 16

Given the figure and table as follows:

To tell the pattern that we notice in the number of squares.

From the figure, we can see that in figure 1, there is 1 square. In figure 2, the pattern changes to 1 + 2 + 1 = 4, in the next figure, it changes to 2 + 3 + 2 = 7, and so on.

Hence, we can notice that the pattern followed is 1 + 1 + 1 = 3 that is the difference between each number of squares is 3.

Therefore, the pattern that we notice in the number of squares is 1 + 1 + 1 = 3.

Page 16   Exercise 1.5   Problem 1

Given:  The numbers 7 thousands and 9 thousands and the equation 7,000 + 9,000.

We have to add the numbers.

Using the place values, we add the numbers.

We are given the numbers 7 thousands and 9 thousands which are both at thousand-place.

Hence, we simply add the numbers 7 + 9 = 16

Hence, we obtain the sum of 7 thousands and 9 thousands is 16 thousands.

Also,7,000 + 9,000 = 16,000.

Therefore, we obtain the equation as 7 thousands + 9 thousands = 16 thousands.

Page 16   Exercise 1.5   Problem 2

Given:  The numbers 23thousands and 14 thousands and the equation 23,000+14,000.

We have to add the numbers.

Using the place values, we add the numbers.

We are given the numbers 23 thousands and 14 thousands which are both at thousand-place.

Hence, we simply add the numbers 23 + 14 = 37.

Hence, we obtain the sum of 23 thousands and 14 thousands is 37 thousands.

Also,23,000 + 14,000 = 37,000.

Therefore, we obtain the equation as 23 thousands + 14 thousands = 37 thousands.

Page 16   Exercise 1.5   Problem 3

Given:  The numbers 29,000&12,000.

We have to add the numbers.

Using the place values, we add the numbers.

We are given the numbers 29,000&12,000 which are both at thousand-place.

Hence, we simply add the numbers 29 + 12 = 41.

Hence, we obtain the sum of 29,000, and 12,000 is 41,000.

Therefore, we obtain the equation as 29,000 + 12,000 = 41,000.

Page 16   Exercise 1.5   Problem 4

Given:  The numbers 3,46,000&24,000.

We have to add the numbers.

Using the place values, we add the numbers.

We are given the numbers 3,46,000&24,000.

We will add the numbers using their place values.

If we add ones, tens, and hundreds place we get 000.

Next, adding thousands place we have 6 + 4 = 10.

Now, adding ten thousands place 4 + 2 = 6.

Hence, we obtain the sum as 3,46,000+24,000=3,70,000.

Therefore, we obtain the equation as 3,46,000 + 24,000 = 3,70,000.

Page 16   Exercise 1.5   Problem 5

Given:  The numbers 538,000&161,000.

We have to add the numbers.

Using the place values, we add the numbers.

We are given the numbers 538,000&161,000.

We will add the numbers using their place values.

If we add ones, tens, and hundreds place we get 000.

Next, adding thousands place we have 8+1=9.

Now, adding ten thousands place 3+6=9 and adding a hundred thousands place is 5 + 1 = 6.

Hence, we obtain the sum as 538,000 + 161,000 = 699,000.

Therefore, we obtain the equation as 538,000 + 161,000 = 699,000.

Page 17   Exercise 1.5  Problem 6 

Given:  The numbers 3 thousands & 2 and the equation 3000×2.

We have to multiply the numbers.

Using the place values, we multiply the numbers.

We are given the numbers 3 thousands & 2.

We multiply each digit of the number with 2.

Hence, we obtain the product of 3 thousands & 2 is 6 thousands as 3 × 2 = 6.

Also,3,000 × 2 = 6,000.

Therefore, we obtain the solution as 3 thousands & 2 = 6  Thousands.

Page 17   Exercise 1.5  Problem 7

Given: We have to multiply 8 thousands×6.

Solving

8 thousands × 6 = 24 thousands

8000 × 6 = 24000

By multiplying we get 8 thousands × 6 = 24 thousands and 8000 × 6 = 24000

Page 17   Exercise 1.5  Problem 8

Given:  We have to multiply 14,000 × 3

Solving

14,000 × 3 = 42,000

On solving the given expression, we get 14,000 × 3 = 42,000

We have to multiply 18,000 × 5.

Solving

18,000 × 5 = 90,000

On solving the given expression, we get 18,000 × 5 = 90,000

Page 17   Exercise 1.5  Problem 9

Given:  We have to multiply 60,000×7.

Solving

60,000 × 7 = 42,000

On solving the given expression, we get 60,000 × 7 = 42,000

Page 17   Exercise 1.5  Problem 10

Given:  We have to divide \(\frac{\text 8 thousands }{4}\)

Solving

\(\frac{\text 8 thousands}{4}\)= 2 thousands and \(\frac{8,000}{4}\)= 2000

On solving the given expression, we get, \(\frac{\text 8 thousands}{4}\) = 2 thousands\(\frac{8,000}{4}\)= 2000

Page 17   Exercise 1.5  Problem 11

Given:  We have to divide \(\frac{\text 72 thousands}{6}\)

Solving

\(\frac{\text 72 thousands}{6}\)= 12 thousands \(\frac{72,000}{6}\)

On solving the given expression, we get: \(\frac{\text 72 thousands}{6}\)= 12 thousands\(\frac{72,000}{6}\)

Page 17   Exercise 1.5  Problem 12

Given:  We have to divide \(\frac{15,000}{5}\)

Solving

\(\frac{15,000}{5}\) = 3,000

On solving the given expression, we get \(\frac{15,000}{5}\) = 3,000

Page 17   Exercise 1.5  Problem 13

Given:  We have to divide \(\frac{96,000}{8}\)

Solving

\(\frac{96,000}{8}\)= 12,000

On solving the given expression, we get \(\frac{96,000}{8}\)= 12,000

Page 17   Exercise 1.5  Problem 14

Given:  We have to divide \(\frac{630,000}{7}\)

Solving

\(\frac{630,000}{7}\) = 90,000

On solving the given expression, we get\(\frac{630,000}{7}\) = 90,000

Page 18  Exercise 1. 5  Problem  15 

Given:  We have to add 120,000 + 340,000

Solving

120,000 + 340,000 = 460,000

On solving the given expression, we get 120,000+340,000=460,000

We have to subtract 120,000 − 34,000.

Solving

120,000 − 34,000 = 86,000

On solving the given expression, we get 120,000−34,000 = 86,000.

Page 18  Exercise 1. 5  Problem  16

Given:  We have to multiply 120,000 × 2.

Solving

120,000 × 2 = 240,000

On solving the given expression, we get 120,000 × 2 = 240,000

Page 18  Exercise 1. 5  Problem  17

Given:  We have to divide \(\frac{120,000}{2}\)

Solving

\(\frac{120,000}{2}\) = 60,000

On solving the given expression, we get \(\frac{120,000}{2}\) = 60,000

Page 18   Exercise 1.5   Problem 18

Given:  We are given an expression 29,000 + n = 41,000.

We have to find the number represented by n.

Bringing n on one side

n = 41,000 − 29,000

n = 12,000

Therefore, the number represented is n=12,000.

Page 18  Exercise 1. 5  Problem  19

Given:  We are given an expression n+24,000=100,000

We have to find the number represented by n

Bringing n on one side

n = 100,000 − 24,0000

n = 76,000

Therefore, the number represented is n = 76,000

Page 18  Exercise 1. 5  Problem  20 

Given:  We are given an expression 254,000 − n = 33,000

We have to find the number represented by n

Bringing n on one side

n = 254,000 − 33,000

n = 221,000

Therefore, the number represented is n = 221,000.

Page 18  Exercise 1. 5  Problem  21

Given:  We are given an expression n−16,000 = 24,000

We have to find the number represented by n

Bringing n on one side

n = 24,000 − 16,000

n = 8,000

Therefore, the number represented is n = 8,000

Page 18  Exercise 1. 5  Problem  22

Given:  We are given an expression 40,000 × n = 120,000

We have to find the number represented by n

Bringing n on one side

n= \(\frac{120,000}{40,000}\)

n = 3

Therefore, the number represented is n = 3.

Page 18  Exercise 1. 5  Problem  23

Given:  We are given an expression n×5 = 40,000

We have to find the number represented by n

Brining n on one side

n = \(\frac{40,000}{5}\)

n = 8,000

Therefore, the number represented is n=8,000

Page 18  Exercise 1. 5  Problem  24

Given:  We are given an expression \(\frac{15,000}{n}\)= 3,000

We have to find the number represented by n

Bringing n on one side

n = 5

Therefore, the number represented is n=5

Page 18  Exercise 1. 5  Problem  25

Given:  We are given an expression \(\frac{n}{8}\)

We have to find the number represented by n

Bringing n on one side

n = 70,000×8

n = 560,000

Therefore, the number represented is n = 560,000

Page 18   Exercise 1.5   Problem 26

Given:  We are given an expression with

LHS = 20,000 + 5,000 + 40 + 6 and

RHS=20,000 + 8,000 + 30 + 9.

We have to compare both sides of the expression, without finding the actual value.

As in the R.H.S 20,000 is added to 8,000, it is higher than in the L.H.S where it is added to 5,000

Therefore

20,000 + 5,000 + 40 + 6<20,000 + 8,000 + 30 + 9

Therefore, by comparing the sides of the expression we get: 20,000 + 5,000 + 40 + 6 < 20,000 + 8,000 + 30 + 9

Page 18   Exercise 1.5   Problem 27

Given:  We are given an expression with

L.HS = 13,100 + 2000 and

R.HS = 13,100 + 200

We have to compare both sides of the expression, without finding the actual value.

As we add 2,000 in the L.H.S , which is higher than 200

Therefore

13,100 + 2000 > 13,100 + 200

Therefore, on comparing we get 13,100 + 2,000>13,100 + 200

Page 18   Exercise 1.5   Problem 28

Given: We are given an expression We have to compare both sides of the expression, without finding the actual value.

As a higher number 7000 is subtracted from 18,151 which is lower as compared to 19,151

18,151−7000<19,151−3,948

Therefore, by comparing we get 18,151−7000<19,151−3,948.

Page 18   Exercise 1.5   Problem 29

Given:  We are given an expression with

L.HS = 700 × 6 and R.HS = 600 × 7

We have to compare both sides of the expression, without finding the actual value.

As 700>600

700×6>600×7

Therefore, by comparing we get 700 × 6>600 × 7

Page 18   Exercise 1.5   Problem 30

Given:  We are given an expression with

L.HS =  \(\frac{56,000}{2}\) and

R.HS = \(\frac{5,600}{2}\)

We have to compare both sides of the expression, without finding the actual value.

As 56,000>5,600 and both are divided by 2 Therefore

\(\frac{56,000}{2}\) > \(\frac{5,600}{2}\)

Therefore, by comparing we get \(\frac{56,000}{2}\) > \(\frac{5,600}{2}\)

Page 19   Exercise 1. 6   Problem 1

Given:  We have to round 297 to the nearest ten.

As 297 is closer to 300 than 290 , 297 is 300 when rounded to the nearest ten.

Therefore, 297 is 300 when rounded to the nearest ten.

Page 19   Exercise 1. 6   Problem 2

Given:  We have to round 1,315 to the nearest ten.

As 1,315 has to be rounded to the nearest ten, the next ten will be 1,320

Therefore, 1,315 is 1,320 when rounded to the nearest ten

Page 19   Exercise 1.6  Problem 3

Given:  We have to round 5,982 to the nearest hundred.

As 6,000 is the nearest hundred to 5,982, 5,892 is 6,000 when rounded to the nearest hundred

Therefore, 5,982 is 6,000 when rounded to the nearest hundred

Page 19   Exercise 1.6  Problem 4

Given:  We have to round 36,250 to the nearest hundred.

As 36,300 is the nearest hundred to 36,250, 36,250 is 36,300 when rounded to the nearest hundred.

Therefore, 36,250 is 36,300 when rounded to the nearest hundred.

Page 19  Exercise 1.6  Problem 5

Given:

We have to round 46,120 to the nearest thousand .

As 46,000 is nearest thousand to 46,120, 46,120 is 46,000 when rounded to the nearest thousand .

Therefore, 46,120 is 46,000 when rounded to the nearest thousand.

Page 19  Exercise 1.6  Problem 6

Given:

We have to round 235,870 to the nearest thousand.

As 236,000 is the nearest thousand to 235,870, 235,870 is 236,000 when rounded to the nearest thousand.

Therefore, 235,870 is 236,000 when rounded to the nearest thousand.

Page 20  Exercise 1. 6   Problem 7

Given:

We have to round 245,230 to the nearest thousand.

As 245,000 is the nearest thousand to 245,230, 245,230 is 245,000 when rounded to the nearest thousand.

Therefore, 245,230 is 245,000 when rounded to the nearest thousand.

Page 20  Exercise 1. 6   Problem 8

Given:

We have to round 247,826 to the nearest thousand .

As 248,000 is the nearest thousand to 247,826 .247,826 is 248,000 when rounded to the nearest thousand .

Therefore, 247,826 is 248,000 when rounded to the nearest thousand.

Page 20  Exercise 1. 6   Problem 9

Given:

We have to round 43,192 to the nearest ten.

As 43,190 is the nearest ten to 43,192, 43,192 is 43,190 when rounded to the nearest ten

Therefore, 49,192 is 43,190 when rounded to the nearest ten.

Page 20  Exercise 1. 6   Problem 10

Given:

We have to round 14,563 to the nearest hundred.

As 14,600 is the nearest hundred.14,563 is 14,600 when rounded to the nearest hundred.

Therefore, 14,563 is 14,600 when rounded to the nearest hundred.

Page 20  Exercise 1. 6   Problem 11

Given:

We have to round 82,926 to the nearest thousand.

As 83,000 is the nearest thousand to 82,926, 82,296 is 83,000 when rounded to the nearest thousand

Therefore, 82,926 is 83,000 when rounded to the nearest thousand.

Page 20  Exercise 1. 6   Problem 12

Given:

We have to round 964,250 to the nearest ten thousand.

As 960,000 is the nearest ten thousand, 964,520 is 960,000 when rounded to the nearest ten thousand.

Therefore, 964,520 is 960,000 when rounded to the nearest ten thousand.

Page 20  Exercise 1. 6   Problem 13

Given:

We have to round 754,000 to the nearest hundred thousand .

As 800,000 is the nearest hundred thousand , 754,000 is 800,000 when rounded to the nearest hundred thousand

Therefore, 754,000 is 800,000 when rounded to the nearest hundred thousand.

Page 20  Exercise 1. 6  Problem 14

Given:

We have to round $438,50 to the nearest ten thousand dollars.

As 440,000 is the nearest ten thousand to 438,50, $438,500 is $ 440,000 when rounded to the nearest ten thousand dollars.

After rounding $438,500 to the nearest ten thousand dollars we get $440,000

Page 20  Exercise 1. 6  Problem 15

Given:

$525,000

To round off the number to nearest ten thousand digit

First, identify the ten thousand digit

Check the previous digit is less than 5 or greater than equal to 5 and apply the rules.

Given the number is $525,000

The ten thousand digit in the number is 2 and the previous digit to it is 5

Since the previous digit to the ten thousand digit is equal to 5 So round up the ten thousand digit.

Therefore, the number $525,000 rounded off to the nearest ten thousand digit as $530,000 The round off answer will be $530,000.

Page 20  Exercise 1. 6  Problem 16

Given: $608,000

To round off the number to nearest ten thousand digit

First identify the ten thousand digit

Check the previous digit is less than 5 or greater than equal to 5 and apply the rules.

Given the number is $608,000

The ten thousand digit in the number is 0 and the previous digit to it is 8

Since the previous digit to the ten thousand digit is equal to 8 So round up the ten thousand digit.

The round-off answer will be $610,000.

Therefore, the number $608,000 rounded off to the nearest ten thousand digit is $610,000.

Page 20  Exercise 1. 6  Problem 17

Given: $974,500

To round off the number to nearest ten thousand digit

First, identify the ten thousand digit

Check the previous digit is less than 5 or greater than equal to 5 and apply the rules.

Given the number is $974,500

The ten thousand digit in the number is 7 and the previous digit to it is 4

Since the previous digit to the ten thousand digit is equal to 4 So round down the ten thousand digit. The round-off answer will be $970,000.

Therefore, the number $974,500rounded off to the nearest ten thousand digit is $970,000.

Page 20  Exercise 1. 6  Problem 18

Given: $990,400

To round off the number to nearest ten thousand digit

First, identify the ten thousand digit

Check the previous digit is less than 5 or greater than equal to 5 and apply the rules.

Given the number is $990,400

The ten thousand digit in the number is 9 and the previous digit to it is 0

Since the previous digit to the ten thousand digit is equal to 0 So round down the ten thousand digit.

The round-off answer will be $990,000.

Therefore, the number $990,400 rounded off to the nearest ten thousand digit is $990,000.

Page 20  Exercise 1. 6  Problem 19

Given: $226,300

To round off the number to nearest ten thousand digit

First, identify the ten thousand digit

Check the previous digit is less than 5 or greater than equal to 5 and apply the rules.

Given the number is $226,300

The ten thousand digit in the number is 2 and the previous digit to it is 6

Since the previous digit to the ten thousand digit is equal to 6 So round up the ten thousand digit.

The round-off answer will be $230,000.

Therefore, the number $226,300 rounded off to the nearest ten thousand digit is $230,000

The post Primary Mathematics Workbook 4A Common Core Edition Solutions Chapter 1 Whole Numbers Exercises 1.1 appeared first on Answer Key for Math.

Glencoe Math Course 2 Volume 1 Chapter 3 Exercise 3.2 Solutions Page 203  Exercise 1 Problem  1

If you would increase the temperature by −5° then you would obtain a temperature of 0°

−5° +5° =0°

Finally, we concluded that the temperature that would make the sum of the two temperatures 0° ⇒ 5°

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 204  Exercise 1  Problem  2

Given:

−5 + ( −7) = _____

To Find – The sum.

Given

​−5 + (−7) = −12

⇒ −12

​−5 + (−7) =  −12
​
Number line:

Finally, we find the sum  ⇒ −12

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Integers Exercise 3.2

Given:

−10 + (−4)=_____

To Find –The sum.

−10 + (−4)= -14

Finally, we find the sum  ⇒ −14

Given:

−14 + (−16) =

To Find – The sum.

Given

−14 + (−16) = −30

⇒ −30

Number line:

Finally, we find the sum ⇒ −30

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Integers Exercise 3.2

Given:

6 + (−7)=_____

To Find-  The sum.

Consider the operation given and simplify
​
​6 + (−7) = −1

⇒ −1

The value of 6 + (−7) is −1

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Integers Exercise 3.2

Given:

−15 + 19 =_____

To Find – The sum.

Given
​
​−15 + 19 = 4

⇒ 4
​
Number line:

Finally, we find the sum  ⇒ 4

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Integers Exercise 3.2

Given:

10 + (−12) =_____

To Find – The sum.

10 + (−12) =−2

= −2

Finally, we find the sum ⇒−2

Given:

−13 + 18 =_____

To Find –  The sum.

−13 + 18 = 5

= 5

Finally, we find the sum ⇒ 5

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Integers Exercise 3.2

Given:

−14 + (−6) + 6 =_____

To Find – The sum.

Consider the operation given and simplify
​
​−14 + (−6) + 6 = −14 + 0

= −14

​

The value of −14+(−6)+6 is−14

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Integers Exercise 3.2

Given:

The temperature is−3°.An hour later it drops 6° and 2 hours later it rises 4°

To Write an additional expression to describe this situation. Then find the sum and explain its meaning

The temperature drops can best be represented by a negative number, while the temperature rise back is positive number.

−3−6 + 4 =−9 + 4 = 5

= 5

Finally we find the sum ⇒ −3− 6 + 4 = 5

Common Core Chapter 3 Integers Exercise 3.2 Answers Glencoe Math Course 2 Page 206  Exercise 1  Problem  3

Given:

−6 + (−8) =_____

To Find –  The sum.

−6 + (−8) = −14

= −14

Finally, we find the sum  ⇒ −14

Step-By-Step Solutions For Exercise 3.2 Chapter 3 Integers In Glencoe Math Course 2 Page 206  Exercise 2  Problem  4

Given:

−3 + 10 =_____

To Find – The sum.

Consider the operation given and simplify

​−3 + 10 =7

⇒ 7

The value of−3 + 10 is  7

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 206  Exercise 3  Problem  5

Given:

−8 + (−4) + 12 =_____

To Find –  The sum.

−8 + (−4) + 12 = − 12 + 12 = 0

⇒  0

Finally, we find the sum ⇒ 0

Page 206 Exercise 4   Problem  6

Given:

Sofia owes her brother $25

She gives her brother the $18

To write an addition expression.

The amount owed can best be represented by a negative number, while the amount paid back is a positive number.

− 25 + 18 − 7

= −7

This means that Sofia still owes her brother

Finally, we write the addition expression ⇒ −25 + 18 = −7

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 206  Exercise 5  Problem  7

The sum is positive

If the absolute value of the negative number is less than the absolute value of the positive number

If both numbers are positive The sum is negative

If the absolute value of the negative number is greater than the absolute value of the positive number

If both numbers are negative The sum is zero

If the absolute value of both integers is equal and if one is positive while the other is negative

Finally, we concluded that we can find a sum is positive, negative or zero without actually adding by the absolute value of the given integers.

Exercise 3.2 Solutions for Chapter 3 Integers Glencoe Math Course 2 Volume 1 Page 207  Exercise 1   Problem  8

Given:

−22 + (−16)

To add both numbers

Given equation is

​− 22 −16

=−38
​
If both numbers has a different sign, add the value and put the greatest value sign

Finally, we conclude the solution using an addition expression and the solution is −38

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 207   Exercise 2  Problem  9

Given:

−10 + (−15)

To add both numbers

If both numbers has a different sign, add the value and put the greatest value sign
​
​−10−15

= −25
​
Finally, we concluded an addition expression to solve the sum and the solution is −25

Examples of problems from Exercise 3.2 Chapter 3 Integers in Glencoe Math Course 2 Page 207  Exercise 3  Problem  10

Given:

6 + 10

To add  both numbers

If both numbers have a + sign, add the value and put a positive sign
​
​6 + 10

= 16
​
Finally, we concluded an addition expression to solve the sum and the solution is 16

Page 207  Exercise 4  Problem  11

Given:

21 + (−21) + (−4)

To add and subtract the numbers

If both numbers has a different sign, add the value and put the greatest value sign

​21 + (−21) + (−4)

= 21−21−4

= −4
​
Finally, we concluded an addition expression to solve the sum and the solution is −4

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 207  Exercise 5  Problem  12

Given:

17 + 20 + (−3)

To add and subtract the given numbers and find the result.

Add the first two numbers, then subtract 3 from their sum.

​17 + 20 + (−3)

= 37 − 3

= 34
​
The value of 17 + 20 + (−3)  is 34

Student Edition Glencoe Math Course 2 Chapter 3 Integers Exercise 3.2 solutions guide Page 207 Exercise 7 Problem  13

Given:

4 + 5

To add both numbers

If both numbers have + sign, add the value and put a positive sign

​4 + 5

= 9
​
Finally, we concluded an addition expression and the solution is 9

Page 207  Exercise 9  Problem  14

Given:

7 + (−11)

To add both numbers

If both numbers has a different sign, add the value and put the greatest value sign
​
​7 + (−11)

=  −4
​
Finally, we concluded the addition expression and the solution is −4

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 207  Exercise 10  Problem  15

Given:

$152 − $20 + $84

To find the sum and explain its meaning

If both numbers has different sign, add the value and put the greatest value sign

​$152 − $20 = $132

=  $132 + $84

=  $216
​
Finally, we concluded an additional expression to represent this situation is  $216

Page 208   Exercise 12  Problem  16

Given:

The given transactions are

Week one  $300

Week two $50

Week three $75

Week four $225

To find the sum and explain its meaning

The withdrawal represents negative (-) and the deposit represents positive (+).

So add the given values, we get

​$300 + (−$50) + (−$75) + $225

= −$125+$525

= $400
​
The total sum using the addition expression is $400

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 208  Exercise 14   Problem  17

The given equation x + (-x) =0 states the property of additive inverse because it has the sum of the number with opposite sides zero

The rule is to change the positive number to a negative number

Finally, we conclude the property as additive inverse property because it has the number wit opposite sides zero.

The given equations x + (-y) = −y + x states the property of commutative, because it allows you to interchange the numbers in a sum

This law simply states that with addition is commutative

Finally, we concluded the property is Commutative property because its interchanges the number in sums.

Step-by-step guide for Exercise 3.2 Chapter 3 Integers in Glencoe Math Course 2 Volume 1 Page 208  Exercise 17 Problem  18

Given:

−9 + m + (−6)

To simplify

Given equation is
​
​−9 + m + (−6)

= −9 + (−6) + m

=(−9 + (−6)) + m

= −15 + m

−9 + m + (−6) = −15 + m

​Finally, we concluded an addition expression to solve the sum and the solution is −15 + m

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 208  Exercise 18  Problem  19

The explanation for correct answer

(A) − 4 + 3

This is the correct answer because the blue line passing on the negative side and stop at −4 and the red line passing on the positive side and stop at 3

(B)−4 + 7

The blue line passing on the negative side and stop at −4 and the red line passing on the positive side and stop at 3, not at 7, so this is the wrong answer

(C) 3 + (−7)

The blue line passing on the negative side and stop at−4, not at 3 and the red line passing on the positive side and stop at 3, not at −7, so this is the wrong answer

(D) 0 + (−7)

The blue line passing on the negative side and stop at −4 not at 0 and the red line passing on the positive side and stop at  3, not at −7, so this is the wrong answer

Finally, we concluded that  (A) −4 + 3 is the correct expression represented by the number line.

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 209   Exercise 19  Problem  20

Given:

18 + (−5)

To add the given value

​18 + (−5)

= 18 + (−5) = 18 − 5

= 13

18 + (−5) = 13
​
Add both the number and the greatest number sign in the result

Finally, we concluded an addition expression to solve the sum and the solution is 13

Page 209   Exercise 20  Problem  21

Given:

−19 + 24

To add the given value

​−19 + 24

= −19 + 24 = 24 + (−19)

= 24 −19

= 5

−19 + 24 = 5
​
Add both the number and the greatest number sign in the result

Finally, we conclude the solution using addition expression and the solution is 5

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 209  Exercise 23  Problem  22

Given:

15 + 9 + (−9)

To add the given value

​15 + 9 + (−9)

= 24 + (−9)

= 24 − 9

= 15

15 + 9 + (−9) = 15
​
Add both the number and the greatest number sign in the result

Finally, we conclude the solution using addition expression and the solution is 15

Page 209  Exercise 24  Problem  23

Given:

−4 + 12 + (−9)

To add the given value

​−4 + 12 + (−9)

=  8 + (−9)

=  8 − 9

= −1

​−4 + 12 + (−9) = −1
​
Add both the number and the greatest number sign in the result

Finally, we conclude the solution using an addition expression and the solution is −1

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 209  Exercise 26  Problem  24

Given:

25 + 3 + (−25)

To add the given value

​25 + 3 + (−25)

=  28 + (−25)

=  28 − 25

=  3

​25 + 3 + (−25) =  3
​
Add both the number and the greatest number sign in the result

Finally, we conclude the solution using addition expression and the solution is 3

Page 209   Exercise 27  Problem  25

Given:

7 + (−19) + (−7)

To add the given value

​7 + (−19) + (−7)

= 7 − 19 − 7

= −12−7

= −19

7 + (−19) + (−7) = −19
​
Add both the number and the greatest number sign in the result

Finally, we conclude the solution using an addition expression and the solution is −19

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 209  Exercise 29  Problem  26

Given:

A quarterback is sacked for a loss of 5 yards

On the next day’s play, his team losses 15 yards.

Then the team gain 12 yards on the third play

Write an additional expression to describe each situation

The gain represents the positive numbers

The losses represent negative numbers

So it can be written as −5 + (−15) + 12

​−5 + (−15) + 12

= −20 + 12

= −8

​−5 + (−15) + 12 = −8
​
Add both the number and the greatest number sign in the result

Finally, we concluded that an addition expression to describe each situation is −5 + (−15) + 12

Page 210   Exercise 31  Problem  27

Given: Temperatures at 8 A.M and 1 P.M

To find – Temperature at 10 P.M

The temperature at  8 A.M was 3°F

The temperature rose at 1 A.M was 14°F

The temperature drops at 10 P.M were

The increase is denoted by a positive number and below zero or drop is denoted by a negative number.

From given

​−3 + 14 + (−12)

=  11 + (−12)

= −1

​−3 + 14 + (−12) = −1

​Thus, the temperature is 1°F  below zero.

The temperature at  10 P.M is 1°F below zero 

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 210  Exercise 32  Problem  28

Given: −8 + 7 + (−3)

To find -The value

Given that −8 + 7 + (−3)

The answer is  (−4)

Explanation:

−8 + 7 + (−3)

​=−1 + (−3)

= −4

−8 + 7 + (−3) = −4
​
The value −8 + 7 + (−3) of is −4 

Page 210  Exercise 30  Problem  29

Given: A bank deposit of $ 75

To find-  Integer format for a given situation.

A bank deposit of $ 75

Explanation:

A bank deposit will increase the money that is in your bank account and thus it is represented by a positive number.

Thus, the integer representation of this situation is 75

The integer representation for bank deposits is 75.

Page 210  Exercise 36  Problem  30

Given:  13°F below zero

To find –  Integer format for a given situation.

Given 13°F below zero.

Explanation:

The temperature below zero is represented by a negative number.

Thus, the integer representation of this situation is = 13°

The integer representation for temperature below zero is −13°

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 210   Exercise 37  Problem  31

Given: A gain of 4 yards

To find- Integer format for a given situation.

A gain of 4 yards

Explanation:

The gain will increase the number that you have and hence it is represented by a positive number.

Thus, the integer representation of this situation is 4.

The integer representation for a gain of yards is 4

Glencoe Math Course 2 Volume 1 Common Core Chapter 3 Page 210  Exercise 38  Problem  32

Given:  Spending of $12.

To find-  Integer format for a given situation.

Spending of $12.

Explanation:

Spending money will decrease the amount that you have and hence it is represented by a negative number.

Thus, the integer representation of this situation is −12.

The integer representation for spending of $12. is  −12

The post Glencoe Math Course 2 Volume 1 Common Core Student Edition Chapter 3 Integers Exercise 3.2 appeared first on Answer Key for Math.

Glencoe Math Course 2 Volume 1 Chapter 4 Rational Numbers Exercise Solutions Page 257  Exercise 1  Problem 1

When we add fractions with the same denominator, we add only the numerators.

To add fractions with unlike denominators, rename the fractions with a common denominator by finding the “Least common multiple”(LCM)

For example:  \(\frac{1}{4}\)+\(\frac{2}{4}\)=\(\frac{3}{4}\)

To subtract fractions with like denominators, subtract the numerators, and write the difference over the denominator.

To subtract fractions with unlike denominators, rename the fractions with a common denominator by finding the “Least common multiple”(LCM)

For example:  \(\frac{4}{2}\)–\(\frac{1}{2}\)=\(\frac{3}{2}\)

The numerators of both fractions are to be multiplied first, followed by the multiplication of the denominators.

Then, the resultant fraction is simplified to its lowest terms, if needed.

For example: \(\frac{1}{2}\)×\(\frac{1}{5}\)=\(\frac{1}{10}\)

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.

The first step to dividing fractions is to find the reciprocal (Reverse the numerator and denominator) of the second fraction.

Next, multiply the two numerators. Then, multiply the two denominators.

For example:  \(\frac{\frac{1}{2}}{\frac{3}{2}}\)=\(\frac{1}{2}\)×\(\frac{2}{3}\)

=  \(\frac{1}{3}\)

When we add, subtract, multiply, or divide fractions, a new fraction is obtained.

Common Core Chapter 4 Rational Numbers Exercise Answers Glencoe Math Course 2 Page 260 Exercise 1 Problem 2

Given: Here it is

∴ \(\frac{24}{36}\)

To find- Write each fraction in simplest form.

Here it is given that

\(\frac{24}{36}\) ÷ 12 = \(\frac{2}{3}\)

= \(\frac{2}{3}\)

∴ \(\frac{24}{36}\) = \(\frac{2}{3}\)

Therefore, The simplest form of the fraction is \(\frac{24}{36}\) = \(\frac{2}{3}\)

Step-By-Step Solutions For Chapter 4 Rational Numbers Exercises In Glencoe Math Course 2 Page 260  Exercise 2  Problem 3

Given: \(\frac{45}{50}\)

To find- Write each fraction in simplest form.

∴ \(\frac{45}{50}\)

Find the Greatest Common Factor (GCF) of 45 and 50, if it exists, and reduce our fraction by dividing both the numerator and denominator by it.

GCF = 5, and getting our simplified answer

\(\frac{45}{50}\) ÷ 5 = \(\frac{9}{10}\)

=  \(\frac{9}{10}\)

Therefore, The simplest form of the fraction is\(\frac{45}{50}\) =  \(\frac{9}{10}\)

Exercise Solutions For Chapter 4 Rational Numbers Glencoe Math Course 2 Volume 1 Page 260  Exercise 3  Problem 4

Given:

To find –  Write each fraction in simplest form.

⇒ \(\frac{88}{121}\)

Find the Greatest Common Factor (GCF) of 88 and 121, if it exists, and reduce our fraction by dividing both the numerator and denominator by it.

GCF = 11 and getting our simplified answer

\(\frac{88}{121}\)÷11= \(\frac{8}{11}\)

= \(\frac{8}{11}\)

Therefore, The simplest form of the fraction is  \(\frac{88}{121}\) = \(\frac{8}{11}\)

Examples Of Problems From Chapter 4 Rational Numbers Exercises In Glencoe Math Course 2 Page 260 Exercise 4 Problem 5

Given:

Graphing graph each fraction or mixed number on the number line below. \(\frac{1}{2}\)

To graph the fraction.

To find – The two whole numbers between which

0<\(\frac{1}{2}\)<1

Since the denominator is 2, divide each space into 2 sections.

Draw a dot at \(\frac{1}{2}\)

Finally, we conclude that the graph has been plotted.

Student Edition Glencoe Math Course 2 Chapter 4 Rational Numbers Exercise Solutions Guide Page 260  Exercise 5  Problem  6

Given:

Graphing graph each fraction or mixed number on the number line below.\(\frac{3}{4}\)

To graph the fraction.

To find – The two whole numbers between which line below.\(\frac{3}{4}\)

0<\(\frac{3}{4}\)<1

Since the denominator is 4, divide each space into 4 sections.

Draw a dot at \(\frac{3}{4}\)

The graph for the given fraction \(\frac{3}{4}\) is

Step-By-Step Guide For Chapter 4 Rational Numbers Exercises In Glencoe Math Course 2 Volume 1 Page 260  Exercise 6  Problem  7

Given:

Graphing graph each fraction or mixed number on the number line below.

1\(\frac{1}{4}\)

To graph the fraction.

To find – The two whole numbers between which 1\(\frac{1}{4}\)

1<1\(\frac{1}{4}\)<2

Since the denominator is 4, divide each space into 4 sections.

Draw a dot at 1\(\frac{1}{4}\)

Finally, we conclude that the graph has been plotted.

Page 260  Exercise 7  Problem  8

Given:

Graphing graph each fraction or mixed number on the number line below.

2\(\frac{1}{2}\)

To graph the fraction

To find –  The two whole numbers between which 2\(\frac{1}{2}\)

2<2\(\frac{1}{2}\)<3

Since the denominator is 2, divide each space into 2 sections.

Draw a dot at 2 \(\frac{1}{2}\)

Finally, we conclude that the graph has been plotted.

Page 262   Exercise 1  Problem  9

Given:

Graph each fraction on a number line. Use a bar diagram if needed. \(\frac{3}{8}\)

To graph the fraction.

To find – The two whole numbers between which −\(\frac{3}{8}\) lies

−1<−\(\frac{3}{8}\)<0

Since the denominator is  8, divide each space into 8 sections.

Draw a dot at −\(\frac{3}{8}\)

Finally, we conclude that the graph has been plotted.

Page 262  Exercise 2  Problem  10

Given:

Graph each fraction on a number line. Use a bar diagram if needed.

−1\(\frac{2}{5}\)

Analyze the given and then graph the fraction.

To find – The two whole numbers between which −1\(\frac{2}{5}\) lies

−2<−1\(\frac{2}{5}\)<−1

Since the denominator is 5, divide each space into 5 sections.

Draw a dot at −1\(\frac{2}{5}\)

The graph for −1\(\frac{2}{5}\) has been plotted:

Page 262  Exercise 3  Problem  11

Given:

Use the number line and complete with < or >,

Plot the given fractions and Analyze the number line to complete the table

To find-  The greater one, first, we have to check whether the denominators are the same.

\(\frac{9}{8}\) and \(\frac{5}{8}\)

Both fractions have the same denominator.

Then check the signs.

In this case, both numbers are positive.

When comparing positive numbers, the larger number is greater.

Hence, it is clear that \(\frac{9}{8}\)> \(\frac{5}{8}\)

The solution is  \(\frac{9}{8}\)> \(\frac{5}{8}\)

Page 262   Exercise 4  Problem  12

Given:

Work with a partner to complete each table. Use the number if needed.

\(\frac{13}{8}\)>\(\frac{3}{8}\)

To complete with < or >.

Hence, it is clear that \(\frac{13}{8}\)>\(\frac{3}{8}\)

Finally, we conclude that the solution is \(\frac{13}{8}\)>\(\frac{3}{8}\)

Page 262  Exercise 5  Problem  13

Given:

Work with a partner to complete each table .Use a number if needed.

\(\frac{15}{8}\)>\(\frac{13}{8}\)

To complete with < or >

Hence, it is clear that  \(\frac{15}{8}\)>\(\frac{13}{8}\)

Finally, we conclude that the solution is  \(\frac{15}{8}\)>\(\frac{13}{8}\)

Page 262   Exercise 6  Problem  14

Given:

Use the number line and complete with < or >.

Plot the given fractions and To find the greater one, first, we have to check whether the denominators are the same.

If we have both denominators.

Then check the signs.

In this case, both numbers are negative.

When comparing negative numbers, the larger number farther from zero is less.

Here −9 is farther from zero. It is less.

Hence, it is clear that −\(\frac{9}{8}\)<−\(\frac{5}{8}\)

The solution is −\(\frac{9}{8}\) <− \(\frac{5}{8}\)

Page 262  Exercise 9 Problem  15

Given:

Identify repeated reasoning compare and contrast the information in the tables.

To compare and contrast

From the table, we can understand that the positive numbers are plotted on the right of the zero on a number line and the negative numbers are plotted on the left side of the zero on a number line.

Also, the positive number values are going on increasing

=  \(\frac{7}{8}\)<\(\frac{9}{8}\)

The negative number values are goes on decreasing,  ⇒ −\(\frac{7}{8}\)>\(\frac{9}{8}\)

Finally, the information in the table has been explained.

Page 262  Exercise 10  Problem  16

Given:

Reason inductively how does graphing −\(\frac{3}{4}\) differ from graphing \(\frac{3}{4}\)

We know that, that the positive numbers are plotted on the right of the zero on a number line and the negative numbers are plotted on the left side of the zero on a number line.

Hence,-\(\frac{3}{4}\)is to be plotted on the left of the zero on a number line.

\(\frac{3}{4}\) is to be plotted on the right of the zero on a number line.

Finally, we can conclude that − \(\frac{3}{4}\) is to be plotted on the left of the zero on a number line. \(\frac{3}{4}\) is to be plotted on the right of the zero on a number line, this is the difference in their graphing.

Page 262  Exercise 11  Problem 17

Given:

How can you graph the negative fractions on the number line? To explain.

The negative numbers are plotted on the left side of the zero on a number line instead of moving right.

For example: To plot −\(\frac{7}{8}\)

Finally, we can conclude that the negative numbers are plotted on the left side of the zero on a number line.

The post Glencoe Math Course 2 Volume 1 Common Core Student Edition Chapter 4 Rational Numbers Exercise appeared first on Answer Key for Math.

Glencoe Math Course 2 Volume 1 Chapter 4 Exercise 4.1 Solutions Page 264 Exercise 1 Problem 1

Given:

To find –  Write each fraction or mixed number as a decimal.

We know that

Use place value to write the equivalent decimal.

\(\frac{3}{10}\) = 0.3

So,\(\frac{3}{10}\) = 0.3

As a decimal, Each fraction or mixed number is \(\frac{3}{10}\) = 0.3 

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Given:

To find-  Write each fraction or mixed number as a decimal.

We know that

Use place value to write the equivalent decimal.

\(\frac{3}{25}\)\(=\frac{3 \times 4}{25 \times 4}\)

⇒  \(\frac{12}{100}\)

⇒  0.12

So, \(\frac{3}{25}\) = 0.12

As a decimal, Each fraction or mixed number is  \(\frac{3}{25}\) = 0.12

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Rational Numbers

Given:

− 6\(\frac{1}{2}\)

To find- Write each fraction or mixed number as a decimal.

We know that

−6\(\frac{1}{2}\)

−6\(\frac{1}{2}\) = −6+ \(\frac{1}{2}\)

⇒ −6 + 0.5

⇒ −5.5

So, -6\(\frac{1}{2}\) =−5.5

As a decimal, Each fraction or mixed number is −6\(\frac{1}{2}\) =−5.5

Given:

−\(\frac{7}{8}\)

To find- Write each fraction or mixed number as a decimal.

We know that −\(\frac{7}{8}\)

Then using long division for 7 divided by 8 and rounding Decimal Places gives us −1.142

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Rational Numbers

Given:

2\(\frac{1}{8}\)

To find- Write each fraction or mixed number as a decimal.

We know that

2\(\frac{1}{8}\)

= 2 + \(\frac{1}{8}\)

= 2 +0.125

= 2.125

2\(\frac{1}{8}\) = 2.125

Then using long division for  2\(\frac{1}{8}\) and rounding Decimal Places gives us 2.125

Given:

− \(\frac{3}{11}\)

To find- Write each fraction or mixed number as a decimal.

We know that

−\(\frac{3}{11}\)

So, −\(\frac{3}{11}\) = 0.273

Then using long division for –\(\frac{3}{11}\) and rounding Decimal Places gives us 0.273.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Rational Numbers

Given:

8\(\frac{1}{3}\)

To find- Write each fraction or mixed number as a decimal.

We know that

8\(\frac{1}{3}\)

=  8 + \(\frac{1}{3}\)

= 8 + 0.333 = 8.333

8\(\frac{1}{3}\) = 8.333

Then using long division for  8\(\frac{1}{3}\) and rounding Decimal Places gives us 8.333

Given: Molly 0.2.

To find-  Write in simplest form

We know that

0.2

0.2 = \(\frac{2}{10}\)

= \(\frac{2}{10}\)

= \(\frac{1}{5}\)

So, \(\frac{1}{5}\) of the fish are Molly

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Rational Numbers

Given: Guppy 0.25

To find-  Write in simplest form

We know that

0.25

0.25 = \(\frac{25}{100}\)

= \(\frac{1}{4}\)

So , \(\frac{1}{4}\) of the fish are Guppy

Given:

Divide 0.4 by 10 as it is in tenth place, then write in simplest form.

We know that

0.4

0.4 = \(\frac{4}{10}\)

= \(\frac{2}{5}\)

The fraction of the aquarium made up by Angelfish is \(\frac{2}{5}\)

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 266  Exercise 1   Problem 2

Given: \(\frac{2}{5}\)

To find- Write each fraction or mixed number as a decimal.

We know that

\(\frac{2}{5}\)

Then using long division for \(\frac{2}{5}\) and rounding Decimal Places gives us 0.4.

Step-By-Step Guide For Exercise 4.1 Chapter 4 Rational Numbers In Glencoe Math Course 2 Page 266  Exercise 2  Problem 3

Given: − \(\frac{9}{10}\)

To find:- Write each fraction or mixed number as a decimal.

We know that

−\(\frac{9}{10}\)

Then using long division for −\(\frac{9}{10}\) and rounding Decimal Places gives us −0.9.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 266  Exercise 3  Problem 4

Given:

To find –  Write each fraction or mixed number as a decimal.

We know that

\(\frac{5}{9}\)

Then using long division for \(\frac{5}{9}\)  and rounding Decimal Places gives us 0.556.

Exercise 4.1 Solutions For Chapter 4 Rational Numbers Glencoe Math Course 2 Volume 1 Page 266  Exercise 4  Problem  5

Given: During a hockey game, an ice resurfacer travels 0.75 miles.

To find – The fraction which represents this distance.

We know that

0.75

​0.75=\(\frac{75}{100}\)

So, 0.75 = \(\frac{3}{4}\)

Finally, we concluded  0.75 = \(\frac{3}{4}\) fraction represents this distance.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267  Exercise 1  Problem 6

Given: \(\frac{1}{2}\)

To find- Write each fraction or mixed number as a decimal.

We know that

So, \(\frac{1}{2}\) = 0.5

Then using long division for 1 divided by 2 and rounding Decimal Places gives us 0.5

Examples of problems from Exercise 4.1 Chapter 4 Rational Numbers in Glencoe Math Course 2 Page 267  Exercise 2  Problem 7

Given: − 4\(\frac{4}{25}\)=

​To find- Write each fraction or mixed number as a decimal.

We know that

−4\(\frac{4}{25}\)

−4\(\frac{4}{25}\) = −4+\(\frac{4}{25}\)

= −4 + 0.16

= − 4.16

So, −4\(\frac{4}{25}\) = − 4.16

Because we know that 25 equals 100 (think quarters to a dollar), converting this fraction to a decimal in the hundredth place will be simple. 4 Times 25 is multiplied by 100 (again, 4 quarters make a dollar).

This means we’d have to multiply 4 by 4 to get \(\frac{16}{100}\)

The decimal for \(\frac{16}{100}\) is 0.16 As a result, 4 equals  \(\frac{4}{25}\) − 4.16

Finally, The decimal for  \(\frac{16}{100}\)  As a result, 4 and equals −4.16.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267  Exercise 3  Problem 8

Given: \(\frac{1}{8}\)

To find- Write each fraction or mixed number as a decimal.

We know that

\(\frac{1}{8}\)

Then using long division for 1 divided by 8 and rounding Decimal Places gives us 0.125.

Common Core Exercise 4.1 Chapter 4 Rational Numbers detailed solutions Glencoe Math Course 2 Page 267  Exercise 4  Problem 9

Given:  \(\frac{3}{16}\)

To find- Write each fraction or mixed number as a decimal.

We know that

\(\frac{3}{16}\)

Then using long division for 3 divided by 16 and rounding Decimal Places gives us 0.188

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267  Exercise 5  Problem 10

Given: −\(\frac{33}{50}\)

To find-  Write each fraction or mixed number as a decimal.

We know that –\(\frac{33}{50}\)

Then using long division for −\(\frac{33}{50}\) and rounding Decimal Places gives us−0.66.

Student Edition Glencoe Math Course 2 Chapter 4 Rational Numbers Exercise 4.1 solutions guide Page 267  Exercise 6  Problem 11

Given: − \(\frac{17}{40}\)

To find- Write each fraction or mixed number as a decimal.

We know that −\(\frac{17}{40}\)

Then using long division for − \(\frac{17}{40}\) and rounding Decimal Places gives us 0.425

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267  Exercise 7  Problem 12

Given:  5\(\frac{7}{8}\)

To find- Write each fraction or mixed number as a decimal.

We know that

5\(\frac{7}{8}\)

Multiply the denominator by the whole number 8 × 5 = 40

Add the answer to the numerator 5\(\frac{7}{8}\)

40 + 7 = 47

Simplified solution

\(=\frac{8 \times 5+7}{8}\)= \(\frac{47}{8}\)

= 5.875

So,5\(\frac{7}{8}\)= 5.875

Then using long division for  5\(\frac{7}{8}\)  and rounding Decimal Places gives us 5.875.

Step-by-step answers for Exercise 4.1 Chapter 4 Rational Numbers in Glencoe Math Course 2 Volume 1 Page 267  Exercise 8  Problem 13

Given:  9\(\frac{3}{8}\)

To find- Write each fraction or mixed number as a decimal.

We know that

9\(\frac{3}{8}\)

9\(\frac{3}{8}\) = 9 + \(\frac{3}{8}\)

=  9.375

So, 9\(\frac{3}{8}\) = 9.37

Then using long division for  9\(\frac{3}{8}\) and rounding Decimal Places gives us 9.37.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267   Exercise 9  Problem 14

Given: −\(\frac{8}{9}\)

We know that

Then using long division for −\(\frac{8}{9}\) and rounding Decimal Places gives us−0.89.

Page 267   Exercise 10  Problem 15

Given:  − \(\frac{1}{6}\)

To find – Using long division write each fraction or mixed number as a decimal.

Given

−\(\frac{1}{6}\)

The decimal form of −\(\frac{1}{6}\) = − 0.1666

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267  Exercise 11 Problem 16

Given: −\(\frac{8}{11}\)

To find- Write each fraction or mixed number as a decimal.

We know that

Then using long division for −\(\frac{8}{11}\)and rounding Decimal Places gives us −0.72.

Page 267  Exercise 12  Problem 17

Given:  2\(\frac{6}{11}\)

To find – Write each fraction or mixed number as a decimal.

We know that

2\(\frac{6}{11}\)

2\(\frac{6}{11}\) = 2 +  \(\frac{6}{11}\)

=  2 + 0.5454

=  2.5454

So, 2\(\frac{6}{11}\) = 2.545

Then using long division for  2\(\frac{6}{11}\)  and rounding Decimal Places gives us 2.545.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267  Exercise 13  Problem 18

Given:−0.2

To find- Write each decimal as a fraction or mixed number in simplest form.

We know that

−0.2

Remove the negative sign from the positive decimal value, convert it to a positive fraction, and then apply the negative sign to the fraction response.

Rewrite the decimal number as a fraction with 1  in the denominator

0.2 = \(\frac{0.2}{1}\)

Multiply to remove 1 decimal place. Here, you multiply top and bottom by 10¹ = 10

\(\frac{0.2}{1}\) × \(\frac{10}{10}\)

=  \(\frac{2}{10}\)

Find the Greatest Common Factor (GCF) of 2 and 10, if it exists, and reduce the fraction by dividing both the numerator and denominator by GCF = 2

Here, we concluded the mixed fraction in simplest form is −0.2 = − \(\frac{1}{5}\)

Page 267  Exercise 14  Problem 19

Given: 0.55

To find- Write each decimal as a fraction or mixed number in simplest form.

We know that0.55 Rewrite the decimal number as a fraction within the denominator

0.55  = \(\frac{0.55}{1}\)

Multiply to remove 2 decimal places. Here, you multiply the top and bottom by 10²

= 1000\(\frac{0.55}{1}\)× \(\frac{100}{100}\) =  \(\frac{55}{100}\)

Find the Greatest Common Factor (GCF) of 55 and 100, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 5

Here, we concluded the mixed fraction in simplest form is 0.55 =\(\frac{11}{20}\)

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 267  Exercise 15  Problem 20

Given: 5.96

To find- Write each decimal as a fraction or mixed number in simplest form.

We know that

5.96

Rewrite the decimal number as a fraction with 1 in the denominator

5.96 = \(\frac{5.96}{1}\)

Multiply to remove 2 decimal places. Here, you multiply top and bottom by 10² = 100\(\frac{5.96}{1}\)×\(\frac{100}{100}\)=\(\frac{596}{100}\)

Find the Greatest Common Factor (GCF) of 596 and 100, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 4

Here, we concluded the mixed fraction in simplest form is 5.96 = 5\(\frac{24}{25}\)

Page 267   Exercise 17   Problem 21

Given: A Praying mantis is an interesting insect that can rotate its head 180 degrees.

Suppose the praying mantis at the right is 10.5 centimeters long.

To find- The mixed number that represents this length.

We know that

Now think about the length you’ve been given 10.5

10.5 = 10 + 0.5

Because 10 is an integer, all we have to do now is convert 0.5 to fractional form to get a mixed number.

0.5 = \(\frac{5}{10}\)=\(\frac{1}{2}\)

Thus, the number is

10 + 0.5 = 10 + \(\frac{1}{2}\)

⇒ 10 \(\frac{1}{2}\)

As a result, the needed mixed number is 10\(\frac{1}{2}\).

Finally,  10\(\frac{1}{2}\)mixed number represents this length.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 268  Exercise 18  Problem 22

Given: Suppose you buy a 1.25− pound package of ham at $5.20 per pound. Find the fraction of the pound bought that is find the portion purchased

⇒ \(\frac{1.25}{1}\)

⇒ \(\frac{125}{100}\)

⇒ \(\frac{5}{4}\)

Finally, \(\frac{5}{4}\) fraction of a pound did you buy.

Given: Suppose you buy a 1.25− pound package of ham at $5.20 per pound.

To find – How much money did you spend?

We know that

The amount of ham purchased in pounds = 1.25

We have a Ham of 1 pound = ​​$​​5.20

The amount spent on ham = The fraction of a pound bought × Price per pound
​
⇒   \(\frac{5}{4}\) × 5.20

⇒  5 × 1.3

⇒   ​​$​​6.5.

​Finally, ​​$​​6.5 amount is spend on ham.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 268  Exercise 19  Problem 23

Given: Write a fraction that is equivalent to a terminating decimal between 0.5and0.75.

To find-  Write a fraction

We have Because both 0.5 and 0.75 are at two places after decimals, we know they are terminating.

Finding the average of a number that is between these two can be done by adding it and then dividing by two.

Adding both value =0.5 + 0.75 = 1.25

When you divide it by two, you get = 1.25  by 2 = 0.625

When we convert it to a fraction, we obtain

​⇒ \(\frac{0.625}{1}\)

​⇒ \(\frac{625}{1000}\)

​⇒ \(\frac{5}{8}\)

Finally, \(\frac{5}{8}\) is the terminating decimal.

Page 268  Exercise 20  Problem 24

Given Fractions in the simplest form that have denominators of2,4,8,16 and 32produce terminating decimals.

Fractions with denominators of  6,12,18, and 24 produce repeating decimals.

To find – The causes of difference.

As you can see, the denominator in  2,4,8,16,32  is of the kind  2¹,2²,2³,2⁴,2⁵.  As a result, the decimal comes to an end.

Consider fractions with denominators of 6,12,18,24.

Now, among all of these
​
6 = 2.3

12 = 2.2.3

18 = 2.3.3

24 = 2.2.2.3
​
All of these integers’ prime factors include a factor other than 2, namely 3.

As previously stated, if the denominator is not in the form of  2^m or 5ⁿ or 2^m⋅5 the decimal is non-terminating.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 268  Exercise 21  Problem 25

Given The value of pi (π)  is 3.1415926…. The mathematician Archimedes believed that π was between 3 \(\frac{1}{7}\) and 3\(\frac{10}{71}\)

Convert the mixed fraction to improper fraction and solve further

Then check whether Archimedes is correct

We know that

π = 3.1415927

3 \(\frac{1}{7}\)

3 \(\frac{1}{7}\) = 3 + \(\frac{1}{7}\)

We know that

Is the same as 1 ÷ 7

Therefore, 3\(\frac{1}{7}\) = 3 + (1÷7)

3 + 0.143 = 3.143

3 \(\frac{10}{71}\)= 3 + \(\frac{10}{71}\)

We know that\(\frac{10}{71}\) Is the same as 10 ÷ 71

Then

3 + \(\frac{10}{71}\)= 3+(10÷71)

3 + 0.141 = 3.141

π = 3.1415927

π value has been rounded to seven decimal digits.

3\(\frac{1}{7}\)  = 3.1428571

Compare these numbers to ensure that pi is contained within the mixed fractions.

3\(\frac{10}{71}\) = 3.1408451

= 3.1408451

It is in this instance.

Finally, we concluded the Archimedes’ statement is correct.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 268  Exercise 22  Problem 26

Given:

Tanya drew a model for the fraction\(\frac{4}{6}\)

Which of the following decimals is equal to \(\frac{4}{6}\)

​A. 0.666

B. 0.6

C. 0.667

D. 0.66777

To find – The decimals.

\(\frac{4}{6}\) = 0.666

Finally, we can conclude that the answer is options A and B.

Page 269   Exercise 23  Problem 27

Given:  \(\frac{4}{5}\)

To find-  Write each decimal as a fraction or mixed number in simplest form.

We know that

We have the equation  4÷5 = 0.80

Then using long division for 4 divided by 5 and rounding Decimal Places gives us 0.80.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 269  Exercise 25  Problem 28

Given: − \(\frac{4}{9}\)

To find− Using long division write each fraction or mixed number as a decimal.

We know that

The decimal form of  − \(\frac{4}{9}\) is − 0.4444

Page 269  Exercise 26   Problem 29

Given:  5\(\frac{1}{3}\)

To find –  Write each decimal as a fraction or mixed number in simplest form.

We know that

Multiply the denominator by the whole number 3 × 5 = 15

Add the answer to the numerator 15 + 1 = 16

Write the answer over the denominator  = \(\frac{16}{3}\)

Simplified Solution

⇒  \(\frac{3×5+1}{3}\) = \(\frac{16}{3}\)

⇒   5.33

Then using long division for   5\(\frac{1}{3}\) rounding Decimal Places gives us 5.33.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 269  Exercise 27  Problem 30

Given: The fraction of a dime that is made up of copper is \(\frac{12}{16}\)

To find- Write this fraction as a decimal

We know that

We have the equation 16 ÷ 12 = 0.750

Then using long division for  \(\frac{12}{16}\)  ,rounding Decimal Places gives us 0.750.

Page 269  Exercise 28  Problem 31

Given:

−0.9

To find- Decimal to a fraction or mixed fraction

Here −0.9

Rewrite the decimal number as a fraction with1 in the denominator

0.9  =  \(\frac{0.9}{1}\)

Multiply to remove 1 decimal place. Here, you multiply top and bottom by 10¹ = 10

\(\frac{0.9}{1}\)×\(\frac{10}{10}\) = \(\frac{9}{10}\)

⇒ −0.9 = −\(\frac{9}{10}\)

Finally, we concluded the value in decimal to fraction −0.9  = −\(\frac{9}{10}\)

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 269  Exercise 29  Problem 32

Given:

0.34

To find- Decimal to a fraction or mixed fraction

​Here it is given that  0.34

Rewrite the decimal number as a fraction with 1 in the denominator

0.34 = \(\frac{0.34}{1}\)

Multiply to remove 2 decimal places. Here, you multiply top and bottom by  10² = 100

\(\frac{0.34}{1}\) × \(\frac{100}{100}\)

=\(\frac{34}{100}\)

Find the Greatest Common Factor (GCF) of 34 and 100, if it exists, reduce the fraction by dividing both numerator and denominator by GCF = 2

Finally, we concluded the value in decimal to fraction   0.34 = \(\frac{17}{50}\)

Page 269   Exercise 30  Problem 33

Given:

2.66

To find- Decimal to a fraction or mixed fraction

Here it is given that

2.66

Rewrite the decimal number as a fraction with 1 in the denominator

2.66 = \(\frac{2.66}{1}\)

Multiply to remove 2 decimal places. Here, you multiply the top and bottom by 10²

\(\frac{2.66}{1}\)×\(\frac{100}{100}\)

=  \(\frac{266}{100}\)

Find the Greatest Common Factor (GCF) of 266 and 100, if it exists, reduce the fraction by dividing both numerator and denominator by GCF= 2

Finally, we concluded the value in decimal to fraction  2.66= 2 \(\frac{33}{50}\)

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 269  Exercise 31  Problem 34

Here an integer is given to us.

−13

We have to convert this into an improper fraction.

Any natural number which has to be converted into a fraction we divided by 1. So now −13 is converted −\(\frac{13}{1}\)

Therefore,  −\(\frac{13}{1}\)is the final answer.

Page 269   Exercise 32   Problem 35

We are given a mixed fraction.

7 \(\frac{1}{3}\)

We have to convert it into an improper fraction.

To convert 7\(\frac{1}{3}\)into an improper fraction

We multiply 7 with 3 and add 1 to the product.

​(7 × 3) + 1 = 22
​
Therefore, 22 is the numerator.

So, the improper fraction is  \(\frac{22}{3}\)

Finally, we conclude the value in an improper fraction  \(\frac{22}{3}\)

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 269  Exercise 33  Problem 36

We are given a negative decimal value

−3.2.

We have to convert it into a negative improper fraction.

Take the decimal −3.2.

Multiply and divide the decimal by 10.

Finally, we conclude, the value of the final answer is − \(\frac{16}{5}\)

Page 269  Exercise 34  Problem 37

Here we are given the time in hours and minutes.

We have to convert it into a decimal.

We are given Nicholas’ time playing the cello as 2 hours and 18 minutes.

First, we convert hours into minutes by multiplying by 60.

2 × 60  = 120 minutes.

Now adding it with the 18-minute

We get 138 minutes.

Now dividing by 60

​⇒  \(\frac{138}{60}\)

​⇒  \(\frac{23}{10}\)

2.3 Hour

Nicholas has been playing the cello for 2.3 hours.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 270  Exercise 35  Problem 38

Given and Find:

We are given fraction and their recurring decimals.

We have to find out which fraction corresponds to 0.88888.

Take option A

1.333333 is not the required answer.

Take option B

0.808080 is not the required answer.

Take option C

0.83333 is not the required answer.

Take option D

0.8888 is the required answer.

The required answer is option D.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 270  Exercise 37  Problem 39

Given:

We are given Zoe’s total bill.

We have to find out which mixed fraction corresponds to the decimal given.

Solution:

We take that

12\(\frac{1}{20}\)

To convert it into improper fractions we multiply 20 with 12 and add 1 to the product.

The improper fraction:

12.05 is the required answer.

Therefore the correct answer is 12.05

Page 270   Exercise 38   Problem 40

Given:

We are given a decimal. 5.69

We have to convert it into the nearest tenths place.

We take the decimal 5.69

We look at 9, which is greater than 5.

So we increase the next number by  1.

Now the decimal is rounded off to 5.7

This has been rounded off to the tenths place.

The rounded-off decimal is 5.7

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 270   Exercise 39  Problem 41

Given:

We are given a decimal. 0.05

We have to convert it into the nearest tenths place.

We take the decimal  0.05

We look at 5, which is greater or equal than 5.

So we increase the next number by 1.

Now the decimal is rounded off to 0.1.

This has been rounded off to the tenths place.

The rounded-off decimal is 0.1.

Page 270  Exercise 40  Problem 42

Given:

We are given a decimal.

98.99

We have to convert it into the nearest tenths place.

We take the decimal 98.99

We look at 9, which is greater or equal than 5.

So we increase the next number by 1.

Now the decimal is rounded off to This has been rounded off to the tenths place 99.0.

The rounded-off decimal is 99.0.

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 270  Exercise 41   Problem 43

Given and Find:

We are given 3 fractions

We have to convert them into decimals and put them onto a number line.

We are given the fraction as \(\frac{1}{2}\)

0.5 is decimal.

Therefore we have shown it on the number line

Page 270  Exercise 42  Problem 44

Given and Find:

We are given 3 fractions

Solution:

We are given the fraction as \(\frac{3}{4}\)

0.75 is a decimal.

Plot these decimals on the number line.

Therefore we have shown it on the number line

Glencoe Math Course 2 Volume 1 Common Core Chapter 4 Page 270 Exercise 43  Problem 45

Given and Find:

We are given 3 fractions

We have to convert them into decimals and put them onto a number line.

We are given the fraction as \(\frac{2}{3}\)

0.66 is given as the fraction.

Plot these decimals on the number line.

Therefore we have shown it on the number line

The post Glencoe Math Course 2 Volume 1 Common Core Student Edition Chapter 4 Rational Numbers Exercise 4.1 appeared first on Answer Key for Math.

Glencoe Math Course 2 Volume 1 Chapter 2 Exercise 2.2 Solutions Page 111   Exercise 1  Problem 1

We need to explain how we can percent help you understand situations

The percentage helps to understand situations involving money

The interest rates are written as percent

Also, find the interest earned on a savings account and the amount of interest charged on bank loans and credit cards.

The sales tax is also indicated in percents.

Hence explained.

Page 111   Exercise 2    Problem 2

Given:

About how many people took lessons at school?

To find – The number of people took lessons at school.

Total number of people surveyed = 200

Number of people took lessons at school =  \(\frac{3}{10}\) 0f 200

\(\frac{3}{10}\)  ×   200 =  60

The number of people took lessons at school = 60

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 111   Exercise 3   Problem 3

Given:

The table shows the survey of 200 people who have learned to play the instrument in different ways.

Sarah estimates the percentage of people who are self-learned in fractions and in percentages.

To find- Compare the number with the actual number and give

To fill in the table of estimated percent and fraction with the actual percent:

Calculation of percentage:

40%

\(\frac{40}{100}\)= \(\frac{4}{10}\)

= \(\frac{2}{5}\)

30%

\(\frac{30}{100}\)= \(\frac{3}{10}\)

25%

\(\frac{25}{100}\)= \(\frac{1}{4}\)

It is less than the actual number.

This is because we rounding the estimated percent as 25 % from the actual percent 26 %.

So it will cause our estimate to be slightly lower than the actual.

It is less than the actual number. Because we are rounding the percentage, the actual percent becomes slightly less than the estimated percent.

Common Core Chapter 2 Percents Exercise 2.2 Answers Glencoe Math Course 2 Page 114   Exercise 1  Problem 4

Given: 52 % of 10 ≈

To find- Estimate the value

Determine the product by rounding the percentage to the nearest tenth:

52 % of 10 ≈ 50

=  \(\frac{1}{2}\)

= 5

Finally, The Value of the estimate is 5.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 114   Exercise 3   Problem 5

Given: 151 % of 70 ≈

To find- Estimate the value

Determine the product by rounding the percentage to the nearest tenth:

151 of 70 ≈ 150

​⇒  1.5 × 70

⇒ 105
​
Finally, The Value of the estimate is 105.

Page 114   Exercise 4   Problem 6

Given: \(\frac{1}{2}\)% of 82 ≈

To find- Estimate the value of the given problem.

Determine the product by rounding the percentage to the nearest tenth:

\(\frac{1}{2}\)% of 82

⇒  \(\frac{1}{2}\)%

= 0.5 %

To find 0.5% of 82

\(\frac{0.5}{100}\) × 82

​=  0.005  ×  82

=  0.41

≈0.4
​
Finally, The Value of the estimate is  \(\frac{1}{2}\)% of  82 ≈ 0.4

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 114  Exercise  5  Problem 7

Given: Of the 78 teenagers at a youth camp, 63 have birthdays in the spring.

To find-  How many teenagers have birthdays in the spring?

Determine the product by rounding the percentage to the nearest tenth:

63 of 78 ≈  60

​⇒  0.6 × 78

⇒   46.8 ≈ 47
​
As a result, approximately c

Finally, We conclude 47 teenagers celebrate their birthdays in the spring.

Step-By-Step Guide For Exercise 2.2 Chapter 2 Percents In Glencoe Math Course 2 Page 114   Exercise 6   Problem 8

Given: About 0.8 of the land in Maine is federally owned. If Maine has 19,847,680 acres, about how many acres are federally owned? (Example 5)

To find- How many acres are federally owned?

Determine the product by rounding the percentage to the nearest tenth:

0.8% of 19,847,680 ≈

\(\frac{0.8}{100}\) × 19847680

0.008 × 19847680

= 158781.44

​Finally, As a result, the feds own approximately 158781 acres.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 114   Exercise 7   Problem 9

Given: Estimation of percentage of a number.

Common Method:

Use percent formulas to figure out percentages and unknowns in equations.

Add or subtract a percentage from a number or solve the equations.

There are many formulas for percentage problems. You can think of the most basic as X/Y = P × 100.

The formulas below are all mathematical variations of this formula.

Let’s explore the three basic percentage problems. X and Y are  number and P is the percentage:

1. Find P percent of  X

2. Find what percent of X is Y.

Example: What is 10% of 150?

Convert the problem to an equation using the percentage formula:

P is 10%, and X is 150, so the equation is 10% × 150 = Y

Convert 10% to a decimal by removing the percent sign and dividing by 100:10/100 = 0.10

Substitute 0.10 for 10% in the equation: 10% × 150 = Y becomes 0.10 × 150 = Y

Do the math: 0.10 × 150 = 15

Y = 15

So 10% of 150 is 15

Double-check your answer with the original question: What is 10% of 150? Multiply 0.10 × 150 = 15

In general, to find n percent of x, we follow these steps:

1. Dividend by 100.

2. Multiply the result by x.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115   Exercise 1  Problem 10

Given:

To convert percentage to a number 47% of 70

Given

47% of 70 ≈ 45

\(\frac{45}{100}\) × 70

=  (0.45)70

=  31.5
​
47% of 70 ≈ 45  = 31.5

The answer for 47 % of 70 ≈ 45 is  31.5

Exercise 2.2 Solutions For Chapter 2 Percents Glencoe Math Course 2 Volume 1 Page 115   Exercise 2  Problem 11

Given:

To convert percentage to a number 39 % of 120

Given

39% of 120 ≈ 40

\(\frac{40}{100}\) × 120

=  (0.40)(120)

=  48

39% of 120 ≈ 40 = 48

The answer for 39 % of 120 ≈ 40 = 48

Page 115  Exercise 3  Problem  12

Given:

To convert percentage to a number 21 % of 90

Given

21%  of  90 ≈ 20

\(\frac{20}{100}\) × 90

=  (0.2)90

=  18
​
21% of 90 ≈ 20 = 18

The answer for 21 % of 90 ≈ 20  = 18

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115   Exercise 4  Problem  13

Given:

To convert percentage to a number 65 % of 152

Given

65 % of 152  ≈ 65

\(\frac{65}{100}\) × 150

=  (0.65)150

=  97.5
​
65 ≈ 65 = 97.5

The answer for 65 % of 152 = 97.5

Common Core Percents Exercise 2.2 Chapter 2 Solutions Glencoe Math Course 2 Page 115   Exercise 5  Problem  14

Given:

To convert percentage to a number 72 % of 238

Given

72 % of 238 ≈ 70

\(\frac{70}{100}\) × 238

=  (0.70)238

=  166.6
​
72 % of 238 ≈ 70 = 166.6

The answer 72 % of 238  = 166.6

Page 115  Exercise 6  Problem  15

Given:

To convert percentage to a number 132% of 54

Given

132 % of  54 ≈ 130

\(\frac{70}{100}\) × 54

= (1.3)54

= 70.2
​
132 % of 54 ≈130=70.2

The answer 132 % of 54 ≈ 130 = 70.2

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115  Exercise 8  Problem  16

Given:

To estimate \(\frac{3}{4}\) % of 168

Given

\(\frac{3}{4}\)% of 168

\(\frac{3}{4}\)% × 168

\(\frac{0.75}{100}\) × 168

=  0.0075 × 168

=  1.26

Therefore, the percentage \(\frac{3}{4}\) of 168 is 1.26.

\(\frac{3}{4}\)% of 168 ≈ 1.3

Examples of problems from Exercise 2.2 Chapter 2 Percents in Glencoe Math Course 2 Page 115   Exercise 9   Problem  17

Given:

To estimate 0.4 % of 510

Given

0.4% of 510

= \(\frac{0.4}{100}\)  ×  510

=  0.004  ×  510

=  2.04

The percentage of 0.4 of 510 is 2.04

Page 115 Exercise 10   Problem  18

Given:

The Financial Literacy Carlie spent $42 at the salon.

Her mother loaned her the money.

Carlie will pay her mother 15% of $42 each week until the loan is repaid.

About how much will Carlie pay each week?

The amount Carlie pay each week is 15 % of $42

=  \(\frac{15}{100}\) × 42

= 0.15 ×  42

= 6.3

Carlie will pay $6.3 amount each week to her mother until the loan is repaid.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 115  Exercise 11   Problem  19

Given:

The United States has 12,383 miles of coastline.

If 0.8 % of the coastline is located in Georgia, about how many miles of coastline are in Georgia?

Given

0.8 % of the coastline in Georgia and 12,383 miles of coastline in United States is
​
=  \(\frac{0.8}{100}\) × 12,383

=  0.008 × 12,383

=  99.064
​
Approximately 99 miles of coastlines are in Georgia.

Student Edition Glencoe Math Course 2 Chapter 2 Percents Exercise 2.2 Guide Page 116  Exercise 14  Problem  20

Given:

Estimate 54% of 76.8 =?

Given

54% of 76.8 = 50% of 76.8

=  \(\frac{50}{100}\) × 76.8

=  \(\frac{1}{2}\)×76.8

=  38.4

So,54%  of  76.8 is approximately 38.4

The percentage 54 % of 76.8is approximately 38.4

Page 116   Exercise 15  Problem  21

Given:

Estimate 10.5% of 238 =?

Given

10.5% of 238 = \(\frac{105}{1000}\) × 238

=  \(\frac{21}{200}\) × 238

=  \(\frac{21}{100}\) × 119

=  \(\frac{2499}{100}\)

≈ 24.99

So,10.5% of 238 is approximately 24

The percentage of 10.5% of 238 is approximately 24

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 116   Exercise 16  Problem  22

Given:

The average white rhinoceros gives birth to a single calf that weight about 3.8% as much as its mother rhinoceros weight 3.75 tons, about how many pounds does its calf weight?

Given

3.8% of 3.75 = \(\frac{3.8}{100}\) × 3.75

=  \(\frac{380}{10000}\) × 375

=  \(\frac{19}{500}\) × 375

=  \(\frac{19}{500}\) × 75

=  14.25t

≈ 0.145t (In pounds)

So, the baby animal weights in 0.145t

Step-by-step answers for Exercise 2.2 Chapter 2 Percents in Glencoe Math Course 2 Volume 1 Page 116  Exercise 18  Problem  23

Given:

Explain how you could find % of $800

By simplifying fraction to omit % symbol and multiplying the values.

\(\frac{3}{8}\)% of 800 =  \(\frac{3}{8}\)×800×\(\frac{1}{100}\)

=  3

\(\frac{3}{8}\)% of 800 =  3

The\(\frac{3}{8}\)% answer of  800 is 3.

Page 116   Exercise 19  Problem 24

Is an estimate for the percent of a number always sometimes or never greater than the actual percent of the number?

Give an example or a counterexample to support your answer

An estimate for the percent of a number is sometimes greater than the actual percent of the number

One estimate for 18% of 40 is, \(\frac{1}{5}\).40 = 8

While ,one estimate for 22% of 60 is ,\(\frac{1}{5}\).60 = 12

While never greater than the actual percent,50% of 30 is

\(\frac{1}{2}\).30 =  15

It is the example for the percent of a number sometimes, or never greater than the actual percent of the number

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 116   Exercise 20  Problem  25

Given:

Cost of bedroom furniture=$1,789.43

Percentage cost of dresser=39.7 of total cost

To find- Cost of the dresser?

Cost of dresser

​=  39.7% of $1,789.43

=   (\(\frac{40}{100}\) × 1,789.43) − (\(\frac{0.3}{100}\) × 1,789.43)

=  715.772 − 5.368

=  710.404 ≈ $720

​Hence, $720 is the best estimate for the cost of the dresser.

Page 117 Exercise 21  Problem 26

Given:

76%of 180 ≈ ?

To find- Evaluate the problem.

76% 180 = 75% of 180 + 1% of 180

=  (\(\frac{75}{100}\) × 180) + (\(\frac{1}{100}\) × 180)
​
=  135 + 1.8

= 136.8 ≈ 137

Therefore by evaluating the equation the percentage of   76% of 180 ≈ 137

Page 117   Exercise 22    Problem 27

Given:

57%of 29 ≈?

To find- Evaluate the problem.

57%  of 29

​=  (\(\frac{60}{100}\) × 29) − (\(\frac{3}{100}\) × 29 )

=  17.4 − 0.87

=  16.53 ≈ 17

​Therefore by evaluating the equation the percentage of 57% of  29 ≈ 17

​

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 117   Exercise 23   Problem  28

Given:

92%of 104 ≈ ?

To find- Evaluate the problem.

Let, 92% of 104

=  (\(\frac{100}{100} × 104\)) − (\(\frac{8}{100} × 104\))
​
=  104 − 8.32

=  95.68 ≈ 96
​
Therefore by evaluating the equation the percentage of 92 of 104 ≈ 96

Page 117   Exercise 25   Problem 29
​
Given:

0.9% of 74 ≈ ?

To find- Evaluate the problem.

0.9% = (1−0.1)%

74 × \(\frac{1-0.1}{100}\) = \(\frac{74×1}{100}\)– \(\frac{74×0.1}{100}\)

=  \(\frac{74}{100}\)–\(\frac{7.4}{100}\)

=  0.74 − 0.074

=  0.666

Therefore by evaluating the equation the percentage of 0.9 of 74 = 0.666

Page 117  Exercise 26  Problem  30

Given:

32% of 89.9 ≈ ?

To find- Evaluate the problem.

30%of89.9 = \(\frac{30}{100}\) ×  89.9

=  26.97

2% of 89.9  = \(\frac{2}{100}\) × 89.9

=  1.798

32 % of 89.9  =  26.97 + 1.798 = 28.768

Therefore by evaluating the equation the percentage of 32 % of 89.9 = 28.8

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 117   Exercise 27   Problem 31

Given:

Total muscles to frown  =  43

Percentage of muscles used to smile = 32%

To find- A number of muscles used to smile?

Number of muscles used to smile

=  32%of43

=   (\(\frac{30}{100}\)) × 43  + (\(\frac{32}{100}\)) × 43
​
=  12.9 + 0.86

=  13.76  ≈ 14

​Therefore,14 muscles are used when using a smile.

Page 117   Exercise 28   Problem  32 

Given:

Coastline of Atlantic coast = 2.069miles

Percentage of coastlines in New Hampshire = \(\frac{6}{10}\)%

To find- Length of coastlines in New Hampshire?

Length of coastlines in New Hampshire = \(\frac{6}{10}\)% of 2.069

\(\frac{0.6}{10}\)%  ×  2.069 =  0.0124

Therefore, the length of coastlines in New Hampshire is 0.0124 miles.

Page 118   Exercise 32   Problem  33

Given:

5n = 120

To find- The value of n =?

The value of n is
​
​5n = 120

n = 120/5

= 24
​
Therefore, the value of n is 24.

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 118   Exercise 33  Problem 34

Given:

1,200 = 4a

To find – Solve each equation show your work

The given equation is 1200 = 4a

Divide both sides by ‘4’, and we get

\(\frac{1200}{4}\) = \(\frac{44}{4}\)

a = 300

By solving the given equation we get a = 300

Page 118   Exercise 34   Problem 35

Given:

6x = 39

To find – Solve each equation show your work

The given equation is 6x = 39

Divide both sides by ‘6’, and we get
​
\(\frac{6x}{6}\) = \(\frac{39}{6}\)

x = 6.5
​
By solving the given equation we get x = 6.5

Page 118   Exercise 36  Problem  36

Given:\(\frac{3}{5}\)

To find – Write three fraction equation

By multiplying the numerator and the denominator by 2,3and 4we get

The equivalent fractional to \(\frac{3}{5}\) is

\(\frac{6}{10}\), \(\frac{9}{15}\), and \(\frac{12}{20}\)

Three fractions equivalent to \(\frac{3}{5}\), \(\frac{6}{10}\),\(\frac{9}{15}\) and \(\frac{12}{20}\)

Glencoe Math Course 2 Volume 1 Common Core Chapter 2 Page 120 Exercise 1 Problem 37

Given:

The bar diagram for eighth and tenth grade.

To find-

Total tickets sold above each bar. We will divide the bar into ten equal sections.

Each section represents ten percent.

Let us divide the bar into 10 equal sections.

The bar for eighth and seventh grade are similar.

By measuring the above bar, we found that 50 % of tickets were sold.

The bar diagrams below show 100% for each grade. Abel Divide each bar into 10 equal sections.

So, each section will represent 10%. The total number of tickets to be sold above each bar is 50 %.

Page 120  Exercise 2  Problem 38

Given:

To Find -The number that belongs in each section.

Then write that Section.

Calculation:
​
​300 ÷ 10 = 30

250 ÷ 10 = 25
​
The number of tickets in each section of eight  & seventh-grade baris 30 and 25 respectively.

Page 120  Exercise 3  Problem 39

Given:

To find- The number of sections to shade for each bar.

Then shade the sections.

Eight

225 ÷ 30 = 7.5

Seventh

200 ÷ 25 = 8

The number of sections to be Shaded in the eighth and seventh-grade bar is 7.5 and 8 respectively.

The eighth grade sold 75 % of their tickets. The seventh grade sold 80% of their tickets.

The Seventh grade sold the greater percent of their tickets.

Hence, the Seventh grade sold a greater percent of their tickets

The post Glencoe Math Course 2 Volume 1 Common Core Student Edition Chapter 2 Percents Exercise 2.2 appeared first on Answer Key for Math.

Glencoe Math Course 2 Volume 1 Chapter 3 Integers Exercise Solutions Page 187   Exercise 1  Problem 1

Because the first three operations close the set of integers:
Add
Subtract
Multiply

These operations will return a set of integers as a result.

When you divide two numbers, though, the result is when the first and second integers are not multiples of each other (In other words,

When the second integer is not a factor of the first integer), Then, rather than an integer, the outcome will be the Ratio between the two integers.

Hence the term “rational” for such numbers.

Finally, we concluded that when we add, subtract, and multiply. will return a set of integers as a result and when we divide integers the outcome will be the Ratio between the two integers. Hence the term “rational” for such numbers.

Page 190   Exercise 2   Problem 2

Given: (10+50) ÷ 5 = ______

To evaluate the expression.

Evaluate the given expression within the brackets first

(10 + 50) = 60

Now, Simplify
​
​60 ÷ (5) =  \(\frac{60}{5}\) = 12

⇒ 12

(10+50) ÷ 5 = 12
​
Finally, we concluded that the result is ⇒ 12

Read and Learn More Glencoe Math Course 2 Volume 1 Common Core Student Edition Solutions

Common Core Chapter 3 Integers Exercise Answers Glencoe Math Course 2 Page 190   Exercise 3  Problem 3

Given:18 + 2(4 − 1) = ______

To evaluate the expression.

Given

18 + 2(4 − 1) =

Evaluate the given expression within the brackets first

(4 − 1) = 3

Then

​2(4 − 1) = 2(3)

⇒ 6

Now, Simplify

​18 + 6 = 24

⇒ 24

18 + 2(4 − 1) = 24
​
We concluded that the 18 + 2(4 − 1) = 24

Step-By-Step Guide For Chapter 3 Integers Exercises In Glencoe Math Course 2 Page 190   Exercise 5    Problem 4

Given: B(2,8)

Graph and label each point on the coordinate grid.

The first number is the number on the x-axis (horizontal) and the second number is the number on the y-axis, i.e., vertical.

The required graph is:

Finally, we plotted Graph and labeled each point on the coordinate grid.

Exercise Solutions For Chapter 3 Integers Glencoe Math Course 2 Volume 1 Page 190   Exercise 6   Problem 5

Given: C(8,1)

Graph and label each point on the coordinate grid.

The first number is the number on the x-axis (horizontal) and the second number is the number on the y-axis, i.e., vertical.

Finally, we Graph and label each point on the coordinate grid.

Common Core Integers Chapter 3 Exercises With Solutions Glencoe Math Course 2 Page 190 Exercise 7 Problem 6

Given: D(3,4)

Graph and label each point on the coordinate grid.

The first number is the number on the x-axis (horizontal) and the second number is the number on the y-axis, i.e., vertical.

The required graph is:

Finally, we plotted Graph and labeled each point on the coordinate grid.

Examples Of Problems From Chapter 3 Integers Exercises In Glencoe Math Course 2 Page 190  Exercise 8  Problem 7

Given: E(1,5)

Graph and label each point on the coordinate grid.

The first number is the number on the x-axis (horizontal) and the second number is the number on the y-axis,i.e., vertical.

The required graph is:

Finally, we plotted Graph and labeled each point on the coordinate grid.

The post Glencoe Math Course 2 Volume 1 Common Core Student Edition Chapter 3 Integers Exercise appeared first on Answer Key for Math.

Vector Integration Applications Gauss Theorem Solved Problems

1. State and prove Gauss’s divergence theorem.

Solution:

Gauss’s divergence theorem: If F is a differentiable vector point function and S is a closed surface enclosing a region V, then \(\int_S\)F. N dS \(\int_V\)  div F dV, where N is the outward drawn unit normal vector to S.

Proof: Let S be a closed surface. Let us choose the coordinate axes so that any line parallel to the axes meets the surface in almost two points. Let R be the projection of S on xy-plane.  Let S₁ and S₂ be the lower and upper parts of S.

Let z=f(x,y) and z = g(x, y) be the equations of S₁ and S₂ which can be put in the form f(x,y)≤z≤ g(x, y)
Let F = F₁i+F₂ j+F₃k where F_1, F₂, and F₃ are scalar point functions.

∴ \(\int_v \text{div} \mathbf{F} d V=\int_V \nabla \cdot \mathbf{F} d V=\iint_V\left(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\right) d V=\int_V \frac{\partial F_1}{\partial x} d V+\int_V \frac{\partial F_2}{\partial y}+\int_V \frac{\partial F_3}{\partial z} d V .\)

Now \(\int_V \frac{\partial F_3}{\partial z} d V=\iiint_V \frac{\partial F_3}{\partial z} d x d y d z=\iiint_R\left[F_3(x, y, z)\right]_f^g d x d y\)

⇒ \(\iint_R\left[F_3(x, y, g)-F_3(x, y, f)\right] d x d y .\)

For the upper part S₂, dx dy = dS cos γ = N . k dS, since the normal to S₂ makes an acute angle y with k.

∴ \(\iint_R F_3(x, y, g) d x d y=\int_{s_2} F_3 \mathbf{N} \cdot \mathbf{k} d S\)

For the lower part S₁ dx dy =- cos γ dS =− N . k dS, since the normal to S₁ makes an obtuse angle y with k.

∴ \(\iint_R F_3(x, y, f) d x d y=-\int_{s_1} F_3 \mathbf{N} . \mathbf{k} d S\)

⇒ \(\int_V \frac{\partial F_3}{\partial z} d V=\int_{s_2} F_3 \mathbf{N} . \mathbf{k} d S+\int_{s_1} F_3 \mathbf{N} \cdot \mathbf{k} d S=\int_S F_3 \mathbf{k} . \mathbf{N} d S\)

Similarly \(\int_V \frac{\partial F_2}{\partial y} d V=\int_S F_2 \mathbf{j} \cdot \mathbf{N} d S \text { and } \int_V \frac{\partial F_1}{\partial x} d V=\int_S F_1 \mathbf{i} \cdot \mathbf{N} d S\)

∴ \(\int_V\left(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}\right) d V=\int_S F_1 \mathbf{i} \cdot \mathbf{N} d S+\int_S F_2 \mathbf{j} \cdot \mathbf{N} d S+\int_S F_3 \mathbf{k} \cdot \mathbf{N} d S\)

⇒ \(\int_V \nabla \cdot \mathbf{F} d V=\int_S\left(F_1 \mathbf{i}+F_2 \mathbf{j}+F_3 \mathbf{k}\right) \cdot \mathbf{N} d S \Rightarrow \int_V \nabla \cdot \mathbf{F} d V=\int_S \mathbf{F} \cdot \mathbf{N} d S\)

2. If F is a continuously differentiable vector point function and S is a closed surface enclosing a region V then prove that \(\int_S\)N×F dS=\(\int_V\) ∇×F dV.

Solution: Let f=a×F where a is any constant vector.

By Gauss’s divergence theorem \(\int_S\)f.N dS= \(\int_V\)∇.f dV.

⇒ \(\int_S(\mathbf{a} \times \mathbf{F}) \cdot \mathbf{N} d S=\int_V \nabla \cdot(\mathbf{a} \times \mathbf{F}) d V \Rightarrow \int_S \mathbf{a} \cdot(\mathbf{F} \times \mathbf{N}) d S=-\int_V \nabla \cdot(\mathbf{F} \times \mathbf{a}) d v\)

⇒ \(-\int_S \mathbf{a} \cdot(\mathbf{N} \times \mathbf{F}) d S=-\int_V(\nabla \times \mathbf{F}) \cdot \mathbf{a} d V \Rightarrow \mathbf{a} \cdot \int_S(\mathbf{N} \times \mathbf{F}) d S=\mathbf{a} \cdot \int_V \nabla \times \mathbf{F} d V\)

⇒ \(\int_S(\mathbf{N} \times \mathbf{F}) d S=\int_V \nabla \times \mathbf{F} d V\) [∵ a is any constant vector]

3. If φ  is a continuously differentiable scalar point function and S is a closed surface enclosing a region V then prove that \(\int_S\)N φ dS= \(\int_V\) ∇φ dV.

Solution: Let f=a φ where a is any constant vector.

By Gauss’s divergence theorem, \(\int_S \mathbf{f} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{f} d V\)

⇒ \(\int_S a \varphi \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{a} \varphi d V \Rightarrow \int_S \mathbf{a} \cdot \varphi \mathbf{N} d S=\int_V \nabla \varphi \cdot \mathbf{a} d V\)

⇒ \(\mathbf{a} \cdot \int_S \varphi \mathbf{N} d S=\mathbf{a} \cdot \int_V \nabla \varphi d V \Rightarrow \int_S \varphi \mathbf{N} d S=\int_V \nabla \varphi d V\) [∵ a is any constant vector]

4. Apply Gauss’s theorem to prove that  \(\int_S\)r. N dS = 3 V.

Solution: \(\int_S\)r.NdS=\(\int_V\)div r dV=\(\int_V\)∇.r dV=\(\int_V\)3 dV=3V, where V is the volume of the region bounded by the closed surface S.

5. Prove that for any closed surface S, \(\iint_S\)N dS = 0.

Solution:\(\iint_S\)N dS=\(\iint_S\)N 1 dS=\(\int_V\)(∇ 1)dV=\(\int_V\)0 dV=0

6. For any closed surface S, prove that \(\iint_S\)Curl F . N dS = 0.

Solution: By Gauss’s  divergence theorem,

\(\iint_S\)F.N dS=\(\iint_S\)(∇×F).N dS=\(\int_V\)div(∇×F) dV=\(\int_V\)0 dV=0.

7. If S is any closed surface enclosing a volume V and F = xi + 2yj+ 3zk, prove that \(\iint_S\)F.N dS=6v.

Solution:  By Gauss’s divergence theorem,

∴ \(\iint_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{2 y\}+\frac{\partial}{\partial z}\{3 z\}\right] d V=\int_V(1+2+3) d V=6 V\)

8. If F = xi- 2yi + 3zk and S is a closed surface enclosing a volume V, show that \(\int_S\)F.N dS=2v.

Solution: 

By Gauss’s divergence theorem,

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \text{div} \mathbf{F} d V=\int_V(\nabla \cdot \mathbf{F}) d V=\int_V\left[\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{-2 y\}+\frac{\partial}{\partial z}\{3 z\}\right] d V\)

⇒ \(\int_V(1-2+3) d V=2 \int_V d V=2 V\)

9. Computed \(\oint_S\)(ax²+by²+cz²) dS over the sphere x² +y² + z² = 1.

Solution: Let φ =x²+y²+z²-1

Normal vector to the surface, ∇φ =\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)

=2xi+2yj+2zk

Unit normal vector, \(\mathbf{N}=\frac{2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}}{\sqrt{4 x^2+4 y^2+4 z^2}}=\frac{2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}}{2 \sqrt{x^2+y^2+z^2}}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

⇒ \(\mathbf{F} \cdot \mathbf{N}=a x^2+b y^2+c z^2 \Rightarrow \mathbf{F} \cdot(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})=a x^2+b y^2+c z^2 \Rightarrow \mathbf{F}=a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k}\)

⇒ \(\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}(a x)+\frac{\partial}{\partial y}(b y)+\frac{\partial}{\partial z}(c z)=a+b+c\)

Volume of the sphere \(x^2+y^2+z^2=1 \text { is } 4 \pi / 3\)

By Gauss’s divergence theorem,

⇒ \(\oint_S\left(a x^2+b y^2+c z^2\right) d S=\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V=\int_V(a+b+c) d V=(a+b+c) V\)

⇒ \((a+b+c) \frac{4}{3} \pi=\frac{4 \pi}{3}(a+b+c)\)

10. If F = axi + byj+ czk and a, b, c are constants, show that ∫F.N dS =\(\frac{4}{3} \pi\) (a + b + c) where S is the surface of the  unit sphere.

Solution:

Given

F = axi + byj+ czk and a, b, c are constants

∴ \(\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}\{a x\}+\frac{\partial}{\partial y}\{b y\}+\frac{\partial}{\partial z}\{c z\}=a+b+c\)

Volume of the sphere, \(V=\frac{4}{3} \pi(1)^3=\frac{4}{3} \pi\)

By Gauss’s divergence theorem, \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V=\int_V(a+b+c) d V\)

⇒ \((a+b+c) \int_V d V=(a+b+c) V=\frac{4}{3} \pi(a+b+c)\)

Stokes Theorem Vector Integration Problems And Solutions

11. Show that \(\iint_S\)(ax dy dz + by dz dx + cz dx dy) = \(\frac{4}{3} \pi\)(a + b + c), where S is the surface of the sphere x² +y² + z²=1. where S is the surface of the sphere.

Solution: 

By Gauss’s divergence theorem,

⇒ \(\iint_S(a x d y d z+b y d z d x+c z d x d y)=\int_s[a x \mathbf{N} \cdot \mathbf{i} d S+\text { by N.j } d S+c z \mathbf{N} \cdot \mathbf{k} d S]\)

⇒ \(\int_s(a x \mathbf{i}+b y \mathbf{j}+c z \mathbf{k}) \cdot \mathbf{N} d S=\int_V\left[\frac{\partial}{\partial x}(a x)+\frac{\partial}{\partial y}(b y)+\frac{\partial}{\partial z}(c z)\right] d V=(a+b+c) \int_V d V\)

⇒ \((a+b+c) V \text { where } V \text { is the volume of } x^2+y^2+z^2 = 1\)

⇒ \((a+b+c) \frac{4}{3} \pi=\frac{4 \pi}{3}(a+b+c) \text {. }\)

12. Apply divergence theorem to evaluate \(\iint_S\) x dy dz+y dz dx + z dx dy where S is the surface x² +y² + z² = 1

Solution: By Gauss’s divergence Theorem,

\(\iint_S\)(x dy dz+y dz dy+z dx dy)= \(\int_S\)[x N.idS+y N.jdS+z N.k dS]

⇒ \(\int_S(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) \cdot \mathbf{N} d S=\int_V\left[\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z)\right] d V=(1+1+1) \int_V d V\)

⇒ \(3 V \text { where } V \text { is the volume of } x^2+y^2+z^2=1=3 \times \frac{4}{3} \pi=4 \pi \text {. }\)

13. Apply Gauss’s divergence theorem to compute the double integral\(\iint_S\) (x+z) dy dz+(y+z) dz dx+(x+y) dxdy where S is the surface of the sphere x² +y² + z² = 4.

Solution: 

By Gauss’s divergence theorem, \(\iint_S(x+z) d y d z+(y+z) d z d x+(x+y) d x d y\)

⇒ \(\int_S(x+z) \mathbf{N} \cdot \mathbf{i} d S+(y+z) \mathbf{N} \cdot \mathbf{j} d S+(x+y) \mathbf{N} \cdot \mathbf{k} d S\)

⇒ \(\int_S[(x+z) \mathbf{i}+(y+z) \mathbf{j}+(x+y) \mathbf{k}] \cdot \mathbf{N} d S\)

⇒ \(\int_v\left[\frac{\partial}{\partial x}(x+z)+\frac{\partial}{\partial y}(y+z)+\frac{\partial}{\partial z}(x+y)\right] d V=\int_V(1+1+0) d V=2 V=2 \frac{4 \pi}{3}(8)=\frac{64 \pi}{3}\)

14. If F = xi−yj+ (z²−1)k find the value of \(\int_S\) F . N dS where S is the closed surface bounded by the planes z = 0, z = 1 and the cylinder x² +y² = 4

Solution: 

Given

F = xi−yj+ (z²−1)k

∴ \(\mathbf{F}=x \mathbf{i}-y \mathbf{j}+\left(z^2-1\right) \mathbf{k}\)

⇒ \(\text{div} \mathbf{F}=\nabla \cdot \mathbf{F}=\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{-y\}+\frac{\partial}{\partial z}\left\{z^2-1\right\}=1-1+2 z=2 z\)

The limits of the region bounded by the given surface are z = 0 to z = 1, \(y=-\sqrt{4-x^2} \text { to } y=\sqrt{4-x^2}\) and x = -2 to x = 2.

By Gauss’s divergence theorem

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V d i v \mathbf{F} d V=\int_V 2 z d V=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} \int_{z=0}^{z=1} 2 z d x d y d z\)

⇒ \(\left.=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} z^2\right]_{z=0}^{z=1} d x d y=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} 1 d x d y\)

⇒ \(\left.\int_{x=-2}^{x=2} y\right] _{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} d x =\int_{x=-2}^{x=2} 2 \sqrt{4-x^2} d x=4 \int_0^2 \sqrt{4-x^2} d x\)

⇒ \(4\left[\frac{x \sqrt{4-x^2}}{2}+\frac{4}{2} \text{Sin}^{-1} \frac{x}{2}\right]_0^2=4\left[2 \frac{\pi}{2}-0\right]=4 \pi\)

15. If F = x i− y j + (z²−1) k, V is the volume of the cylinder bounded by z = 0, z = 1 and x² +y² = a² and S is the surface of the cylinder, show that \(\int_S\) F . N dS = π a²

Solution: Given=x i− y j + (z²−1)k, Now F₁=x,F₂=−y,F₃=z−1

∴ \(\frac{\partial F_1}{\partial x}=1, \frac{\partial F_2}{\partial y}=-1, \frac{\partial F_3}{\partial z}=2 z . \text{div} \mathbf{F}=\nabla \cdot \mathbf{F}=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=1-1+2 z=2 z\)

By the Gauss divergence theorem,

⇒ \(\int_S \mathbf{F} \mathbf{N} d S=\int_v \text{div} \mathbf{F} d V=\int_V 2 z d V=\int_{x=-a}^{x=a} \int_{y=\sqrt{a^2-x^2}}^{y=-\sqrt{a^2-x^2}} \int_{z=0}^{z=1} 2 z d x d y d z\)

⇒ \(\left.\int_{x=-a}^{x=a} \int_{y=-\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}} z^2\right]_{0}^{1} d x d y=\int_{x=-a}^{x=a}\int_{-y=\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}} d x d y \left.= \int_{x=-a}^{x=a} y\right]_{y=-\sqrt{a^2-x^2}}^{y=\sqrt{a^2-x^2}} dx
\)

⇒ \(\left.\int_{x=-a}^{x=a} 2 \sqrt{a^2-x^2} d x=x \sqrt{a^2-x^2}+a^2 \text{Sin}^{-1}(x / a)\right]_{-a}^a=a^2[\pi / 2-(-\pi / 2)]=\pi a^2\)

Applications Of Gauss Theorem In-Plane Solved Examples

16. By transforming into triple integral, evaluate \(\int_S\) ( x³dy dz + x²y dz dx + x²z dx dy) where S is the closed surface consisting of the cylinder x²+y² = a² and the circular disc z = 0 and z = b.

Solution:

Let \(F_1=x^3, F_2=x^2 y, F_3=x^2 z. \quad \frac{\partial F_1}{\partial x}=3 x^2, \frac{\partial F_2}{\partial y}=x^2, \frac{\partial F_3}{\partial z}=x^2\)

⇒ \(\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=3 x^2+x^2+x^2=5 x^2\)

By Gauss’s divergence theorem \(\iint_S\left(x^3 d y d z+x^2 y d z d x+x^2 z d x d y\right)\)

⇒ \(\iiint_V 5 x^2 d x d y d z=4 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}} \int_{z=0}^{z=b} 5 x^2 d x d y d z\)

⇒ \(20 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}}\left[x^2 z\right]_{z=0}^{z=b} d x d y=20 b \int_{x=0}^a \int_{y=0}^{\sqrt{a^2-x^2}} x^2 d x d y\)

⇒ \(\left.20 b \int_0^a x^2 y\right]_0^{\sqrt{a^2-x^2}}=20 b \int_0^a x^2 \sqrt{a^2-x^2} d x\)

Put x = a sin θ

∴ dx = a cos θ dθ

x = 0 ⇒ θ = 0

x = a ⇒ θ = π/2

⇒ \(20 b \int_0^{\pi / 2} a^2 \sin ^2 \theta \sqrt{a^2-a^2 \sin ^2 \theta} a \cos \theta d \theta=20 a^4 b \int_0^{\pi / 2} \sin ^2 \theta \cos ^2 \theta d \theta\)

⇒ \(20 a^4 b \int_0^{\pi / 2} \sin ^2 \theta\left(1-\sin ^2 \theta\right) d \theta=20 a^4 b\left[\int_0^{\pi / 2} \sin ^2 \theta-\int_0^{\pi / 2} \sin ^4 \theta d \theta\right]\)

⇒ \(20 a^4 b\left[\frac{1}{2} \cdot \frac{\pi}{2}-\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right]=20 a^4 b \cdot \frac{\pi}{16}=5 a^4 b \frac{\pi}{4}\)

17. If F = 2xyi +yz²j + xzk and S is a rectangular parallelopiped bounded by x = 0, y = 0,z = 0,x = 2,y= 1 and z = 3 verily Gauss’s divergence theorem.

Solution:  Consider the six faces of the rectangular parallelopiped bounded by x=o,y=0,z=0,x=2,y=1, and z=3.

Case (1): For the face OADB, the outward normal

N = -k, z = 0, dS = dx dy.

∴ \(\int_{S_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{y=0}^{y=1}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{k}) d x d y=\int_{x=0}^{x=2} \int_{y=0}^{y=1}-x z d x d y=0 .\)

Case (2): For the face OBEC, the outward normal, N = -i, x = 0, dS = dx dz.

∴ \(\int_{S_2} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 2 x y d y d z=0\)

Case (3): For the face OCFA, the outward normal, N = -j, y = 0, dS = dz dx

∴ \(\int_{S_3} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+z x \mathbf{k}\right) \cdot(-\mathbf{j}) d x d z=\int_{x=0}^{x=2} \int_{z=0}^{z=3}-y z^2 d x d z=0\)

Case (4): For the face ADGF, the outward normal, N = I, x = 2, dS = dy dz

⇒ \(\int_{S_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot \mathbf{i} d y d z=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 2 x y d y d z\)

⇒ \(\left.\left.=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 4 y d y d z=\int_{y=0}^{y=1} 4 y z\right]_{z=0}^{z=3} d y=\int_0^1 12 y d y=6 y^2\right]_0^1=6\)

Case (5): For the faces BDGE, the outward normal, N = j, y = 1, dS = dz dx

⇒ \(\int_{S_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(\mathbf{j}) d z d x\)

⇒ \(\left.\left.=\int_{x=0}^{x=2} \int_{z=0}^{z=3} y z^2 d z d x=\int_{x=0}^{x=2} \int_{z=0}^{z=3} z^2 d z d x=\int_{x=0}^{x=2} z^3 / 3\right]_0^3 d x=\int_{x=0}^{x=2} 9 d x=9 x\right]_0^2=18\)

Case (6):  For the faces CEGF, the outward normal, N = k, z = 3, dS = dx dy

⇒ \(\int_{S_6} \mathbf{F} \cdot \mathbf{N} d S=\int_{x=0}^{x=2} \int_{y=0}^{y=1}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+z x \mathbf{k}\right) \cdot \mathbf{k} d x d y\)

⇒ \(\left.\left.\int_{x=0}^{x=2} \int_{y=0}^{y=1} x z d x d y=\int_{x=0}^{x=2} \int_{y=0}^{y=1} 3 x d x d y=\int_{x=0}^{x=2} 3 x y\right]_{y=0}^{y=1} d x=\int_{x=0}^{x=2} 3 x d x=3 x^2 / 2\right]_0^2=6 .\)

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S\) sum of the six integrals over the six faces

⇒ 0 + 0 + 0 + 6 + 18 + 6 = 30.

⇒ \(\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}(2 x y)+\frac{\partial}{\partial y}\left(y z^2\right)+\frac{\partial}{\partial z}(x z)\right] d V=\int_V\left(2 y+z^2+x\right) d V\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(x+2 y+z^2\right) d x d y d z=\int_{x=0}^{x=2} \int_{y=0}^{y=1}\left[x z+2 y z+\frac{z^3}{3}\right]_{z=1}^{z=3} d x d y\)

⇒ \(\left.\int_{x=0}^{x=2} \int_{y=0}^{y=1}(3 x+6 y+9) d x d y=\int_{x=0}^{x=2}\left[3 x y+3 y^2+9 y\right]\right]_{y=0}^{y=1} d x=\int_0^2(3 x+12) d x\)

∴ \(\left[\frac{3 x^2}{2}+12 x\right]_0^2=6+24=30\) ∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V\)

∴ Gauss’s divergence theorem is verified.

18. Evaluate \(\int_S\)F.N dS where F = 2xy i +yz² +xzk and S is the surface of the parallelopiped formed by x=0 ,  y = 0, z = 0  x = 2,y = 1, z = 3

Solution: Consider the parallelopiped O A B C P Q R S surrounded by x=0,y=0, z=0, x=2,y=1, and z=3.

1. For the face PQAR, I is the outward normal.

N = i, x = 2, dS = dy dz.

∴ \(\int_{R_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot \mathbf{i} d y d z\)

⇒ \(=\int_{y=0}^{y=1} \int_{z=0}^{z=3} 2 x y d y d z=\int_{y=0}^{y=1}[4 y z]_{z=0}^{x=3} d y=\int_{y=0}^{y=1} 12 y d y=\left[6 y^2\right]_0^1=6\)

2. For the faces OBSC, -i is the outward normal. N = -i, x = 0, dS = dy dz

∴ \(\int_{R_2} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_2}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z=-\iint_R 2 x y d y d z=0 .\)

3. For the face BQPS, j is the outward normal. ∴ N = j, y = 1 and dS = dx dz.

∴ \(\int_{R_3} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_3}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot \mathbf{j} d x d z=\iint_{R_3} y z^2 d x d z\)

⇒ \(=\int_{x=0}^{x=2} \int_{z=0}^{z=3} z^2 d x d z=\int_{x=0}^{x=2}\left[\frac{z^3}{3}\right]_{z=0}^{z=3} d x=\int_{x=0}^{x=2} 9 d x=[9 x]_0^2=18\)

4. For the face OARC, -j is the outward normal. ∴ N = -j, y = 0 and dS = dx dz

∴ \(\int_{R_4} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_4}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{j}) d x d z=-\iint_{R_4} y z^2 d x d z=0\)

5. For the face PRCS, k is the outward normal. ∴ N = k, z = 3 and dS = dx dy.

∴ \(\int_{R_5} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_5}\left(2 x y \mathbf{i}+y z^2 \mathbf{J}+x z \mathbf{k}\right) \cdot \mathbf{k} d x d y=\iint_{R_5} x z d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=1} 3 x d x d y=\int_{x=0}^{x=2}[3 x y]_{y=0}^{y=1} d x=\int_{x=0}^{x=2} 3 x d x=\left[\frac{3 x^2}{2}\right]_0^2=6\)

6. For the face OAQB, -k is the outward normal. N = -k, z = 0 and dS = dx dy

∴ \(\int_{R_6} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_6}\left(2 x y \mathbf{i}+y z^2 \mathbf{j}+x z \mathbf{k}\right) \cdot(-\mathbf{k}) d x d y=-\iint_{R_6} x z d x d y=0\)

∴\(\int_s \mathbf{F} \cdot \mathbf{N} d S=6+0+18+0+6+0=30\)

Exercise 5 Gauss Theorem And Stokes Theorem Step-By-Step Solutions

19. Evaluate\(\iint_S\) (x dydz +y dzdx + z dxdy) taken over the outer surface of the cube[0,a;0,a;0,a].

Solution:  By Gauss’s Divergence theorem,

⇒ \(\iint_S(x d y d z+y d z d x+z d x d y)=\iiint_v\left[\frac{\partial}{\partial x}\{x\}+\frac{\partial}{\partial y}\{y\}+\frac{\partial}{\partial z}\{z\}\right] d x d y d z\)

⇒ \(\iiint_V(1+1+1) d x d y d z=3 \int_{x=0}^{x=a} \int_{y=0}^{y=a} \int_{z=0}^{z=a} d x d y d z=3 \int_{x=0}^{x=a} \int_{y=0}^{y=a}[z]_{z=0}^{z=a} d x d y\)

∴ \(3 \int_{x=0}^{x=a} \int_{y=0}^{y=a} a d x d y=3 \int_{x=0}^{x=a}[a y]_{y=0}^{y=a} d x=3 \int_{x=0}^{x=a} a^2 d x\left[3 a^2 x\right]_{x=0}^{x=a}=3 a^3\)

Gauss Theorem Vector Integration Practice ProblemsGauss Theorem Vector Integration Practice Problems

20. Evaluate \(\int_S\)F . N dS where F = 2x²y i -y²j + 4xz²k taken over the region in the first octant bounded by y² + z²= 9 and x = 2.

Solution:  \(\int_S\)F . N dS=\(\int_V\)∇.F dV=\(\int_V\)∇.(2x²yi-y²j+4xz² k)dv

⇒ \(\int_v\left[\frac{\partial}{\partial x}\left(2 x^2 y\right)+\frac{\partial}{\partial y}\left(-y^2\right)+\frac{\partial}{\partial z}\left(4 x z^2\right)\right] d V=\int_V(4 x y-2 y+8 x z) d V\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=3} \int_{z=0}^{z=\sqrt{9-y^2}}(4 x y-2 y+8 x z) d x d y d z\)

= \(\int_{x=0}^{x=2} \int_{y=0}^{y=3}\left[4 x y z-2 y z+4 x z^2\right]_{=0}^{z=\sqrt{9-y^2}} d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=3}(2 x-1) 2 y \sqrt{9-y^2}+4 x\left(9-y^2\right) d x d y\)

⇒ \(\int_{x=0}^{x=2}\left[(2 x-1) \frac{\left(9-y^2\right)^{3 / 2}}{-3 / 2}+4 x\left(9 y-\frac{y^3}{3}\right)\right]_{y=0}^{y=3} d x=\int_{x=0}^{x=2}\left[4 x(27-9)+\frac{2(2 x-1)}{3} \times 27\right] d x\)

∴ \(\int_0^2[72 x+36 x-18] d x=\int_0^2(108 x-18) d x=\left[54 x^2-18 x\right]_0^2=216-36=180\)

21. Evaluate by Gauss divergence theorem for \(\iint_S\) 4xz dy dz -y² dz dx+yz dx dy where S is the surface of the cube bounded by the planes  x=0,x=1,y=0,y=1,z=0,z=1

Solution:  

Let \(F_1=4 x z, F_2=-y^2, F_3=y z\)

⇒ \(\frac{\partial F_1}{\partial x}=4 z, \frac{\partial F_2}{\partial y}=-2 y, \frac{\partial F_3}{\partial z}=y \cdot \frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=4 z-2 y+y=4 z-y\)

By Gauss’s divergence theorem, \(\iint_S 4 x z d y d z-y^2 d z d x+y z d x d y\)

⇒ \(\left.\iiint_V(4 z-y) d x d y d z=\int_{x=0}^{x=1} \int_{y=0}^{y=1} \int_{z=0}^{z=1}(4 z-y) d x d y d z=\int_{x=0}^{x=1} \int_{y=0}^{y=1}\left[2 z^2-y z\right]\right]_{z=0}^{z=1} d x d y\)

∴ \(\int_{x=0}^{x=1} \int_{y=0}^{y=1}(2-y) d x d y=\int_{x=0}^{x=1}\left[2 y-\frac{y^2}{2}\right]_{y=0}^{y=1} d x=\int_{x=0}^{x=1} \frac{3}{2} d x=\left[\frac{3 x}{2}\right]_{x=0}^{x=1}=\frac{3}{2} .\)

22. Find the value of \(\int_S\)(F x ∇φ) .N dS, where F = x² i +y²j + z²k,  φ= xy+yz + zx, S is the surface bounded by x = ± 1 ,y = ± 1, z = ± 1.

Solution:

Given \(\mathbf{F}=x^2 \mathbf{i}+y^2 \mathbf{j}+z^2 \mathbf{k}, \varphi=x y+y z+z x . \quad \nabla \varphi=(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k}\)

⇒ \(\mathbf{F} \times \nabla \varphi=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
x^2 & y^2 & z^2 \\
y+z & z+x & x+y
\end{array}\right|\)

⇒ \(\mathbf{i}\left(x y^2+y^3-z^3-x z^2\right)-\mathbf{j}\left(x^3+x^2 y-y z^2-z^3\right)+\mathbf{k}\left(x^2 z+x^3-y^3-y^2 z\right)\)

∴ \(\nabla \cdot(F \times \nabla \varphi)=\frac{\partial}{\partial x}\left(x y^2+y^3-z^3-x z^2\right)-\frac{\partial}{\partial y}\left(x^3+x^2 y-y z^2-z^3\right)[latex]

+[latex]\frac{\partial}{\partial z}\left(x^2 z+x^3-y^3-y^2 z\right)\)

= \(y^2-z^2-x^2+z^2+x^2-y^2=0\)

By Gauss’s divergence theorem;\(\int_S\)(F×∇φ).N dS=\(\int_V\)∇.(F×∇φ)dV=0.

23. Verify Gauss divergence theorem for F = 4xzi -y²j  +yzk taken over the curve bounded by x = 0, x- 1,y = 0,y= 1,z = 0,z= 1.

Solution:  Consider the cube OABCPQRS surrounded by the following faces.

1. For the face PQAR, I is the outward normal.

N = i, x = 1, ds = dy dz.

∴ \(\int_{R_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=1} \int_{z=0}^{z=1}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{i} d y d z\)

⇒ \(=\int_{y=0}^{y=1} \int_{z=0}^{z=1} 4 x z d y d z=\int_{y=0}^{y=1} \int_{z=0}^{z=1} 4 z d y d z=\int_{y=0}^{y=1}\left[2 z^2\right]_{z=0}^{z=1} d y=\int_{y=0}^{y=1} 2 d y=[2 y]_0^1=2\)

2. For the face OBSC, -i is the outward normal. ∴ N = -i, x = 0, and dS = dy dz

∴ \(\int_{R_2} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_2}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z=-\iint_{R_2} 4 x z d y d z=0\)

3. For the face BQPS, j is the outward normal. ∴ N = j, y = 1, and dS = dx dz

∴ \(\int_{R_3} \mathbf{F} \cdot \mathbf{N} d S=\iint_{R_3}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{j} d x d z\)

= \(-\iint_{R_3} y^2 d x d z=-\int_{x=0}^{x=1} \int_{z=0}^{z=1} d x d z=-[x]_0^1[z]_0^1=-1\)

4. For the face OARC, -j is the outward normal. ∴ N = -j, y = 0 and dS = dx dz

∴ \(\int_{R_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{R_4}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{j}) d S=\iint_{R_4} y^2 d x d z=0\)

5. For the face PRCS, k is the outward normal. ∴ N = k, z = 1, and dS = dx dy

∴ \(\int_{R_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{R_5}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{k} d S\)

= \(\iint_{R_5} y z \cdot d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1} y d x d y=[x]_0^1 \cdot\left[\frac{y^2}{2}\right]_0^1=\frac{1}{2} .\)

⇒ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=2+0-1+0+\frac{1}{2}+0=\frac{3}{2}\)

⇒ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=2+0-1+0+\frac{1}{2}+0=\frac{3}{2}\)

⇒ \(\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}(4 x z)+\frac{\partial}{\partial y}\left(-y^2\right)+\frac{\partial}{\partial z}(y z)\right] d V=\int_V(4 z-2 y+y) d V\)

⇒ \(=\int_{x=0}^{x=1} \int_{y=0}^{y=1} \int_{z=0}^{z=1}(4 z-y) d x d y d z=\int_{x=0}^{x=1} \int_{y=0}^{y=1}\left[2 z^2-y z\right]_{z=0}^{z=1} d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1}(2-y) d x d y\)

⇒ \(=\int_{x=0}^{x=1}\left[2 y-\frac{y^2}{2}\right]_{y=0}^{y=1} d x=\int_{x=0}^{x=1}\left(2-\frac{1}{2}\right) d x=\left[3 \frac{x}{2}\right]_{x=0}^{x=1}=\frac{3}{2}\) ∴ \(\int_S F \cdot \mathbf{N} d S=\int_V \nabla \cdot \mathbf{F} d V\)

∴ Gauss’s divergence theorem is verified.

Solved Problems On Vector Integration Using Stokes Theorem

24. Verify Gauss’s divergence theorem to evaluate \(\int_S\){(x³ −yz) i- 2x²yi + zk }. N dS over the surface of a cube bounded by the coordinate planes x=y=z=a.

Solution:

∴ \(\mathbf{F}=\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k} . \quad \text{div} \mathbf{F}=\nabla \cdot \mathbf{F}=3 x^2-2 x^2+1=x^2+1\)

⇒ \(\int_V \text{div} \mathbf{F} d V=\int_V\left(x^2+1\right) d V=\int_{x=0}^{x=a} \int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(x^2+1\right) d x d y d z\)

⇒ \(=\int_{x=0}^{x=a} \int_{y=0}^{y=a}\left[\left(x^2+1\right) z\right]{ }_{z=0}^{z=a} d x d y=\int_{x=0}^{x=a} \int_{y=0}^{y=a} a\left(x^2+1\right) d x d y=\int_{x=0}^{x=a}\left[a\left(x^2+1\right) y\right]_{y=0}^{y=a} d x\)

⇒ \(\left.=a^2 \int_{x=0}^{x=a}\left(x^2+1\right) d x=a^2\left(\frac{x^3}{3}+x\right)\right]_0^a=a^2\left(\frac{a^3}{3}+a\right)=\frac{a^5}{3}+a^3\)

Case (1): For the face OAQB, the outward normal, N = -k, dS = dx dy, z=0

∴ \(\int_{s_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{s_1}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] \cdot(-\mathbf{k}) d S=-\int_{s_1} z d S=0\)

Case (2): For the face CSPR, the outward normal, N =k, dS = dx dy, z = a

∴ \(\int_{s_2} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_2}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] \cdot \mathbf{k} d S\)

⇒ \(=\int_{S_2} z d S=a \int_{x=0}^{x=a} \int_{y=0}^{y=a} d x d y=a \int_{x=0}^{x=a} [y]_{y=0}^{y=a} d x=a^2 \int_0^a d x=\left[a^2 x\right]_{x=0}^{x=a}=a^3\)

Case (3): For the face OASC, the outward normal, N = -j, dS = dx dy, y = 0

∴ \(\int_{s_3} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_3}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right) \cdot(-\mathbf{j}) d S=\int_{s_3}\left(2 x^2 y\right) d S=0\)

Case (4): For the face BQPR, the outward normal, N = j, dS = dx dz, y = a

∴ \(\int_{s_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{s_4}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right) \cdot \mathbf{j} d S=\int_{s_4}-2 x^2 y d S\)

⇒ \(\left.-2 a \int_{x=0}^{x=a} \int_{z=0}^{z=a} x^2 d x d z=-2 a \int_{x=0}^{x=a}\left[x^2 z\right]_{z=0}^{z=a} d x=-2 a^2 \int_0^a x^2 d x=-2 a^2 \frac{x^{3}}{3}\right]_{0}^{a}=-\frac{2 a^5}{3}\)

Case (5): For the face OBRC, the outward normal, N = -i, dS = dy dz, x = 0

∴ \(\int_{s_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_5}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right) \cdot(-\mathbf{i}) d S=-\int_{S_5}\left(x^3-y z\right) d S\)

⇒ \(\left.\int_{y=0}^{y=a} \int_{z=0}^{z=a} y z d y d z=\int_{y=0}^{y=a}\left[y z^2 / 2\right]\right]_{z=0}^{z=a} d y=\frac{a^2}{2} \int_{y=0}^{y=a} y d y=\left[\frac{a^2}{2} \cdot \frac{y^2}{2}\right]_{y=0}^{y=a}=\frac{a^4}{4}\)

Case (6): For the face AQPS, the outward normal, N = i, dS = dy dz, x = a

∴ \(\int_{s_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{s_5}\left[\left(x^3-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+z \mathbf{k}\right] \cdot \mathbf{i} d S\)

= \(\int_{s_5}\left(x^3-y z\right) d S=\int_{y=0}^{y=0} \int_{z=0}^{y=a}\left(a^3-y z\right) d y d z\)

⇒ \(\int_{y=0}^{y=a}\left[a^3 z-y z^2 / 2\right]_{z=0}^{z=a} d y=\int_{y=0}^{y=a}\left[a^4-\frac{a^2 y}{2}\right] d y=\left[a^4 y-\frac{a^2 y^2}{4}\right]_{y=0}^{y=a}=a^5-\frac{a^4}{4}\)

∴ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=\int_{s_1} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_2} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_3} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_4} \mathbf{F} \cdot \mathbf{N} d S\)

+\(\int_{s_5} \mathbf{F} \cdot \mathbf{N} d S+\int_{s_6} \mathbf{F} \cdot \mathbf{N} d S\)

= \(0+a^3+0-\frac{2 a^5}{3}+\frac{a^4}{4}+a^5-\frac{a^4}{4}=a^3+\frac{a^5}{3}\)

∴ \(\int_V \text{div} \mathbf{F} d V=\int_s \mathbf{F} \cdot \mathbf{N} d S\)

∴ Gauss’s divergence theorem is verified.

25. Verify Gauss’s divergence theorem for F = (x² -yz) i- 2x² y J + 2k taken over the cube bounded by the planes x = 0, x= a, y = 0, y =a, z = 0, z = a.

Solution: 

∴ \(\int_V \text{div} \mathbf{F} d V=\int_V \nabla \cdot \mathbf{F} d V=\int_V\left[\frac{\partial}{\partial x}\left(x^2-y z\right)+\frac{\partial}{\partial y}\left(-2 x^2 y\right)+\frac{\partial}{\partial z}(2)\right] d V\)

⇒ \(\int_V\left(2 x-2 x^2\right) d V=\int_{x=0}^{x=a} \int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(2 x-2 x^2\right) d x d y d z\)

⇒ \(\int_{x=0}^{x=a} \int_{y=0}^{y=a}\left(2 x-2 x^2\right) d x d y [z]_{z=0}^{z=a}=\int_{x=0}^{x=a} \int_{y=0}^{y=a} a\left(2 x-2 x^2\right) d x d y\)

⇒ \(\left.\int_{x=0}^{x=a} a\left(2 x-2 x^2\right) d x \quad y\right]_{y=0}^{y=a}\quad\)

= \(\int_{x=0}^{x=a} a^2\left(2 x-2 x^2\right) d x=a^2\left[x^2-2 x^3 / 3\right]_0^a=a^2\left[a^2-2 a^3 / 3\right]\)

⇒ \(a^4-2 a^5 / 3\)

Case (1): For the face ADGF, N = i, dS = dy dz and x = a

∴ \(\int_{S_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_1}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot \mathbf{i} d S=\int_{S_1}\left(x^2-y z\right) d S\)

⇒ \(\int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(a^2-y z\right) d y d z=\int_{y=0}^{y=a}\left[a^2 z-y z^2 / 2\right]_{z=0}^{z=a}dy\)

⇒ \(\left.\int_0^a\left(a^3-a^2 y / 2\right) d y=a^3 y-a^2 y^2 / 4\right]_0^a=a^4-a^4 / 4=3 a^4 / 4\)

Case (2): For the face OBEC, N = -i, dS = dy dz and x = 0.

∴ \(\int_{S_2} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_2}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot(-\mathbf{i}) d S=\int_{y=0}^{y=a} \int_{z=0}^{z=a}(0-y z)(-1) d y d z\)

⇒ \(\left.\left.\int_{y=0}^{y=a} \int_{z=0}^{z=a} y z d y d z=\int_{y=0}^{y=a} y z^2 / 2\right]_{z=0}^{z=a} d y=\int_{y=0}^{y=a}\left[a^2 y / 2\right] d y=\frac{a^2 y^2}{4}\right]_0^a=\frac{a^4}{4}\)

Case (3): For the face BEGD, N = I, dS = dx dz and y = a

∴ \(\int_{S_3} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_3}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot(\mathbf{j}) d S=\int_{x=0}^{x=a} \int_{z=0}^{z=a}-2 a x^2 d x d z\)

⇒ \(\left.\int_{x=0}^{x=a}\left[-2 a x^2 z\right]_{z=0}^{z=a} d x=\int_0^a-2 a^2 x^2 d x=-2 a^2 x^3 / 3\right]_0^a=-2 a^5 / 3 .\)

Case (4): For the face OCFA, N = -j, dS = dx dz, y = 0

∴ \(\int_{S_4} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_4}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right] \cdot(-\mathbf{j}) d S=0\)

Case (5): For the face CFGE, N = k, dS = dx dy, z = a

∴\(\int_{S_5} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_5}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right) \cdot \mathbf{k} d S\)

⇒ \(\left.\left.\int_{x=0}^{x=a} \int_{y=0}^{y=a} 2 d x d y=\int_{x=0}^{x=a} 2 y\right]_{y=0}^{y=a} d x=\int_{x=0}^{x=a} 2 a d x=2 a x\right]{ }_0^a=2 a^2\)

Case (6): for this face OADB, N = -k, dS = dx dy, z = 0.

∴ \(\int_{S_6} \mathbf{F} \cdot \mathbf{N} d S=\int_{S_6}\left[\left(x^2-y z\right) \mathbf{i}-2 x^2 y \mathbf{j}+2 \mathbf{k}\right) \cdot(-\mathbf{k}) d S\)

⇒ \(\left.\left.\int_{x=0}^{x=a} \int_{y=0}^{y=a}(-2) d x d y=\int_{x=0}^{x=a}(-2 y)\right]_{y=0}^{y=a} d x=\int_{x=0}^{x=a}-2 a d x=-2 a x\right]_0^a=-2 a^2\)

∴ \(\mathbf{F} \cdot \mathbf{N} d S=\int_{S_1} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_2} \mathbf{F} \cdot \mathbf{N} d S\)

+ \(\int_{S_3} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_4} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_5} \mathbf{F} \cdot \mathbf{N} d S+\int_{S_6} \mathbf{F} \cdot \mathbf{N} d S\)

= \(3 a^4 / 4+a^4 / 4-2 a^5 / 3+0+2 a^2-2 a^2=a^4-2 a^5 / 3\)

∴ \(\int_V \text{div} \mathbf{F} d V=\int_S \mathbf{F} \cdot \mathbf{N} d S .\)

∴ Gauss’s divergence theorem was verified.

26. State and prove Green’s theorem in a plane.

Solution:  

Green’s theorem

Let S be a closed region in the plane enclosed by a curve C if P and Q are continuous and differentiable scalar functions of x and y in S, then\(\int_C\) P dx + Q dy = \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\)dx dy,  the line integral being taken along the entire boundary C of S such that S is on the left as one advance along C

Proof: Let any line parallel to either co-ordinate axes cut C in at most two points.

Let S  lie between the lines x=a, x=b,and y=c y=d.

Let y=f(x) be the curve C₁ (AEB) and y=g(x)  be the curve C₂ (ADB) where f(x)≤ g(x)

Consider \(\iint_S \frac{\partial P}{\partial y} d x d y=\int_{x=a}^{x=b} \int_{y=f(x)}^{y=g(x)}\left(\frac{\partial P}{\partial y} d y\right) d x\)

= \(=\int_{x=a}^{x=b}[P(x, y)]_{y=f(x)}^{y=g(x)} d x=\int_a^b[P(x, g(x))-P(x, f(x))] d x\)

= \(\int_a^b P(x, g) d x-\int_a^b P(x, f) d x=-\int_{C_1} P(x, y) d x-\int_{C_2} P(x, y) d x=-\int_{c_3} P d x\)

Similarly , we can prove that \(\iint_S\)\(\left(\frac{\partial Q}{\partial x}\right)\)dx dy=\(\int_C\)dy.

∴\(\int_C\) P dx + Q dy = \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\)dx dy

27. Evaluate \(\oint_C\)(x dy-y dx) around the circle C where C is x² +y²=1.

Solution:

By Green’s theorem, \(\int_c P d x+Q d y=\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

Put P = -y, Q = x. Then \(\frac{\partial Q}{\partial x}=1, \frac{\partial P}{\partial y}=-1\)

∴ \(\int_C x d y-y d x=\iint_S[1-(-1)] d x d y=2 \iint_S d x d y\)

= \(2 \text { (Area of the surface } S \text { ) }=2 \pi(1)^2=2 \pi\)

28. If f and g are two continuous and differentiable scalar point functions over the region V enclosed by the surface S, then prove that

1. \(\int_V\left[f \nabla^2 g+\nabla f \cdot \nabla g\right] d V\)=\(=\int_S(f \nabla g) \cdot N d S\)
2. \(\int_V\left(f \nabla^2 g-g \nabla^2 f\right) d V\)=\(=\int_S(f \nabla g-g \nabla f) \cdot \mathbf{N} d S\)

Solution:

1. Let F = f ∇g.

Then \(\nabla \cdot \mathbf{F}=\nabla \cdot(f \nabla g)=f(\nabla \cdot \nabla g)+\nabla f \cdot \nabla g=f \nabla^2 g+\nabla f \cdot \nabla g\)

By Gauss’s divergence theorem, \(\int_V \nabla \cdot \mathbf{F} d V=\int_S \mathbf{F} \cdot \mathbf{N} d S\)

⇒ \(\int_V\left[f \nabla^2 g+\nabla f \cdot \nabla g\right] d V=\int_S(f \nabla g) \cdot \mathbf{N} d S\)

2. From (1); \(\int_V\left(f \nabla^2 g+\nabla f \cdot \nabla g\right) d V=\int_S(f \nabla g) \cdot \mathbf{N} d S\) → (1)

Interchanging f and g in (1), we get \(\int_V\left(g \nabla^2 f+\nabla g \cdot \nabla f\right) d V=\int_s(g \nabla f) \cdot \mathbf{N} d S\) → (2)

(1) – (2) ⇒ \(\int_V\left(f \nabla^2 g-g \nabla^2 f\right) d V=\int_S(f \nabla g-g \nabla f) \cdot \mathbf{N} d S\)

Solved Examples Of Vector Integration In Plane Using Gauss Theorem

29. Show that the area bounded by a simple closed curve C is given by ½ \(\oint_C\) x dy-y dx and hence find area of ellipse x=a cos θ, y= b sin θ ,0≤θ≤2π.

Solution:

Here P = -y, Q = x. Then \(\frac{\partial Q}{\partial x}=1, \frac{\partial P}{\partial y}=-1\)

By Green’s theorem \(\frac{1}{2} \int_C x d y-y d x=\frac{1}{2} \int_C P d x+Q d y=\frac{1}{2} \iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

= \(\frac{1}{2} \iint_S[1-(-1)] d x d y=\iint_S d x d y\) = Area of the surface bounded by the curve C.

Equations of ellipse are x = a cos θ, y = b sin θ, 0 ≤ θ ≤ 2π

∴ \(\frac{d x}{d \theta}=-a \sin \theta, \frac{d y}{d \theta}=b \cos \theta\)

∴ Area of the ellipse = \(\frac{1}{2} \int_c x d y-y d x=\frac{1}{2} \int_0^{2 \pi}[(a \cos \theta)(b \cos \theta) {-}(b \sin \theta)(-a \sin \theta)] d \theta\)

= \(\frac{1}{2} \int_0^{2 \pi}\left(a b \cos ^2 \theta+a b \sin ^2 \theta\right) d \theta=\frac{a b}{2} \int_0^{2 \pi} d \theta=\frac{a b}{2}(2 \pi)=\pi a b \text { sq.unit }\)

30. \(\oint_C\) (cos x sin y- xy) dx + sin x cos y dy, by Green’s theorem, where C is the circle x²+y²=1

Solution:  By Green’s theorem\(\int_C\) P dx + Q dy = \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\)dx dy

Let P = cos x sin y -xy, Q = sin x cos y. \(\frac{\partial P}{\partial y}=\cos x \cos y-x, \frac{\partial Q}{\partial x}=\cos x \cos y\).

The limits of the surface of the circle x² + y² = 1 are x = -1 to x = 1 and \(y=-\sqrt{1-x^2} \text { to } y=\sqrt{1-x^2} \text {. }\)

∴\(\int_c(\cos x \sin y-x y) d x+\sin x \cos y d y=\iint_s(\cos x \cos y-\cos x \cos y+x) d x d y\)

= \(\iint_S x d x d y=\int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} x d x d y=\int_{x=-1}^{x=1} 2 x d x \int_0^{y=\sqrt{1-x^2}} d y=\int_{x=-1}^{x=1} 2 x \sqrt{1-x^2} d x=0\)

31. Evaluate by Green’s theorem \(\oint_C\)(x²– cosh y) + (y + sinx) dy, where C is the rectangle with vertices (0, 0), (π, 0), (π, 1), (0, 1).

Solution:

By theorem \(\int_c P d x+Q d y=\iint_s\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

Let \(P=x^2-\cosh y, Q=y+\sin x \text {.. Then } \frac{\partial P}{\partial y}=-\sinh y, \frac{\partial Q}{\partial x}=\cos x\)

The limits of the surface of integration are x = 0 to x = π and y = 0 to y = 1.

∴ \(\int_C\left(x^2-\cosh y\right) d x+(y+\sin x) d y=\int P d x+Q d y\)

= \(\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\int_{x=0}^{x=\pi} \int_{y=0}^{y=1}(\cos x+\sinh y) d x d y\)

= \(\int_{x=0}^{x=\pi}[y \cos x+\cosh y]_{y=0}^{y=1} d x=\int_0^{\pi}(\cos x+\cosh 1-1) d x\)

= \([\sin x+x \cosh 1-x]_{x=0}^{x=\pi}=\pi \cosh 1 – \pi=\pi(\cosh 1-1)=\pi\left(\frac{e+e^{-1}}{2}-1\right)\)

32. Using Green’s theorem, evaluate \(\oint_C\)(x² +y²)dx + 3xy² dy where C is the circle x²+y²=4

Solution:

Here \(P=x^2+y^2, Q=3 x y^2\) ∴ \(\frac{\partial P}{\partial y}=2 y, \frac{\partial Q}{\partial x}=3 y^2\). The limits of the surface of integration are x = -2 to x = 2 and \(y=-\sqrt{4-x^2} \text { to } y=\sqrt{4-x^2}\)

By Green’s theorem, \(\int_C\left(x^2+y^2\right) d x+3 x y^2 d y=\iint_S\left(3 y^2-2 y\right) d x d y\)

= \(\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}\left(3 y^2-2 y\right) d x d y=\int_{x=-2}^{x=2}\left[y^3-y^2\right]_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}} d x=\int_{x=-2}^{x=2} 2\left(4-x^2\right)^{3 / 2} d x\)

= \(4 \int_0^2\left(4-x^2\right)^{3 / 2} d x=4 \int_0^{\pi / 2}\left(4-4 \sin ^2 \theta\right)^{3 / 2} 2 \cos \theta d \theta \text { where } x=2 \sin \theta\)

= \(64 \int_0^{\pi / 2} \cos ^4 \theta d \theta=64 \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}=12 \pi\)

33. Evaluate \(\oint_C\) (3x + 4y) dx + (2x− 3y) dy where ‘C’ is the circle  x²+y²= 4.

Solution:

Here P = 3x + 4y, Q = 2x – 3y. \(\frac{\partial P}{\partial y}=4, \frac{\partial Q}{\partial x}=2\). The limits of the surface of integration are x = -2 to x = 2 and \(y=-\sqrt{4-x^2} \text { to } y=\sqrt{4-x^2}\).

By Green’s theorem, \(\int_C(3 x+4 y) d x+(2 x-3 y) d y\)

= \(\iint_y(2-4) d x d y=\int_{x=-2}^{x=2} \int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}(-2) d x d y=\int_{x=-2}^{x=2}[-2 y]_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}\)

= \(\int_{-2}^2-4 \sqrt{4-x^2} d x=-8 \int_0^2 \sqrt{4-x^2} d x=-8\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \text{Sin}^{-1} \frac{x}{2}\right]_{0}^{2}=-8\left[2 \frac{\pi}{2}\right]=-8 \pi\)

Step-By-Step Guide To Vector Integration Using Stokes Theorem

34. Evaluate by Green’s theorem \(\oint(y-\sin x) d x\)+ cos x dy where C is the triangle enclosed by the lines x=0, x=π/2,  πy=2x.

Solution:

Here P = y – sin x, Q = cos x. ∴ \(\frac{\partial P}{\partial y}=1, \frac{\partial Q}{\partial x}=-\sin x\)

The limits of the surface of integration are x = 0 to \(x=\frac{\pi}{2} ; y=0 \text { to } y=\frac{2 x}{\pi}\)

By Green’s theorem \(\int_C(y-\sin x) d x+\cos x d y=\int_C P d x+Q d y\)

= \(\iint_s\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\int_{x=0}^{x=\pi / 2} \int_{y=0}^{y=2 x / \pi}(-\sin x-1) d x d y\)

=\(\int_{x=0}^{x=\pi / 2}[(-\sin x-1) y]_{y=0}^{y=2 x / \pi} d x\)

= \(\int_0^{\pi / 2}(-\sin x-1) \frac{2 x}{\pi} d x=-\frac{2}{\pi} \int_0^{\pi / 2} x(1+\sin x) d x\)

=\(-\frac{2}{\pi}[x(x-\cos x)]_{0}^{\pi /2}-\int_0^{\pi / 2}(x-\cos x) d x\)

= \(-\frac{2}{\pi}\left[\frac{\pi}{2}\left(\frac{\pi}{2}\right)-\left(\frac{x^2}{2}-\sin x\right)_0^{\pi / 2} \right]=-\frac{2}{\pi}\left[\frac{\pi^2}{4}-\frac{\pi^2}{8}+1\right]=-\frac{\pi}{4}-\frac{2}{\pi}\)

35. Compute \(\oint_C\) (x²− 2xy) dx + (x²y + 3) dy around the boundary C of the region defined by y² = 8x and x = 2 by applying Green’s theorem.

Solution:

Here \(P=x^2-2 x y, Q=x^2 y+3\)  ∴ \(\frac{\partial P}{\partial y}=-2 x, \frac{\partial Q}{\partial x}=2 x y\)

The limits of the surface of integration are x = 0 to x = 2 and y = 0 to \(\sqrt{8 x}\).

By Green’s Theorem, \(\oint_c\left(x^2-2 x y\right) d x+\left(x^2 y+3\right) d y=\int_c P d x+Q d y\)

= \(\iint_S\left[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right] d x d y=\int_{x=0}^{x=2} \int_{y=-\sqrt{8 x}}^{y=\sqrt{8 x}}(2 x y+2 x) d x d y=\int_{x=0}^{x=2}\left[x y^2+2 x y\right]_{y=-\sqrt{8 x}}^{y=\sqrt{8 x}} d x\)

= \(\int_{x=0}^{x=2}(4 x \sqrt{8 x}) d x=\left[8 \sqrt{2} \frac{x^{5 / 2}}{5 / 2}\right]_{x=0}^{x=2}=\frac{128}{5}\)

36. Find the area bounded by x^2/3 +y^2/3 = a^2/3  using Green’s theorem.

Solution:

The parametric equation of \(x^{2 / 3}+y^{2 / 3}=a^{2 / 3} \text { is } x=a \cos ^3 \theta, y=a \sin ^3\) θ and θ variation from 0 to 2π.

By green’s theorem, Area = \(\frac{1}{2} \int_C(x d y-y d x)\)

= \(\frac{1}{2} \int_{\theta=0}^{\theta=2 \pi} a\cos ^3 \theta 3 a \sin ^2 \theta \cos \theta d \theta-a \sin ^3 \theta\left(3 a \cos ^2 \theta\right)(-\sin \theta) d \theta\)

= \(\frac{1}{2} \int_{\theta=0}^{2 \pi} 3 a^2 \cos ^2 \theta \sin ^2 \theta\left(\cos ^2 \theta+\sin ^2 \theta\right) d \theta=\frac{1}{2} \int_0^{2 \pi} 3 a^2 \cos ^2 \theta \sin ^2 \theta d \theta\)

= \(6 a^2 \int_0^{\pi / 2} \sin ^2 \theta \cos ^2 \theta d \theta=6 a^2 \times \frac{1}{4} \times \frac{1}{2}=\frac{3 a^2}{4} \times \frac{\pi}{2}=\frac{3 \pi a^2}{8} \text { sq.unit }\)

37. Verify Green’s theorem in the plane for \(\oint_C\)(3x²– 8y²) dx + (4y- 6xy) dy where C is the region bounded by y = \(\sqrt{x}\)and y = x².

Solution:

P = \(P=3 x^2-8 y^2, Q=4 y-6 x y . \quad \frac{\partial P}{\partial y}=-16 y ; \frac{\partial Q}{\partial x}=-6 y .\)

⇒ \(\int_C\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y=\iint_s \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

= \(\iint_s(-6 y+16 y) d x d y=10 \int_{x=0}^{x=1}\left(\int_{y=x^2}^{y=\sqrt{x}} y d y\right) d x\)

= \(10 \int_{x=0}^{x=1}\left[\frac{y^2}{2}\right]_{y=x^2}^{y=\sqrt{x}} d x \quad=5 \int_0^1\left(x-x^4\right) d x=5\left[\frac{x^2}{2}-\frac{x^5}{5}\right]_0^1=5\left[\frac{1}{2}-\frac{1}{5}\right]=5\left[\frac{5-2}{10}\right]=\frac{3}{2}\)

∴ \(\int_c\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

= \(\int_{C_1}\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y+\int_{C_2}\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

Where c₁ is the curve y = x² from O to A and C₂ is the curve = √x from A to O

= \(\int_0^1\left(3 x^2-8 x^4\right) d x+\left(4 x^2-6 x^3\right) 2 x d x+\int_1^0\left(3 x^2-8 x\right) d x+(4 \sqrt{x}-6 x \sqrt{x}) \frac{d x}{2 \sqrt{x}}\)

= \(\int_0^1\left(3 x^2+8 x^3-20 x^4\right) d x-\int_0^1\left(3 x^2-8 x+2-3 x\right) d x=\int_0^1\left(8 x^3-20 x^4+11 x-2\right) d x\)

= \(\left[2 x^4-4 x^5+\frac{11 x^2}{2}-2 x\right]_0^1=2-4+\frac{11}{2}-2=\frac{3}{2}\)  ∴ Green’s theorem is verified.

38. Verify Green’s theorem in the plane for\(\oint_C\)(xy + y²) dx + x²dy where C is the closed curve of the region bounded by y = x andy = x².

Solution:

Here P = xy + y², Q = x² ∴ \(\frac{\partial P}{\partial y}=x+2 y, \frac{\partial Q}{\partial x}=2 x\)

By Green’s theorem, \(\int_c\left(x y+y^2\right) d x+x^2 d y=\int_c P d x+Q d y=\iint_s\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y\)

= \(\iint_s[2 x-(x+2 y)] d x d y=\iint_s(x-2 y) d x d y=\int_{x=0}^{x=1} \int_{y=x^2}^{y=x}(x-2 y) d x d y\)

= \(\int_{x=0}^{x=1}\left[x y-y^2\right]_{y=x^2}^{y=x} d x=\int_0^1\left[\left(x^2-x^2\right)-\left(x^3-x^4\right)\right] d x=\int_0^1\left(x^4-x^3\right) d x\)

= \(\left[\frac{x^5}{5}-\frac{x^4}{4}\right]_0^1=\frac{1}{5}-\frac{1}{4}=-\frac{1}{20}\)

= Line integral along y = x² (from O to A) + line integral along y = x (from A to O)

= \(=\int_0^1\left[x\left(x^2\right)+x^4\right] d x+x^2 \cdot 2 x d x+\int_1^0\left(x^2+x^2\right) d x+x^2 d x\)

= \(\int_0^1\left(3 x^3+x^4\right) d x-\int_0^1 3 x^2 d x=\left[3 \frac{x^4}{4}+\frac{x^5}{5}-x^3\right]_{0}^{1}=\frac{3}{4}+\frac{1}{5}-1=\frac{15+4-20}{20}=-\frac{1}{20}\)

39. Verify Green’s theorem in the plane for \(\oint_C\)(2xy − x²)dx + (x +y²) dy where C is the boundary of the region enclosed by y = x² and y²=x described in the positive sense.

Solution:

Here P = 2xy – x², Q = x² + y².

∴ \(\frac{\partial P}{\partial y}=2 x, \frac{\partial Q}{\partial x}=2 x\)

⇒ \(\iint_S\left[\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right] d x d y=\iint_S(2 x-2 x) d x d y=0\)

⇒ \(\int_c P d x+Q d y\) = Line integral along y = x² (from O to A) + Line integral along y² = x (from A to O) = I₁ + I₂

Now \(I_1=\int_{x=0}^{x=1} P d x+Q d y=\int_0^1\left[2 x\left(x^2\right)-x^2\right] d x+\left(x^2+x^4\right) 2 x d x\)

= \(\left.\int_0^1\left(2 x^3-x^2+2 x^3+2 x^5\right) d x=\int_0^1\left(2 x^5+4 x^3-x^2\right) d x=\frac{x^6}{3}+x^4-\frac{x^3}{3}\right]_0^1=1\)

⇒ \(I_2=\int_{x=1}^{x=0} P d x+Q d y=\int_1^0\left(2 x \sqrt{x}-x^2\right) d x+\left(x^2+x\right) \frac{1}{2 \sqrt{x}} d x\)

= \(\int_1^0\left[2 x \sqrt{x}-x^2+x \sqrt{x} / 2+\sqrt{x} / 2\right] d x=\int_1^0\left[5 x^{3 / 2} / 2-x^2+x^{1 / 2} / 2\right] d x\)

= \(\left.\frac{5}{2} \times \frac{x^{5 / 2}}{5 / 2}-\frac{x^3}{3}+\frac{1}{2} \times \frac{x^{3 / 2}}{3 / 2}\right]_1^0=-1+\frac{1}{3}-\frac{1}{3}=-1\)

∴ \(\int_c P d x+Q d y=I_1+I_2=1-1=0\)

∴ Green’s theorem is verified.

Vector Integration Solved Problems With Applications Of Gauss Theorem

40. Verify Green’s theorem, \(\oint_C\)(3x²– 8y²) dx + (4y- 6xy)dy where C is the boundary enclosed by x = 0,y = x+y= 1.

Solution:

Here P = 3x² – 8y², Q = 4y – 6xy. ∴ \(\frac{\partial P}{\partial y}=-16 y, \frac{\partial Q}{\partial x}=-6 y\)

The limits o the surface of integration are x = 0 to x = 1 and y = 0 to y = 1 – x

By Green’s theorem, \(\oint\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

= \(\int_c P d x+Q d y=\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1-x}(-6 y+16 y) d x d y\)

= \(\int_{x=0}^{x=1} \int_{y=0}^{y=1-x} 10 y d x d y=\int_{x=0}^{x=1}\left(5-10 x+5 x^2\right) d x=\int_{x=0}^{x=1}\left[5 y^2\right]_{x=0}^{y=1-x} d x\)

=\(\int_{x=0}^{x=1} 5(1-x)^2 d x\)

= \(\left[5 x-5 x^2+\frac{5 x^3}{3}\right]_{x=0}^{x=1}=5-5+\frac{5}{3}=\frac{5}{3}\)

Given planes x = 0, y = 0 and x + y = 1 from a triangle in xy-plane with vertices O(0,0), A(1,0) and B(0,1).

∴ \(\int_c\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y\)

= Line integral along \(\overrightarrow{O A}\) + Line along \(\overrightarrow{A B}\) + Line integral along \(\overrightarrow{B O}\).

1. Line integral along \(\left.\overrightarrow{O A}=\int_0^1 3 x^2 d x=x^3\right]_0^1=1\)

2. Line integral along \(\overrightarrow{A B}\). Here x + y = 1 ⇒ y = 1 – x varies from 1 to 0.

Line integral along \(\overrightarrow{A B}=\int_1^0\left[3 x^2-8(1-x)^2\right] d x+[4(1-x)-6 x(1-x)](-d x)\)

= \(\int_1^0\left(3 x^2-8-8 x^2+16 x\right) d x-\left(4-4 x-6 x+6 x^2\right) d x=\int_1^0\left(-11 x^2+26 x-12\right) d x\)

= \(\left[-\frac{11 x^3}{3}+13 x^2-12 x\right]_1^0=\frac{11}{3}-13+12=\frac{8}{3} .\)

3. Line integral along \(\left.\overrightarrow{B O}=\int_1^0 4 y^2 d y=2 y^2\right]_1^0=-2\)

∴ \(\int_C\left(3 x^2-8 y^2\right) d x+(4 y-6 x y) d y=1+\frac{8}{3}-2=\frac{5}{3}\)

∴ Green’s theorem is verified.

41. State and prove Stake’s theorem.

Solution: 

Stake’s theorem

Let S be a surface bounded by a closed non-intersecting curve C. If F is any differentiable vector point function, then

\(\int_C\)F.dr= \(\int_S\) curl F.Nds, where N is the outward drawn unit normal vector to S and C is traversed in the positive direction.

proof: Let S be a surface which is such that its projection on xy,yz,zx or y=h(z,x) where f,g,h are simple valued continuous and differentiable functions. Let F=F₁i+F₂j+F₃k.

Then curl F =∇×F=∇ × (F₁i + F₂j + F₃k)

=∇ × F₁i × F₂j × F₃k

Now \(\nabla \times F_1 \mathbf{I}=\left|\begin{array}{ccc}
\mathbf{I} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_1 & 0 & 0
\end{array}\right|=\frac{\partial F_1}{\partial z} \mathbf{J}-\frac{\partial F_1}{\partial y} \mathbf{k}\)

⇒ \(\left(\nabla \times F_1 \mathbf{I}\right) \cdot \mathbf{N}=\left[\frac{\partial F_1}{\partial z}(\mathbf{j} \cdot \mathbf{N})-\frac{\partial F_1}{\partial y}(\mathbf{k} \cdot \mathbf{N})\right]\) (1)

Let z = f(x,y) be the equation of S.

For any point in S, r = xi + yj + zk = xi + yi + f(x,y)k…

∴ \(\frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}+\frac{\partial z}{\partial y} \mathbf{k} \text {. Since } \frac{\partial \mathbf{r}}{\partial y} \text { is the tangent vector to } S, \mathbf{N} \cdot \frac{\partial \mathbf{r}}{\partial y}=0\)

⇒ \(\mathbf{N} \cdot \mathbf{J}+(\mathbf{N} \cdot \mathbf{k}) \frac{\partial z}{\partial y}=0 \Rightarrow \mathbf{N} \cdot \mathbf{j}=-(\mathbf{N} \cdot \mathbf{k}) \frac{\partial z}{\partial y}\)

From (1); \(\left(\nabla \times F_1 \mathbf{i}\right) \cdot \mathbf{N}=-(\mathbf{N} \cdot \mathbf{k}) \frac{\partial F_1}{\partial z} \cdot \frac{\partial z}{\partial y}-(\mathbf{N} \cdot \mathbf{k}) \frac{\partial F_1}{\partial y}\)

∴ \(\left[\left(\nabla \times F_1 \mathbf{i}\right) \cdot \mathbf{N}\right] d S=-\left(\frac{\partial F_1}{\partial y}+\frac{\partial F_1}{\partial z} \cdot \frac{\partial z}{\partial y}\right)(\mathbf{N} \cdot \mathbf{k}) d S\)

= \(-\frac{\partial}{\partial y} F_1(x, y, z) \cos \gamma d S=-\frac{\partial F_1}{\partial y} d x d y\)

Let R be the projection of S on xy-plane. Then

∴ \(\int_S\left(\nabla \times F_1 \mathbf{i}\right) \cdot \mathbf{N} d S=\int_R \int-\frac{\partial F_1}{\partial y} d x d y=\int_C F_1 d x\), by green’s theorem

Similarly \(\int_S\left(\nabla \times F_2 \mathbf{j}\right) \cdot \mathbf{N} d S=\int_c F_2 d y \text { and } \int_S\left(\nabla \times F_3 \mathbf{k}\right) \cdot \mathbf{N} d S=\int_c F_3 d z\)

∴ \(\int_S \nabla \times\left(F_1 \mathbf{i}+F_2 \mathbf{j}+F_3 \mathbf{k}\right) \cdot \mathbf{N} d S=\int_c F_1 d x+F_2 d y+F_3 d z\)

∴ \(\int_S(\nabla \times F) \cdot \mathbf{N} d S=\int_c F \cdot d r\)

42. Prove by Stake’s theorem curl grad φ = 0.

Solution:  Let be a  surface enclosed by a simple closed curve C.

∴ By stoke’s theorem, \(\int_s(\text{curl} \text{grad} \varphi) \cdot \mathbf{N} d S\)

= \(\int_s[\nabla \times(\nabla \varphi)] \cdot \mathbf{N} d S=\int_c \nabla \varphi \cdot d \mathbf{r}=\int_c\left(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\right) \cdot(\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d y)\)

= \(\left.\int_C\left(\frac{\partial \varphi}{\partial x} d x+\frac{\partial \varphi}{\partial y} d y+\frac{\partial \varphi}{\partial z} d z\right)=\int_C d \varphi=\varphi\right]_P^P=0\)

∴ \(\int_S(\text{curl} \text{grad} \varphi) \cdot \mathbf{N} d S=0 \Rightarrow \text{curl}(\text{grad} \varphi)=0\)

43. Find \(\int_C\)T .dr where T is the unit tangent vector and C is the unit circle in the xy-plane with centre at the origin.

Solution:  By stokes theorem ,\(\int_C\)T.dr=\(\int_S\)(curl T).N dS=\(\int_S\)(∇×T).N dS=\(\int_S\)0 dS=0.

44. Prove that \(\oint_C\)r.dr = 0.

Solution:  By stokes theorem ,\(\int_C\)T.dr=\(\int_S\)(curl r).N dS=\(\int_S\)0. N dS=0.

45. By Stoke’s theorem prove that div curl F = 0.

Solution:  Let S be the surface enclosed by a simple closed curve C

∴ \(\int_s \text{div} \text{curl} \mathbf{F} d S=\int \nabla \cdot \nabla \times \mathbf{F} d S=\int_s \Sigma \mathbf{i} \frac{\partial}{\partial x}(\nabla \times \mathbf{F}) d S=\int_S \Sigma \frac{\partial}{\partial x}[\mathbf{i} \cdot(\nabla \times \mathbf{F})] d S\)

= \(\Sigma \frac{\partial}{\partial x} \int_S(\nabla \times F) \cdot i d S=\Sigma \frac{\partial}{\partial x} \int_C\left(F_2 d y+F_3 d z\right)=0\)

∴ div curl F = 0.

46. Verify Stoke’s theorem to evaluate \(\int_C\)xy dx + xy² dy, where C is the square in the  xy-plane with vertices (1, 0), (- 1, 0), (0, 1), (0,- 1)

.Solution:

Let F = \(x y \mathbf{i}+x y^2 \mathbf{j} . \text{curl} \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x y & x y^2 & 0
\end{array}\right|=\left(y^2-x\right) \mathbf{k}\)

⇒ \(\int_c x y d x+x y^2 d y=\oint_C \mathbf{F} \cdot d \mathbf{r}=\int_s \nabla \times \mathbf{F} \cdot \mathbf{N} d S=\int_S\left(y^2-x\right) \mathbf{k} \cdot \mathbf{N} d S\)

Since k. N ds = dx dy and R is the region ABCD in xy-plane,

We have \(\int_S\left(y^2-x\right) \mathbf{k} \cdot \mathbf{N} d S=\iint_R\left(y^2-x\right) d x d y\)

Equation to \(\stackrel{\leftrightarrow}{A B} \text { is } \frac{x}{1}+\frac{y}{1}=1 \Rightarrow y=1-x\)

Equation to \(\stackrel{\leftrightarrow}{B C} \text { is } \frac{x}{-1}+\frac{y}{1}=1 \Rightarrow y=1+x\)

Equation to \(\stackrel{\leftrightarrow}{C D} \text { is } \frac{x}{-1}+\frac{y}{-1}=1 \Rightarrow y=-1-x\)

Equation to \(\stackrel{\leftrightarrow}{D A} \text { is } \frac{x}{1}+\frac{y}{-1}=1 \Rightarrow y=x-1\)

⇒ \(\iint_R\left(y^2-x\right) d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=1-x}\left(y^2-x\right) d x d y+\int_{x=-1}^{x=0} \int_{y=0}^{y=1+x}\left(y^2-x\right) d x d y\)

+ \(\int_{x=-1}^{x=0} \int_{y=-1-x}^{y=0}\left(y^2-x\right) d x d y+\int_{x=0}^{x=1} \int_{y=x-1}^{y=0}\left(y^2-x\right) d x d y\)

= \(\int_{x=0}^{x=1}\left[\frac{y^3}{3}-x y\right]_{y=0}^{y=1-x} d x+\int_{x=-1}^{x=0}\left[\frac{y^3}{3}-x y\right]_{y=0}^{y=1+x} d x\)

+\(\int_{x=-1}^{x=0}\left[\frac{y^3}{3}-x y\right]_{y=-1-x}^{y=0} d x+\int_{x=0}^{x=1}\left[\frac{y^3}{3}-x y\right]_{y=x-1}^{y=0} d x\)

= \(\int_0^1\left[\frac{(1-x)^3}{3}-x(1-x)\right] d x+\int_{-1}^0\left[\frac{(1+x)^3}{3}-x(1+x)\right] d x\)

–\(\int_{-1}^0\left[\frac{(-1-x)^3}{3}-x(-1-x)\right] d x – \int_0^1\left[\frac{(x-1)^3}{3}-x(x-1)\right] d x\)

= \(\int_0^1\left[\frac{(1-x)^3}{3}-x+x^2\right] d x+\int_{-1}^0\left[\frac{(1+x)^3}{3}-x-x^2\right] d x\)

+\(\int_{-1}^0\left[\frac{(1+x)^3}{3}-x-x^2\right] d x-\int_0^1\left[\frac{(x-1)^3}{3}-x^2+x\right] d x\)

= \(\left[\frac{(1-x)^4}{-12}-\frac{x^2}{2}+\frac{x^3}{3}\right]_0^1+\left[\frac{(1+x)^4}{12}-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^0\)

+\(\left[\frac{(1+x)^4}{12}-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-1}^0-\left[\frac{(x-1)^4}{12}-\frac{x^3}{2}+\frac{x^2}{2}\right]_0^1\)

= \(\left[0-\frac{1}{2}+\frac{1}{3}+\frac{1}{12}\right]+\left[\frac{1}{12}-0-0-0+\frac{1}{2}+\frac{1}{3}\right]\)

+\(\left[\frac{1}{12}-0-0-0+\frac{1}{2}-\frac{1}{3}\right]-\left[0-\frac{1}{3}+\frac{1}{2}-\frac{1}{12}+0-0\right]\)

= \(\frac{-6+4+1}{12}+\frac{1+6-4}{12}+\frac{1+6-4}{12}-\frac{-4+6-1}{12}=-\frac{1}{12}+\frac{3}{12}+\frac{3}{12}-\frac{1}{12}=\frac{4}{12}=\frac{1}{3}\).

Case (1): Line integral along AB: y = 1-x, dy = -dx, x varies from 1 to 0.

∴ \(\int_{C_1} x y d x+x y^2 d y=\int_1^0 x(1-x) d x+x(1-x)^2(-d x)=\int_1^0\left(x-x^2-x+2 x^2-x^3\right) d x\)

= \(\int_1^0\left(x^2-x^3\right) d x=\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_1^0=-\left[\frac{1}{3}-\frac{1}{4}\right]=-\frac{1}{12}\)

Case (2): Line integral along BC: y = 1+x, dy = dx, x varies from 0 to -1.

∴ \(\int_{C_2} x y d x+x y^2 d y=\int_0^{-1} x(1+x) d x+x(1+x)^2 d x=\int_0^{-1}\left(x+x^2+x+2 x^2+x^3\right) d x\)

= \(\int_0^{-1}\left(2 x+3 x^2+x^3\right) d x=\left[x^2+x^3+\frac{x^4}{4}\right]_0^1=1-1+\frac{1}{4}=\frac{1}{4}\)

Case (3): Line integral along CD: y = -1, dy = -dx, x varies from -1 to 0.

∴ \(\int_{C_3} x y d x+x y^2 d y=\int_{-1}^0 x(-1-x) d x+x(-1-x)^2(-d x)\)

=\(\int_{-1}^0\left(-x-x^2-x-2 x^2-x^3\right) d x\)

= \(-\int_{-1}^0\left(2 x+3 x^2+x^3\right) d x=-\left[x^2+x^3+\frac{x^4}{4}\right]_0^{-1}=1-1+\frac{1}{4}=\frac{1}{4}\)

Case (4): Line integral along DA: y = x-1, dy = dx, x varies from 0 to 1.

∴ \(\int_{C_4} x y d x+x y^2 d y=\int_0^{1} x(x-1) d x+x(x-1)^2 d x=\int_0^{1}\left(x^2-x+x^3-2 x^2+x\right) d x\)

= \(\int_0^1\left(x^3-x^2\right) d x=\left[\frac{x^4}{4}-\frac{x^3}{3}\right]_0^1=\frac{1}{4}-\frac{1}{3}=-\frac{1}{12}\)

∴ \(\int_S\)xy dx+xy² dy=1/12+1/4+1/4−1/12=1/3

∴  Stoke’s theorem is verified.

47. Verify Stoke’s theorem for the function F =x²i + xy j integrated round the square in the plane z= 0 whose sides are along the line x = 0, y = 0, x = a, y = a.

Solution:

∴ \({F}=x^2 \mathbf{i}+x y \mathbf{j} . \quad \text{curl} \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x^2 & x y & 0
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(y-0)=y \mathbf{k}\)

F. dr = (x2i + xyj) . (dx i + dy j + dz k) = x2 dx + xy dy

Let N be the unit normal vector to the surface of the square.

By Stoke’s theorem \(\int_S \boldsymbol{F} \cdot d \mathbf{r}=\int_S \text{cur} l \boldsymbol{F} \cdot \mathbf{N} d S\)

Since the surface of the square lies in the xy-plane, N = k.

⇒ \(\int_S \text{curl} \mathbf{F} \cdot \mathbf{N}dS =\int_S y k \cdot{k} d S=\int_S y d S=\int_0^a \int_0^a y d x d y\)

=\(\int_0^a\left[\frac{y^2}{2} \right]_0^a d x=\int_0^a \frac{a^2}{2} d x=\left[\frac{a^2 x}{2}\right]_0^a=\frac{a^3}{2}.\)

Case (1): Along the side OA: y = 0 and x varies from 0 to a.

Case (2): Along the side AB: x = a, dx = 0 and y varies from 0 to a.

Case (3): Along the side BC: y = a, dy = 0 and x varies from a to 0.

Case (4): Along the side CO: x = 0, y varies from a to 0.

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_{c_1} \mathbf{F} \cdot d \mathbf{F}+\int_{c_2} \mathbf{F} \cdot d \mathbf{r}+\int_{c_3} \mathbf{F} \cdot d \mathbf{r}+\int_{c_4} \mathbf{F} \cdot d \mathbf{r}=\frac{a^3}{3}+\frac{a^3}{2}-\frac{a^3}{3}+0=\frac{a^3}{2}\)

=\(\int_S\) curl F.NdS

∴ Stoke’s theorem is verified

48. Verify Stoke’s theorem to evaluate \(\oint_C\) F .dr where F =y²i + x² j- (x+ z)k and C is the boundary of the triangle with vertices (0, 0, 0), (1, 0, 0), (1, 1, 0).

Solution:

F = y² i + x² j – (x + z) k.

∴ \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y^2 & x^2 & -x-z
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(-1-0)+\mathbf{k}(2 x-2 y)=\mathbf{j}+(2 x-2 y) \mathbf{k}\)

Let N be the unit normal vector to the surface of the triangle.

Since the triangle lies in xy-plane, N=k, and dS=dx dy.

In xy -plane, the vertices of the triangle are O(0,0), A(1,0), and B(1,1).

By Stokes theorem,

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S=\int_S[\mathbf{j}+(2 x-2 y) \mathbf{k}] \cdot \mathbf{k} d S\)

= \(\iint_R(2 x-2 y) d x d y=\int_{x=0}^{x=1} \int_{y=0}^{y=x}(2 x-2 y) d x d y\) [∵ Along OB, x = y]

= \(\left.=\int_{x=0}^{x=1}\left[2 x y-y^2\right]{ }_{y=0}^{y=x} d x=\int_0^1\left(2 x^2-x^2\right) d x=\int_0^1 x^2 d x=\frac{x^3}{3}\right]_0^1=\frac{1}{3}\)

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_{O A} \mathbf{F} \cdot d \mathbf{r}+\int_{A B} \mathbf{F} \cdot d \mathbf{r}+\int_{B O} \mathbf{F} \cdot d \mathbf{r}\)

1. Along OA, y = 0, z = 0, dy = 0, dz = 0, x varies from 0 to 1

∴ \(\int_{O A} \mathbf{F} \cdot d \mathbf{r}=\int_0^{1} y^2 d x=0\)

2. Along AB, x = 1, z = 0, dx = 0, dz = 0, y varies from 0 to 1

∴ \(\left.\int_{A B} \mathbf{F} \cdot d \mathbf{r}=\int_0^1 x^2 d y=\int_0^1 d y=y\right]_0^1=1\)

3. Along BO, x = y, z = 0, dx = dy and x varies from 1 to 0.

∴ \(\left.\int_{B O} \mathbf{F} \cdot d \mathbf{r}=\int_1^0 y^2 d x+x^2 d y=\int_1 x^2 d x+x^2 d x=\int_1 2 x^2 d x=\frac{2 x^{30}}{3}\right]_1=-\frac{2}{3}\)

∴ \(\int_C F \cdot d \mathbf{F}=0+1-\frac{2}{3}=\frac{1}{3}\)

∴ Stoke’s theorem is verified.

49. Verify Stoke’s theorem for \(\oint_C\) F =- y³i + x³j,  where S is the circular disc x² +y² ≤ 1, z= 0.

Solution:  F=−y³i+x³j.

The boundary of C of S is a circle in xy -plane, x² +y²= 1, z=0

The parametric equations are x=cos θ,y=sin θ,z=0 where 0≤ θ≤2π.

∴ \(\int_c \mathbf{F} \cdot d \mathbf{r}=\int_c-y^3 d x+x^3 d y=\int_{\theta=0}^{\theta=2 \pi}-\sin ^3 \theta(-a \sin \theta) d \theta+\cos ^3 \theta(a \cos \theta) d \theta\)

Put x = sin θ

∴ dx = cos θ dθ

x = 0 ⇒ θ = 0

x = 1 ⇒ θ = π/2

= \(\int_0^{2 \pi}\left(\cos ^4 \theta+\sin ^4 \theta\right) d \theta=\int_0^{2 \pi}\left[\left(\cos ^2 \theta+\sin ^2 \theta\right)^2-2 \cos ^2 \theta \sin ^2 \theta\right] d \theta\)

= \(\int_0^{2 \pi}\left[1-\frac{1}{2} \sin ^2 2 \theta\right] d \theta=\int_0^{2 \pi}\left[1-\frac{1-\cos 4 \theta}{4}\right] d \theta\)

= \(\int_0^{2 \pi}\left[\frac{3+\cos 4 \theta}{4}\right] d \theta=\left[\frac{3 \theta}{4}+\frac{\sin 4 \theta}{16}\right]_0^{2 \pi}=\left[\frac{3 \pi}{2}+0\right]=\frac{3 \pi}{2}\)

∴ \(\nabla \times \mathbf{F}=\left[\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
-y^3 & x^3 & 0
\end{array}\right]=\mathbf{i}(0-0)-\mathbf{j}(0+0)+\mathbf{k}\left(3 x^2+3 y^2\right)=3\left(x^2+y^2\right) \mathbf{k}\)

Let R be the projection of S in the xy-plane.

∴ \(\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S=\iint_R 3\left(x^2+y^2\right) \mathbf{k} \cdot \mathbf{k} d x d y=3 \iint_R\left(x^2+y^2\right) d x d y\)

= \(3 \int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}}\left(x^2+y^2\right) d x d y=12 \int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}}\left(x^2+y^2\right) d x d y\)

= \(12 \int_{x=0}^{x=1}\left[x^2 y+\frac{y^3}{3}\right]_0^{\sqrt{1-x^2}} d x=12 \int_0^1\left[x^2 \sqrt{1-x^2}+\frac{1-x^2}{3} \sqrt{1-x^2}\right] d x\)

= \(4 \int_0^1\left(1+2 x^2\right) \sqrt{1-x^2} d x=4 \int_0^{\pi / 2}\left(1+2 \sin ^2 \theta\right) \cos \theta \cos \theta d \theta\)

= \(4 \int_0^{\pi / 2}\left(\cos ^2 \theta+2 \sin ^2 \theta \cos ^2 \theta\right) d \theta=4\left[\frac{1}{2} \cdot \frac{\pi}{2}+2 \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right]=4\left[\frac{\pi}{4}+\frac{\pi}{8}\right]=\frac{3 \pi}{2}\)

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S\)

∴ Stoke’s theorem is verified.

50. Verify Stoke’s theorem for F = (2x-y) i -yz² j -y²zk, where S is the upper half surface of the sphere x² +y² +z²= 1 and C is its boundary.

Solution: The boundary C os S is a circle in xy-plane, i.e…,  x² +y² +z²= 1 z=0.

The parametric equations are x=cos t,y= sin t, 0≤t≤2π

F=(2x-y)i-yz²j+y²zk

r=xi+yj+zk⇒dr=dxi+dyj+dzk

F.dr=(2x-y)dx-yz² dy+y²z dz=(2cos t-sin t) (-sin t)(dt)

=(sin² t-2 cos t sin t)dt=(sin² t-sin 2t)dt.

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_0^{2 \pi}\left(\sin ^2 t-\sin 2 t\right) d t=4 \int_0^{\pi / 2} \sin ^2 t d t+\left[\frac{\cos 2 t}{2}\right]_0^{2 \pi}\)

= \(4 \cdot \frac{1}{2} \cdot \frac{\pi}{2}+\frac{1}{2}[\cos 4 \pi-\cos 0]=\pi+\frac{1}{2}(1-1)=\pi\)

Also \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x-y & -y z^2 & -y^2 z
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(0+1)=\mathbf{k}\)

Let R be the projection of S in the xy-plane. Then k. N dS = dx dy.

∴ \(\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S=\int_S \mathbf{k} \cdot \mathbf{N} d S=\iint_R d x d y\)

= \(4 \int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}} d x d y=4 \int_{x=0}^{x=1}[y]{ }_{y=0}^{y=\sqrt{1-x^2}} d x=4 \int_0^1 \sqrt{\left(1-x^2\right)} d x\)

= \(4\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \text{Sin}^{-1} x\right]_0^1=4\left[\frac{1}{2} \times \frac{\pi}{2}\right]=\pi\)

∴ \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S\)

∴ Stoke’s theorem is verified.

51. Verify Stoke’s theorem for A = 2 yi + 3xj− z² k where S is the upper half surface of the sphere x²+y² + z² = 9 and C is its boundary.

Solution: The boundary Cof S is a circle in xy-plane  is the circle  x²+y²=9,z=0

The parametric equations are x=3 cos θ,y=3 sin θ , z=0

⇒dx =− 3 sin θ dθ ,dy =3 cos θ dθ,dz=0

∴ \(\int_C\)A.dr=\(\int_C\)(2yi+3xj−z²k).(dxi+dyj+dzk)

= \(\int_c 2 y d x+3 x d y-z d z=\int_0^{2 \pi} 2(3 \sin \theta)(-3 \sin \theta d \theta)+3(3 \cos \theta)(3 \cos \theta d \theta)-0\)

= \(\int_0^{2 \pi}\left[-18 \sin ^2 \theta+27 \cos ^2 \theta\right] d \theta=\int_0^{2 \pi}\left(27-45 \sin ^2 \theta\right) d \theta\)

= \(27(2 \pi-0)-45 \times 4 \times \frac{1}{2} \times \frac{\pi}{2}=54 \pi-45 \pi=9 \pi\)

∴ \(\nabla \times \mathbf{A}=\left|\begin{array}{lll}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 y & 3 x & -z^2
\end{array}\right|=\mathbf{i}(0-0)-\mathbf{j}(0-0)+\mathbf{k}(3-2)=\mathbf{k}\)

Let R be the projection of S in the xy-plane.

∴ \(\int_S(\nabla \times \mathbf{A}) \cdot \mathbf{N} d S=\iint_R(\mathbf{k} \cdot \mathbf{N}) d S=\iint_R d x d y\)

= \(\int_{x=-3}^{x=3} \int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}} d x d y=4 \int_{x=0}^{x=3} \int_{y=0}^{y=\sqrt{9-x^2}} d x d y\)

= \(4 \int_{x=0}^{x=3}\left[y\right]_{y=0}^{\sqrt{=9-x}} d x=4 \int_{0}^3 \sqrt{9-x^2} d x=4\left[\frac{x}{2} \sqrt{9-x^2}+\frac{9}{2} \text{Sin}^{-1} \frac{x}{3}\right]_0^3=4\left[\frac{9}{2} \times \frac{\pi}{2}\right]=9 \pi\)

∴ \(\int_S(\nabla \times \mathbf{A}) \cdot \mathbf{N} d S=\int_c \mathbf{A} \cdot d \mathbf{r}\)

∴ Stoke’s theorem is verified.

52. Apply Stoke’s theorem to evaluate\(\int_C\) (y dx + z dy+x dz) where C is the curve of intersection of x²+ y² + z²= a² and x + z =a.

Solution:  

For the sphere x² + y² + z² = a², Centre O = (0, 0, 0), radius = a.

Let A be the centre and r be the radius of the circle of intersection of x² + y² + z² = a² and x + z = a.

∴ \(O A=\left|\frac{0+0-a}{\sqrt{2}}\right|=\frac{a}{\sqrt{2}}\)

⇒ \(r^2=a^2-O A^2=a^2-\frac{a^2}{2}=\frac{a^2}{2} \Rightarrow r=\frac{a}{\sqrt{2}}\)

Since the plane of the circle is perpendicular to the y-axis, the vector normal to the plane is j.  ∴ N = j.

Let F = yi + zj + xk

∴ \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y & z & x
\end{array}\right|=\mathbf{i}(0-1)-\mathbf{j}(1-0)+\mathbf{k}(0-1)\)

= – i – j – k

By stoke’s theorem, \(\int_C(y d x+z d y+x d z)=\int_C \mathbf{F} \cdot d \mathbf{r}=\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d S\)

= \(\int_S(-\mathbf{i}-\mathbf{j}-\mathbf{k}) \cdot \mathbf{j} d S=\int_S(-1) d S=-S=-(\text { Area of the circle })={-}\left(\pi a^2 / 2\right)\)

53.Evaluate \(\int_S\)∇xF.N dS using Stoke’s theorem, where F = (2x -y) i -yz²-y²z k and S is tlie upper halfof the sphere x² +y² +z² = 1 and C is its boundary.

Solution:  The boundary C of S is a circle in xy-plane, i.e x² +y² =1,z=0.

The parametric equations are x=cos t, y=sint, 0≤t ≤2π.

F = (2x – y)i – y z j + y z k.

r = xi + yj + zk ⇒ dr = dxi + dyj + dzk.

= \(\mathbf{F} \cdot d \mathbf{r}=(2 x-y) d x-y z^2 d y+y^2 z d z=(2 \cos t-\sin t)(-\sin t)(d t)\)

= \(\left(\sin ^2 t-2 \cos t \sin t\right) d t=\left(\sin ^2 t-\sin 2 t\right) d t\)

By Stoke’s Theorem,

∴ \(\int_S(\nabla \times \mathbf{F}) \cdot \mathbf{N} d s=\int_C \mathbf{F} \cdot d \mathbf{r}=\int_0^{2 \pi}\left(\sin ^2 t-\sin 2 t\right) d t=4 \int_0^{\pi / 2} \sin ^2 t d t+\left[\frac{\cos 2 t}{2}\right]_0^{2 \pi}\)

= \(4 \cdot \frac{1}{2} \cdot \frac{\pi}{2}+\frac{1}{2}[\cos 4 \pi-\cos 0]=\pi+\frac{1}{2}(1-1)=\pi\)

54. Evaluate  \(\int_S\)(∇x F). N dS where F =yi + (x- 2xz) j -xy k and S is the surface S  of the sphere x² +y² + z² = a², above the xy-plane.

Solution:  The boundary C of S is a circle in xy-plane, i.e. x² +y² =1,z=0.

The parametric equations are x=a cos t, y= a sin t,z=0, 0≤t ≤2π. And dx =-a isn’t dt, dy= a cos t dt, dz=0.

∴ \(\int_S \nabla \times \mathbf{F} \cdot \mathbf{N} d S=\int_C \mathbf{F} \cdot d \mathbf{r}=\int_C[y \mathbf{i}+(x-2 x z) \mathbf{j}-x y \mathbf{k}] \cdot d \mathbf{r}\)

= \(\int_C y d x+(x-2 x z) d y-x y d z=\int_{t=0}^{t=2 \pi} a \sin t(-a \sin t) d t+(a \cos t-0) a \cos t d t\)

= \(a^2 \int_{t=0}^{t=2 \pi}\left(\cos ^2 t-\sin ^2 t\right) d t\)

= \(a^2 \int_{t=0}^{t=2 \pi} \cos 2 t d t=\left[\frac{a^2}{2} \sin 2 t\right]_{t=0}^{t=2 \pi}=0\)

55. Prove that \(\oint_C\) f∇g .dr  \(\oint_C\)(∇fx ∇g) .N dS.

Solution: By Stoke’s theorem,

∴ \(\oint_c(f \nabla g) \cdot d \mathbf{r}=\int_S[\nabla \times(f \nabla g)] \cdot \mathbf{N} d S=\int_s[\nabla f \times \nabla g+f \text { curl } \text{grad} g] \cdot \mathbf{N} d S\)

= \(\int_S(\nabla f \times \nabla g) \cdot \mathbf{N} d S\) ∵ curl grad g = 0.

56. Prove that \(\int_S\) curlφ f .dS = \(\int_C\) φ f .dr− \(\int_S\) ∇φ x f.dS.

Solution: Applying Stoke’s theorem to the function φ f.

∴ \(\oint_c(f \nabla g) \cdot d \mathbf{r}=\int_s[\nabla \times(f \nabla g)] \cdot \mathbf{N} d S=\int_s[\nabla f \times \nabla g+f \mathrm{curl} \text{grad} g] \cdot \mathbf{N} d S\)

∴ \(\int_s(\nabla f \times \nabla g) \cdot \mathbf{N} d S .\)

∵ curl grad g = 0.

57.Prove that \(\oint_C\)f∇f.dr = 0.

Solution:

By Stoke’s theorem, \(\int_C(f \nabla f) \cdot d \mathbf{r}=\int_S(c u r l f \nabla f) \cdot \mathbf{N} d S\)

= \(\int_S \phi \text{curl} \mathbf{f} \cdot d \mathbf{S}=\int_C \phi \mathbf{f} \cdot d \mathbf{r}-\int_S \nabla \phi \times \mathbf{f} \cdot d \mathbf{S} .\)

The post Vector Integration Applications Gauss Theorem And Applications Gauss Theorem In Plane And Applications Stokes Theorem And Applications Solved Problems Exercise 5 appeared first on Answer Key for Math.

Solved Problems On Line Surface And Volume Integrals

1. Evaluate \(\int_0^1\)(e^t + e^-2t j + Z k) dt.

Solution: 

Given

⇒ \(\left.\left.\left.\int_0^1\left(e^t \mathbf{i}+e^{-2 t} \mathbf{j}+t \mathbf{k}\right) d t=\mathbf{i} \int_0^1 e^t d t+\mathbf{j} \int_0^1 e^{-2 t} d t+\mathbf{k} \int_0^1 t d t=\mathbf{i} e^t\right]_0^1+\mathbf{j} \frac{e^{-2 t}}{-2}\right]_0^1+\mathbf{k} \frac{t^2}{2}\right]_0^1\)

= \(\mathbf{i}(e-1)+\mathbf{j}\left(\frac{e^{-2}}{-2}+\frac{1}{2}\right)+\frac{1}{2} \mathbf{k}=\mathbf{i}(e-1)+\mathbf{j}\left(\frac{1-e^{-2}}{2}\right)+\frac{1}{2} \mathbf{k}\)

2. IfF(Z) = ti + (t²– 2t) j + (3t² + 3t³) k then find \(\int_0^1\)f(t) dt.

Solution:

⇒ \(\int_0^1 \mathbf{f}(t) d t=\int_0^1\left[t \mathbf{i}+\left(t^2-2 t\right) \mathbf{j}+\left(3 t^2+3 t^3\right) \mathbf{k}\right] d t\)

= \(\int_0^1 t d t+\mathbf{j} \int_0^1\left(t^2-2 t\right) d t+\mathbf{k} \int_0^1\left(3 t^2+3 t^3\right) d t\)

= \(i\left[\frac{t^2}{2}\right]_0^1+\mathbf{j}\left[\frac{t^3}{3}-t^2\right]_0^1+\mathbf{k}\left[t^3+\frac{3 t^4}{4}\right]_0^1=\frac{1}{2} \mathbf{i}-\frac{2}{3} \mathbf{j}+\frac{7}{4} \mathbf{k} . .\)

3. If f(t) = (t−t²) i + 2t³ j- 3k, then find \(\int_1^2\)f(t) dt.

Solution:

⇒ \(\int_1^2 \mathbf{f}(t) d t=\int_1^2\left[\left(t-t^2\right) \mathbf{i}+2 t^3 \mathbf{j}-3 \mathbf{k}\right] d t=\mathbf{i} \int_1^2\left(t-t^2\right) d t+\mathbf{j} \int_1^2 2 t^3 d t-\mathbf{k} \int_1^2 3 a\)

= \(\mathbf{i}\left[\frac{t^2}{2}-\frac{t^3}{3}\right]_1^2+\mathbf{j}\left[\frac{t^4}{2}\right]_1^2-\mathbf{k}[3 t]=\mathbf{i}\left[2-\frac{8}{3}-\frac{1}{2}+\frac{1}{3}\right]+\mathbf{j}\left[8-\frac{1}{2}\right]-\mathbf{k}[6-3]\)

= \(-\frac{5}{6} i+\frac{15}{2} j-3 k\)

4. If f (t) = 2i- J + 2k and  f(3) = 4i- 2j + 3k then find\(\int_2^ 3\)f.\(\frac{d f}{d t}\) dt.

Solution:

∴ \(\left.\int_2^3\left(\mathbf{f} \cdot \frac{d \mathbf{f}}{d t}\right) d t=\frac{1}{2} \mathbf{f}(t)^2\right]_2^3=\frac{1}{2}\left[\mathbf{f}(3)^2-\mathbf{f}(2)^2\right]=\frac{1}{2}\left[(4 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})^2-(2 \mathbf{i}-\mathbf{j}+2 \mathbf{k})^2\right]\)

= \(\frac{1}{2}[(16+4+9)-(4+1+4)]=\frac{1}{2}(29-9)=10\)

5. lf f(t) = 5t²i + tj-t³k. find\(\left(f \times \frac{d^2 \mathbf{f}}{d t^2}\right)\)dt.

Solution:

∴ \(\int_1^2\left(\mathbf{f} \times \frac{d^2 \mathbf{f}}{d t^2}\right) d t=\left[\mathbf{f} \times \frac{d \mathbf{f}}{d t}\right]_1^2=\left[-2 t^3 \mathbf{i}+5 t^4 \mathbf{j}-5 t^2 \mathbf{k}\right]_1^2=-14 \mathbf{i}+75 \mathbf{j}-15 \mathbf{k} .\)

6. If \(\frac{d^2 \mathbf{r}}{d t^2}\)= 6ti- 24t² + 4 sin t k, find r given that r = 2i + j and  \(\frac{d r}{d t}\)=−i−3k at t=0

Solution:

At \(t=0, \frac{d \mathbf{r}}{d t}=-\mathbf{i}-3 \mathbf{k} \Rightarrow-4 \mathbf{k}+\mathbf{c}_1=-\mathbf{i}-3 \mathbf{k} \Rightarrow \mathbf{c}_1=-\mathbf{i}+\mathbf{k}\)

∴ \(\frac{d \mathbf{r}}{d t}=3 t^2 \mathbf{i}-8 t^3 \mathbf{j}-4 \cos t \mathbf{k}-\mathbf{i}+\mathbf{k}=\left(3 t^2-1\right) \mathbf{i}-8 t^3 \mathbf{j}+(1-4 \cos t)\)

r = \(\left.\int \frac{d \mathbf{r}}{d t} d t=\int\left[\left(3 t^2-1\right) \mathbf{i}-8 t^3 \mathbf{j}+(1-4 \cos t) \mathbf{k}\right)\right] d t=\left(t^3-t\right) \mathbf{i}-2 t^4 \mathbf{j}+(t-4 \sin t) \mathbf{k}+\mathbf{c}_2\)

At \(t=0, \mathbf{r}=2 \mathbf{i}+\mathbf{j} \Rightarrow \mathbf{c}_2=2 \mathbf{i}+\mathbf{j}\)

∴ \(r=\left(t^3-t\right) \mathbf{i}-2 t^4 \mathbf{j}+(t-4 \sin t) \mathbf{k}+2 \mathbf{i}+\mathbf{j}=\left(t^3-t+2\right) \mathbf{i}+\left(1-2 t^4\right) \mathbf{j}+(t-4 \sin t) \mathbf{k}\)

7. lf\(\frac{d^2 \mathbf{r}}{d t^2}\)=−k²r, show that \(\left(\frac{d r}{d t}\right)^2\) = c −k²r².

Solution:

Given that \(\frac{d^2 \mathbf{r}}{d t^2}=-k^2 \mathbf{r} \Rightarrow 2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}=- k^2\left(2 \mathbf{r} \cdot \frac{d \mathbf{r}}{d t}\right)\)

⇒ \(\int\left(2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}\right) d t=-k^2 \int\left(2 \mathbf{r} \cdot \frac{d \mathbf{r}}{d t}\right) d t \Rightarrow\left(\frac{d \mathbf{r}}{d t}\right)^2=-k^2 \mathbf{r}^2+\mathbf{c} \Rightarrow\left(\frac{d \mathbf{r}}{d t}\right)^2=\mathbf{c}-k^2 \mathbf{r}^2\)

Vector Integration Line Integral Exercises

8. IfA = ti− t²j+ (t- 1)k, B = 2 t² + 6tk, find (a)\(\int_1^2\)(A.B) dt   (b) \(\int_0^2\) (A x B) dt

Solution:

1. \(\mathbf{A} \cdot \mathbf{B}=\left[t \mathbf{i}-t^2 \mathbf{j}+(t-1) \mathbf{k}\right] \cdot\left[2 t^2 \mathbf{i}+6 t \mathbf{k}\right]=2 t^3+6 t(t-1)=2 t^3+6 t^2-6 t^2\)

∴ \(\int_0^2(\mathbf{A} \cdot \mathbf{B}) d t=\int_0^2\left(2 t^3+6 t^2-6 t\right) d t=\left[\frac{4 t^4}{4}+\frac{6 t^3}{3}-\frac{6 t^2}{2}\right]_0^2=16+16-12=20\)

2. \(\mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
t & -t^2 & t-1 \\
2 t^2 & 0 & 6 t
\end{array}\right|=\mathbf{i}\left(-6 t^3-0\right)-\mathbf{j}\left(6 t^2-2 t^3+2 t^2\right)+\mathbf{k}\left(0+2 t^4\right)\)

= \(-6 t^3 \mathbf{i}+\left(2 t^3-8 t^2\right) \mathbf{j}+2 t^4 \mathbf{k}\)

⇒ \(\int_0^2(\mathbf{A} \times \mathbf{B}) d t=\int_0^2\left[-6 t^3 \mathbf{i}+\left(2 t^3-8 t^2\right) \mathbf{j}+2 t^4 \mathbf{k}\right] d t=\left[-\frac{6 t^4}{4} \mathbf{i}+\left(\frac{2 t^4}{4}-\frac{8 t^3}{3}\right) \mathbf{J}+\frac{2 t^5}{5} \mathbf{k}\right]_0^2\)

= \(-24 \mathbf{i}+\left(8-\frac{64}{3}\right) \mathbf{j}+\frac{64}{5} \mathbf{k}=-24 \mathbf{i}-\frac{40}{3} \mathbf{j}+\frac{64}{5} \mathbf{k}\)

9. Evaluate \(\int_c\)(x² +y²) dx, where C is the arc of the parabola y²= 4ax between (0, 0)and (a, 2a).

Solution:

⇒ \(\int_C\left(x^2+y^2\right) d x\)

⇒ \(=\int_{x=0}^{x=a}\left(x^2+4 a x\right) d x\)

⇒ \(=\left[\frac{x^3}{3}+2 a x^2\right]_{x=0}^{x=a}\)

⇒ \(=\frac{a^3}{3}+2 a^3\)

∴ \(\frac{7 a^3}{3}\)

10. Evaluate \(\int_c \frac{d x}{(x+y)}\) , where  C is the curve x=at²,y=2at,0≤t≥2

Solution:

⇒ \(\int_C \frac{d x}{x+y}\)

⇒ \(=\int_{t=0}^{t=2} \frac{d\left(a t^2\right)}{a t^2+2 a t}\)

⇒ \(=\int_{t=0}^{t=2} \frac{2 a t d t}{a t^2+2 a t}\)

∴ \(=2 \int_0^2 \frac{d t}{t+2}\)=\(2 \log (t+2)]_0^2\)

11. Show that \(\int_c\)[(x-y)³ dx + (x-y)³ dy] = 3πa⁴ taken along the circle x²+y² = a² in    the counter clockwise sense.

Solution:

The parametric equations circle x+y = a are x=a cos θ, y= a sin θ.

dx= -a sin θ dθ, dy= a cos θ dθ and θ varies from 0 to 2π.

⇒ \(\int_C\left[(x-y)^3 d x+(x-y)^3 d y\right]\)

⇒ \(=\int_0^{2 \pi}(a \cos \theta-a \sin \theta)^3(-a \sin \theta d \theta)\)+ \((a \cos \theta-a \sin \theta)^3(a \cos \theta d \theta)\)

⇒ \(a^4 \int_0^{2 \pi}(\cos \theta-\sin \theta)^4 d \theta=a^4 \int_0^{2 \pi}\left(\cos ^4 \theta-4 \cos ^3 \theta \sin \theta+6 \cos ^2 \theta \sin ^2 \theta-4 \cos \theta \sin ^4 \theta+\sin ^4 \theta\right) d \theta\)

⇒ \(a^4 4 \int_0^{\pi / 2}\left(\cos ^4 \theta-0+6 \cos ^2 \theta \sin ^2 \theta-0+\sin ^4 \theta\right) d \theta\)

⇒ \(4 a^4 \int_0^{\pi / 2}\left[\left(\cos ^2 \theta+\sin ^2 \theta\right)^2+4 \cos ^2 \theta \sin ^2 \theta\right] d \theta\)

∴ \(4 a^4 \int_0^{\pi / 2}\left(1+4 \cos ^2 \theta \sin ^2 \theta\right) d \theta=4 a^4\left[\frac{\pi}{2}+4 \times \frac{1}{4} \times \frac{1}{2} \times \frac{\pi}{2}\right]=3 \pi a^4\).

12. Define line integral and explain the Cartesian form of line integral.

Line integral:

An integral which is to be evaluated along a curve is called a ” Line Integral. Suppose r=xi+yj+zk defines a piecewise smooth curve C joining two points A and B. Suppose F is a vector point function defined and continuous along C. If s denotes the arc length of the curve C then \(\frac{d r}{d s}\)= T is a unit vector along the tangent to the curve C at the point r.

The component of the vector F along the tangent is F.\(\frac{d r}{d s}\). The integral of F.\(\frac{d r}{d s}\) along from A to B written as \(\int_A^B\left(F \cdot \frac{d r}{d s}\right)\) ds \(=\int_A^B \mathbf{F} \cdot d \boldsymbol{r}\)  \(=\int_C F \cdot d r\) . is an example of a line integral. it is called tangent line integral of F along C.

Cartesian form: If F = F₁i +F₂ j + F₃k then

F.dr =(F₁i +F₂ j + F₃k). (dxi+dyj+dzk)= F₁dx+F₂dy+F₃dz

∴ \(\int_C\)F.dr=\(\int_C\)F₁dx+F₂dy+F₃dz

If the parametric equation of the curve C are x=x(t),y(t), z=z(t) and if t₁ at A, t=t₂ at  B then \(\int_C\)F.dr=\(=\int_{t_1}^{t_2}\left[F_1 \frac{d x}{d t}+F_2 \frac{d y}{d t}+F_3 \frac{d z}{d t}\right]\)dt

13. If F- 3xyi- 5zj + 10xk, evaluate \(\int_c\)F.dr along the curve, x = t² + 1, y = 2t², z= t³  from t = 1 to t= 2.

Solution:

Given

F- 3xyi- 5zj + 10xk

Let \(\mathbf{r}=x \mathbf{i}+y \mathbf{J}+z \mathbf{k}=\left(t^2+1\right) \mathbf{i}+2 t^2 \mathbf{j}+t^3 \mathbf{k} \Rightarrow \frac{d \mathbf{r}}{d t}=2 t \mathbf{l}+4 t \mathbf{j}+3 t^2 \mathbf{k} …\)

⇒ \(\mathbf{F}=3 x y \mathbf{i}-5 z \mathbf{j}+10 x \mathbf{k}=3\left(t^2+1\right) 2 t^2 \mathbf{i}-5 t^3 \mathbf{j}+10\left(t^2+1\right) \mathbf{k}\)

⇒ \(6 t^2\left(t^2+1\right) \mathbf{i}-5 t^3 \mathbf{J}+10\left(t^2+1\right) \mathbf{k}\)

⇒ \(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}=6 t^2\left(t^2+1\right) \cdot 2 t-5 t^3 \cdot 4 t+10\left(t^2+1\right) \cdot 3 t^2\)

⇒ \(12 t^5+12 t^3-20 t^4+30 t^4+30 t^2=12 t^5+10 t^4+12 t^3+30 t^2\)

∴ \(\left.\int \mathbf{F} \cdot d \mathbf{r}=\int\left(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\right) d t=\int_1^2\left(12 t^5+10 t^4+12 t^3+30 t^2\right) d t=12 t^6+2 t^5+3 t^4+10 t^3\right]_1^2\)

= 138 + 64 + 48 + 80 – 2 – 2 – 3 – 10 = 320 – 17 = 303.

Volume Integrals Solved Examples

14. If F- 3xyi- 5zj + l0xk, evaluate \(\int_c\)F.dr along the curve, x =t² , y = 2t², from z = t³  to t= 1 t=2

Solution:

Given

F- 3xyi- 5zj + l0xk

Let r=xi+yj+zk=t²i+2t²j+t³k ⇒\(\frac{d r}{d t}\)=2ti+4tj+3t²k

F=3xyi-5zj+10xk=3t²(2t²)i-5t³j+10t²k= 6t⁴i-5t³j+10t²k

⇒ \(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}=\left(6 t^4 \mathbf{i}-5 t^3 \mathbf{j}+10 t^2 \mathbf{k}\right) \cdot\left(2 t \mathbf{i}+4 t \mathbf{j}+3 t^2 \mathbf{k}\right)\)

⇒ \(\left(6 t^4\right)(2 t)+\left(-5 t^3\right)(4 t)+\left(10 t^2\right)\left(3 t^2\right)=12 t^5-20 t^4+30 t^4=12 t^5+10 t^4\)

∴ \(\left.\int \mathbf{F} \cdot d \mathbf{r}=\int \mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\right) d t=\int\left(12 t^5+10 t^4\right) d t=\left[2 t^6+2 t^5\right]_1^2=(128+64)-(2+2)=188\).

15. Find ∫F .dr where F = xyi+yzj + zxk over the curve Curve r = ti + t²j + t³k, t varying from −1 to 1 .

Solution:

⇒ \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+3 \mathbf{k}=t \mathbf{i}+t^2 \mathbf{j}+t^3 \mathbf{k} \Rightarrow x=t, y=t^2, z=t^3 \text { and } \frac{d \mathbf{r}}{d t}=\mathbf{i}+2 t \mathbf{j}+3 t^2 \mathbf{k}\)

⇒ \(\mathbf{F}=x y \mathbf{i}+y z \mathbf{j}+z x \mathbf{k}=t^3 \mathbf{i}+t^{\mathbf{5}} \mathbf{j}+t^4 \mathbf{k}\)

∴ \(\int \mathbf{F} \cdot d \mathbf{r}=\int\left(\mathbf{F} \cdot \frac{d \mathbf{r}}{d t}\right) d t=\int_{-1}^1\left(t^3+2 t^5+3 t^5\right) d t=\int_{-1}^1\left(t^3+5 t^5\right) d t=\left[\frac{t^4}{4}+5 \frac{t^7}{7}\right]_{-1}^1\)

⇒ \(\left(\frac{1}{4}+\frac{5}{7}\right)-\left(\frac{1}{4}-\frac{5}{7}\right)=\frac{10}{7}\)

16. Find \(\int_c\)F.dr where C is the arc of y- x² in xy-plane from (0, 0) to (1, 1) and F=x²i +y²j.

Solution:  The equation of the given curve is y= x²⇒ 2x dx. Given F=xi+yj+zk the curve lies in xy plane from (0,0) t0 (1,2) the limits of integration are  x=0 to  x=1

∴ \(\int_c F \cdot d r\)

⇒ \(=\int_C x^2 d x+y^2 d y\)

⇒ \(=\int_0^1 x^2 d x+x^4(2 x d x)\)

⇒ \(=\int_0^1\left(x^2+2 x^3\right) d x\) \(=\left[\frac{x^3}{3}+\frac{2 x^6}{6}\right]_0^1\)

∴ \(=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)

17. If F = 3xy i -y² j, evaluate  \(\int_c\)Fdr where C is the curve y = 2x² in the xy-plane from (0, 0) to (1, 2).

Solution:

The equation of given curve C is y= x²⇒ 4x dx. The integration was performed in xy- plane along C from (0,0) to (1,2).

∴ x varies from 0 to 1.

⇒ \(\int_C F \cdot d r\)

⇒ \(\int_C\left(3 x y \mathbf{i}-y^2 \mathbf{j}\right) \cdot(d x \mathbf{i}+d y \mathbf{j}+d z \mathbf{k})\)

⇒ \(=\int_C\left(3 x y d x-y^2 d y\right)\)

⇒ \(=\int_{x=0}^{x=1} 3 x \cdot\left(2 x^2\right) d x-4 x^4 \cdot 4 x d x\)

⇒ \(=\int_0^1\left(6 x^3-16 x^3\right) d x\) \(\left.=6 \frac{x^4}{4}-16 \frac{x^6}{6}\right]_0^1\)

⇒ \(=\frac{3}{2}-\frac{8}{3}=\frac{9-16}{6}\)

∴ \(-\frac{7}{6}\).

18. Evaluate   \(\int_c\)F.dr, where F = x²y² i +yj and the curve C is y= 4x in the xy-plane c from (0, 0) to (4, 4).

Solution:

The equation of curve C is y²=4x⇒2y dy =4dx⇒ ydy =2dx.

∴ x varies from 0 to 4.

⇒ \(\int_C F \cdot d r\)

⇒ \(\int_C x^2 y^2 d x+y d y\)

⇒ \(\int_0^4 x^2 \cdot 4 x d x+2 d x\)

⇒ \(\int_0^4\left(4 x^3+2\right) d x\) \(\left.=x^4+2 x\right]_0^4\)

=256+8=264.

19. Evaluate \(\int_c\)F.dr where F = 3x²i + (2xz -y)j + zk along the straight line C from(0,0,0) to (1,2).

Solution:

Equation of the joining (0,0,0) to (2,1,3) are  \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}=t\) (say).

∴ x = 2t, y = t, z = 3t, t varies from 0 to 1. Also \(\frac{d x}{d t}=2, \frac{d y}{d t}=1, \frac{d z}{d t}=3\)

⇒ \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \Rightarrow d \mathbf{r}=d x \mathbf{i}+d y \mathbf{j}+d z \mathbf{k}=\left(\frac{d x}{d t} \mathbf{i}+\frac{d y}{d t} \mathbf{j}+\frac{d z}{d t} \mathbf{k}\right) d t=(2 \mathbf{i}+\mathbf{j}+3 \mathbf{k}) d t\)

⇒ \(\mathbf{F}=3 x^2 \mathbf{i}+(2 x z-y) \mathbf{j}+z \mathbf{k}=12 t^2 \mathbf{i}+\left(12 t^2-t\right) \mathbf{j}+3 t \mathbf{k}\)

⇒ \(\int_C \mathbf{F}. d \mathbf{r}=\int_C\left[12 t^2 \mathbf{i}+\left(12 t^2-t\right) \mathbf{j}+3 t \mathbf{k}\right] \cdot(2 \mathbf{i}+\mathbf{j}+3 \mathbf{k}) d t=\int_C\left(24 t^2+12 t^2-t+9 t\right) d t\)

∴ \(\left.\int_0^1\left(36 t^2+8 t\right) d t=12 t^3+4 t^2\right]_0^1=12+4=16\)

Exercises On-Line, Surface, And Volume Integrals With Answers

20. Evaluate \(\int_c\)A.dr where C is the line joining (0,0,0) and (2,1,1), given A = (2y + 3)i + xzj + (yz−x)k.

Solution:

The equations of the line joining (0,0,0) and (2,1,1) are  \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}\) =t.

Then along the line C,x=2t,y=t,z=t.

∴ At (0,0,0,) , t=0 and at (2,1,1,) , t=1.

r=xi+yj+zk=2ti+rj=tk.

∴ dr=(2i+j+k)dt.

∫A.dr=\(\int_C[2(2 y+3)+1(x z)+1(y z-x)] d t\)

=\(\int_0^1[2(2 t+3)+(2 t \cdot t)+(t \cdot t-2 t)] d t\)

∴ \(\left.=\int_0^1\left(2 t^2+2 t+6\right) d t=t^3+t^2+6 t\right]_0^1\)=8.

21. Evaluate \(\oint_c\)F .dr where C is the circle x²+y² = 1, z = 0 and F =yi + zj +xk.

Solution:

The equation of the circle is x+y+=1,z=0

∴ dz=0.

In parametric form, x=cos θ,y=sin θ, z=0 and θ varies from 0 to 2π.

∴ \(\int_C F \cdot d r\)=\(\int_C(y \mathbf{i}+z \mathbf{j}+x \mathbf{k}) \cdot(\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d z)\)

⇒ \(\int_C(y d x+z d y+x d z)\)=\(\int_C y d x\)

⇒ \(=\int_{\theta=0}^{2 \pi} \sin \theta(-\sin \theta) d \theta\)

⇒ \(\theta=-4 \int_0^{\pi / 2} \sin ^2 \theta d \theta\)

=4(1/2)(π/2)=−π

22. If F = (3x² + 6y)i- 14yzi + 20xz² k, evaluate \(\int_c\)F.dralong the straight line joining (0, 0, 0) to (1, 0, 0) to(l, 1, 0) to (1, 1, 1)5

Solution:

(1) Line integral along the line from (0, 0, 0) to (1, 0, 0):  Here y = 0, z = 0, x varies from 0 to 1, dy = 0, dz = 0.

\(\int_{c_1} \mathbf{F} \cdot d r\)=\(\int_0^1 3 x^2 d x\)

=1.

(2) Line integral along the line from (1,0,0) to (1,1,0):

Here x=1, y varies from 0 to 1, z=0; dx=0, dz=0.

∴ \(\int_{c_2} \mathbf{F} \cdot d r\)=\(\int_0^1\)=0

(3) Line integral along the line from (1,1,0) t0 (1,1,1):

Here x=1,y=1, and z varies from 0 to 1. dx=0, dy=0.

⇒ \(\int_{c_3} \mathbf{F} \cdot d r\)=\(\int_0^1 20 z^2 d z\)=\(\frac{20 z^3}{3}\)\(]_0^1\)

⇒ \(\frac{20}{3}\)

∴ \(\int_{c} \mathbf{F} \cdot d r\)\(\)=\(\int_{c_1} \mathbf{F} \cdot d r\)+\(\int_{c_2} \mathbf{F} \cdot d r\)+\(\int_{c_3} \mathbf{F} \cdot d r\)=1+0+\(\frac{20}{3}\)=\(\frac{23}{3}\).

Step-By-Step Solutions To Vector Integration Problems

23. If F =(x² +y²)i− 2xy j, evaluate \(\oint_c\)F .dr where the curve C is the rectangle in the xy-plane bounded by y = 0,y = b, x = 0, x = a.

Solution:

Since the integration takes place in xy -plane (z=0),

∴ \(\oint_c\)F.dr= \(\oint_c\)F₁dx+F₂ dy=\(\oint_c\)(x²+y²)dx-2xy dy

(1) Line integral along OP: Here y=0, dy=0 and x varies from o to a. \(\int_{O P} F \cdot d r\)=\(\int_0^a x^2 d x\)=\(\frac{a^3}{3}\)

(2)  Line integral along PQ: Here x=a, dx=0, and y changes from 0 to b.

∴ \(\int_{P Q} F \cdot d r\)=\(\int_a^b(-2 a y) d y\)=\(\left.-a y^2\right]_0^b\)=−ab²

(3) Line integral along QR: Here y=b, dy=0, and x changes from a to 0.

∴\(\int_{Q R} F \cdot d r\)=\(\int_b^0 0\)=0

∴\(=\int_a^0\left(x^2+b^2\right) d x\)

=\(\left[\frac{x^3}{3}+b^2 x\right]_a^0\)

=\(-\frac{1}{3} a^3-a b^2\)

(4) Line integral along RO: Here x=0, dx=0, and y varies from b to 0.

∴ \(\int_{R O} F \cdot d r\)=\(\int_b^0 0\)

∴ \(\oint_c \boldsymbol{F} \cdot d \boldsymbol{r}\)=\(\int_{O P} F \cdot d r\)+ \(\int_{P Q} F \cdot d r\)+\(\int_{Q R} F \cdot d r\)+\(\int_{R O} F \cdot d r\)=\(\frac{a^3}{3}-a b^2-\frac{1}{3} a^3-a b^2\)=-2ab²

24. Find \(\int_c\)y²dx−x²dy, where C the curve represents sides of ΔABC, where A =(1, 0), B = (0, 1), C = (- 1, 0).

Solution:

Equation of \(\overleftrightarrow{A B}\) is \(\frac{y-0}{1-0}\)=\(=\frac{x-1}{0-1}\)⇒ y=1−x

Equation  of \(\overleftrightarrow{B C}\) is \(\frac{y-1}{0-1}\)\(=\frac{x-0}{-1-0}\) ⇒ y=1+x.

Equation of \(\overleftrightarrow{C A}\) is y=0.

Case(1): Line integral along \(\overleftrightarrow{A B}\): y=1-x, dy =−dx, x varies from 1 to 0

∴ \(\int_{c_1} y^2 d x-x^2 d y=\int_1^0(1-x)^2 d x-x^2(-d x)=\int_1^0\left(1+x^2-2 x+x^2\right) d x\)

= \(\int_1^0\left(1-2 x+2 x^2\right) d x=\left[x-x^2+\frac{2 x^2}{3}\right]_1^0=-\left(1-1+\frac{2}{3}\right)=-\frac{2}{3}\)

Case(2): Line integral aling \(\overleftrightarrow{B C}\): y=1+x,dy=dx, x varies from 0 to −1

∴ \(\int_{c_2} y^2 d x-x^2 d y=\int_0^{-1}(1+x)^2 d x-x^2 d x=\int_0^{-1}\left(1+2 x+x^2-x^2\right) d x\)

= \(\int_0^{-1}(1+2 x) d x=\left[x+x^2\right]_0^{-1}=-1+1=0\)

Case(3): Line integral along \(\overleftrightarrow{C A}\): y=0,dy=0, x varies from −1 to 1.

∴ \(\int_{c_3} y^2 d x-x^2 d y=\int_{-1}^1 0=0\)

∴ \(\int_C y^2 d x-x^2 d y=\int_{c_1} y^2 d x-x^2 d y+\int_{c_2} y^2 d x-x^2 d y+\int_{c_3} y^2 d x-x^2 d y\)

= \(-\frac{2}{3}+0+0=-\frac{2}{3}\)

25. Find the work done when a force F = (x² − y² + x) i− (2xy+y)j moves a particle in xy-plane from (0, 0) to (1, 1) along the parabola y²= x.

Solution:

Given curve is y²=x⇒ 2y dy=dx. Work done by F is \(\int_C F \cdot d r\). the integration is performed in xy-plane and y varies from 0 to 1.

⇒ \(\int_c \mathbf{F} \cdot d \mathbf{r}=\int_c\left(x^2-y^2+x\right) d x-(2 x y+y) d y=\int_0^1\left(y^4-y^2+y^2\right) 2 y d y-\left(2 y^3+y\right) d y\)

⇒ \(\int_0^1\left(2 y^5-2 y^3-y\right) d y=\left[\frac{y^6}{3}-\frac{y^4}{2}-\frac{y^2}{2}\right]_0^1=\frac{1}{3}-\frac{1}{2}-\frac{1}{2}=\frac{2-3-3}{6}=\frac{-4}{6}=\frac{-2}{3}\)

26. If F = (x +y²) i−2xj+ 2yzk, evaluate \(\int_S\) F . N dS where S is the surface of plane 2x +y + 2z = 6 in the first octant.

Solution: Let φ =2x+y=2z-6.

The vector normal to the surfaces S is ∇φ=\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2i+j+2k.

Unit normal vector, N \(=\frac{2 i+j+2 k}{\sqrt{4+1+4}}\)=\(=\frac{1}{3}(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})\).

Let R be the projection of S on xy-plane.

Now R is bounded by the x-axis, y-axis, and the line 2x+y=6z= 0

⇒ \(\mathbf{F} \cdot \mathbf{N}=\left[\left(x+y^2\right) \mathbf{i}-2 x \mathbf{j}+2 y z \mathbf{k}\right] \cdot \frac{1}{3} (2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})=\frac{1}{3}\left[2 x+2 y^2-2 x+4 y z\right]=\frac{2}{3}\left(y^2+2 y z\right)\)

⇒ \(\mathbf{N} \cdot \mathbf{k}=\frac{1}{3}(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k}) \cdot \mathbf{k}=\frac{2}{3}\)

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\iint_R \frac{2}{3}\left(y^2+2 y z\right) \frac{d x d y}{(2 / 3)}=\iint_R\left(y^2+2 y z\right) d x d y\)

⇒ \(\iint_R\left[y^2+y(6-2 x-y)\right] d x d y=2 \int_{x=0}^{x=3} \int_{y=0}^{y=6-2 x} y(3-x) d x d y\)

⇒ \(2 \int_{x=0}^{x=3}\left[\frac{y^2}{2}(3-x)\right]_{y=0}^{y=6-2 x} d x=\int_{x=0}^{x=3}(3-x)(6-2 x)^2 d x\)

⇒ \(\int_0^3(3-x)\left(36-24 x+4 x^2\right) d x=\int_0^3\left(108-108 x+36 x^2-4 x^3\right) d x\)

∴ \(\left[108 x-54 x^2+12 x^3-x^4\right]_0^3=324-486+324-81=81\)

Vector Integration Solved Problems For Students

27. Evaluate \(\int_S\)F.N dS where F = xy i- x²j + (x + z) k and S is the surface of the planes 2x + 2y + z=6 in the first octant.

Solution:

Let φ =2x+2y=z-6.

The vector normal to the surfaces S is  ∇φ=\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2i+j+2k.

Unit normal vector, N \(=\frac{2 i+2 j+k}{\sqrt{4+4+1}}\)=\(\frac{1}{3}\)(2i+2j+k).

Let R be the projection of S on xy-plane.

Now R is bounded by x-axis, y-axis, and the line 2x+y=6,z= 0

F.N =[(xyi-x²j+(x+z)K].1/3(2i+2j+k)=\(=\frac{1}{3}\left(2 x y-2 x^2+x+z\right)\).

⇒ \(\mathbf{N} \cdot \mathbf{k}=\frac{1}{3}(2 \mathbf{i}+2 \mathbf{j}+\mathbf{k}) \cdot \mathbf{k}=\frac{1}{3}\)

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\iint_R \frac{1}{3}\left(2 x y-2 x^2+x+z\right) \frac{d x d y}{(1 / 3)}\)

⇒ \(\iint_R\left(2 x y-2 x^2+x+6-2 x-2 y\right) d x d y=\int_{x=0}^{x=3} \int_{y=0}^{y=3-x}\left(2 x y-2 x^2-x-2 y+6\right) d x d y\)

⇒ \(\int_{x=0}^{x=3}\left[x y^2-2 x^2 y-x y-y^2+6 y\right]_{y=0}^{y=3-x} d x\)

⇒ \(\int_{x=0}^{x=3}\left[x(3-x)^2-2 x^2(3-x)-x(3-x)-(3-x)^2+6(3-x)\right] d x\)

⇒ \(\int_0^3\left(9 x-6 x^2+x^3-6 x^2+2 x^3-3 x+x^2-9+6 x-x^2+18-6 x\right) d x\)

⇒ \(\int_0^3\left(3 x^3-12 x^2+6 x+9\right) d x=\left[\frac{3 x^4}{4}-4 x^3+3 x^2+9 x\right]_0^3=\frac{243}{4}-108+27+27\)

∴ \(\frac{243}{4}-54=\frac{27}{4}\)

28.Evaluate\(\int_S\) F. N dS where F= 18zi- 12J + 3yk and S is the part of the planes 2x + 3y + 6z=12 located in the first octant.

Solution: Let φ =2x+3y+6z-12. The vector normal to the plane is ∇φ=2i+3j+6k.

Unit normal vector, N=\(\frac{2 i+3 j+6 k}{\sqrt{4+9+36}}\)=\(\frac{2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}}{7}\).

Let R be the projection of S on xy-plane.

∴ \(\mathbf{F} \cdot \mathbf{N}=(18 z \mathbf{i}-12 \mathbf{j}+3 y \mathbf{k}) \cdot\left(\frac{2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}}{7}\right)=\frac{1}{7}(36 z-36+18 y)=\frac{6}{7}(6 z-6+3 y)\)

⇒ \(\mathbf{N} \cdot \mathbf{k}=\frac{1}{7}(2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}) \cdot \mathbf{k}=\frac{6}{7} \cdot \quad d S=\frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\frac{d x d y}{6 / 7}\)

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\iint_R \frac{6}{7}(6 z-6+3 y) \frac{d x d y}{(6 / 7)}=\iint_R(6 z-6+3 y) d x d y\)

⇒ \(\iint_R(12-2 x-3 y-6+3 y) d x d y=\int_{x=0}^{x=6} \int_{y=0}^{y=(12-2 x) / 3}(6-2 x) d x d y\)

⇒ \(\int_{x=0}^{x=6}[(6-2 x) y]_{y=0}^{y=(12-2 x) / 3} d x=\int_{x=0}^{x=6} \frac{4}{3}(3-x)(6-x) d x=\frac{4}{3} \int_0^6\left(x^2-9 x+18\right) d x\)

∴ \(\frac{4}{3}\left[\frac{x^3}{3}-1 \frac{9 x^2}{2}+18 x\right]_0^6=\frac{4}{3}[72-162+108]=24\)

29. Evaluate \( \int_S\)F.N dS where F =yi + 2xj -zk and S is the surface of the plane s 2x+y = 6 in the first octant, cut of f by the plane z = 4.

Solution: Let φ =2x+y-6.

The vector to the surfaces S is ∇φ =\(i \frac{\partial \varphi}{\partial x}+j \frac{\partial \varphi}{\partial y}+k \frac{\partial \varphi}{\partial z}=\)=2i+j

Unit normal vector  to th surface is N \(=\frac{\nabla \varphi}{|\nabla \varphi|}\)=\(\frac{2 \mathbf{i}+\mathbf{j}}{\sqrt{5}}\).

Let r be the projection of S over xz plane.

The boundaries of R is x=0 to x=3 and z-0 to z=4.

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R(y \mathbf{i}+2 x \mathbf{j}-z \mathbf{k}) \cdot \frac{2 \mathbf{i}+\mathbf{j}}{\sqrt{5}} \frac{d x d z}{|\mathbf{j} \cdot \mathbf{n}|}\)

⇒ \(\iint_R \frac{2 y+2 x}{\sqrt{5}} \frac{d x d z}{1 / \sqrt{5}}=\iint_R[2(6-2 x)+2 x] d x d z=\int_{x=0}^{x=3} \int_{z=0}^{z=4}(12-2 x) d x d z\)

⇒ \(\left.\int_{x=0}^{x=3}(12-2 x) z\right]{ }_{z=0}^{z=4} d x=\int_{x=0}^{x=3}(12-2 x) 4 d x=4\left[12 x-x^2\right]_0^3=4[36-9]=108\)

Practice Problems Online, Surface, And Volume Integrals

30. Evaluate Evaluate \(\int_S\)F.N dS where F = zi +xj−  3y² zk and S is the surfaces x² +y² = 16 included in the first octant between z = 0 and z = 5.

Solution: Let φ = x²+y²-16

The normal to the surfaces S is grad  φ = ∇φ=\(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2xi+2yj

Unit normal vector, N=\(\frac{2 x i+2 y i}{\sqrt{4 x^2+4 y^2}}\)

=\(\frac{x \mathbf{i}+y \mathbf{j}}{\sqrt{x^2+y^2}}\)

=\(\frac{x \mathbf{i}+y \mathbf{j}}{4}\)

Let R be the projection of S on xy-plane .

In yz- plane, for the surface y varies from 0 to 4 and z varies from 0 to 5.

Then \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}=\int_{y=0}^{y=4} \int_{z=0}^{z=5} \frac{x z+x y}{4} \cdot \frac{d y d z}{x / 4}\)

⇒ \(\int_{y=0}^{y=4} \int_{z=0}^{z=5}(z+y) d y d z=\int_{y=0}^{y=4}\left[z^2 / 2+y z\right]{ }_{z=0}^{z=5} d y=\int_0^4\left[\frac{25}{2}+5 y\right] d y=\left[\frac{25 y}{2}+\frac{5 y^2}{2}\right]_0^4\)

⇒ 50 + 40 = 90.

31. Evaluate \(\int_S\)F.N dS where F = 6z i + (2x +y) j- x k and S is the surface of the region bounded by x² + z² = 9, x == 0,y = 0, z = 0 and y = 8.

Solution: Let φ x² +y²+ z² − 9

The normal to the surfaces S is ∇φ \(\mathbf{i} \frac{\partial \varphi}{\partial x}+\mathbf{j} \frac{\partial \varphi}{\partial y}+\mathbf{k} \frac{\partial \varphi}{\partial z}\)=2xi+2zk

Unit normal vector, N=\(\frac{2 x i+2 y i}{\sqrt{4 x^2+4 y^2}}\)

=\(\frac{x \mathbf{i}+z \mathbf{k}}{\sqrt{x^2+z^2}}\)

=\(=\frac{1}{3}(x \mathbf{i}+z \mathbf{k})\)

F.N =(6zi+(2x+y)j-xk).\(\frac{1}{3}\) (xi+zk)=1/3 (6xz-xz)=\(\frac{5}{3}\) xz

N.K=\(\frac{1}{3}\) (xi+zk)k. =z/3. Let R be the projection of S on xy-plane.

In xy-plane, for the surfaces x varies from 0 to 3 and y varies from 0 to 8.

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{k}|}=\int_{x=0}^{x=3} \int_{y=0}^{y=8}\left(\frac{5}{3} x z\right) \frac{d x d y}{(z / 3)}\)

\(=\int_{x=0}^{x=3} \int_{y=0}^{y=8} 5 x d x d y\)=\(\int_{x=0}^{x=3} 40 x d x=\left[20 x^{2}\right]_{0}^{3}=180\)

32. Evaluate \(\int_S\)F.N dS, where F =yzi + zxj + xyk. and S is the part ofthe sphere x² +y² + z² = 1 which lies in the first octant.

Solution: Let φ =x² +y²+ z²−1

Normal vector to the surfaces is ∇φ=2(xi+yj+zk)

Unit normal vector, N=\(\frac{2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{\sqrt{4 x^2+4 y^2+4 z^2}}\)

=\(\frac{x \mathbf{i}+y \mathbf{j}+\mathbf{k}}{\sqrt{x^2+y^2+z^2}}\)

=xi+yj+zk.

F.N (yzi+zxj+xyk).(xi+yj+zk)=xyz+xyz+xyz=3xyz, N.i=x.

Let R be the projection of S on yz-plane.

Then \(\int_S F \cdot N d S\)

=\(\iint_R F \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \boldsymbol{i}|}\)

=\(\iint_R 3 x y z d y d z / x\)

=\(3 \iint_R y z d y d z\)

In yz-plane x=0, the equation of the surface becomes y²+ z²=1

∴ y varies from o to 1 and z varies from o to \(\sqrt{1-y^2}\)

∴ \(\int_s \mathbf{F} \cdot \mathbf{N} d S=3 \int_{y=0}^{y=1} \int_{z=0}^{z=\sqrt{1-y^2}} y z d y d z=3 \int_{y=0}^{y=1}\left[\frac{z^2}{2}\right]_{z=0}^{z=\sqrt{1-y^2}} y d y\)

⇒ \(\frac{3}{2} \int_0^1 y\left(1-y^2\right) d y=\frac{3}{2}\left[\frac{y^2}{2}-\frac{y^4}{4}\right]=\frac{3}{2}\left[\frac{1}{2}-\frac{1}{4}\right]=\frac{3}{8}\)

33. Evaluate \(\int_S\)F.N dS where F =y²z² i + z²x² j + x²y² k and the surfaces x² +y² + z² = 1 above xy-plane.

Solution: Let φ= x² +y²+ z²−1

Normal vector to the surfaces is ∇φ=2(xi+yj+zk)

Unit normal vector, N=\(=\frac{2(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{\sqrt{4 x^2+4 y^2+4 z^2}}\)

=\(\frac{x+y+z k}{\sqrt{x^2+y^2+z^2}}=\)

=xi+yj+zk.

F.N =( y² z² i+z² x² j+x² y² k).(xi+yj+zk)=xy² z² +x² yz² +x² y² z

Let R be the projection of S in xy−plane.

Then \(\int_S F \cdot N d S\)

=\(\iint_R F \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}\)

=\(\iint_R \frac{\left(x y^2 z^2+x^2 y z^2+x^2 y^2 z\right)}{z} d x d y\)

=\(\iint_R\left(x y^2 z+x^2 y z+x^2 y^2\right) d x d y\)

In xy-plane , z=0 and the equation of the surfaces becomes x+y=1.

x varies from -1 to 1 and z varies from

=\(\sqrt{1-x^2}\)  to \(\sqrt{1-x^2}\)

∴ \(\int_S F \cdot N d S\)

=\(\iint_R\left(x y^2 z+x^2 y z+x^2 y^2\right) d x d y\)

=\(\iint_R x^2 y^2 d x d y\)

=\(y\int_{x=-1}^{x=1} \int_{y=-\sqrt{1-x^2}}^{y=\sqrt{1-x^2}} x^2 y^2 d x d y\)

⇒ \(4 \int_{x=0}^{x=1} \int_{y=0}^{y=\sqrt{1-x^2}} x^2 y^2 d x d y=4 \int_{x=0}^{x=1}\left[\frac{x^2 y^3}{3}\right]_0^{\sqrt{1-x^2}} d x=\frac{4}{3} \int_{x=0}^{x=1} x^2\left(1-x^2\right)^{3 / 2} d x\)

Put x = sin θ.

Then dx = cos θ dθ

x = 0, 1 ⇒ θ = 0, π/2

⇒ \(\frac{4}{3} \int_{\theta=0}^{0=\pi / 2} \sin ^2 \theta\left(1-\sin ^2 \theta\right)^{3 / 2} \cos \theta d \theta\)

⇒ \(\frac{4}{3} \int_0^{\pi / 2} \sin ^2 \theta \cos ^4 \theta d \theta=\frac{4}{3} \times \frac{1}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}=\frac{\pi}{24}\)

34. Find the surface area of the sphere given by x = a sin θ cos φ, y = a sin θ sin φ , z= a cos θ,0≤θ≤π,0≤φ≤2π.

Solution: x² +y² +z² = a² sin² θ cos² φ+a² sin² θ sin² φ +a² cos²  θ

= a² sin² θ (cos² φ+sin² φ)a² cos²  θ=a² sin² θ  + a² cos²  θ= a²( sin² θ + cos²  θ) =a²

Volume Integrals Practice Problems With Step-By-Step Guidance

35. If F = 4xzi -y² j+yz k, evaluate ∫F . N dS where S is the surface of the cube bounded by x = 0, x = a ,y = 0, y=a, z = 0, z = a.

Solution:

Consider the cube OABCPQRS surrounded by the following faces:

(1) For the faces PQAR, i is the outward normal:

∴ N=i, x=a, dS=dy dz.

∴ \(\int_{R_1} \mathbf{F} \cdot \mathbf{N} d S=\int_{y=0}^{y=a} \int_{z=0}^{z=a}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{i} d y d z\)

⇒ \(\int_{y=0}^{y=a} \int_{z=0}^{z=a} 4 x z d y d z=\int_{y=0}^{y=a} \int_{z=0}^{z=a} 4 a z d y d z\)

⇒ \(\int_{y=0}^{y=a}\left[2 a z^2\right]_{z=0}^{z=a} d y=\int_{y=0}^{y=a} 2 a^3 d y=\left[2 a^3 y\right]_0^a=2 a^4\)

(2) For the faces OBSC,−i is the outward normal:

∴ N=-i,x=0 , and dS =dy dz

∴ \(\int_{R_2} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\iint_{R_2}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z\)

=\(-\iint_{R_2} 4 x z d y d z\)=0

(3) For the face BQPS, j is the outward normal:

∴N=j,y=a, and dS=dx dz

∴\(\int_{R_3} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\iint_{R_3}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right)\).j dx dz

=\(-\iint_{R_3} y^2 d x d z\)

=\(-a^2 \int_{x=0}^{x=a} \int_{z=0}^{z=a} d x d z\)

=\(-a^2[x]_0^a[z]_0^a\)

=-a⁴

(4) For the OARC, -j is the  outward normal:

∴ N=-j, y=0 and dS =dx dz

∴ \(\int_{R_4} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\int_{R_4}\)(4xzi-y²j+yzk).(-j) dS

=\(\int_{R_4}\)∫y² dx dz=0

(5) For the face PRCS, k is the outward normal:

∴ N=k, z=a, and dS= dx dy

∴ \(\int_{R_5} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\int_{R_5}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot \mathbf{k} d S\)

=\(\iint_{R_5} y z \cdot d x d y\)

=\(\int_{x=0}^{x=a} \int_{y=0}^{y=a} a y d x d y\)

=\(a[x]_0^a \cdot\left[\frac{y^2}{2}\right]_0^a\)

∴ \(=\frac{a^4}{2}\)

(6) For the face OAQB, -k is the outward normal:

∴ N=-k, z=0 and dS= dx dy

∴ \(\int_{R_6} \mathbf{F} \cdot \mathbf{N} d S\)

=\(\int_{R_6}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{k}) d S\)

=\(-\iint_{R_6} y z d x d y\)=0

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d S\)=2a⁴+0-a⁴+0+½ a⁴+0=3/2a⁴.

36. If F = 2xzi−xj+yk, evaluate ∫F dV where V is the region bounded by the surfaces x = 0, x = 2, y = 0, y = 6, z=x², z = 4.

Solution:

⇒ \(\int_V \mathbf{F} d V=\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4}\left(2 x z \mathbf{i}-x \mathbf{j}+y^2 \mathbf{k}\right) d x d y d z\)

⇒ \(i\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} 2 x z d x d y d z-\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} x d x d y d z+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} y^2 d x d y d z\)

⇒ \(\left.\left.\left.=\mathbf{i} \int_{x=0}^{x=2} \int_{y=0}^{y=6} x z^2\right]_{z=x^2}^{z=4} d x d y-\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6} x z\right]_{z=x^2}^{z=4} d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6} y^2 z\right]_{z=x^2}^{z=4} d x d y\)

⇒ \(\text { i } \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(16 x-x^5\right) d x d y-\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4 x-x^3\right) d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4 y^2-x^2 y^2\right) d x d y\)

⇒ \(\left.\left.=\mathbf{i} \int_{x=0}^{x=2}\left(16 x-x^5\right) y\right]_{y=0}^{y=6} d x-\mathbf{j} \int_{x=0}^{x=2}\left(4 x-x^3\right) y\right]_{y=0}^{y=6} d x+\mathbf{k} \int_{x=0}^{x=2}\left[4 y^3 / 3-x^2 y^3 / 3\right]_{y=0}^{y=6} d x\)

⇒ \(\mathbf{i} \int_0^2\left(96 x-6 x^5\right) d x-\mathbf{j} \int_0^2\left(24 x-6 x^3\right) d x+\mathbf{k} \int_0^2\left(288-72 x^2\right) d x\)

∴ \(\mathbf{i}\left[48 x^2-x^6\right]_0^2-\mathbf{j}\left[12 x^2-3 x^4 / 2\right]_0^2+\mathbf{k}\left[288 x-24 x^3\right]_0^2=128 \mathbf{i}-24 \mathbf{j}+384 \mathbf{k}\)

Solved Examples Of Vector Calculus Integrals

37. Evaluate \(\int_V\)F dV where F =xi +yj + zk and V is the region bounded by x = 0, x =2,y = 0, y = 6, z = 4 and z=x² .

Solution:

⇒ \(\int_V F d V=\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) d x d y d z\)

⇒ \(i\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} x d x d y d z+j\int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} y d x d y d z+k \int_{x=0}^{x=2} \int_{y=0}^{y=6} \int_{z=x^2}^{z=4} z d x d y d z\)

⇒ \(i \int_{x=0}^{x=2} \int_{y=0}^{y=6}[x z]_{z=x^2}^4 d x d y+\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6}[y z]_{z=x^2}^{z=4} d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left[\frac{z^2}{2}\right]_{z=4}^{x^2} d x d y\)

⇒ \(i \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4 x-x^3\right) d x d y+\mathbf{j} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left(4-x^2\right) y d x d y+\mathbf{k} \int_{x=0}^{x=2} \int_{y=0}^{y=6}\left[8-\frac{x^4}{2}\right] d x d y\)

⇒ \(\mathbf{i} \int_{x=0}^{x=2}\left[\left(4 x-{x}^3\right) y\right]_{y=0}^{y=6} d x+\mathbf{j} \int_{x=0}^{x=2}\left[\left(4 x-x^2\right) \frac{y^2}{2}\right]_{y=0}^{y=6} d x+\mathbf{k} \int_{x=0}^{x=2}\left[\left(8-\frac{x^4}{2}\right) y\right]_{y=0}^{y=6} d x\)

⇒ \(\mathbf{i} \int_{x=0}^{x=2} 6\left(4 x-x^3\right) d x+\mathbf{j} \int_{x=0}^{x=2} 18\left(4-x^2\right) d x+\mathbf{k} \int_{x=0}^{x=2} 6\left(8-\frac{x^4}{2}\right) d x\)

⇒ \(\mathbf{i}\left[12 x^2-\frac{6 x^4}{4}\right]_{x=0}^{x=2}+\mathbf{j}\left[18\left(4 x-\frac{x^3}{3}\right)\right]_{x=0}^{x=2}+\mathbf{k}\left[6\left(8 x-\frac{x^5}{10}\right)\right]_{x=0}^{x=2}\)

⇒ \(\mathbf{i}(48-24)+\mathbf{j}\left[18\left(8-\frac{8}{3}\right)\right]+\mathbf{k}\left[6\left(16-\frac{32}{10}\right)\right]=24 \mathbf{i}+96 \mathbf{j}+\frac{384}{5} \mathbf{k}\)

38. If F = (2x²– 3z) i- 2xy j- 4x k, then evaluate ∫\(\int_V\)∫∇.F dV where V is the closed region bounded by the planes x = 0,y = 0, z = 0 and 2x + 2y + z = 4. Also, Evaluate ∫\(\int_V\)∫∇×F dV.

Solution:

∴ \(\nabla \cdot \mathbf{F}=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) \cdot\left[\left(2 x^2-3 z\right) \mathbf{i}-2 x y \mathbf{j}-4 x \mathbf{k}\right]\)

⇒ \(\frac{\partial}{\partial x}\left(2 x^2-3 z\right)+\frac{\partial}{\partial y}(-2 x y)+\frac{\partial}{\partial z}(-4 x)=4 x-2 x=2 x\)

∴ \(\iiint_V \nabla \cdot \mathbf{F} d V=\iiint_V 2 x d x d y d z=2 \int_{x=0}^{x=2} \int_{y=0}^{y=2-x} \int_{z=0}^{z=4-2 x-2 x} x d x d y d z\)

⇒ \(\left.2 \int_{x=0}^{x=2} \int_{y=0}^{y=2-x} x[z]\right]_{z=0}^{z=4-2 x-2 y} d x d y=2 \int_{x=0}^{x=2} \int_{y=0}^{y=2-x} x(4-2 x-2 y) d x d y\)

⇒ \(\left.2 \int_{x=0}^{x=2}\left[4 x y-2 x^2 y-x y^2\right]\right]_{y=0}^{y=2-x} d x=2 \int_0^2\left[4 x(2-x)-2 x^2(2-x)-x(2-x)^2\right] d x\)

⇒ \(2 \int_0^2\left[x^3-4 x^2+4 x\right] d x=2\left[\frac{1}{4} x^4-\frac{4}{3} x^3+2 x^2\right]_0^2=2\left[4-\frac{32}{3}+8\right]=\frac{8}{3}\)

We have \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x^2-3 z & -2 x y & -4 x
\end{array}\right|\)

⇒ \(\left[\frac{\partial}{\partial y}(-4 x)-\frac{\partial}{\partial z}(-2 x y)\right] \mathbf{i}-\left[\frac{\partial}{\partial x}(-4 x)-\frac{\partial}{\partial z}\left(2 x^2-3 z\right)\right] \mathbf{j}+\left[\frac{\partial}{\partial x}(-2 x y)-\frac{\partial}{\partial y}\left(2 x^2-3 z\right)\right] \mathbf{k}\)

⇒ \(0 \mathbf{i}-(-4+3) \mathbf{j}+(-2 y) \mathbf{k}=\mathbf{j}-2 y \mathbf{k}\)

∴ \(\iiint_V \nabla \times \mathbf{F} d V=\iiint_V(\mathbf{j}-2 y \mathbf{k}) d x d y d z\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=2-x z=4} \int_{z=0}^{-2 x-2 y}(\mathbf{j}-2 y \mathbf{k})dx dy dz =\int_{x=0}^{x=2} \int_{y=0}^{y=2-x}(\mathbf{j}-2 y \mathbf{k})(4-2 x-2 y) d x d y.\)

⇒ \(\int_{x=0}^{x=2}\left[\mathbf{j}\left(4 y-2 x y-y^2\right)-2 \mathbf{k}\left(2 y^2-x y^2-\frac{2}{3} y^3\right)\right]_{y=0}^{y=2-x} d x\)

⇒ \(=\int_{x=0}^{x=2}\left[\mathbf{j}(2-x)(4-2 x-2+x)-2 \mathbf{k}(2-x)^2\left\{2-x-\frac{2}{3}(2-x)\right\}\right] d x\)

⇒ \(\int_0^2\left[(2-x)^2 \mathbf{j}-\frac{2}{3}(2-x)^3 \mathbf{k}\right] d x=\int_0^2\left[(x-2)^2 \mathbf{j}+\frac{2}{3}(x-2)^3 \mathbf{k}\right] d x\)

⇒ \(\left[\frac{(x-2)^3}{3}\right]_0^2 \mathbf{j}+\left[\frac{2}{3} \frac{(x-2)^4}{4}\right]_0^2 \mathbf{k}=\frac{8}{3} \mathbf{j}-\frac{8}{3} \mathbf{k}=\frac{8}{3}(\mathbf{j}-\mathbf{k})\)

39. Evaluate ∫∫∫(2x+y)dV where V is closed region bounded by the cylinder z = 4−x² and the planes x = 0,y = 0 and y = 2, z = 0.

Solution:

∴ \(\iiint_V(2 x+y) d V=\int_{x=0}^{x=2} \int_{y=0}^{y=2} \int_{z=0}^{z=4-x^2}(2 x+y) d x d y d z=\int_{x=0}^{x=2} \int_{y=0}^{y=2}[(2 x+y) z]\int_{z=0}^{z=4-x^2} d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=2}(2 x+y)\left(4-x^2\right) d x d y=\int_{x=0}^{x=2}\left[\left(4-x^2\right)\left(2 x y+\frac{y^2}{2}\right)\right]_{y=0}^{y=2} d x\)

⇒ \(\int_{x=0}^{x=2}\left(4-x^2\right)(4 x+2) d x=\int_0^2\left(8+16 x-2 x^2-4 x^3\right) d x\)

⇒ \(\left[8 x+8 x^2-\frac{2 x^3}{3}-x^4\right]_0^2=16+32-\frac{16}{3}-16=\frac{80}{3}\)

40. If φ = 45x²y, evaluate ∫\(\int_V\)∫φ dV where V is the closed region bounded by the plane 4x + 2y + z = 8,x = 0,y = 0,z = 0.

Solution:

∴ \(\iiint_V \varphi d V=\iiint_V 45 x^2 y d x d y d z=\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x} \int_{z=0}^{z=8-4 x-2 y} 45 x^2 y d x d y d z\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x}\left[45 x^2 y z\right]_{z=0}^{z=8-4 x-2 y} d x d y=\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x}\left[45 x^2 y(8-4 x-2 y)\right] d x d y\)

⇒ \(\int_{x=0}^{x=2} \int_{y=0}^{y=4-2 x} 45 x^2\left(8 y-4 x y-2 y^2\right) d x d y=\int_{x=0}^{x=2} 45 x^2\left[4 y^2-2 x y^2-\frac{2}{3} y^3\right]_{y=0}^{y=4-2 x} d x\)

⇒ \(\int_{x=0}^{x=2}\left[720 x^2(2-x)^2-360 x^3(2-x)^2-240 x^2(2-x)^3\right] d x\)

⇒ \(120 \int_{x=0}^{x=2}\left[6 x^2\left(4-4 x+x^2\right)-3 x^3\left(4-4 x+x^2\right)-2 x^2\left(8-12 x+6 x^2-x^3\right)\right] d x\)

⇒ \(120 \int_{x=0}^{x=2}\left(24 x^2-24 x^3+6 x^4-12 x^3+12 x^4-3 x^5-16 x^2+24 x^3-12 x^4+2 x^5\right) d x\)

⇒ \(120 \int_{x=0}^{x=2}\left(8 x^2-12 x^3+6 x^4-x^5\right) d x=120\left[\frac{8 x^3}{3}-3 x^4+\frac{6 x^5}{5}-\frac{x^6}{6}\right]_0^2\)

= 2560 – 5760 + 4608 – 1280 = 128.

41. Evaluate I=∫∫∫\sqrt{\left(a^2 b^2 c^2-b^2 c^2 x^2-c^2 a^2 y^2-a^2 b^2 z^2\right)} dx dy dz taken throughout the domain \(\left\{(x, y, z): \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1\right\}\)

Solution: The limits of integration are ;x varies from -a to a; y varies from -b\(\sqrt{1-x^2 / a^2}\) to b\(\sqrt{1-x^2 / a^2}\) and z varies from -c \(\sqrt{1-x^2 / a^2-y^2 / b^2}\) to c \(\sqrt{1-x^2 / a^2-y^2 / b^2}\)

∴ \(I=\iiint \sqrt{\left(a^2 b^2 c^2-b^2 c^2 x^2-c^2 a^2 y^2-a^2 b^2 z^2\right)} d x d y d z\)

⇒ \(a b c \int_{x=-a}^{x=a} \int_{y=-b \sqrt{1-x^2 / a^2}}^{y=b \sqrt{1-x^2 / a^2}} \quad \int_{z=-c \sqrt{1-x^2 / a^2-y^2 / b^2}}^{z=c \sqrt{1-x^2 / a^2-y^2 / b^2}} \sqrt{\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)} d x d y d z\)

⇒ \(8a b c \int_{x=-a}^{x=a} \int_{y=0}^{y=b \sqrt{y=1-x^2 / a^2}} \quad \int_{z=0}^{z=c \sqrt{z=1-x^2 / a^2-y^2 / b^2}} \sqrt{\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)} d x d y d z\)

⇒ \(=8 a b c^2 \int_{x=0}^{x=a} \int_{y=0}^{y=b \sqrt{y=1-x^2 / a^2}}\left[0+\frac{1}{2}\left(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)\left(\frac{\pi}{2}-0\right)\right] d x d y\)

⇒ \(8 a b c^2 \frac{\pi}{4} \int_{x=0}^{x=a} \int_{y=0}^{y=b \sqrt{1-x^2 / a^2}}\left[\left(1-\frac{x^2}{a^2}\right)-\frac{y^2}{b^2}\right] d x d y\)

⇒ \(2 a b c^2 \pi \int_{x=0}^{x=a}\left[\left(1-\frac{x^2}{a^2}\right) y-\frac{y^3}{3 b^2}\right]_{y=0}^{y=b \sqrt{1-x^2 / a^2}} d x\)

⇒ \(2 a b c^2 \pi \int_0^a b\left(1-\frac{x^2}{a^2}\right)^{3 / 2}-\frac{1}{3 b^2} b^3\left(1-\frac{x^2}{a^2}\right)^{3 / 2} d x\)

Put x = a sin θ.

Then dx = a cos θ dθ.

x = 0, a ⇒ θ = 0, π/2

⇒ \(2 a b c^2 \pi \int_0^a \frac{2 b}{3}\left(1-\frac{x^2}{a^2}\right)^{3 / 2} d x=\frac{4 a b^2 c^2 \pi}{3} \int_0^{\pi / 2}\left(1-\sin ^2 \theta\right)^{3 / 2}(a \cos \theta d \theta)\)

∴ \(\frac{4 a^2 b^2 c^2 \pi}{3} \int_0^{\pi / 2} \cos ^4 \theta d \theta=\frac{4 a^2 b^2 c^2 \pi}{3} \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2}=\frac{a^2 b^2 c^2 \pi^2}{4}\)

42. Find the volume of a sphere of radius ‘a’.

Solution: The equation of the sphere with the center origin and radius a is x²+y²+z²=a²

The volume of the sphere V \(=\int_V d V\)

The limits of integration are x=± a,y=±\(\sqrt{a^2-x^2}\),z=±\(\sqrt{a^2-x^2-y^2}\)

∴ \(V=\int_{v}dv=\int_{x=-a}^{x=a} \int_{y=- \sqrt{a^2 – x^2}}^{y= \sqrt{a^2+x^2}} \int_{z=- \sqrt{a^2-x^2-y^2}}^{z= \sqrt{a^2-x^2-y^2}}d x d y d z\)

⇒ \(8 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}} \int_{z=0}^{z=\sqrt{a^2-x^2-y^2}} d x d y d z=8 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}}[z]_{z=0}^{z=\sqrt{a^2-x^2-y^2}} d x d y\)

⇒ \(8 \int_{x=0}^{x=a} \int_{y=0}^{y=\sqrt{a^2-x^2}} \sqrt{a^2-x^2-y^2} d x d y\)

⇒ \(8 \int_{x=0}^{x=a}\left[\frac{y}{2} \sqrt{a^2-x^2-y^2}+\frac{a^2-x^2}{2} \text{Sin}^{-1} \frac{y}{\sqrt{a^2-x^2}}\right]_{y=0}^{y=\sqrt{a^2-x^2}} d x\)

⇒ \(8 \int_{x=0}^{x=a}\left(\frac{a^2-x^2}{2}\right) \frac{\pi}{2} d x=2 \pi\left[a^2 x-\frac{x^3}{3}\right]_0^a=2 \pi\left[a^3-\frac{a^3}{3}\right]=\frac{4 \pi a^3}{3}\)

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