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Rings, Integral Domains And Fields

Examples Of Rings In Mathematics With Solutions

The ring is the second algebraic system of the subject of Modern Algebra. The abstract concept of rings has its origin from the set of integers. Even though integers, real numbers, integers modulo − n, and Matrices are endowed with two binary operations, when dealt them in Groups we have considered only one binary operation ignoring the other.

The concept of Ring will take into account both addition and multiplication. The algebra of rings will follow the pattern already laid out for groups.

Definition. (Ring.) Let R be a non-empty set and +, · be two binary operations in R. (R, +, ·) is said to be a ring if, for a,b,c ∈ R;

R₁ · a + b = b + a

R₂· (a + b) + c = a + (b + c)

R₃. there exists 0 ∈ R such that a + 0 = a for a ∈ R.

R₄. there exists −a ∈ R such that a + (−a) = 0 for a ∈ R.

R₅· (a · b) · c = a · (b · c) and

R₆· a · (b + c) = a · b + a · c and (b + c) · a = b · a + c · a.

Note

1. The operation ’+’ is called the addition and the operation ’ · ’ is called the multiplication in the ring (R, +, •).

2. The ring (R, +, ·) is also called the ring R.

3. We write a.b as ab.

4. The properties R₁,R₂,R₃,R₄ merely state that (R, +) is a commutative group. Thus (R, +) is called the additive group of the ring R.

5. The identity element ’ 0 ’ in (R, +) is called the zero of the ring R. The zero of a ring should not be confused with the zero of the numbers.

6. By R₃, 0 + 0 = 0 for 0 ∈ R.

7. The properties R₅, and R₆ may be respectively called the Associative and Distributive laws.

In view of Note (4) and Note (7) a ring may also be defined as follows:

Definition. (Ring.) Le t.R be a non-empty set and +, · b e two binary operations in R.(R, +, ·) is said to be a ring, if

(1) (R, +) is a commutative group,

(2) (R, ·) is a semigroup and

(3) Distributive laws hold.

Definition. (Unity Element.) In a ring (R, +, ·) if there exists 1 ∈ R such that a.1 = 1. a = a for every a ∈ R then we say that R is a ring with unity element or identity element.

Note 1. If R is a ring with identity element, by R₄, we have −1 ∈ R so that

1 + (−1) = 0.

2. A ring with unity element contains at least two elements 0 and 1 if R 6= {0}.

Definition. In a ring (R, +, ·) if a.b = b.a for a, b ∈ R then we say that R is a commutative ring.

Imp. A ring R (1) need not be commutative under multiplication and (2) need not have an identity (unity) element under multiplication, unless or otherwise stated.

Example. 1. Let R = {0} and +, · be the operations defined by 0 + 0 = 0 and 0 · 0 = 0. Then (R, +, ·) is clearly a ring called the Null ring or Zero ring.

Integral Domains Definition And Examples Step-By-Step

Example 2. The set Z of integers w.r.t. usual addition and multiplication is a commutative ring with unity element.

For (1) (Z, +) is a commutative group (2) Multiplication is associative in Z and (3) Multiplication is distributive over addition.

Example. 3. The set N of natural numbers is not a ring w.r.t. usual addition and multiplication, because, (N, +) is not a group.

Example. 4. The sets Q, R, and C are rings under the usual addition and multiplication of numbers.

Example 5. The set of integers mod under the addition and multiplication mode is a ring.

Example. 6. The set of irrational numbers under addition and multiplication is not a ring as there is no zero element.

Let (R, +, ·) be a commutative ring with unity element. Then (R, +) is a commutative group and (R, ·) is a semi-group with identity. element 1. So we have the following results which are obvious from the theory of groups.

1. The zero element of R is unique and a + 0 = a for every element ’ a ’ in R.

2. For a ∈ R the additive inverse −a ∈ R is unique and a + (−a) = 0.

3. The identity element 1 ∈ R is unique and a · 1 = 1.a = a for every a ∈ R.

4. For a ∈ R, −(−a) = a. 5. For 0 ∈ R, −0 = 0. 6. For a, b ∈ R, −(a + b) = −a − b.

5. For a, b, c ∈ R, a + b = a + c ⇒ b = c and b + a = c + a ⇒ b = c

6. For a, b, x ∈ R, the equations a+x = b and x+a = b have unique solutions.

7. For 1 ∈ R, the identity element, −(−1) = 1.

8. For a, b₁, b₂, . . . . . . .b_n ∈ R, from R₆, we have a (b₁ + b₂ + . . . .. + b_n) = ab₁ + ab₂ + . . . + ab_n and (b₁ + b₂ + . . . .. + b_n) a = b₁a + b₂a + . . . . + b_na.

Notation. 1. If R is a ring and a, b ∈ R then a + (−b) ∈ R, a + (−b) is written as a − b.

9. If R is a ring and a ∈ R then a + a ∈ R and a + a is written as 2a.

10. If R is a ring and a ∈ R then a · a ∈ R and a · a is written as a2.

Rings, Integral Domains, And Fields Some Basic Properties Of Rings

Theorem 1. If R is a ring and 0, a, b ∈ R, then

1. 0a = a0 = 0,
2. a(−b) = (−a)b = −(ab)
3. (−a)(−b) = ab
4. a(b − c) = ab − ac. )

Proof. (1) 0a = (0 + 0)a ⇒ 0 + 0a = 0a + 0a (By R₃, R₆)

∴ 0 = 0a (By right cancellation law of (R, +) )

Similarly, we can prove that a0 = 0. Hence 0a = a0 = 0

(2) To prove that a(−b) = −(ab) we have to show that a(−b) + (ab) = 0.

a(−b) + ab = a{(−b) + b} = a0 = 0 ( By R6, R4) ⇒ a(−b) = −(ab)

Similarly, we can prove that (−a)b = −(ab).

Hence, a(−b) = (−a)b =−(ab).

(3) (−a)(−b) = −{(−a)b} = −{−(ab)} = ab ( by (2 ))[ ∵ (R, +) is a group ]

(4) a(b − c) = a[b + (−c)] = ab + a(−c) = ab − ac (By R6 ) [By theorem (1), (2)]

Similarly, we can prove that (b − c)a = ba − ca.

Solved Problems On Rings, Integral Domains, And Fields

Theorem. 2. If (R, +, ·) is a ring with unity then this unity 1 is the only multiplicative identity.

Proof. Suppose that there exist 1, 10 ∈ R such that 1.x = x.1 = x and 10 · x = x · 10 = x∀x ∈ R

Regarding 1 as identity, 1.10 = 10. Regarding 10 as identity, 1.10 = 1

Thus 10 = 1.10 = 1. ∴ 1 is the only multiplicative identity.

Theorem, 3. If R is a ring with unity element 1 and a ∈ R then

1. (−1)a = −a
2. (−1)(−1) = 1

Proof. (1) (−1)a+a = (−1)a+1a = {(−1)+1}a = 0a = 0 ( ∵ a = 1a, R6, R4)

∴ (−1)a = −a

(2) For a ∈ R we have (−1)a = −a. . Taking a = −1, (−1)(−1) = −(−1) =1

Rings, Integral Domains, And Fields Boolean Ring

Definition. In a ring R if a₂ = a∀a ∈ R then R is called a Boolean ring.

Theorem 1. If R is a Boolean ring then (1) a + a = 0∀a ∈ R (2) a + b = 0 ⇒ a = b and (3) R is commutative under multiplication. Or, Every Boolean ring is abelian.

Proof. (1) a ∈ R ⇒ a + a ∈ R.

Since a₂ = a∀a ∈ R, we have (a + a)₂ = a + a ⇒ (a + a)(a + a) = a + a ⇒ a(a + a) + a(a + a) = a + a ⇒ a₂ + a₂ + a₂ + a₂ = a + a (By R₆) ⇒

(a+a)+(a+a) = a+a ( ∵ R is Boolean ) ⇒ (a+a)+(a+a) = (a+a)+0 (By R₃ ) ⇒ a + a = 0 [By left cancellation law of group (R, +)]

(2) For a, b ∈ R, a + b = 0 ⇒ a + b = a + a ⇒ b = a [By (1)]

(3) a, b ∈ R ⇒ a + b˙ ∈ R ⇒ (a + b)₂ = a + b

(∵ R is Boolean) ⇒ (a + b)(a + b) = a + b ⇒ a(a + b) + b(a + b) = a + b (By R₆) ⇒ a₂ + ab + ba + b₂ = a + b (By R₆ ) ⇒ (a + ab) + (ba + b) =

a + b. (∵ R is Boolean) ⇒ (a + b) + (ab + ba) = a + b [ ∵ (R, +) is a group ] ⇒ (a + b) + (ab + ba) = (a + b) + 0 ⇒ ab + ba = 0 ⇒ ab = ba (By (2))

Rings, Integral Domains And Fields Solved Problems

Example. 1. If R is a ring with identity element 1 and 1 = 0 then R = {0}.

Solution. x ∈ R ⇒ x = 1x ⇒ x = 0x ⇒ x = 0

[By Theorem 1(1)]

∴ R = {0}

Thus a ring R with unity has at least two elements if R₆ = {0}.

Rings And Their Properties With Detailed Examples

Example. 2. Prove that the set of even integers is a ring, commutative without unity under the usual addition and multiplication of integers.

Solution. Let R = the set of even integers. Then R = {2x | x ∈ Z}.

a, b, c ∈ R ⇒ a = 2m, b = 2n, c = 2p where m, n, p ∈ Z.

(R, +) is a commutative group. (see ex. in groups)

a · b = (2m)(2n) = 2l where l = 2mn ∈ Z

∴ Multiplication ( · of integers is a binary operation in R.

(a · b) · c = (2m · 2n) · 2p = 8mnp and a · (b · c) = 2m · (−2n · 2p) = 8mnp

∴ (a.b).c = a.(b.c) ⇒ Multiplication (•) is associative in R.

a. (b + c) = 2m(2n + 2p) = 2m · 2n + 2m · 2p = a · b + a · c

Similarly, (b + c) · a = b · a + c · a

∴ Distributive laws hold in R. Hence (R, +, ·) is a ring.

Since ’ 1 ’ is not an even integer; 1 ∈/ R and hence R has no unity element.

Example. 3. (R, +) is an abelian group. Show that (R, +, ·) is a ring if multiplication (·) is defined as a.b=0 ∀a, b ∈ R.

Solution. To prove that (R, +, ·) is a ring we have to show that (R, ·) is semigroup and distributive laws hold.

∀a, b ∈ R, a.b = 0 where 0 ∈ R is the zero element in the group.

∴ multiplication ‘0 ’ is a binary operation in R.

Let a, b, c ∈ R. Then (a · b) · c = 0.c = 0; a.(b · c) = a · 0 = 0 (By Def.)

∴ (a · b) · c = a · (b, c)∀a, b, c ∈ R ∴ (R, ·) is a semi group.

Let a, b, c ∈ R. a ∈ R, b + c ∈ R ⇒ a · (b + c) = 0

a ∈ R, b ∈ R ⇒ ab = 0; a ∈ R, c ∈ R ⇒ ac = 0 ⇒ ab + ac = 0 + 0 = 0.

Hence a · (b + c) = a · b + a · c

Similarly, we can prove that (b + c) · a = b · a + c· a.

∴ Distributive laws hold.

Example. 4. Prove that Z _m  = {0, 1, 2, . . . ., m− 1} is a ring with respect to addition and multiplication modulo m.

Solution. We denote addition modulo m by +_m and multiplication modulo m by × _m

We also know that a+_m b = a + b(modm) = r where r is the remainder when a + b is divided by m. a × m b = ab(modm) = s where s is the remainder when ab is divided by m.

Let a, b, c ∈ Z_m. a+_m  b = a + b( mod m) ∈ Z_m ⇒ +_m  is a binary operation in Z_m . a +_m b = a + b(modm) = b + a(modm) = b +m a ⇒ +_m  is commutative in Z_m

(a+_m b) + mc = (a + b) + c(modm) = a + (b + c)(modm) = a +m (b +_m  c)

∴ +m is associative in Z_m

There exists 0 ∈ Z_m

such that 0 +m a = 0 + a(modm) = a(modm) = a.

⇒ 0 is the zero element.

For 0 ∈ Z_m , we have 0 +m 0 = 0(modm) ⇒ additive inverse of 0 = 0.

For a 6= 0 ∈ Z_m we have 0 < a < m ⇒ 0 < m − a < m ⇒ m − a ∈ Z_m

a +_m (m − a) = a + (m − a)(modm) = m(modm) = 0(modm)

∴ inverse of a 6= 0 ∈ Z_m is m − a ∈ Z_m

Hence ( Z_m, +) is an abelian group.

Let a, b, c ∈ Z_m. a× _m b = ab(modm) ∈ Z_m ⇒ × _m is a binary operation in Z_m

(a × _m b) × m c = (ab)c(modm) = a(bc)(modm) = a × _m (b × _m c)

∴ × _m  is associative in Z_m.

a× m (b+_m c) = a(b+ c)( mod m) = ab+ ac( mod m) = (a × _m b)+m (a × _m c) and (b +_m c) × m a = (b × _m a) +m (c × _m a) so that distributive laws hold.

∴ ( Z_m ,+_m, × _m ) is a ring.

Note. Put m = 6 in the above proof to prove that Z₆  is a ring.

Applications Of Rings, Integral Domains, And Fields In Mathematics

Example. 5. Prove that the set R = {a, b} with addition ( +) and multiplication (·) defined as follows is a ring

and

Solution. From the above tables, clearly +,· are binary operations in R.

1. (a + a) + b = a + b = b; a + (a + b) = a + b = b ⇒ (a + a) + b = a + (a + b) (a + b) + a = b + a = b; a + (b + a) = a + b = b ⇒ (a + b) + a = a + (b + a), etc,

∴ Associativity is true.

2. a ∈ R is the zero element because a + a = a, b + a = b

3. a + b = b = b + a ⇒ commutativity is true.

4. a + a = a ⇒ additive inverse of a = a and b + b = a ⇒ additive inverse of b = b.

5. a · (a · b) = a · a = a; (a · a) · b = a · b = a ⇒ a · (a · b) = (a · a) · b, etc

∴ Associativity is true.

6. a · (b + a) = a · b = a; a · b + a · a = a + a = a ⇒ a · (b + a) = a · b + a · a (b + a) · a = b · a = a; b · a + a · a = a + a = a ⇒ (b + a) · a = b · a + a · a, etc.

∴ Distributive laws are true.

Hence (R, +, ·) is a ring.

Example. 6. If R is a ring and a, b, c, d ∈ R then prove that

1. (a + b)(c + d) = ac + ad + bc + bd, and
2. a + b = c + d ⇔ a − c = d − b

Solution.

(1) (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd ( By R₆)

(2) a + b = c + d ⇔ (a + b) + (−b) = (c + d) + (−b)

⇔ a + (b + (−b)) = (c + d) + (−b)

⇔ a + 0 = (c + d) + (−b)

⇔ a + (−c) = (−c) + {(c + d) + (−b)}

⇔ a − c = ((−c) + c) + {d + (−b)}

⇔ a − c = 0 + (d − b) ⇔ a − c = d − b (By R4 )

Rings, Integral Domains And Fields Exercise 1

1( a )
1. If R is a ring and a, b, c ∈ R prove that (a − b) − c = (a-c) − b

2. If R is a ring and a, b ∈ R then prove that the equation a + x = b has unique solution in R.

3. In a ring R if ’ a ’ commutes with ’ b ’ prove that ’ a ’ commutes with ’ −b0 ’ where a, b ∈ R.

4. If R is a ring with unity element ’ 1 ’ and R₆= {0} prove that 1 6= 0 where 0 ∈ R is the zero element.

5. R is a Boolean ring and for a ∈ R, 2a = 0 ⇒ a = 0 then prove that R = {0}. 6. If R is a commutative ring prove that (a + b)² = a² + 2ab + b²∀a, b ∈ R.

6. If R is a ring and a, b, c, d ∈ R evaluate (a − b)(c − d).

7. If R = {a√2 | a ∈ Q} is (R, +, ·) under ordinary addition and multiplication, a ring ?

8. Is the set of all pure imaginary numbers = {iy | y ∈ R} a ring with respect to addition and multiplication of complex numbers?

9. If Z = the set of all integers. and ’ n ’ is a fixed integer prove that the set nZ = {nx | x ∈ Z} is a ring under ordinary addition and multiplication of integers.

Answers

2. b − a 7. ac + bd − ad − bc 8. Not a ring 9. Not a ring

Rings, Integral Domains, And Fields Zero Divisors Of A Ring

Though rings are a generalisation of the number system some algebraic properties of the number system need not hold in general rings.

The product of two numbers can only be zero if at least one of them is zero, whereas in any ring it may not be true.

For example, in the ring (Z₆, +, ·) of modulo −6, we have 2.3 = 0 with neither 2 = 0 nor 3 = 0.

Definition. (Zero Divisors). Two non-zero elements a, and b of a ring R are said to be zero divisors (divisors of zero) if ab = 0, where 0 ∈ R is the zero element.

In particular,’ a ’ is the left zero divisor, and ’ b ’ is the right zero divisor.

Definition. (Zero Divisor). a 6= 0 ∈ R is a zero divisor if there exists b 6= 0 ∈ R such that ab = 0.

Note. 1. In a commutative ring there is no distinction between left and right zero divisors.

2. A ring R has no zero divisors ⇔ a, b ∈ R and ab = 0 ⇒ a = 0 or b = 0

Example.1. The ring of integers Z has no zero divisors.

Step-By-Step Guide To Understanding Rings, Integral Domains, Fields

Example. 2. In the ring (Z₁₂, +, ·), the elements 2, 3, 4, 6, 8, 9, 10 are zero divisors.

For 2 · 6 = 0, 3 : 4 = 0, 3 · 8 = 0, 4 · 6 = 0, 4 · 9 = 0, 6 · 10 = 0

Observe that the G. C. D of any of {2, 3, 4, 6, 8, 9, 10} and 12 6= 1.

Example3. The ring (M₂, +, ·) of 2 × 2 matrices whose elements are in Z, has zero divisors.

For, we have A = 10006= O, B =  00106= O, where O is a zero matrix, such that AB = O.

Example 4. The ring (Z₃, +, ·) of modulo −3 has no zero divisors.

Example 5. The ring Z × Z = {(a, b) | a, b ∈ Z} has zero divisors.

For, (0, 1), (1, 0) ∈ Z × Z ⇒ (0, 1) · (1, 0) = (0, 0) = zero element in Z × Z.

Imp. In the ring of integers Z, all the solutions of x² − 4x + 3 = 0 are obtained by factoring as x² − 4x + 3 = (x − 1)(x − 3) and equating each factor to zero. While doing so, we are using the fact that Z is an Integral Domain, so that it has no zero divisors.

But if we want to find all solutions of an equation in a ring R which has zero divisors, we can do so, by trying every element in the Ring by substitution in the product (x − 1)(x − 3) for zero.

Example 6. In the ring Z of integers, the equation x² +2x+4 = 0 i.e. (x+1) 2 +3 = 0 has no solution as (x + 1)² + 3 ≥ 3∀x ∈ Z.

But, in the ring Z₆ = {0, 1, 2, 3, 4, 5}, (x+1)² +3 takes respectively the values 4, 1, 0, 1, 4, 3 for x = 0, 1, 2, 3, 4, 5 ∈ Z₆.

∴ x² + 2x + 4 = 0 has 2 ∈ Z₆ as solution.

Rings, Integral Domains, And Fields Cancellation Laws In A Ring

If (R, +, ·) is a ring, then (R, +) is an abelian group. So, cancellation laws with respect to addition are true in R. Now, we are concerned about the cancellation laws in R, namely ab = ac ⇒ b = c, ba = ca ⇒ b = c for a, b, c ∈ R with respect to multiplication.

Definition. (Cancellation laws). In a ring R, for a, b, c ∈ R if a₆ = 0, ab = ac ⇒ b = c and a₆ = 0, ba = ca ⇒ b = c then we say that cancellation laws hold in R.

Theorem. A ring R has no zero divisors if and only if the cancellation laws hold in R.

Proof. Let the ring have zero divisors. We prove that cancellation laws hold in R.

a, b, c ∈ R and a₆ = 0, ab = ac ⇒ ab − ac = 0

⇒ a(b − c) = 0 ⇒ b − c = 0 ( ∵ a₆ = 0) ⇒ b = c

Similarly we can prove a₆ = 0, ba = ca ⇒ b = c

Conversely, let the cancellation laws hold in R. We prove that R has no zero divisors. If possible, suppose that there exist a, b ∈ R such that a₆ = 0, b₆ = 0, and ab = 0.

ab = 0 ⇒ ab = a0 ⇒ b = 0 (By cancellation law)

This is a contradiction.

∴ a₆= 0, b₆ = 0 and ab = 0 is not true in R.

∴ R has no zero divisors.

Note. The importance of having no zero divisors in a ring R, is, that an equation ax = b where a₆ = 0, b ∈ R can have at most one solution in R.

For x₁, x₂ ∈ R if ax₁ = b and ax₂ = b then ax₁ = ax₂ ⇒ x₁ = x₂ (By cancellation law)

If a₆ = 0 ∈ R has a multiplicative inverse, say, a− 1 ∈ R then the solution is a− 1b ∈ R.

Rings, Integral Domains, And Fields Solved Problems

Example. 1. Find the zero-divisors of Z₁₂, the ring of residue classes modulo – 12.

Solution. Z₁₂ = {0, 1, . . . . . . .11}.

For a¯ 6= 0 ∈ Z₁₂ there should exist ¯b ∈ Z₁₂ such that a¯ × ¯b ≡ 0(mod12)

We have 2 × 6 = 0, 3 × 4 = 0, 4 × 3 = 0, 6 × 2 = 0, 8 × 3 = 0, 8 × 6 = 0, 8 × 9 = 0, 10 × 6 = 0

∴ 2, 3, 4, 6, 8, 9, 10 are zero divisors.

Worked Examples Of Integral Domains In Abstract Algebra

Example. 2. Solve the equation x₂ − 5x + 6 = 0 in the ring Z₁₂.

Solution.

Given

x₂ − 5x + 6 = 0

In the ring of integers Z, which has no zero divisors, x₂ − 5x + 6 = 0 ≡ (x − 2)(x − 3) = 0 has two solutions 2, 3 ∈ Z.

But in Z₁₂; for x = 6, (x − 2)(x − 3) = (4)(3) = 12 = 0 and for x = 11, (x − 2)(x − 3) = (9)(8) = 72 = 0.

∴ the given equation has 4 solutions, namely, 2,3,6, and 11 in the ring Z₁₂.

Example. 3. In the ring Z_n, show that the zero divisors are precisely those elements that are not relatively prime to n. (or) show that every non-zero element of Zn is a unit or zero divisor:

Solution. Let m ∈ Z_n = {0, 1, 2, . . . , n − 1} and m 6= 0. Let m be not relatively prime to n.

Then G. C.D of m, n = (m, n) 6= 1.

Let (m, n) = d.

We have (m, n) = d ⇒ md , nd  = 1 ⇒ md , nd ∈ Zn and md 6= 0, nd 6= 0

∴ m nd  = md  n = 0(modn). Thus m 6= 0, nd 6= 0 ⇒ m nd  = 0

⇒ m is a zero divisor.

∴ Every m ∈ Zn which is not relatively prime to n is a zero divisor.

Let m ∈ Zn be relatively prime to n.

Then (m, n) = 1

∴ Let mr = 0 for some r ∈ Zn.

We have mr = 0(modn) ⇒ n| mr ⇒ n| r ( ∵ (m, n) = 1) ⇒ r = 0 (0 ≤ r < n − 1)

∴ If m ∈ Zn is relatively prime to n then m is not a zero divisor.

Note. If p is a prime, then Z_p ring has no zero divisors.

Theorems And Properties Of Rings And Fields With Examples

1. (g) Some Special Types Of Rings

Definition.  (Integral Domain)

A commutative ring with unity containing no zero divisors is an Integral Domain.

Note. 1. Some authors define integral domain without a unity element.

2. For “Integral Domain” we simply use the word “Domain” and denote by the symbol D. Imp. D is an integral domain ⇔

1. D is a ring,
2. D is commutative,
3. D has a unity element and
4. D has no zero divisors.

Example 1. The ring of integers Z is naturally an integral domain.

1 ∈ Z is the unity element and ∀a, b ∈ Z we have ab = ba (commutativity) and ab = 0 ⇒ a = 0 or b = 0 (no zero divisors).

Example 2. (Z₆, +, ·) where Z₆ = {0, 1, 2, 3, 4, 5}, the set of integers under the modulo −6 system, is a ring. 1 ∈ Z₆ is the unity element and ∀a, b ∈ Z₆ we have ab(mod6) = ba(mod6)( commutative ) But, for 2 6= 0, 3 6= 0(mod6), 2.3 = 6(mod6) = 0 and hence Z₆ has zero divisors. Therefore, Z₆ is not an integral domain.

Example. 3. If Q = the set of all rational numbers and R = the set of all real numbers then (Q, +, ·) and (R, +, ·) are integral domains.

Example 4. -7 = {0, 1, 2, 3, 4, 5, 6}, the set of all integers under modulo −7 is an integral domain with respect to addition and multiplication modulo −7.

Example 5. The ring (M₂, +, ·) of 2× 2 matrices is not an integral domain because it is not commutative and has zero divisors.

Example 6. Z × Z = {(a, b) | a, b ∈ Z} is not an Integral Domain under the addition and multiplication of components.

Theorem. 1. In an integral domain, cancellation laws hold.(Write the proof of (1) part of a theorem in Art. 1.6)

Theorem. 2. A commutative ring with unity is an integral domain if and only if the cancellation laws hold. (Write the proof of a theorem in Art. 1.6)

Definition. (Multiplicative Inverse). Let R be a ring with the unity element ’

1 ’. A non-zero element a ∈ R˙ is said to be invertible under multiplication, if there exists b ∈ R such that ab = ba = 1, b ∈ R is called the multiplicative inverse of a ∈ R.

From the theory of groups, the multiplicative inverse of a 6= 0 ∈ R, if exists, is unique. It is denoted by a − 1. Also aa− 1 = a− 1a = 1.

Definition. (Unit of a Ring). Let R be a ring with unity. An element u ∈ R is said to be a unit of R if it has a multiplicative inverse in R.

Note.

1. The zero element of a ring is not a unit.

2. The unity element of a ring and the unit of a ring R is different. The unity element is the multiplicative identity while a unit of a ring is an element of the ring having a multiplicative inverse in the ring. Of course, the unity element is a unit.

Integral Domains And Fields Classification With Solved Examples

Theorem. 3. In a ring R with unity, if a (6= 0) ∈ R has a multiplicative inverse, then it is unique.

Proof. Suppose that there exist b, b₀ ∈ R such that ab = ba = 1 and ab₀ = b₀a = 1.

Then ab = ab₀ = 1.

By the definition of cancellation law, b = b₀.

Example 1. It is a ring with a unity element = 1. We have 1.1 = 1 and (−1)(−1) = 1 for −1, 1 ∈ Z.

If a 6= ±1 ∈ Z then there exists no b ∈ Z such that ab = ba = 1.

Therefore, −1;1 are the only units in the ring Z.

Observe that any element is also a unit.

Example 2. Consider the ring Z7 = {0, 1, 2, 3, 4, 5, 6} under addition and multiplication modulo −7.

It has a unity element = 1 which is also a unit.

Further 24 = 4.2 = 1(mod7), 3.5 = 5.3 = 1(mod7) and 6.6 = 1(mod7). Thus every non-zero element is a unit.

Example 3. Consider the ring Z × Z = {(m, n); m, n ∈ Z}.

The unity element = (1, 1) which is also a unit: Also, (1, −1)(1, −1) = (1.1, (−1)(−1)) = (1, 1); (−1, 1)(−1, 1) = (1, 1) and (−1, −1)(−1, −1) = (1, 1)

Thus (1, 1), (1, −1), (−1, 1) and (−1, −1) are units in Z × Z.

Definition. (Division Ring or Skew Field) Let R be a ring with a unity element.

If every non-zero element of R is a unit then R is a Division Ring.  (R, +,)isaDivisionring ⇔ (1)R is a ring, (2) R has a unity element and (3) every non-zero element in R is invertible under multiplication.

Example 1. ( Z, +,) is not a division ring, for, 2 6= 0 ∈ Z has no multiplicative inverse in Z.

Example 2. (Q, +, ·) and (R, +, ·) are division rings.

Example 3. The ring (M₂, +, ·) of non-singular 2 × 2 matrices is a division ring.

Definition. (Field) Let R be a commutative ring with a unity element. If every nonzero element of R is invertible under multiplication then R is a field.

Another Definition. A commutative ring with unity is called a field if every nonzero element is a unit.

(R, +, ·) is a field ⇔ (1)R is a ring, (2) R is commutative (3) R has a unity element and (4) every non-zero element of R is a unit. Usually, a field is denoted by the symbol F.

Note. 1 . A division ring which is also commutative is a field.

2. In a field, the zero element and the unity element are different. Therefore, a field has at least two elements.

Example 1. We know that (Q, +) where Q = the set of all rational numbers is an additive group and (Q − {0}, ·) is a multiplicative group. Further distributive laws hold. Therefore (Q, +, ·) is a field.

Example  2. (Z, +, ·) where Z = the set of all integers is not a field, because all non-zero elements of Z are not units.

Example  3. (Z₇, +, ·) where Z₇ = the set of integers under modulo −7 is a field.

Theorem. 4. A field has no zero – divisors.

Proof. Let (F, +, ·) be a field.

Let a, b ∈ F and a₆= 0.

a₆= 0 ∈ F, F is a field ⇒ there exists a−1 ∈ F such that aa−1 = a−1a = 1.

ab = 0. ⇒ a−1(ab) = a−10 ⇒ a−1a b = 0 ⇒ 1b = 0 ⇒ b = 0

Thus a, b ∈ R, a₆ = 0 and ab = 0 ⇒ b = 0.

Similarly, we can prove that, a, b ∈ R, b₆ = 0 and ab = 0 ⇒ a = 0.

∴ F has no zero divisors.

Note. A division ring has no zero divisors. (Write the proof of the theorem (3))

Theorem. 5. Every field is an integral domain.

Proof. Let (F, +, ·) be a field. Then the ring F is a commutative ring with unity and having every non-zero element as a unit.

But an integral domain is a commutative ring with unity and having no zero divisors. So, we have to prove that F has no zero divisors. (Write the proof of the above Theorem (3))

Note. The converse of the above theorem need not be true. However, an integral domain with a finite number of elements can become a field.

Theorem 6. Every finite integral domain is a field.

Proof. Let 0, 1, a₁, a₂, . . . . . . , a_n be all the elements of the integral domain D.

Then D has n + 2 elements which is finite. Integral domain D is a commutative ring with unity and having no zero divisors.

So, we have to prove that every non-zero element of D has a multiplicative inverse in D.

Let a ∈ D and a₆ = 0

Now consider the n + 1 products a₁, aa₁, aa₂ , . . . ., aan.

If possible, suppose that aai = aaj for i 6= j.

Since a₆ = 0, by cancellation law, we have ai = aj.

This is a contradiction since i 6= j.

Therefore a₁, aa₁, aa₂, . . . . . . . . . , aan are (n + 1) distinct elements in D.

Since D has no zero divisors, none of these (n +1) elements is zero element.

Hence, by counting ;

a₁, aa₁, aa₂, . . . . . . .., aan are the (n+1) elements 1, a₁, a₂, . . . . . . , a_n in some order.

∴ a₁ = 1 or a = 1 or aai = 1 for some i.

For a₆ = 0 ∈ D there exists b = ai ∈ D such that ab. = 1

⇒ a₆ = 0 ∈ D has a multiplicative inverse in D.

∴ D is a field.

Theorem 7. If p is a prime then Z_p, the ring of integers modulo p, is a field.

Proof. In Example 4 on page 191, we proved that (Z_p, +, ·) is a ring.

Since Z

p = {0, 1, 2, . . . . . . , p − 1} has p distinct elements, and Z_p is a finite ring.

We prove now that Z_p is an integral domain.

Clearly, 1 ∈ Z_p is the unity element.

For a, b ∈ Z_p, ab(modp) ≡ ba(modp) ⇒ ab = ba and hence Zp is commutative.

For a, b ∈ Z_p and ab = 0 ⇒ ab ≡ 0(modp) ⇒ p|ab ⇒ p|a or p | b ( ∵ p is prime)

⇒ a ≡ 0(modp) or b ≡ 0(modp) ⇒ a = 0 or b = 0.

∴ Z_p has no zero divisors. Thus (Z_p, +, ·) is a finite integral domain .

∴ Z_p is a field.

Theorem. 8. If (Zn, +, ·) is a field then n is a prime number.

Proof. If possible let m be a divisor of n.

∴ there exists q ∈ Z such that n = mq. Clearly 1 ≤ m, q ≤ n.

mq = n ⇒ mq ≡ 0(modn). Since Zn is a field, Zn has no zero divisors.

∴ mq ≡ 0(modn) ⇒ m = 0(modn) or q = 0(modn)

⇒ m = n or q = n ⇒ m = n or m = 1( ∵ mq = n). ∴ n is a prime number.

Theorem. 9. Z p = {0, 1, 2, . . . , p − 1} is a field if and only if p is a prime number.

Proof. Write the proofs of Theorem 7 and Theorem 8.

Note. In the field Z_p= {0, 1, 2, . . . . . . , p − 1} where p is a prime, 1 and p − 1

are the only elements that are their own multiplicative inverses.

Rings, Integral Domains, And Fields Solved Problems

Example. 4. Find all solutions of x₂ − x + 2 = 0 over Z₃[1].

Solutions.

Given

x₂ − x + 2 = 0

We have Z₃ = {0, 1, 2} under modulo – 3 system. Z₃[1] = {a + ib | a, b ∈ Z3 and i₂ = −1 = {0, 1, 2, i, 1 + i, 2 + i, 2i, 1 + 2i, 2 + 2i}, containing 9 elements.

Let P(x) = x₂ − x+2. Then P(0) 6= 0, P(1) 6= 0, P(2) 6= 0, P(i) = −1− i+2 6= 0

P(1 + i) = (1 − 1 + 2i) − 1 − i + 2 6= 0, P(2 + i) = (4 − 1 + 4i) − (2 + i) + 2 6= 0, P(2i) = −4 6= 0

P(1+2i) = (1−4+4i)−(1+2i)+2 6= 0, P(2+2i) = (4−4+8i)−(2+2i)+2 6= 0

∴ x₂ − x + 2 = 0 has no solution over Z3[1].

Example. 5. Show that 1, p − 1 are the only elements of the field Z_p, p is prime, that are their own multiplicative inverses.

Solution. Observe that, in the Z_p field, x2 − 1 = 0 has only two solutions.

x₂ − 1 = 0 ⇒ x₂ = 1 ⇒ x = 1 ⇒ Multiplicate inverse of x = x.

So, we have to prove that 1, p − 1 are solutions of x₂ − 1 = 0 in Z_p.

1 ∈ Z

p ⇒ 12 − 1 = 1 − 1 = 0

p − 1 ∈ Zp ⇒ (p − 1) 2 − 1 = p2 − 2p + 1 − 1 = p2 − 2p

= p(p − 2) = 0(p − 2) = 0 ( ∵ p = 0(modp))

Example. 6. In a ring R with unity if a ∈ R has multiplicative inverse then a ∈ R is not a zero divisor.

Solution. a ∈ R has multiplicative inverse

⇒ There exists a−1 ∈ R, such that aa−1 = a−1a = 1, where 1 ∈ R is the unity element, To prove that a ∈ R is not a zero divisor we have to prove that

for b ∈ R so that ab = 0 or ba = 0 ⇒ b = 0 only.

ab = 0 ⇒ a−1(ab) = a−10 ⇒ 1b = 0 ⇒ b = 0; ba = 0 ⇒ (ba)a−1 = 0a−1 ⇒ b1 = 0 ⇒ b = 0

∴ a ∈ R is not a zero divisor.

Example. 7. Construct a field of two elements.

Solution. Let F = {0, 1} and addition ( +), multiplication ( ) in F be defined as follows :

Clearly, + and – are binary operations in F.

We have 0 + 1 = 1 + 0 and 0.1 = 1.0 and hence +, · are commutative.

The two operations are associative.

0 ∈ F is the zero elements and 1 ∈ F is the unity element.

Clearly, distributivity is also true.

Additive inverse of 0 = 0, additive inverse of 1 = 1.

The multiplicative inverse of 16= 0 ∈ F is 1 .

Hence ({0, 1}, +, ·) is a field.

Example. 8. Show that the set R of all real-valued continuous functions defined on [0, 1] is a commutative ring with unity, with respect to addition (+) and multiplication ( •) of functions defined as (f + g)(x) = f(x) + g(x) and (f · g)(x) = f(x) · g(x)∀x ∈ [0, 1] and f, g ∈ R.

Solution. f, g are real-valued continuous functions on [0, 1] ⇒ (1)f + g and f.g are real-valued continuous functions on [0, 1] and (2) f(x), g(x) are real numbers for x ∈ [0, 1].

∴ The addition and multiplication of functions are binary operations in R.

Let f, g, h ∈ R ∀x ∈ [0, 1], ((f + g) + h)(x) = (f + g)(x) + h(x) = (f(x) + g(x)) + h(x)

= f(x) + (g(x) + h(x)) = f(x) + (g + h)(x) = (f + (g + h))(x)

∴ (f + g) + h = f + (g + h)∀f, g, h ∈ R

If O(x) = 0∀x ∈ [0, 1] then O is a real-valued continuous function.

Therefore there exists O ∈ R so that (f + O)(x) = f(x) + O(x) = f(x)∀x ∈ [0, 1] and f ∈ R.

If f is a real-valued continuous function on [0, 1] then ’ −f0 ’ is also a real-valued continuous function so that (−f)(x) = −f(x)∀x ∈ [0, 1].

Therefore for f ∈ R there exists −f ∈ R so that (f + (−f))(x) = f(x) − f(x) = 0 = 0(x)∀x ∈ [0, 1]

That is, additive inverse exists ∀f ∈ R. ∴ (R, +) is a commutative group.

∀x ∈ [0, 1]; ((fg)h)(x) = (fg)(x)h(x) = (f(x)g(x))h(x) = f(x)(g(x)h(x)) = f(x)(gh)(x) = (f(gh))(x)

∴ (fg)h˙ = f(gh)∀f, g, h ∈ R

∀x ∈ [0, 1]; (f(g + h))(x) = f(x)(g + h)(x) = f(x)(g(x) + h(x)) = f(x)g(x) + f(x)h(x) = (fg)(x) + (fh)(x) = (fg + fh)(x)

∴ f(g + h) = fg + fh∀f, g, h ∈ R

Similarly (g + h)f = gf + hf ∀f, g, h ∈ R. Hence (R, +, ·) is a ring.

∀x ∈ [0, 1], (fg)(x) = f(x)g(x) = g(x)f(x) = (gf)(x)

∴ fg = gf∀f, g ∈ R.

∴ R is a commutative ring.

The constant function e(x) = 1∀x ∈ [0, 1] is real-valued and continuous.

Also e ∈ R is such that (ef)(x) = e(x)f(x) = f(x)∀x ∈ [0, 1]

∴ e ∈ R defined as above is the unity element.

Example 9. Prove that the set Z[1] = a + bi | a, b ∈ Z, i ² = −1 of Gaussian integers is an integral domain with respect to the addition and multiplication of numbers. Is it a field?

Solution. Let Z(1) = {a + bi | a,b ∈ Z}.

Let x, y ∈ Z (i) so that x = a + bi,y = c + di where a,b,c,d ∈ Z

x + y = (a + c) + (b + d)i = a₁ + b₁ i where a₁ = a + c, b₁ = b + d ∈ Z

x,y = (ac − bd) + (ad + bc)i = a₂ + b₂ i where a₂ = ac − bd,b₂ = ad + bc ∈ Z

∴ +, · are binary operations in Z(1).

Since the elements of Z (1) are complex numbers we have that

1. Addition and multiplication are commutative in Z(1),
2. Addition and multiplication are associative in Z (1) and
3. Multiplication is distributive over addition in Z(1).

Clearly zero element = 0 + 0i = 0 and unity element = 1 + 0i = 1.

Further, for every x = a + ib ∈ Z(i) we have −x = (−a) + i(−b) ∈ Z(i) so that x + (−x) = {a + (−a)} + i{b + (−b)} = 0 + i0 = 0 ⇒ Additive inverse exists.

∴ Z(i) is a commutative ring with a unity element.

For x,y ∈ Z(i),x.y = 0 ⇒ x = 0 or y = 0 since x,y are complex numbers.

Hence Z(i) is an integral domain with a unity element.

For α = 3 + 4i 6= 0 ∈ Z(i) we have β = 253 − i 254 so that α · β˙ = 259 + 1625  + i −2512 + 1225  = 1 + i0 = 1. But β /∈ Z(i) as 253 , − 254 ∈/ Z.

So, every non-zero element of Z(i) is not invertible. ∴ Z(i) is not a field.

Example. 10. Prove that Q[√2] = {a + b√2 | a,b ∈ Q} is a field with respect to ordinary addition and multiplication of numbers.

Solution. Let x,y,z ∈ Q[√2] so that

x = (a₁ + b₁)√2,y = (a₂ + b₂)√2,z = (a₃ + b₃)√2 where a₁,b₁,a₂,b₂,a₃,b₃ ∈ Q

x+y = (a₁ + a₂)+(b₁ + b₂) √2 = a+b√2 where a₁ +a₂ = a,b₁ +b₂ = b ∈ Q

x · y = (a₁a₂ + 2b₁b₂)+(a₁b₂ + a₂b₁) √2 = c+d√2 where c = a₁a₂ +2b₁b₂ ∈ Q and d = a₁b₂ + a₂b₁ ∈ Q

∴ Addition (+) and multiplication ( ) of numbers are binary operations in Q[√2].

x + y = (a₁ + a₂) + (b₁ + b₂) √2 = (a₂ + a₁) + (b₂ + b₁) √2 = (a₂ + b₂)√2+ (a₁ + b₁)√2 = y + x ⇒ Addition is commutative.

(x + y) + z = (a₁ + a₂ + a₃) + (b₁ + b₂ + b₃) √2 and x + (y + z) = (a₁ + a₂ + a₃) + (b₁ + b₂ + b₃) √2 ⇒ (x + y) + z = x + (y + z) ⇒ Addition is associative.

For 0 ∈ Q we have 0 + 0√2 = 0 ∈ Q[√2] so that x + 0 = x for x ∈ Q[√2] ⇒ 0 ∈ Q[√2] is the zero element.

For x = (a₁ + b₁)√2 ∈ Q[√2] we have −x = (−a₁) + (−b₁) √2 ∈ Q[√2] so that x + (−x) = 0 ⇒ Additive inverse exists.

∴ (Q[√2], +) is a commutative group.

x · y = (a₁ + b₁)√2 · (a₂ + b₂)√2 = (a₁a₂ + 2b₁b₂) + (a₁b₂ + a₂b₁) √2  = (a₂a₁ + 2b₂b₁)+(a₂b₁ + b₂a₁) √2 = y·x ⇒ Multiplication is commutative.

(x · y) · z = (a₁a₂ + 2b₁b₂ + a₁b₂ + a₂b₁)√2 · (a₃ + b₃)√2 = (a₁a₂a₃ + 2b₁b₂ a₃ + 2a₁b₂b₃ + 2a₃b₁b₃)+(a₁a₂b₃ + 2b₁b₂b₃ + a₁a₃b₂ + a₂a₃b₁) √2 and x · (y · z)

= (a₁ + b₁)√2 (a₂a₃ + 2b₂b₃ + a₂b₃ + a₃b₂)√2 = (a₁a₂a₃ + 2a₁b₂b₃ + 2a₂b₁b₃ + 2a 3b₁b₂)+(a₁a₂b₃ + a₁a₃b₂ + a₂a₃b₁ + 2b₁b₂b₃) √2

∴ (x, y), z = x.(y, z) ⇒ Multiplication is associative.

x · (y + z) = (a₁ + b₂)√2 (a₂ + a₃ + b₂ + b₃)√2 = (a₁a₂ + a₁a₃ + 2b₁b₂ + 2b₁b₃) + (a₁b₂ + a₁b₃ + a₂b₁ + a₃b₁) √2 and x·y+x, z = (a₁a₂ + 2b₁b₂ + a₁b₂ + a₂b₁)√2 +(a₁a₃ + 2b₁b₃ + a₁b₃ + a₃b₁)√2 = (a₁a₂ + 2b₁b₂ + a₁a₃ + 2b₁b₃) + (a₁b₂ + a₂b₁ + a₁b 3 + a₃b₁) √2

∴ x · (y + z) = x · y + x · z ⇒ Distributivity is true.

Hence (Q[√2], +, ·) is a ring.

1 = 1 + 0√2 ∈ Q[√2] so that x₁ = (a₁ + b₁)√2 (1 + 0√2) = x∀x ∈ Q[√2].

∴ Q[√2] is a commutative ring with a unity element.

To show that Q[√2] is a field we have to prove further every non-zero element in Q[√2] has a multiplicative inverse.

Let a + b√2 ∈ Q[√2] and a 6= 0 or b 6= 0

⇒ Then \(\frac{1}{a+b \sqrt{2}}=\frac{a-b \sqrt{2}}{a^2-2 b^2}=\left(\frac{a}{a^2-2 b^2}\right)+\left(\frac{-b}{a^2-2 b^2}\right) \sqrt{2}\)

⇒ since \(a^2-2 b^2 \neq 0 for a \neq 0 or b \neq 0. a, b \in Q \Rightarrow \frac{a}{a^2-2 b^2}, \frac{-b}{a^2-2 b^2} \in Q\)

⇒ For \(a+b \sqrt{2} \neq 0 \in Q[\sqrt{2}]\)

there exists \(\left(\frac{a}{a^2-2 b^2}\right)+\left(\frac{-b}{a^2-2 b^2}\right) \sqrt{2} \in Q[\sqrt{2}]\)

⇒ such \(t(a+b \sqrt{2})\left[\left(\frac{a}{a^2-2 b^2}\right)+\left(\frac{-b}{a^2-2 b^2}\right) \sqrt{2}\right]=1=1+0 \sqrt{2}\)

∴ Every non-zero element of Q[√2] is invertible.

Hence Q[√2] is a field. Ex.

Example11. If Z = the set of integers then prove that the set z × z = {(m, n) | m, n ∈ z} with respect to addition (+) and multiplication (•) defined as  (m₁, n₁) + (m₂, n₂) = (m₁ + m₂, n₁ + n₂) and (m₁, n₁) · (m₂, n₂) = (m₁m₂, n₁n₂) ∀ (m₁, n₁) , (m₂, n₂) ∈ z × z is a ring and not an integral domain.
Solution.

Let x = (m₁, n₁) , y = (m₂, n₂) , z = (m₃, n₃) ∈ z×z so that m₁, n₁, m₂, n₂, m₃, n₃ ∈ Z

(1) x + y = (m₁, n₁) + (m₂ + n₂) = (m₁ + m₂, n₁ + n₂) ∈ z × z x · y = (m₁, n₁) · (m₂, n₂) = (m₁m₂, n₁n₂) ∈ z × z as m₁ + m₂, n₁ + n₂, m₁m₂, n₁n₂ ∈ z

∴ + and · are binary operations in z × z

(2) x + y = (m₁ + m₂, n₁ + n₂) = (m₂ + m₁, n₂ + n₁) = y + x and x.y = (m₁m₂, n₁n₂) = (m₂m₁, n₂n₁) = yx  ⇒ + and · are commutative.

(3) (x + y) + z = (m₁ + m₂, n₁ + n₂) + (m₃, n₃) = (m₁ + m₂ + m₃, n₁ + n₂ + n₃) = (m₁ + m₂ + m₃, n₁ + n₂ + n₃) = x + (y + z) (x · y) · z = (m₁m₂, n₁n₂) · (m₃, n₃) = ((m₁m₂) m₃, (n₁, n₂) n₃) = (m₁ (m₂m₂) , n₁ (n₂n₃)) = x.(y.z). ⇒ + and – are associative.

(4) x · (y + z) = (m₁, n₁) · (m₂ + m₃, n₂ + n₃) = (m₁ (m₂ + m₃) , n₁ (n₂ + n₃)) = (m₁m₂ + m₁m₃, n₁n₂ + n₁n₃) = (m₁m₂, n₁n₂) + (m₁m₃, n₁n₃) = x · y + x · z

Since multiplication is commutative, (y + z) · x = y · x + z · x

∴ Distributivity is true.

(5) For 0 ∈ z we have (0, 0) ∈ z×z and (m, n)+(0, 0) = (m+0, n+0) = (m, n)

∴ (0, 0) ∈ z × z is the zero elements.

(6) For 1 ∈ z we have (1, 1) ∈ z × z and (m, n).(1, 1) = (m · 1, n · 1) = (m, n)

∴ (1, 1) ∈ z × z is the unity element. Hence z × z is a commutative ring with unity.

But we have, (0, 1), (1, 0) ∈ z × z and (0, 1) 6= (0, 0), (1, 0) 6= (0, 0) such that (0, 1) · (1, 0) = (0 · 1, 1 · 0) = (0, 0)

∴ (0, 1), (1, 0) are zero divisors in z× z. Hence z× z is not an integral domain.

Rings, Integral Domains, And Fields Exercise 2

1. List all zero divisors in the ring Z₂₀. Also, find the units in Z₂₀. Is there any relationship between zero divisors and units?

2. Solve the equation 3x = 2 in (a) Z₇ (b)Z₂₃

3.

1. Find all solutions of x³ − 2x² − 3x = 0 in Z_12.
   
2. Find all solutions of x² + x − 6 = 0 in Z₁₄ .

4. Describe all units in (a) Z₄ (b) Z₅. Prove that Z₂ × Z₂  = {(0, 0), (0, 1), (1, 0), (1, 1)} under componentwise addition and multiplication is a Boolean ring.

5. Find all solutions of a₂ + b₂ = 0 inZ₇.

6. Write the multiplication table for Z₃[1] = {0, 1, 2,i, 1+i, 2+i, 2i, 1+2i, 2+ 2i}.

7. R is a set of real numbers. Show that R × R forms a field under addition and multiplication defined by (a,b) + (c,d) = (a + c,b + d) and (a,b) : (c,d) = (ac − bd,ad + bc) is a field. (Hint. R × R = C = a + ib | a,b ∈ R,i2 = −1 )

9. If Z is the set of all integers and addition ⊕, multiplication (x) is defined in Z as a ⊕ b = a + b − 1 and a × b = a + b − ab∀a,b ∈ Z then prove that (Z, ⊕, ×) is a commutative ring.

10. Let (R, +) be an abelian group. If multiplication (·) in R is defined as a · b = 0, 0000 is the zero element in R, ∀a,b ∈ R then prove that (R, +, ·) is a ring.

11. If R = {0, 1, 2, 3, 4} prove that (R, +5, x5) under addition and multiplication modulo – 5 is a field.

12. Give examples of (1) a commutative ring with unity (2) an integral domain and (3) a Division ring.

13. If R = the set of all even integers and (+) is ordinary addition and multiplication (∗) is defined as a ∗ b = ab₂ ∀a,b ∈ R then prove that (R, +, ∗) is
a commutative ring.

14. S is a non-empty set containing n elements. Prove that P(S) forms finite Boolean ring w.r.t ’+’ and ’ ’ ’defined as A + B = (A ∩ B) − (A ∪ B) and
A · B = A ∩ B∀A, B ∈ P(S). Find addition and multiplication tables when S = {a,b}.

15. If R₁, R₂,……, R_n are rings, then prove that R₁ × R₂ × ……. R_n = {(r₁,r₂,……, r_n) | ri ∈ Ri} forms a ring under componentwise addition and multiplication, that is, (a₁,a₂,……, a_n)+(b₁,b₂,……, b_n) = (a₁ + b₁,a₂ + b₂,…, an + bn) and (a₁,a₂,……, a_n) · (b₁,b₂,……, b_n) = (a₁b₁,a₂b₂,…., a_n b_n)

Rings, Integral Domains & Fields Integral Multiples And Integral Powers Of An Element

Integral multiples: Let (R, +, ·) be a ring and a ∈ R. We define 0a = O where ’ 0 ’ is the integer and O is the zero element of the ring

If n ∈ N we define na = a + a + . . . + a, (n terms ).

If n is negative integer, na = (−a) + (−a) + . . . + (−a), (−n terms ).

(−n)a = (−a) + (−a) + . . . + (−a), n terms = n(−a) = −(na) where n ∈ N.

The set {na | n ∈ Z, a ∈ R} is called the set of integral multiples of an element ’ a ’. It may be noted that for n ∈ Z, a ∈ R we have na ∈ R. Theorem.

If m, n ∈ Z and a, b ∈ R, a ring, then (1)(m + n)a = ma + an, (2) m(na) = (mn)a, (3) m(a + b) = ma + mb and (iv)m(ab) = (ma)b. (Proof is left as an exercise )

Note 1. If the ring R has unity element then for n ∈ Z and a ∈ R we have na = (n1)a = (n1)a

2. If m, n ∈ Z and a, b ∈ R, a ring then we have (ma)(nb) = m{a(nb)} = m{(na)b} = m{n(ab)} = (mn)(ab).

Integral powers: Let (R, +, ) be a ring and a ∈ R.

For n ∈ N we write an = a.a . . . a(n times ).

It may be noted that an = an − 1.a.

Theorem. If m, n ∈ N and a, b ∈ R, a ring then

am · an = am +n and

(am)n = amn (Proof is left as an exercise )

Rings, Integral Domains, And Fields Idempotent Element And Nilpotent Element Of A Ring

Definition. In a ring R, if a 2 = a for a ∈ R then ’ a ’ is called an idempotent element of R with respect to multiplication.

Theorem. 1. If a 6= 0 is an idempotent element of an integral domain with unity then a = 1.
Proof. Let (R, +, ·) be an integral domain.

a₆ = 0 ∈ R is an idempotent element ⇒ a 2 = a

∵ a₂ = a₁ ( ∵ al = a)

⇒ a₂ − a₁ = 0 ⇒ a(a − 1) = 0 (’ 0 ’ is the zero element)

⇒ a − 1 = 0 since R has no zero divisors ⇒ a = 1.

Note. 1. An integral domain with unity contains only two idempotent elements ’ 0 ’ and ’1’.

2. A division ring contains exactly two idempotent elements.

3. The product of two idempotent elements in a commutative ring R is idempotent.

For, (ab)² = (ab)(ab) = a(ba)b = a(ab)b = (aa)(bb) = a²b² = ab for a, b ∈ R which are idempotent:

Definition. Let R be a ring and a 6= 0 ∈ R. If there exists n ∈ N such that an = 0 then ’ a ’ is called a nilpotent element of R.

Theorem. 2. An integral domain has no nilpotent element other than zero.

Proof. Let R be an integral domain and a₆ = 0 ∈ R.

we have a 1 = a₆ = 0, a₂ = a · a₆ = 0 since R has no zero divisors.

Let an ≠ 0 for n ∈ N.

Then an +1 = an. a₆ = 0, since R has no zero divisors.

∴ By induction, a₆ = 0 for every n ∈ N.

Hence a₆ = 0 ∈ R is not a nilpotent element.

Example1. In the ring (Z₆, +, ·), 3 and 4 and idempotent elements, for 32 = 3 and 42 = 4.

Example 2. In the ring (Z₈, +, ·), there are no idempotent elements.

Example 3. In the ring (Z₈, +, ·), 2 and 4 are nilpotent elements, for 23 = 0 and 42 = 0.

Example 4. In the ring Z₆, +˙, there are no nilpotent elements.

Example. 1. If a,b are nilpotent elements in a commutative ring R then prove that a + b, a · b are also nilpotent elements.

Solution.

Given

a,b are nilpotent elements in a commutative ring R

a,b ∈ R are nilpotent elements

⇒ there exists m,n ∈ N such that am = 0,bn = 0. We have

⇒ (a + b)m+n = am+n + (m + n)C1 · am+n−1 · b + … + (m + n)Cn · am · bn + … + … + bm+n = am an + (m + n)C1an−1 · b + (m + n)C2 · an−2 · b2 + … + (m + n)Cn · bn + (m + n)Cn+1 · am−1 · b + (m + n)Cn+2 · am−2 · b2 + … + bm bn = 0 ( ∵ am = 0 = bn.)

Also, (ab)mn = amn · bmn = (am)n (bn)m = 0 ( ∵ R is commutative ).

∴ a + b and a · b are nilpotent elements.

Rings, Integral Domains And Fields Characteristic Of A Ring

Definition. The characteristic of a ring R is defined as the least positive integer p such that pa = 0 for all a ∈ R. In case such a positive integer p does not exist then we say that the characteristic of R is zero or infinite.

Note.

1. If R is a ring and Z = {n ∈ N | na = 0∀a ∈ R} 6= φ then the least element in Z is the characteristic of R.

2. If the ring R has characteristic zero then ma = 0 ’ where a 6= 0 can hold only if m = 0.

3. If the characteristic of a ring R is not zero then we say that the characteristic of R is finite.

4. As the integral domain, division ring and field are also ring characteristics that have meaning for these structures. Imp. If for some a ∈ R, pa 6= 0 then characteristic of R 6= p. Characteristic of a ring R = p ⇒ pa = 0∀a ∈ R.

Example 1 . R = {0, 1, 2, 3, 4, 5, 6} = Z7 is a ring under addition and multiplication modulo 7. Zero element of R = 0.

∀a ∈ R we have 7a ≡ 0(mod7) ⇒ 7a = 0∀a ∈ R. Further for 1 ∈ R, p(1) = p 6= 0 where p 6= 0 and 0 < p < 7.

∴ 7 is the least positive integer so that 7a = 0∀a ∈ R ⇒ Characteristic of R = 7.

Example. 2. The characteristic of the ring (Z, +, ·) is zero. For, there is no positive integer n so na = 0 for all a ∈ Z.

Example3. If R 6= {0} and the characteristic of R is not zero then the characteristic of R > 1.

Characteristic of R = 1 ⇒ 1a = 0∀a ∈ R ⇒ a = 0∀a ∈ R ⇒ R = {0}.

Example 4. For any element x ∈ Z3[1] ring, we have 3x = 0∀x ∈ Z3[1] ⇒ characteristic of Z3[i] = 3.

Example. 5. In the ring, R = {0, 3, 6, 9} ⊂ Z12, 4x = 0∀x ∈ R, and ’ 4 ’ is the least positive integer.

∴ Characteristic of R = {0, 3, 6, 9} = 0

Theorem 1. If R is a ring with a unity element, then R has characteristic p > 0 if and only if p is the least positive integer such that p₁ = 0.
Proof.

Let characteristic of R = p(> 0)

By definition, pa = 0∀a ∈ R. In particular p₁ = 0.

Conversely, let p be the least positive integer such that p₁ = 0.

∴ q < p and q ∈ N ⇒ q₁ 6= 0. Then for any a ∈ R we have

p · a = a + a + . . . + a(p terms ) = a(1 + 1 + . . . + 1) = a(p₁) = a0 = 0.

∴ p is the least positive integer so that p.a = 0∀a ∈ R. ∴ Characteristic of R = p.

Theorem 2. The characteristic of a ring with a unity element is the order of the unity element regarded as a member of the additive group.

Proof. Let (R, +, ·) be a ring so that (R, +) is its additive group.

Case 1. Let O(1) = 0 when the unity, element 1 is regarded as an element of (R, +). By the definition of the order of an element in a group, there exists no positive integer n so that n₁ = 0.

∴ Characteristic of R = 0.

Case 2. Let O(1) = p(6= 0).

By the definition of the order of elements in a group, p is the least positive integer,

so p₁ = 0. For any a ∈ R, pa = p(1a) = (pl)a = 0a = 0

∴ Characteristic of R = p.

For example., For the commutative ring Z× Z, the zero element = (0, 0) and the unity element = (1, 1). By the definition of the order of an element in the additive group Z × Z, there exists no positive integer m such that m(1, 1) = (m, m) = (0, 0).

Therefore, the characteristic of Z × Z is zero.

Theorem 3. The characteristic of an integral domain is either a prime or zero. 

Proof. Let (R, +, ·) be an integral domain. Let the characteristic of R = p(6= 0).

If possible, suppose that p is not a prime.

Then p = mn where 1 < m, n < p. a 6= 0 ∈ R ⇒ a · a = a2 ∈ R and a2 6= 0 ( ∵ R is integral domain )

pa 2 = 0 ⇒ (mn)a 2 = 0 ⇒ (ma)(na) = 0 ⇒ ma = 0 or na = 0 ( ∵ R is integral domain )

Let ma = 0. For any x ∈ R, (ma)x = 0 ⇒ a(mx) = 0 ⇒ mx = 0 ( ∵ a 6= 0)

This is absurd, as 1 < m < p and characteristic of R = p.

∴ ma 6= 0. Similarly, we can prove that na 6= 0.

This is a contradiction and hence p is a prime.

Theorem 4. The characteristic of a field is either a prime or zero.

Proof. Since every field is an integral domain, by the above theorem the characteristic of a field is either a prime or zero.

Note.

1. The characteristic of a division ring is either a prime or zero.
2. The characteristic of Z p, where p is a prime, is p.

Rings, Integral Domains & Fields Solved Problems

Example. 2. The characteristic of an integral domain ( R, +, ·) is zero or a positive integer according as the order of any non-zero element of R regarded as a member of the group ( R, +).

Solution. Let a ∈ R and a 6= 0.

Case (1). Let O(a) = 0 when ’ a ’ is regarded as a member of (R, +).

By the definition of order, there exists no positive integer n so that na = 0.

∴ Characteristic of R = 0.

Case (2). Let O(a) = p.

By the definition of order, p is the least positive integer, so that pa = 0.

For any x ∈ R, pa = 0 ⇒ (pa)x = 0x ⇒ a(px) = 0 ⇒ px = 0 since a 6= 0.

∴ p is the least positive integer so that px = 0∀x ∈ R.

Hence characteristic of R = p.

Example. 3. If R is a non-zero ring so that a 2 = a∀a ∈ R proves that the characteristic of R = 2 or proves that the characteristic of a Boolean ring is 2.

Solution. Since a₂ = a∀a ∈ R, we have (a + a)₂ = a + a

⇒ (a + a)(a + a) = a + a ⇒ a(a + a) + a(a + a) = a + a

⇒ a₂ + a₂ + a₂ + a₂ = a + a ⇒ (a + a) + (a + a) = (a + a) + 0 ⇒ a + a = 0 ⇒ 2a = 0.

∴ for every a ∈ R, we have 2a = 0. Further for a 6= 0, 1a = a 6= 0.

∴ 2 is the least positive integer so 2a = 0∀a ∈ R.

Hence characteristic of R = 2.

Example. 4. Find the characteristic of the ring Z₃ × Z₄.
Solution.

The characteristic of the ring Z₃ × Z₄

We have Z₃ = {0, 1, 2}, Z₄ = {0, 1, 2, 3} Z₃ × Z₄= {(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3)} contains 12 ordered pairs as elements. Zero element (0, 0) and unity element = (1, 1)

We have, 1(1, 1) = (1, 1) ≠ (0, 0); 2(1, 1) = (2, 2) ≠ (0, 0);

3(1, 1) = (3, 3) = (0, 3)≠  (0, 0); 4(1, 1) = (4, 4) = (1, 0) ≠  (0, 0);

5(1, 1) = (5, 5) = (2, 1)≠ (0, 0); 6(1, 1) = (6, 6) = (0, 2) ≠  (0, 0);

7(1, 1) = (7, 7) = (1, 3) ≠  (0, 0); 8(1, 1) = (8, 8) = (2, 0)≠  (0, 0);

9(1, 1) = (9, 9) = (0, 1)≠  (0, 0); 10(1, 1) = (10, 10) = (1, 2)≠  (0, 0);

11(1, 1) = (11, 11) = (2, 3) ≠  (0, 0); 12(1, 1) = (12, 12) = (0, 0);

L Least positive integer = 12. Hence characteristic of Z₃ × Z₄ = 12.

Also, G. C. D of 3, 4 = (3, 4) = 1 ⇒ the additive group Z₃ × Z₄ is isomorphic with Z₁₂

∴ Characteristic of Z₃ × Z₄ = Characteristic of Z₁₂ = 12.

 Example. 5. If the characteristic of a ring is 2 and the elements a, b of the ring commute prove that (a + b)₂ = a₂ + b₂ = (a − b)₂.

Solution. Since characteristic of the ring R = 2 ⇒ 2x = 0∀x ∈ R.

a, b ∈ R commute ⇒ ab = ba.

(a+ b)₂ = (a+ b)(a+ b) = a(a+ b)+ b(a+ b) = a₂ + ab+ ba+ b₂ = a₂ +2ab+ b₂

a, b ∈ R ⇒ ab ∈ R and 2(ab) = 0. ( ∵ characteristic of R = 2)

(a + b)₂ = a₂ + 0 + b₂ = a₂ + b₂.

Similarly, we can prove that (a − b)₂ = a₂+ b₂.

Example. 6. If R is a commutative ring with unity of characteristic = 3 then prove that (a + b)₃ = a₃ + b3∀a, b ∈ R

Solution. R is a ring with characteristic = 3 ⇒ 3x = 0, zero elements of R∀x ∈ R.

Since R is a commutative ring, by Binomial Theorem, (a + b)₃ = a₃ +3a₂b + 3ab₂ + b₃

a, b ∈ R ⇒ a2b, ab₂ ∈ R ⇒ 3a₂b = 0, 3ab₂ = 0. ∴ (a + b)₃ = a₃ + b₃.

Rings, Integral Domains, And Fields Exercise 3

1. Prove that the characteristic of the ring Zn = {0, 1, 2, . . . , n − 1} under addition and multiplication modulo n, is n.

2. Prove that the characteristic of a field is either prime or zero.

3. Prove that the characteristic of a finite integral domain is finite.

4. Prove that any two non-zero elements of an integral domain regarded as the members of its additive group are of the same order.

5. Give examples of a field with zero characteristics and a field with characteristics.

6. Find the characteristics of the rings (1) 2Z (2) Z × Z

7. If R is a commutative ring with unity of characteristic = 4 then simplify (a + b) 4 for all a, b ∈ R.

8. If R is a commutative ring with unity of characteristic = 3 compute and simplify (1)(x + y) 6(2)(x + y) 9∀x, y ∈ R

Answers

5. Zn ring, Z5 ring
6. (1) 0 (2) 0 7. a4 + a3b + ab3 + b4
7. (1)x6 + 2x3y3 + y6 (2) x9 + y9

Rings, Integral Domains & Fields Divisibility Units, Associates And Primes In A Ring.

Definition. (Divisor or Factor) Let R be a commutative ring and a 6= 0, b ∈ R. If there exists q ∈ R such that b = aq then a’ is said to divide ’ b ’.

Notation. ’ a ’ divides ’ b ’ is denoted by a | b and a ’ does not divide ’ b ’ is denoted by aχb.

Note.

1. If ’ a ’ divides ’ b ’ then we say that a ’ is a divisor or factor of ’ b ’.

2. For a₆ = 0, 0 ∈ R we have a · 0 = 0 and hence every non-zero element of a ring R is a divisor of 0 ’ 0 ’ Zero element of R.

3. a₆ = 0, b ∈ R and a | b ⇔ b = aq for some q ∈ R.

For example., 1. In the ring Z of integers, 3115 and 317.

For example., 2. In the ring Q of rational numbers, 3/7 because there exists (7/3) ∈ Q such that 7 = 3.(7/3),

For example., 3. In a field F, two non-zero elements are divisors to each other.

For example., 4. In the ring Z₆, 4 | 2, In the ring Z₈, 3 | 7 and in the ring Z₁₅, 9 | 12.

For example., 5. The unit of a ring R divides every element of the ring if a ∈ R is a unit then aa −1 = a−1a = 1 where a−1 ∈ R.

For any b ∈ R we have b = 1b = aa−1b = a a−1b ⇒ a | b.

Theorem. 1. If R is a commutative ring with unity and a b, c ∈ R then

1. a | a (2) a | b and b|c → a|c
2. a|b → a|bx∀x ∈ R
3. a | b and a|c ⇒ a|bx + cy∀x, y ∈ R.

Proof.

(1) If 1 ∈ R is the unity element in R then a = a.1 which implies that a | a.

(2) a | b ⇒ b = aq₁ for some q₁ ∈ R; b | c ⇒ c = bq₂ for some q₂ ∈ R.

Now c = bq₂ = (aq₁) q₂ = a (q₁ · q₂) = aq where q = q₁q₂ ∈ R ⇒ a | c.

(3) a | b ⇒ b = aq for some q ∈ R.

Now bx = (aq)x = a(qx) = aq₁ where q₁ = qx ∈ R ⇒ a | bx.

(4) a|b ⇒ a|bx∀x ∈ R; a|c ⇒ a|cy∀y ∈ R

a | bx ⇒ bx = aq₁ for some q₁ ∈ R; a | cy ⇒ cy = aq₂ for some q₂ ∈ R.

∴ bx+cy = aq₁+aq₂ = a (q₁ + q₂) = aq where q = q₁+q₂ ∈ R ⇒ a | (bx+cy).

Note. a | b and a|c ⇒ a|b± c. Definition. (Greatest Common Divisor G.C.D)

Let R be a commutative ring and a, b ∈ R, d ∈ R is said to be the greatest common divisor of ’ a ’ and ’ b ’ if

1. d | a and d | b and
2. whenever c | a and c | b

where c ∈ R then c | d. Notation. If ’ d ’ is a greatest common divisor (G. C. D) of ’ a ’ and ’ b ’ then we write d = (a, b).

Definition. (Unit) Let R be a commutative ring with unity. An element a ∈ R is said to be a unit in R if there exists an element b ∈ R such that ab = 1 in R. However, the unity element ’ 1 ’ is also a unit because 1, 1 = 1.

2. In a ring, the unity element is unique, while, units may be more than one.

3. If ab = 1 then a− 1 = b. So, a unit in a ring R is an element of the ring so its multiplicative inverse is also in the ring. That is a ∈ R is a unit of R means that the element d a is invertible.

4. Units of a ring are in fact the two divisors of the unity element in the ring.

5. ’ a ’ is a unit in R ⇒ ab = 1 for some b ∈ R ⇒ ’ b ’ is also a unit of R. example. In a field F, every non-zero element has a multiplicative inverse. So, every non-zero element in a field is a unit.

Theorem 2. Let D be an integral domain. For a, b ∈ D, if both a | b and b | a are true then a = ub where u is a unit in D.

Proof. a | b ⇒ b = aq₁ for some q₁ ∈ D; b | a ⇒ a = bq₂ for some q₂ ∈ D.

b = aq₁ = (bq₂) q₁ = b (q₂q₁) ⇒ 1 = q₂q₁, by using cancellation property in integral domain.

∴ q₂q₁ = 1 ⇒ q₂ is a unit in D.

Hence a = bq₂ where q₂ is a unit in D.

Definition. (Associates) Let R be a commutative ring with unity. Two elements ’ a ’ and ’ b ’ in R are said to be associates if b = ua for some unit u in R.

Note. The relation of being associated in a ring R is an equivalence relation in R.

For example., 1. If ’ 1 ’ is the unity element in the ring R then ’ 1 ’ is a unit in R. For a(6= 0) ∈ R we have a = 1. a and a, a are associates in R.

For example., 2. In the ring Z of integers, the units are 1 and −1 only. For a₆ = 0 ∈ Z, we have a = a : 1 and a = (−a)(−1) only. Therefore, a ∈ Z has only two associates, namely, a, −a.

For example., 3. In the ring Z₆ = {0, 1, 2, 3, 4, 5} of integers modulo −6, the units are 1,5 only.

For 2 ∈ Z₆ ; 2 ≡ 2.1(mod6) and 2 ≡ 4.5(mod6)

∴ 2 has two associates 2,4.

Theorem. 3. In an integral domain D, two non-zero elements a, b ∈ D are associates iff a | b and b | a.

Proof. From Theorem (2) we see that a | b and b | a

⇒ there exists unit u ∈ D such that a = ub ⇒ a, b are associates. a, b are associates in D ⇒ there exists unit u in D such that a = ub ⇒ b | a. u is unit in D ⇒ there exists unit v ∈ D such that uv = 1.

Now a = ub ⇒ va = v(ub) ⇒ va = (vu)b ⇒ va = (1) b ⇒ b = va ⇒ a | b.

Hence a | b and b | a.

Definition. (Trivial Divisors and Proper Divisors)

Let a 6= 0 be an element in the integral domain D. The units in D and the associates of ’ a ’ are divisors of ’ a ’. These divisors of ’ a ’ is called Trivial divisors of ’ a ’. The remaining divisors of ’ a ’ are called the proper divisors of ’ a ’.

For example., Consider the integral domain (Z, +, ·). The units in Z are 1 and −1 only.

For a₆ = 0 ∈ Z, the trivial divisors are 1, −1, a, −a only. The remaining divisors of ’ a ’ are proper divisors.

3 ∈ Z has only trivial divisors ±1, ±3 and no proper divisors.

6 ∈ Z has trivial divisors ±1, ±6 and also proper divisors ±2, ±3.

Definition. (Prime and Composite elements)

Let ’ a ’ be a non-zero and non-unit element in an integral domain D. If ’ a ’ has no proper divisors in D then ’ a ’ is called a prime element in D. If ’ a ’ has proper divisors in D then ’ a ’ is called Composite element in D.

Note. a ∈ D is a prime element and a = bc then one of b or c is a unit in D.

For example., 1. In the integral domain (Z, +, ·);

5 ∈ Z is a prime element and 6 ∈ Z is a composite element.

For example., 2. In the integral domain (Q, +, ·); 6 ∈ Q is the prime element since all its divisors are units.

Rings, Integral Domains, And Fields Solved Problems

Example.1. Find all the units of Z₁₂ the ring of residue classes modulo 12.
Solutions.

We have Z₁₂  = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.

Clearly, the unity element = 1 is a unit. a¯ ∈ Z₁₂  is a unit if there exists ¯b ∈ Z₁₂  such that a¯¯b = 1.

For ~a = even there is no ¯b so that a¯¯b ≡ 1(mod12), as a¯¯b, is even. So we have to verify for a¯ = odd.

For 5 ∈ Z₁₂  we have 5 × 5 = 1; 7 ∈ Z₁₂ we have 7 × 7 = 1 and 11 ∈ Z₁₂ . we have 11 × 11 = 1. ∴ 1, 5, 7, and 11 are the units in Z₁₂

Example.2. Prove that ±1, ±i are the only four units in the domain of Gaussian integers.

Solution. Z[1] =a + ib | a,b ∈ Z, i² = −1 is the integral domain of Gaussian integers. 1 + 0i = 1 is the unity element. Let x + iy ∈ Z[i] be a unit. By the definition, there exists u + iv ∈ Z[1] such that (x + iy)(u + iv) = 1 ⇒ |(x + iy)(u + iv)| = 1 ⇒ x² + y² u² + v² = 1

⇒ x² = 1,y² = 0 or x² = 0 or y² = 1 ⇒ x = ±1,y = 0 or x = 0,y = ±1.

∴ ±1 + 0i, 0 ± 1 i i.e., ±1; ±i are the possible units.

Example.3. Find all the associates of (2 − i) in the ring of Gaussian integers. (N. U. 97).

Solution. We have 2 − i = (2 − i) · 1; 2 − i = (−2 + i) · (−1); (2 − i) = (−2i − 1) · I and 2 − i = (2i + 1) · (−i).

∴ 2 − i, −2 + i, −2i − 1 and 2i + 1 are the associates.

Example 4. In the domain of Gaussian integers, prove that the associates of a + ib are a + ib, −a − ib, ia − b, −i a + b.
Solution.

Since ±1 and ±i are the four units of Z[1], a + ib = (a + ib).1; a + ib = (−a − ib) · (−1); a + ib = (ia − b) · (−i) and a + ib = (−i a + b) · i

∴ a + ib, −a − ib, i a − b, and −i a + b are the associates of a + ib.

Example. 5. If D is an integral domain and U is a collection of units in D, Prove that (U, ·) is a group.
Solution. (Left to the reader)

Example. 6. Find all units of Z₁₄.
Solution.

Z14 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} Unity element = 1 is a unit. Since 14 is even, the number in Z14 cannot be a unit.

For 3 ∈ Z₁₄. we have 3.5 = 15 = 1 ⇒ 3, 5 are units.

For 9 ∈ Z_14, we have 9.11 = 99 = 1 ⇒ 9, 11 are units.

For 13 ∈ Z₁₄ we have 13.13 = 169 = 1 ⇒ 13 is a unit.

Example. 7. Find all the units in the matrix ring M₂ (Z₂) 
Solution.

We have Z₂ = {0, 1} so that 0 + 0 = 0, 0 + 1 = 1 + 0 = 1 and 1 + 1 = 0.

⇒ \(\mathrm{M}_2\left(\mathrm{Z}_2\right)=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right) where a, b, c, d \in\{0,1\}\)

⇒ Number of elements in \(M_2\left(Z_2\right)=2^4=16\)

⇒ Clearly \(\mathrm{I}_2=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\) is the unity element and hence an unit.

A ∈ M₂ (Z₂) is a unit in M₂ if there exists a B ∈ M₂ such that AB = I2 the unity element. AB = I2 happens when A is non-singular and B = A−1.

Hence the · · · its of M₂ (Z₂) are all the non-singular matrices. Matrices having only one ’ 0 ’ and three ’ 1 ’s are :

⇒ \(\left(\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right)\)

which are non-singular Hence the above 4 matrices are units. Matrices having two ‘ 0 ‘s and two 1 “s are

⇒ \(\left(\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}
0 & 0 \\
1 & 1
\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}
1 & 0 \\
1 & 0
\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}
0 & 1 \\
0 & 1
\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\)

Among the above six matrices of

⇒ \(\mathrm{M}_2 ;\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right)\)

⇒ \(\left(\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right)\)

are only nonsingular. Hence these two matrices are units.

Matrices having three ‘0’s and one ‘1 are:

⇒ \(\left(\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right)\)

which are all singular.

The zero matrix

⇒ \(\mathrm{O}=\left(\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right)\)

and the matrix having all ‘l’s

⇒ \(\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)\)

are both singular. Hence the units of

⇒ \(\mathrm{M}_2\left(\mathrm{Z}_2\right) \)are

⇒ \(\left(\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right),\)

⇒ \(\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)\)

⇒ \(\text { and }\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right) \text { which are six in number. }\)

Rings, Integral Domains, And Fields Some Noncommutative Examples.

There are many rings that are not Commutative under multiplication. We study three non-commutative rings, namely, the ring of square matrices over a field, the ring of endomorphisms of an abelian group, and the Quaternions.

Rings, Integral Domains And Fields Solve Problems

Example 1. Prove that the set of all 2 × 2 matrices over the field of Complex numbers is a ring with unity under addition and multiplication of matrices.

Solution. Let \(R=\left\{\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]: a, b, c, d \in \mathrm{C}\right\} be the set of 2 \times 2 matrices over \mathrm{C}\)
Let A=

⇒ \(\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[a_{i j}\right]_{2 \times 2}, B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right]=\left[b_{i j}\right]_{2 \times 2} and C=\)

⇒ \([\left[\begin{array}{ll}
c_{11} & c_{12} \\
c_{21} & c_{22}
\end{array}\right]=\left[c_{i j}\right]_{2 \times 2} \)

be three elements in \mathrm{R}.

(1) A + B = [aij] 2× 2 + [bij] 2× 2 = [aij + bij] 2× 2 and

A + B = [bij + aij] 2× 2 = B + A ( ∵ aij,bij ∈ C)

∴ Addition is a binary operation and also Commutative. (2) (A + B) + C =  [aij + bij] 2× 2 + [cij] 2× 2 = [(aij + bij) + cij] 2× 2  = [aij + (bij + cij)] 2× 2 = A + (B + C) ( ∵ aij,bij,cij ∈ C)

∴ Addition is associative.

⇒ (3) We have O = \(\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=[0]_{2 \times 2} \in \mathrm{R} such that A+\mathrm{O}=\)

⇒ \(\left[a_{i j}+0\right]_{2 \times 2}= \left.a_{i j}\right]_{2 \times 2} = A\)

= Therefore  O = \(\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\) is the Zero element.

(4) For A − [aij] 2× 2 ,aij ∈ C ⇒ −aij ∈ C so that aij + (−aij) = o ∈ C.

∴ there exists −A = [−aij] 2× 2 ∈ R such that A+(−A) = [aij + (−aij)] 2× 2 = [0] 2× 2 = O.

∴ (R, +) is an abelian group.

(5) Let A = [aij] 2× 2 ,B = [bjk] 2× 2 ,C = [ckl] 2× 2 ∈ R.

From the definition of multiplication; AB = [aij] [bjk] = [uik] 2× 2 where uik = P2j=1 aij bjk = ai1b1k + ai2b2k ∈ C

∴ Multiplication is a binary operation.

(6) \((A B) C=\left[\sum_{j=1}^2 a_{i j} b_{j k}\right]_{2 \times 2}\left[c_{k l}\right]_{2 \times 2}\)

=\(\left[\sum_{k=1}^2\left(\sum_{j=1}^2 a_{i j} b_{j k}\right) c_{k l}\right]\)

=\(\left[\sum_{j=1}^2 a_{i j}\left(\sum_{k=1}^2 b_{j k} c_{k l}\right)\right]\)

=A(B C) therefore Multiplication is associative.

(7) \(A(B+C)=\left[a_{i j}\right]_{2 \times 2}\left[b_{j k}+c_{j k}\right]_{2 \times 2}\)

=\(\left[\sum_{j=1}^2 a_{i j}\left(b_{j k}+c_{j k}\right)\right]\)

= \(\left[\sum_{j=1}^2\left(a_{i j} b_{j k}+a_{i j} c_{j k}\right)\right]\)

= \(\left[\sum_{j=1}^2 a_{i j} b_{j k}\right]+\left[\sum_{j=1}^2 a_{i j} c_{j k}\right]=A B+A C\)

Similarly, we can prove that (B + C)A = BA + CA. ∴ Distributive laws hold.

Hence (R, +, ·) is a ring

⇒ \(\text { Since } 1 \in \mathrm{C}, I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \in \mathbb{R} \text {. For } A\)

⇒ \(=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right]_{2 \times 2}\)

⇒ \(A I=\left[\begin{array}{ll}
a_{11}+0 & 0+a_{12} \\
a_{21}+0 & 0+a_{22}
\end{array}\right]_{2 \times 2}=\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right] \) = A. Also IA = A.

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \)

⇒ \(
\text { Let } A=\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right] \text { and } B=\left[\begin{array}{ll}
2 & 0 \\
0 & 1
\end{array}\right] \text {. }\) Then AB =

⇒ \(
\left[\begin{array}{ll}
2+0 & 0+2 \\
2+0 & 0+2
\end{array}\right]=\)

⇒ \(
\left[\begin{array}{ll}
2 & 2 \\
6 & 4
\end{array}\right]
\)

and BA =\(
\left[\begin{array}{ll}
2+0 & 4+0 \\
0+4 & 0+4
\end{array}\right]=\left[\begin{array}{ll}
2 & 4 \\
4 & 4
\end{array}\right] \)

so that AB 6= BA

Hence (R, +, ·) is not a commutative ring.

Notation. The ring of all 2 × 2 matrices over the field of complex numbers C is denoted by M₂(C). If F is a field the ring of all n × n matrices over F is denoted M n( F). The zero element in M_n( F) is denoted by O·n× n and the unity element by I

Zero divisors in M₂(C) : A =

⇒ \(
\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right] \neq \mathrm{O} \text { and } B=\left[\begin{array}{ll}
0 & 0 \\
0 & 1
\end{array}\right] \neq \mathrm{O}\)

Then AB = \(
\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right] \neq \mathrm{O} \text { and } B A=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\mathrm{O}
\)

We observe that AB 6= BA and A₆ = O, B₆ = O ⇒ BA = O. Therefore there exist Zero divisors in M₂(F) where F is a field.

Nilpotent element in M₂(C) :

For A = \(
\left[\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right]
\)

we have A₂ = \(
\left[\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 2 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\mathrm{O} \)

Therefore A is a nilpotent matrix in M₂(C).

⇒ \( B=\left[\begin{array}{cc}
1 & 1 \\
-1 & -1
\end{array}\right]\) is also a nilpotent matrix element in M₂(C).

Example.2. The set of 2 × 2 matrices of the form

⇒ \(
\left[\begin{array}{cc}
x & y \\
-\bar{y} & \bar{x}
\end{array}\right]
\)

where x, y are complex numbers and x,¯ y¯ denote the complex conjugates of xy; is a skew field for compositions of matrix addition and multiplication.

Solution. Let M =\(
\left\{\left[\begin{array}{cc}
\frac{x}{-y} & \bar{x}
\end{array}\right]: x=a+i b, y=c+i d ; a, b, c, d \in \mathrm{R}\right\} \) be the set
of 2 × 2 matrices.

⇒ \(
\text { Let } A=\left[\begin{array}{cc}
x_1 & y_1 \\
-y_1 & \bar{x}_1
\end{array}\right], B=\left[\begin{array}{cc}
x_2 & y_2 \\
-y_2 & \bar{x}_2
\end{array}\right], C=\left[\begin{array}{cc}
x_3 & y_3 \\
-y_3 & \overline{x_3}
\end{array}\right] \in \mathrm{M} \)

⇒ \(
\text { (1) } A+B=\left[\begin{array}{cc}
x_1+x_2 & y_1+y_2 \\
-\bar{y}_1-\bar{y}_2 & \overline{x_1}+\bar{x}_2
\end{array}\right]=\left[\begin{array}{cc}
x_1+x_2 & y_1+y_2 \\
-\overline{y_1+y_2} & \underline{x_1+x_2}
\end{array}\right] \in \mathrm{M} \text {, since }
\)

¯Z₁ ± Z₂ = Z¯₁ ± Z¯₂ for Z¯₁, Z¯₂ ∈ C

⇒ \(
\text { A. } B=\left[\begin{array}{ll}
x_1 & y_1 \\
-y_1 & \bar{x}_1
\end{array}\right]\left[\begin{array}{cc}
x_2 & y_2 \\
-y_2 & -x_2
\end{array}\right]=\left[\begin{array}{cc}
x_1 x_2-y_1 y_2 & x_1 y_2+y_1 x_2 \\
-y_1 x_2-x_1 y_2 & -y_1 y_2+x_1 x_2
\end{array}\right]\)

If u = x₁x₂ − y₁y¯₂ and ν = x₁y₂ + y₁x¯₂ then u = ¯x₁x¯₂ − y¯₁y₂ and v¯ = x¯₁y¯₂ + ¯y₁x₂

A,B =\(
\left[\begin{array}{cc}
u & v \\
-\bar{v} & \bar{u}
\end{array}\right] \in \mathrm{M}
\)

Hence addition ( + ) and multiplication

(1) are binary operations.

(2) Clearly A + B = B + A for any A,B ∈ M.

(3) Clearly (A + B) + C = A + (B + C) and (A,B) · C = A(B,C) for any A,B,C ∈ M because addition and multiplication of matrices are associative.

(4) There exists O =

⇒ \(
\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=\left[\begin{array}{cc}
0+i 0 & 0+i 0 \\
-0+i 0 & 0+i 0
\end{array}\right] \in
\)

M so that A +O = A for any A ∈ M.

(5) For A =\(
\left[\begin{array}{ll}
x & y \\
-y & \bar{x}
\end{array}\right] \text { there exists }-A=\left[\begin{array}{ll}
-x & -y \\
\bar{y} & -x
\end{array}\right] \text { so that } A+(-A)=
\)

⇒ \(
\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0
\)

= 0 (Zero matrix)

(6) For any A,B,C ∈ M, distributive laws, namely, A· (B+C) = A·B+A·C

and (B + C) · A = B · A + C · A are clearly true. Hence (M, +, −) is a ring

⇒ \(
\text { (7) We have } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1+i 0 & 0+i 0 \\
-0+i 0 & \frac{1+i \cdot 0}{1+}
\end{array}\right] \in \text { so that } A \cdot I=I \cdot A=A \text { for any } A \in \mathrm{M} \text {. }
\)

∴ the ring M has unity element I.

(8) Let A 6= 0 ∈ M so that A =\(
\left[\begin{array}{cc}
x & y \\
-y & \bar{x}
\end{array}\right]=\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]
\)

where a,b,c,d are not all zero.

Det A = (a + ib)(a − ib) − (c + id)(−c + id) = a₂ + b₂ + c₂ + d₂ 6= 0

Since det A 6= 0, A 6= O is invertible. Hence (M, +, ·) is a skew field.

Note. The matrix  \(
\left[\begin{array}{cc}
x & y \\
-y & \bar{x}
\end{array}\right] \text { is also given as }\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]
\) in the problem.

Rings, Integral Domains & Fields Ring Of Quaternions

Example 3. Prove that the set of Quaternions is a skew field.

Solution. Let Q = R × R × R × R = {α₀ + α₁i + α₂j + α₃k | α₀,α₁,α₂,α₃ ∈ R} where i,j,k are quaternion units satisfying the relations :

i ²  = j²  = k ²   = i · jk = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. Let

X, Y, Z ∈ Q so that X = α₀ + α₁i + α₂j + α₃k, Y = β₀ + β₁i + β₂j + β₃k and Z = γ₀ + γ₁i + γ₂j + γ₃k where α_t, β_t, γ_t for t = 0, 1, 2, 3 are real numbers.

We define X = Y ⇔ α_t = βt for t = 0, 1, 2, 3.

We define addition (+) as X + Y = (α₀ + β₀) + (α₁ + β1) i + (α2 + β2) j +

(α₃ + β₃) k and multiplication (·) as X.Y = ( α₀β₀− α₁β₁ − α₂β₂ − α₃β₃) + (α₀β₁ + α₁β₀ + α₂β₃ − α₃β₂) i

+ (α₀β₂ + α₂β₀ + α₃β₁ − α₁β₃) j + (α₀β₃ + α₃β₀ + α₁β₂ − α₂β₁) k.

(1) ∀X, Y ∈ Q; X + Y = (α₀+ β₀) + (α₁+ β₁) i + (α₂ + β₂) j + (α₃ + β₃) k

As αt + βt for t = 0, 1, 2, 3 ∈ R, X + Y ∈ Q. ∴ addition (+) is a binary operation.

(2) ∀X, Y ∈ Q; X + Y = (α₀+ β₀) + (α₁+ β₁) i + (α₂ + β₂) j + (α₃ + β₃) k

= (β0 + α₀) + (β₁ + α₁) i + (β₂ + α₂) j + (β₃ + α₃) k˙ = Y + X

∴ addition is commutative:

( ∵ α_t + β_t = β_t + α_t for t = 0, 1, 2, 3) .

(3) ∀X, Y, Z ∈ Q

(X + Y) + Z = {(α₀ + β₀) + (α₁ + β₁) i + (α₂ + β₂) j + (α₃ + β₃) k} + (γ₀ + γ₁i + γ₂j + γ₃k)

= [(α₀ + β₀) + γ₀] + [(α₁ + β₁) + γ₁] i + [(α₂ + β₂) + γ₂] j + [(α₃ + β₃) + γ₃] k

= [α₀ + (β₀ + γ₀)] + [α₁ + (β₁ + γ₁)] i + [α₂ + (β₂ + γ₂)] j + [α₃ + (β₃ + γ₃)] k

= X + (Y + Z). ( ∵ (α_t + βt) + γ_t = α_t + (β_t + γ_t) for t = 0, 1, 2, 3) .

∴ addition is associative.

(4) For O = 0 + 0i + 0j + 0k ∈ Q and X = α₀ + α₁i + α₂j + α₃k we have

O + X = (0 + α₀) + (0 + α₁) i + (0 + α₂) j + (0 + α₃) k = α₀ + α₁i + α₂j + α₃k =

X = X + O

∴ O = 0 + 0i + 0j + 0k is the additive identity.

(5) For X = α₀ + α₁i + α₂j + α₃k there exists −X = (−α₀) + (−α₁) i +

(−α₂) j + (−α₃) k ∈ Q such that X + (−X) = [α₀ + (−α₀)] + [α₁ + (−α₁)] i +

[α₂ + (−α₂)] j + [α₃ + (−α₃)] k = 0 + 0i + 0j + 0k = 0, the additive identity.

∴ every element has an additive inverse.

Hence, from (1), (2), (3), (4) and (5) : (Q, +) is abelian group.

(6) X.Y = b₀ + b₁i + b₂j + b₃k where b₀ = α₀β₀ − α₁β₁ − α₂β₂ − α₃β₃

b₁ = α₀β₁ + α₁β₀ + α₂β₃ − α₃β₂, b₂ = α₀β₂ + α₂β₀ + α₃β₁ − α₁β₃

and b3 = α0β3 + α3β0 + α1β2 − α2β1 are real numbers.

∴ multiplication (·) is a binary operation.

(7) (X : Y) · Z = (b₀ + b₁i + b₂j + b₃k) · (γ₀ + γ₁i + γ₂j + γ₃k)

= (b₀γ₀ − b₁γ₁ − b₂γ₂ − b₃γ₃) + (b₀γ₁ + b₁γ₀ + b₂γ₃ − b₃γ₂) i + (b₀γ₂ + b₂γ₀ + b₃γ₁ − b₁γ₃) j

+ (b₀γ₃ + b₃γ₀ + b₁γ₂ − b₂γ₁) k

Y.Z = c₀ + c₁i + c₂j + c₃k where c₀ = β₀γ₀ − β₁γ₁ − β₂γ₂ − β₃γ₃,

c₁ = β₀γ₁ + β₁γ₀ + β₂γ₃ − β₃γ₂,

c₂ = β₀γ₂ + β₂γ₀ + β₃γ₁ − β₁γ₃,

c₃ = β₀γ₃ + β₃γ₀ + β₁γ2 − β2γ₁.

X.(Y.Z) = (α₀ + α₁i + α₂j + α₃k) . (c₀ + c₁i + c₂j + c₃k)

= (α₀c₀ − α₁c₁ − α₂c₂ − α₃c₃) + (α₀c₁ + α₁c₀ + α₂c₃ − α₃c₂) i

+ (α₀c₂ + α₂c₀ + α₃c₁ − α₁c₃) j + (α₀c₃ + α₃c₀ + α₁c₂ − α₂c₁) k

Since the corresponding terms of (X.Y). Z and X.(Y, Z) are equal we have

∴ multiplication is associative.

(8) Both the distributive laws, namely, X.(Y + Z) = X.Y + X.Z and (Y + Z) · X = Y · X + Z · X can be proved to be true.

From the truth of the above 8 properties, we establish that (Q+, ·) is a ring.

(9) There exists 1 = 1 + 0i + 0j + 0k ∈ Q such that

∀X ∈ Q we have 1.X = (1 + 0i + 0j + 0k) · α0˙ + α₁i + α₂j + α₃k

= (1.α₀ − 0.α₁ − 0.α₂ − 0.α₃) + (1.α₁ + α₀ · 0 + 0.α₃ − 0.α₂) i

+ (1.α₂ + 0.α₀ + 0.α₁ − α₁ · 0) j + (1 · α₃ + 0.α₀ + 0 · α₂ − 0.α₁) k

= α₀ + α₁i + α₂j + α₃k = X. Also X.1= X.

∴ 1 = 1 + 0i + 0j + 0k ∈ Q is the unity element.

(10) Let X 6= 0, the zero element. Then not all α0, α1, α2, α3 are zero ∈ R.

∴ α²₀ + α² ₁+ α²₂ + α²₃ = ∆ ≠0 ∈ R.

For the real numbers α ₀

⇒ \(
\frac{\alpha_0}{\Delta}, \frac{-\alpha_1}{\Delta}, \frac{-\alpha_2}{\Delta}, \frac{-\alpha_3}{\Delta}
\) there exists

⇒ \(
\mathrm{X}^1=\frac{\alpha_0}{\Delta}-\frac{\alpha_1}{\Delta} i-\frac{\alpha_2}{\Delta} j-\frac{\alpha_3}{\Delta} k \in \mathrm{Q} . \quad \text { Further } \mathrm{X} \cdot \mathrm{X}^1=\left(\frac{\alpha_0^2}{\Delta}+\frac{\alpha_1^2}{\Delta}+\frac{\alpha_2^2}{\Delta}+\frac{\alpha_3^2}{\Delta}\right)
\)

⇒ \(
+\left(\frac{-\alpha_0 \alpha_1}{\Delta}+\frac{\alpha_0 \alpha_1}{\Delta}-\frac{\alpha_2 \alpha_3}{\Delta}+\frac{\alpha_3 \alpha_2}{\Delta}\right) i+\left(\frac{-\alpha_0 \alpha_2}{\Delta}+\frac{\alpha_2 \alpha_0}{\Delta}-\frac{\alpha_3 \alpha_1}{\Delta}+\frac{\alpha_1 \alpha_3}{\Delta}\right) j
\)

⇒ \(
+\left(\frac{-\alpha_0 \alpha_3}{\Delta}+\frac{\alpha_3 \alpha_0}{\Delta}-\frac{\alpha_1 \alpha_2}{\Delta}+\frac{\alpha_2 \alpha_1}{\Delta}\right)
\)

k = 1 + 0i + 0j + 0k = 1, the unity element.

Similarly, we can prove that X1 · X = 1.

We have X.Y = (α₀β₀ − α₁β₁ − α₂β₂ − α₃β₃)+(α₀β₀ + α₁β₀ + α₂β₃ − α₃β₂) i

+ (α₀β₂ + α₂β₀ + α₃β₁ − α₁β₃) j + (α₀β₃ + α₃β₀ + α₁β₂ − α₂β₁) k

= b₀ + b₁i + b₂j + b₃k and

∴ every non-zero element of Q has a multiplicative inverse.

Y.X = (β₀α₀ − β₁α₁ − β₂α₂ − β₃α₃) + (β₀α₁ + β₁α₀ + β₂α₃ − β₃α₂) i

+ (β₀α₂ + β₂α₀ + β₃α₁ − β₁α₃) j + (β₀α₃ + β₃α₀ + β₁α₂ − β₂α₁) k = a₀ + a₁i + a₂j + a₃k.

we observe that  b ₀ = a ₀,b ₁ ≠ a ₁,b ₂ ≠a ₂,_{b 3} ≠a ₃.

∴ multiplication is not commutative.

Hence (Q+, ·) is a Division ring or Skew field.

Note. We can take 1 = (1, 0, 0, 0),i = (0, 1, 0, 0),j = (0, 0, 1, 0) and k = (0, 0, 0, 1)

2. The set G = {±1, ±i, ±j, ±k} form a nonabelian group of order 8 under multiplication (·) defined as follows:

i ₂  = j  ₂  = k  ₂  = i,k = −1,ij = −ji = k; jk = −kj = i and ki = −ik = j

Rings, Integral Domains, And Fields Ring Of Endomorphisms Of An Abelian Group.

Let G be an abelian group. A homomorphism of G into itself is an endomorphism of G. The set of all endomorphisms of G is denoted by Hom (G, G) or

Hom (G).

For f,g ∈ Hom(G, G) if we define addition (+) and multiplication (·) of two

endomorphisms as (f + g)(x) = f(x) + g(x) and (f,g)(x) = f(g(x))∀x ∈ G then < Hom(G, G), +, > is a ring.

Note. Hom (G, G) is not commutative as the composition of functions is not commutative.

Integration Of Vectors

Integration is the inverse operation of differentiation. Let F (t) be a differentiable vector function of a scalar variable t and

Let \(\frac{d}{d t} \mathbf{F}(t)\)=\(\mathbf{f}(t)\). Then \(\int \mathbf{f}(t) d t\)= F(t)

F (t) is called the primitive of f (t)

The set of all primitives of f (t), that is\(\int \mathbf{f}(t) d t\) = F (t) + C where C is any arbitrary constant vector, is called the indefinite integral of f (t).

Hence, the indefinite integral of F (t) is not unique.

 Note. If f(t)= f₁(t)i + f₂(t)j + f₃(t)k , then

∫f(t)dt  =  i∫f₁(t)dt +j∫f₂(t)dt + k ∫f₃(t)dt + C

Integration Of Vectors Examples And Solutions

Integration Of Vectors Definite Integral

Let  ∫f(t)dt  = F (t) . Then F (b)- F (a) is called the definite integral of F (t) between

The limits, t = a and t = b. This is denoted\(\int_a^b \mathbf{f}(t) dt \)=\([\mathbf{F}(t)]_a^b\) =F(b)-F(a)

This integral can also be defined as a limit of a sum in a manner analogous to that of elementary integral calculus.

Note. Let f(t)=f₁(t)i + f₂(t)j + f₃(t)k .

Then \(\int_a^b \mathbf{f}(t) d t\)=\(\mathbf{i} \int_a^b f_1(t) d t+\mathbf{j} \int_a^b f_2(t) d t+\mathbf{k} \int_a^b f_3(t) d t\)

Standard Results

1. We have \(\frac{d}{d t}(\mathbf{r} \cdot \mathbf{s})\)=\(\frac{d \mathbf{r}}{d t} \cdot \mathbf{s}+\mathbf{r} \cdot \frac{d \mathbf{s}}{d t}\)

∴ \(\int\left(\frac{d \mathbf{r}}{d t} \cdot \mathbf{s}+\mathbf{r} \cdot \frac{d \mathbf{s}}{d t}\right) d t\) = r.s + c

Here c is a scalar constant, since the integrand is a scalar

We have \(\frac{d}{d t}\left(\mathbf{r}^2\right)\)=\(2 \mathbf{r} \cdot \frac{d \mathbf{r}}{d t}\)

∴ \(\int\left(2 \mathbf{r} \cdot \frac{d \mathbf{r}}{d t}\right) d t\) = r²+c

2.\(\frac{d}{d t}\left(\frac{d \mathbf{r}}{d t}\right)^2\)=\(2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}\)

∴\(\int\left(2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}\right) d t\)=\(\left(\frac{d \mathbf{r}}{d t}\right)^2+c\)

where c is a scalar constant

3.We have\(\frac{d}{d t}(\mathbf{r} \times \mathbf{s})\)=\(\frac{d \mathbf{r}}{d t} \times \mathbf{s}+\mathbf{r} \times \frac{d \mathbf{s}}{d t}\)

∴\(\int\left(\frac{d \mathbf{r}}{d t} \times \mathbf{s}+\mathbf{r} \times \frac{d \mathbf{s}}{d t}\right) d t\)=\(\mathbf{r} \times \mathbf{s}+\mathbf{c}\)

Here the constant c is a vector quantity since the integrand is also a vector quantity.

4. If an Inconstant vector, we have \(\frac{d}{d t}(\mathbf{a} \times \mathbf{r})\)=\(\mathbf{a} \times \frac{d \mathbf{r}}{d t}\)

∴  \(\int\left(\mathbf{a} \times \frac{d \mathbf{r}}{d t}\right) d t=\mathbf{a} \times \int \frac{d \mathbf{r}}{d t} d t\)=\(\mathbf{a} \times \mathbf{r}+\mathbf{c}\)

5. Now \(\frac{d}{d t}\left(\mathbf{r} \times \frac{d \mathbf{r}}{d t}\right)\)=\(\frac{d \mathbf{r}}{d t} \times \frac{d \mathbf{r}}{d t}+\mathbf{r} \times \frac{d^2 \mathbf{r}}{d t^2}\)=\(\mathbf{r} \times \frac{d^2 \mathbf{r}}{d t^2}\)

∴ \(\int\left(\mathbf{r} \times \frac{d^2 \mathbf{r}}{d t^2}\right) d t\)=\(\mathbf{r} \times \frac{d \mathbf{r}}{d t}+\mathbf{c}\)

6. Ifr = |r|, then \(\frac{d}{d t}\left(\frac{\mathbf{r}}{r}\right)\)=\(\frac{1}{r} \frac{d \mathbf{r}}{d t}-\frac{1}{\dot{r}^2} \frac{d r}{d t} \mathbf{r}\)

∴ \(\int\left(\frac{1}{r} \frac{d \mathbf{r}}{d t}-\frac{1}{r^2} \frac{d r}{d t} \mathbf{r}\right) d t\)=\(\frac{\mathbf{r}}{r}+\mathbf{c}\)

7. If c is a scalar constant then c is a scalar constant then∫ cr dt=c∫r dt

8. If r and s are any two vector functions of a scalar t, then ∫(r+s) dt= ∫r dt+ ∫ s dt

Integration Of Vectors Solved Problems

Exmple.1. Evaluate \(\int_0^1\left(e^t \mathbf{i}+e^{-2 t} \mathbf{j}+t \mathbf{k}\right) d t\)

Solution:

Given

= \(\mathbf{i}\left[e^t\right]_0^1+\mathbf{j}\left[-\frac{1}{2} e^{-2 t}\right]_0^1+\mathbf{k}\left[\frac{t^2}{2}\right]_0^1=(e-1) \mathbf{i}-\frac{1}{2}\left(e^{-2}-1\right) \mathbf{j}+\frac{1}{2} \mathbf{k}\)

Solved Problems On Vector Integration Step-By-Step

Example. 2. Evaluate \(\int_2^3 \mathbf{f} \cdot \frac{d \mathbf{f}}{d t} d t\) if f(2) = 2i – i+2k and f(3)= 4i-2j+3k

Solution:

Given

(2) = 2i – i+2k and f(3)= 4i-2j+3k

We know that \(\int\left(2 \mathbf{f} \cdot \frac{d \mathbf{f}}{d t}\right) d t=\mathbf{f}^2+c\)

∴ \(\int_2^3(\mathrm{f} \cdot \frac{d \mathrm{f}}{d t}) d t=\frac{1}{2}[\mathrm{f}^2{ }_2^3=\frac{1}{2}[\mathrm{f}^2(3)-\mathrm{f}^2(2)]\)

= \(\frac{1}{2}\left[(4 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})^2-(2 \mathbf{i}-\mathbf{j}+2 \mathbf{k})^2\right]=\frac{1}{2}[(16+4+3)-(4+1+4)]=10\)

Example. 3. Find the value of \(\frac{d \mathbf{r}}{d t}\) by integrating \(\frac{d^2 \mathbf{r}}{d t^2}\) = -n²r

Solution:

The given equation is \(\frac{d^2 \mathbf{r}}{d t^2}=-n^2 \mathbf{r}\)

Taking the dot product with \(2 \frac{d \mathrm{r}}{d t}\) both sides

and integrating we have \(\int\left(2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}\right) d t=-n^2 \int\left(2 r \cdot \frac{d \mathbf{r}}{d t}\right) d t\)

⇒ \(\left(\frac{d \mathbf{r}}{d t}\right)^2=n^2 \mathbf{r}^2+\mathbf{c}\)

where c is any constant vector.

Exercises On Line, Surface, And Volume Integrals With Solutions

Example. 4. If f(t) =5t² i+tj-t³k find \(\int_1^2\left(\bar{f} \times \frac{d^2 \tilde{f}}{d t^2}\right) d t\)

Solution: \(\int_1^2\left(\mathbf{f} \times \frac{d^2 \mathbf{f}}{d t^2}\right\} d t=\left[\mathbf{f} \times \frac{d \mathbf{f}}{d t}\right]_1^2\)

Given \(\mathbf{f}(t)=5 t^2 \mathbf{i}+t \mathbf{j}-t^3 \mathbf{k}\)

∴ \(\frac{d \mathbf{f}}{d t}=10 t \mathbf{i}+\mathbf{j}-3 t^2 \mathbf{k}\)

f x \(\frac{d \mathbf{f}}{d t}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
5 t^2 & t & -t^3 \\
10 t & 1 & -3 t^2
\end{array}\right|\)

= \(-2 t^3 \mathbf{i}+5 t^4 \mathbf{j}-5 t^2 \mathbf{k}\)

∴ \(\left[\mathbf{f} \times \frac{d \mathbf{f}}{d t}\right]_1^2=\left[-2 t^3 \mathbf{i}+5 t^4 \mathbf{j}-5 t^2 \mathbf{k}\right]_1^2=-14 \mathbf{i}+75 \mathbf{j}-15 \mathbf{k}\)

∴ \(\int_1^2\left(\mathbf{f} \times \frac{d^2 \mathbf{f}}{d t^2}\right) d t=-14 \mathbf{i}+75 \mathbf{j}-15 \mathbf{k}\)

Integration Of Vectors Exercise 6(a)

1. If F(t) = (t-t²)i+2t³j-3k find \(\int_1^2 \mathbf{F}(t) d t\)
Solution:

Given

F(t) = (t-t²)i+2t³j-3k

⇒ \(\int_1^2\left[\left(t-t^2\right) \mathrm{i}+2 t^3 \mathrm{j}-3 \mathrm{k}\right] d t\)

= \(\left[\left(\frac{t^2}{2}-\frac{t^3}{3}\right) \mathrm{i}+\frac{2 t^4 \mathrm{j}}{4}-3 t \mathrm{k}\right]_1^2=-\frac{5}{6} \mathrm{i}+\frac{15}{2} \mathrm{j}-3 \mathrm{k}\)

Surface Integrals Of Vectors Examples And Solutions

2.If F(t) =ti+(t²-2t)j+(3t²+3t³)k find \(\int_0^1 f(t) d t\)
Solution:

Given

F(t) =ti+(t²-2t)j+(3t²+3t³)k

⇒ \(\int_0^1 f(t) d t=\int_0^1\left[t \mathbf{i}+\left(t^2-2 t\right) \mathbf{j}+\left(3 t^2+3 t^3\right) \mathbf{k}\right] d t\)

= \(\left[\frac{t^2}{2} \mathbf{i}+\left(\frac{t^3}{3}-t^2\right) \mathbf{j}+\left(t^3+\frac{3}{4} t^4\right) \mathbf{k}\right]_0^1=\frac{1}{2} \mathbf{i}-\frac{2}{3} \mathbf{j}+\frac{7}{4} \mathbf{k}\)

= \(\frac{1}{12}[6 \mathbf{i}-8 \mathbf{j}+21 \mathbf{k}]\)

3. If A=ti-t²j+(t-1)k and B= 2t²i+6tK find \(\int_1^2(\mathbf{A} \cdot \mathbf{B}) d t\)
Solution:

Given

A=ti-t²j+(t-1)k and B= 2t²i+6tK

A . B = \(t\left(2 t^3\right)+6 t(t-1)\)

∴ \(\int_0^2(\mathbf{A} \cdot \mathbf{B}) d t=\int_0^2\left(2 t^3+6 t^2-6 t\right) d t=12\)

4. If a=ti-3j+2tk : B= i-2j+2k : C= 3i+tj-k find
(1)\(\int_1^2[\mathbf{A B C}] d t\)
(2) \(\int_1^2[\mathrm{~A} \times(\mathrm{B} \times \mathrm{C})]dt\)
Solution:

Given

a=ti-3j+2tk : B= i-2j+2k : C= 3i+tj-k

(1) [А B C] = \(\left|\begin{array}{ccc}
t & -3 & 2 t \\
1 & -2 & 2 \\
3 & t & -1
\end{array}\right|\) =14 t-21

∴ \(\int_1^2[\mathbf{A ~ B ~ C}] d t=\int_1^2(14 t-21) d t=0\)

(2) Use \(\mathbf{A} \times(\mathbf{B} \times \mathbf{C})=(\mathbf{A} \cdot \mathbf{C}) \mathbf{B}-(\mathbf{A} \cdot \mathbf{B}) \mathbf{C}\) and then integrate.

5. If \(\frac{d^2 \mathbf{r}}{d t^2}\)=6ti-24t²j+4 sin tk find r given thatr= 2i+j and \(\frac{d \mathbf{r}}{d t}\)=-i-3k at t=0
Solution:

Given

\(\frac{d^2 \mathbf{r}}{d t^2}\)=6ti-24t²j+4 sin tk

∫ \(\frac{d^2 \mathbf{r}}{d t^2} \cdot d t=\int\left(6 t \mathbf{i}-24 t^2 \mathbf{j}+4 \sin t \mathbf{k}\right) d t\)

⇒ \(\frac{d \mathbf{r}}{d t}=3 t^2 \mathbf{i}-8 t^3 \mathbf{j}-4 \cos t \mathbf{k}+\mathbf{A}\) when t = \(0 \quad \frac{d \mathbf{r}}{d t}=-\mathbf{i}-3 \mathbf{k}\)

–\(\mathbf{i}-3 \mathbf{k}=-4 \mathbf{k}+\mathbf{A} \Rightarrow \mathbf{A}=-\mathbf{i}+\mathbf{k}\)

Hence \(\frac{d \mathrm{r}}{d t}=\left(3 t^2-1\right) \mathbf{i}-8 t^3 \mathbf{j}+(1-4 \cos t) \mathbf{k}\)

Again integrating \(\mathbf{r}=\int \frac{d \mathbf{r}}{d t} d t=\left(t^3-t\right) \mathbf{i}-2 t^4 \mathbf{j}+(t-4 \sin t) \mathbf{k}+\mathbf{B}\) when t=0,2 i+j=B

∴ \(\mathbf{r}=\left(t^2-t+2\right) \mathbf{i}+\left(1-2 t^4\right) \mathbf{j}+(t-4 \sin t) \mathbf{k}\)

Volume Integrals Solved Problems With Step-By-Step Explanations

6. Find F(t), given \(\frac{d \mathbf{F}}{d t}\) =12 cos 2ti-8 sin 2tj+2tk and F(0)=0
Solution:

F(t) = \(\int d \mathbf{F}=\int(12 \cos 2 t \mathbf{i}-8 \sin 2 t \mathbf{j}+2 t \mathbf{k}) d t\)

F(t) = \(6 \sin 2 t \mathbf{i}+4 \cos 2 t \mathbf{j}+t^2 \mathbf{k}+\mathbf{C}\)

Given \(\mathbf{F}(0)=0 \Rightarrow 4 \mathbf{j}+\mathbf{C}=0 \Rightarrow \mathbf{C}=-4 \mathbf{j}\)

∴ \(\mathrm{F}(t)=6 \sin 2 t \mathrm{i}+(4 \cos 2 t-4) \mathrm{j}+t^2 \mathrm{k}\)

7. If a,b, and n are constants and r= a cos nt+b sin nt. Show that \(\frac{d^2 \mathbf{r}}{d t^2}+n^2 \mathbf{r}\)=0
Solution:

Given

a,b, and n are constants and r= a cos nt+b sin nt

⇒ \(\frac{d \mathbf{r}}{d t}=-\mathbf{a} n \sin n t+\mathbf{b} n \cos n t\)

⇒ \(\frac{d^2 \mathbf{r}}{d t^2}=-\mathbf{a} n^2 \cos n t-\mathbf{b} n^2 \sin n t=-n^2 \mathbf{r}\)

∴ \(\frac{d^2 \mathbf{r}}{d t^2}+n^2 \mathbf{r}=0\)

Applications Of Vector Integration In Physics And Engineering

8. Given \(\frac{d^2 \mathbf{r}}{d t^2}\)=-k²r show that \(\left(\frac{d \mathbf{r}}{d t}\right)^2\)=c-k²r²
Solution:

The given equation is \(\frac{d^2 \mathbf{r}}{d t^2}=-n^2 \mathbf{r}\)

Taking the dot product with \(2 \frac{d \mathbf{r}}{d t}\) both sides and integrating we have \(\int\left(2 \frac{d \mathbf{r}}{d t} \cdot \frac{d^2 \mathbf{r}}{d t^2}\right) d t=-n^2 \int\left(2 r \cdot \frac{d \mathbf{r}}{d t}\right) d t \quad\left(\frac{d \mathbf{r}}{d t}\right)^2=n^2 \mathbf{r}^2+\mathbf{c}\)

where c is any constant vector.

Integration Of Vectors Line, Surface, Volume, Integrals Oriented Curve

Let C be a curve in space. A be the initial point and B be the terminal point of curve C . When the direction along C oriented
from A to B is positive then the direction B to A is then called the negative direction. If the two points coincide, curve C is called the closed curve.

Integration Of Vectors Smooth Curve

A curve r = F (t) is called a smooth curve if F (t) is continuously differentiable. A curve C is said to be piecewise smooth if it consists of a finite number of smooth curves. The curve of is composed of three smooth curves C₁, C_2, and C₃.

Integration Of Vectors Line Integrals

Let r = f (t) define a smooth curve C joining points A and B. Let us be the differential of arc length at P e C.

Then \(\frac{d \mathbf{r}}{d s}\)=T, is the unit vector along the tangent to the curve C at P.

Let F (r) be a vector point, a function defined and continuous along C. The component of F (r) along the tangent at P is F (r). T.

The integral ∫F.Tds taken along the curve C is called the Linear integral of F along

C. This is written as\(\int_A^B \mathbf{F} \cdot \mathbf{T} d s\)=\(\int_C\left(\mathbf{F} \cdot \frac{d \mathbf{r}}{d s}\right) d s\)=\(\int_C \mathbf{F} \cdot d \mathbf{r}\)

This is also called the tangential line integral of F along C.

Note. The other types of line integrals are \(\int_C \mathbf{F} \times d \mathbf{r} \text { and } \int_C \phi d \mathbf{r}\)

where F is a continuous vector and Φ  a continuous scalar point function. In general, any integral which is to be evaluated along a curve is called a line integral. Such integrals can be defined in terms of limits of sums as are the integrals of elementary calculus.

Integration Of Vectors Circulation

Let C be a simple closed curve (i.e. a curve that does not intersect itself anywhere). The line integral of F along C is called the circulation of F along C. It is often denoted by

∴ \(\oint_{\mathrm{C}} \mathrm{F} \cdot d \bar{r}\)=\(\oint_{\mathrm{C}}\left(\mathrm{F}_1 d x+\mathrm{F}_2 d y+\mathrm{F}_3 d z\right)\)

Integration Of Vectors Cartesian Form

LetF(r) = F₁i+F₂j+F₃k   (r) = xi+yj+zk

Then \(\oint_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}\)=\(\oint_{\mathbf{C}}\left(\mathrm{F}_1 d x+\mathrm{F}_2 d y+\mathrm{F}_3 d z\right) \cdot(\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d z)\)

⇒ \(\oint_C\left(\mathrm{~F}_1 d x+\mathrm{F}_2 d y+\mathrm{F}_3 d z\right)\) is the line integral in cartesian form. If, x,y, and z are functions of t, then

⇒ \(\oint_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}\)=\(\oint_{\mathbf{C}} \mathrm{F}_1 d x+\mathrm{F}_2 d y+\mathrm{F}_3 d z\)

= \(\oint_{\mathbf{C}}\left(\mathrm{F}_1 \frac{d x}{d t}+\mathrm{F}_2 \frac{d y}{d t}+\mathrm{F}_3 \frac{d z}{d t}\right) d t\)

Integration Of Vectors Work Done By A Force

If A is the force F acting on a particle moving along C, then \(\int_C \mathbf{A} d \mathbf{r}\) denotes the work done by the force.

Integration Of Vectors Solved  Problems

Example. 1. Find  \(\int_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}\) where F= xyi+yzj+ zx+k and the curve C is r= ti+t²j+t³k , t varying from -1 to 1.

Solution:

Given \(\mathbf{r}=t \mathbf{i}+t^2 \mathbf{j}+t^3 \mathbf{k}\), for the curve \(\mathbf{C}\)

∴ \(\frac{d \mathbf{r}}{d t}=\mathbf{i}+2 t \mathbf{j}+3 t^2 \mathbf{k}\)

Also \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

∴ x=t, y=t^2, z = \(t^3\), is the curve C

Again F = \(x y \mathbf{i}+y z \mathbf{j}+z x \mathbf{k}\)

∴ F over the curve C is F = \(t\left(t^2\right) \mathbf{i}+\left(t^2\right)\left(t^3\right) \mathbf{j}+\left(t^3\right)(t) \mathbf{k}=t^3 \mathbf{i}+t^5 \mathbf{j}+t^4 \mathbf{k}\)

∴ \(\mathrm{F} \cdot \frac{d \mathrm{r}}{d t}=\left(t^3 \mathrm{i}+t^5 \mathrm{j}+t^4 \mathrm{k}\right) \cdot\left(\mathrm{i}+2 t \mathrm{j}+3 t^2 \mathrm{k}\right)=t^3+2 t^6+3 t^6=t^3+5 t^6\)

∴ \(\oint_{\mathrm{C}} \mathrm{F} \cdot d \mathrm{r}=\int_C\left(\mathrm{~F} \cdot \frac{d \mathrm{r}}{d t}\right) d t=\int_{t=-1}^1\left(t^3+5 t^6\right) d t=\left[\frac{t^4}{4}+\frac{5 t^7}{7}\right]_{-1}^1=\frac{10}{7}\)

Step-By-Step Guide To Vector Surface Integrals

Example.2. Evaluate \(\int_C \mathbf{A} d \mathbf{r}\)  where F= 3x²i=92xy-y)j=zk along the straight line C from (0,0,0) to (2,1,3)

Solution:

The equation to the line joining (0,0,0) and (2,1,3) is \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}=t\)

Then along the line C

x=2 t, y=t, z=3 t

Also \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}=2 t \mathbf{i}+t \mathbf{j}+3 t \mathbf{k}\)

∴ \(d \mathbf{r}=(2 \mathbf{i}+\mathbf{j}+3 \mathbf{k}) d t\)

Given \(\mathrm{F}=3 x^2 \mathrm{i}+(2 x z-y) \mathrm{j}+z \mathrm{k}\)

And along C

F = \(3(2 t)^2 \mathbf{i}+[2(2 t)(3 t)-t] \mathbf{j}+3 t \mathbf{k}=12 t^2 \mathbf{i}+\left(12 t^2-t\right) \mathbf{i}+3 t \mathbf{k}\)

F. dr = \(\left[24 t^2+\left(12 t^2-t\right)+9 t\right] d t=\left(36 t^2+8 t\right) d t\)

At (0,0,0) and at (2,1,3), t=1.

∴ \(\int_{\mathbf{C}} \mathbf{F} . d \mathbf{r}=\int_{t=0}^1\left(36 t^2+8 t\right) d t=\left[12 t^3+4 t^2\right]_0^1=16\)

Example.3. If F= (3x²+6y)i-14yz j +20 xz² k calculate \(\int_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}\) along the lines from (0,0,0), to (1,0,0,) , then to (1,1,0) , and then to (1,1,1).

Solution:

Given F = \(\left(3 x^2+6 y\right) \mathrm{i}-14 y z \mathrm{j}+20 x z^2 \mathrm{k}\)

r = \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \quad d \mathbf{r}=\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d z\)

∴ \(\int_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}=\int_{\mathbf{C}}\left(3 x^2+6 y\right) d x-14 y z d y+20 x z^2 d z\)

1. Along the line from (0,0,0) to (1,0,0)

y = \(0, z=0 \Rightarrow \quad d y=0, d z=0\)

x varies from 0 to 1

The integral over this part of the path

∴ \(\int_{\mathrm{C}_1} \mathrm{~F} \cdot d \bar{r}=\int_{x=0}^1 3 x^2 d x=\left[x^3\right]_0=1\)

2. Along the line (1,0,0) to (1,1,0) x=1, z=0 ⇒ dx=0, dz=0

y varies from 0 to 1. The integral over this part of the path

∴ \(\int_{C_2} \mathbf{F} \cdot d \mathbf{r}=\int_{y=0}^1(0-0+0) d y=0\)

3. Along the line (1,1,0) to (1,1,1), x=1, y=1

dx=0, dy=0

z varies from 0 to 1

The integral over this part of the path

∴ \(\int_{\mathbf{C}_3} \mathbf{F} \cdot d \mathbf{r}=\int_{z=0}^1 20 z^2 d z=\left[\frac{20 z^3}{3}\right]_0^1=\frac{20}{3}\)

∴ Adding \(\int_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}=1+0+\frac{20}{3}=\frac{23}{3}\)

Example.4. If F = 3xyi-y²j evaluates \(\int_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}\) where C is curved y= 2x² in xy plane from (0,0) to (1,2).

Solution:

Given

F = 3xyi-y²j

The equation of the curve C is y=\(2 x^2\)

dy = 4x d x

Given \(\mathbf{F}=3 x y \mathbf{i}-y^2 \mathbf{j}\)

Since the integration is performed in the xy plane (z=0) and x varies from 0 to 1.

∴ \(\int_{\mathbf{C}} \mathbf{F} . d \mathbf{r}=\int_{z=0}^1 \mathbf{F}_1 d x+\mathbf{F}_2 d y=\int_{\mathbf{C}} 3 x y d x-y^2 d y\)

= \(\int_{x=0}^1 3 x\left(2 x^2\right) d x-4 x^4(4 x) d x\)

= \(\int_0^1\left(6 x^3-16 x^5\right) d x=\left[\frac{6 x^4}{4}-\frac{16 x^6}{6}\right]_0^1=-\frac{7}{6}\)

Example.5 If F(x²+y²)i-2xyz evaluate \(\int_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}\) where the curve C is the rectangle in the xy plane bounded by y=0, y=b, x=0, x=a

Solution:

Equation to the curve C \(x^2+y^2=1, z=0\) ∴ dz=0

In parametric form, \(x=\cos \theta, y=\sin \theta, z=0\)

The circulation of F = \(y \mathbf{i}+z \mathbf{j}+x \mathbf{k}\), along C is \(\oint_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}=\oint_{\mathbf{C}} \mathbf{F}_1 d x+\mathbf{F}_2 d y+\mathbf{F}_3 d z=\oint_{\mathbf{C}} y d x+z d y+x d z=\oint_{\mathbf{C}} y d x\)

For the circle \(\theta\) varies from 0 to \(2 \pi\)

∴ \(\oint \mathbf{F} \cdot d \mathbf{r}=\int_0^{2 \pi} \sin \theta(-\sin \theta) d \theta=-4 \int_0^{\pi / 2} \sin ^2 \theta d \theta=-4\left(\frac{1}{2}\right) \frac{\pi}{2}=-\pi\)

Worked Examples Of Vector Line And Volume Integrals

Example.6. If F=yi+ zj+xk find the circulation of f around the curve C where C is the circle x²+y²=1, z=0

Solution:

Equation to the curve C \(x^2+y^2=1, z=0\) ∴ dz=0

in parametric form, \(x=\cos \theta, y=\sin \theta, z=0\)

The circulation of \(\mathbf{F}=y \mathbf{i}+z \mathbf{j}+x \mathbf{k}\), along \(\mathbf{C}\) is \(\oint_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}=\oint_{\mathbf{C}} \mathbf{F}_1 d x+\mathbf{F}_2 d y+\mathbf{F}_3 d z=\oint_{\mathbf{C}} y d x+z d y+x d z=\oint_{\mathbf{C}} y d x\)

For the circle \(\theta\) varies from 0 to \(2 \pi\)

∴ \(\oint \mathbf{F} . d \mathbf{r}=\int_0^{2 \pi} \sin \theta(-\sin \theta) d \theta=-4 \int_0^{\pi / 2} \sin ^2 \theta d \theta=-4\left(\frac{1}{2}\right) \frac{\pi}{2}=-\pi\)

Example.7. Find the work done in moving a particle force field F=3xi+(2xz-y)j+zk along the straight line form (0,0,0) to (2,1,3)

Solution:

Given curve is \(\mathbf{F}=3 x^2 \mathbf{i}+(2 x z-y) \mathbf{j}+z \mathbf{k}\)

Let \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+3 \mathbf{k} \Rightarrow d \mathbf{r}=d x \mathbf{i}+d y \mathbf{j}+d \mathbf{k}\)

F. \(d \mathbf{r}=3 x^2 d x+(2 x z-y) d y+z d z\)

The Cartesian equation of line through the points (0,0,0),(2,1,3) is \(\frac{x-0}{2-0}=\frac{y-0}{1-0}=\frac{z-0}{3-0}\) i.e. \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3}=t\) (say)

The parametric equations are, x=2t, y=t, z=3t.

dx = 2 dt, dy=dt, dz=3 dt on C, t varies from 0 to 1.

The work done by particles in the force field is

= \(\int_C \mathbf{F} \cdot d \mathbf{r}=\int_{t=0}^1 3\left(2 t^2\right)^2 2 d t+(2.2 t .3 t-t) d t+(3 t) 3 d t\)

= \(\left.\left.\int_0^1\left(24 t^2+12 t^2-t+9 t\right) d t=\int_0^1\left(3 \dot{6} \dot{t}^2+8 t\right) d t\)

= \(36 \frac{t^3}{3}+\frac{8 t^2}{2}\right]_0^1=12 t^3+4 t^2\right]_0^1=1^7+4=16\)

Integration Of Vectors Exercise6 ( b )

1. Evaluate \(\int_{\mathbf{C}} \mathbf{F} \cdot d \mathbf{r}\)where F = x²y²i + yj and the curve C is y² = 4x in the xy plane from (0,0) to (4,4).

Solution:

The equation of the curve C is \(4 x=y^2\) ∴4 dx=2y dy

Given F = \(x^2 y^2 \mathbf{i}+y \mathbf{j}\)

The integration is performed in xy – plane (z=0) and y varies from 0 to 4.

∴ \(\int_c \mathbf{F} . d \mathbf{r}=\int_c \mathbf{F}_1 d x+\mathbf{F}_2 d y=\int_c x^2 y^2 d x+y d y\)

= \(\int_{y=0}^4\left(\frac{y^2}{4}\right)^2 y^2\left(\frac{1}{2} y\right) d y+y d y=\int_0^4\left(\frac{y^7}{32}+y\right) d y=\left[\frac{y^8}{256}+\frac{y^2}{2}\right]_0^4=264\)

2. If F = (3x² + 6y) i- 14yzj + 20xz²k evaluate ∫F.dr along the straight line joining (0,0,0) and (1,1,1).
Solution:

Equation to the line joining (0,0,0) and (1,1,1) is \(\frac{x}{1}=\frac{y}{1}=\frac{z}{1}=t\).

∴ \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}=(\overline{\mathbf{i}}+\mathbf{j}+\mathbf{k}) t\)

d \(\mathbf{r}=(\mathbf{i}+\mathbf{j}+\mathbf{k}) d t\)

Given \(\mathrm{F}=\left(3 x^2+6 y\right) \mathbf{i}-14 y z \mathbf{j}+20 x z^2 \mathbf{k}\)

Along the straight line x=y=z=t

F = \(\left(3 t^2+6 t\right) \mathbf{i}-14 t^2 \mathbf{j}+20 t^3 \mathbf{k}\)

At (0,0,0), t=0 and at (1,1,1), t=1

⇒ \(\overline{\mathrm{F}} \cdot d \bar{r}=\left[\left(3 t^2+6 t\right)-14 t^2+20 t^3\right] d t\)

∴ \(\left.\int_C \overline{\mathrm{F}} \cdot d \bar{r}=3 t^2-\frac{11 t^3}{3}+5 t^4\right]_0^1=\frac{13}{3}\)

Examples Of Vector Integration In Electromagnetism

3. If F = 3xyi- 5zj+ l0xk evaluate ∫F.dr  along the curve x = t²+l, y=2t², z = t³ from t=1 to t=2
Solution:

Given x = \(t^2+1, y=2 t^2, z=t^3\)

r = \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}=\left(t^2+1\right) \mathbf{i}+2 t^2 \mathbf{j}+t^3 \mathbf{k}\)

dr = \(\left(2 t \mathbf{i}+4 t \mathbf{j}+3 t^2\right) d t\)

Along the given curve C

4. If F = (2y + 3)i + xzj + (yz – x)k evaluate ∫F.dr along the curve C given x = t², y = t, z =t³ from t = 0 to t = 1 (or) c is the line joining (0,0,0) and (2,1,1)
Solution:

F.dr = \([(2 y+3) \mathbf{i}+x z \mathbf{j}+(y z-x) \mathbf{k}] \cdot(\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d z)\)

= \((2 y+3) d x+x z d y+(y z-x) d x\)

= \((2 t+3)(4 t)+\left(2 t^2\right)\left(t^3\right)+\left[(t) t^3-2 t^2\right]\left(3 t^2\right)\)

= \(3 t^6+2 t^5-6 t^4+8 t^2+12 t\)

∴ \(\int_0^1 \mathbf{F} \cdot d \mathbf{r}=\int_0^1\left[3 t^6+2 t^5-6 t^4+8 t^2+12 t\right] d t\)

= \(\left[\frac{3}{7} \boldsymbol{t}^7+\frac{2}{6} \boldsymbol{t}^6-\frac{\mathbf{6}}{5} \boldsymbol{t}^5+\frac{8}{3} \boldsymbol{t}^3+\mathbf{6} t^2\right]_0^1=\frac{288}{35}\)

5. Find the line integral of the vector F = zi + xj+ yk over the curve C given by  x = a cost, y = a sin t,z = from z = t/2π  from z=0 to z = 1.

Solution:

F = \(z \mathbf{i}+x \mathbf{j}+y \mathbf{k}\)

Given x = \(a \cos t, y=a \sin t, z=\frac{t}{2 \pi}\)

∴ dx = \(-a \sin t d t, d y=a \cos t d t, d z=\frac{1}{2 \pi} d t\)

z = \(0 \Rightarrow t=0\) and \(z=1 \Rightarrow t=2 \pi\)

∴ \(\int \mathbf{F} \cdot d \mathbf{r}=\int z d x+x d y+y d z\)

= \(\frac{a}{2 \pi} \int_0^{2 \pi}\left(\frac{t}{2 \pi}(-a \sin t)+a \cos t-\sin t\right) t \cdot d t+a^2 \int_0^{2 \pi} \cos ^2 t \cdot d t\)

= \(\frac{a}{2 \pi}(t \sin t+\cos t)-\frac{a}{2 \pi}[-\cos t+\sin t]_0^{2 \pi}+4 a^2 \int_0^{2 \pi} \cos ^2 t \cdot d t\)

= \(\frac{a}{2 \pi}(2 \pi)-0+4 a^2=\left(\frac{1}{2} \cdot \frac{\pi}{2}\right)=a+\pi a^2\)

6. Evaluate \(\int_{\mathbf{C}}\)(x² + xy) dx + (x² + y²)dy where C is the square formed by the lines  x = ± 1, y = ± 1

Solution:

Given F. dr = \(\int\left(x^2+x y\right) d x+\left(x^2+y^2\right) d y\)

(1) Along \(\mathbf{A B}, y=1, d y=0\)

and x changes from 1 to -1.

∴ \(\int \mathbf{F} \cdot d \mathbf{r}=\int_1^{-1}\left(x^2+x\right) d x\)

= \(\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_1^{-1}=-\frac{2}{3} .\)

(2) Along B C, x=-1, d x=0, changes from 1 to -1

∴ \(\int \mathbf{F} \cdot d \mathbf{r}=\int_1^{-1}\left(1+y^2\right) d y=\left[y+\frac{y^3}{3}\right]_1^{-1}=-\frac{8}{3}\)

(3) Along CD, \(y=-1, d y=0\) and x changes from -1 to 1 .

∴ \(\int \mathbf{F} . d \mathrm{r}=\int_{-1}^1\left(x^2-x\right) d x=\left[\frac{x^3}{3}+\frac{y^2}{2}\right]_{-1}^1=\frac{2}{3}\)

(4) Along D A, \(x=1, d x=0\) and y changes from -1 to 1 .

∴ \(\int \mathbf{F} . d \mathbf{r}=\int_{-1}^1\left(1+y^2\right) d y=\left[y+\frac{y^3}{3}\right]_{-1}^1=\frac{8}{3}\)

∴ \(\int \mathbf{F} . d \mathbf{r}\) along \(x= \pm, y= \pm 1\) is \(\left[-\frac{2}{3}-\frac{8}{3}+\frac{2}{3}+\frac{8}{3}\right]=0\)

Properties Of Vector Integrals With Solved Problems

7. Find the circulation of F = (2x- y + 2z) i + (x + y- z) j + (3x- 2y- 5z)k along the circle x²+ y² = 4, z = 0 . [Hint: Take x = 2 cos t, y = 2sint  and t from 0 to 2π ].

Solution:

The equation to the curve C \(x^3+y^3=4, z=0\)

In parametric form \(z=0, x=2 \cos \theta_1, y=2 \sin \theta\)

For the circle \(\theta\) varies from 0 to \(2 \pi\)

∴ The circulation of \(\mathbf{F}\) along C is \(\oint_C \mathbf{F} \cdot d \mathbf{r}=\oint_C \mathbf{F}_1 d x+\mathbf{F}_2 d y+\mathbf{F}_3 d z\)

= \(\oint_c(2 x-y+2 z) d x+(x+y-z) d y+(3 x-2 y-5 z) d z\)

= \(\int_0^{2 \pi}(4 \cos \theta-2 \sin \theta)(-2 \sin \theta) d \theta+(2 \cos \theta+2 \sin \theta) 2 \cos \theta\)

= \(\int_0^{2 \pi}(4-\sin 2 \theta) d \theta=4(2 \pi)-\int_0^{2 \pi} \sin \theta=8 \pi-0=8 \pi\)

8. Evaluate \(\int_{\mathbf{C}}\)F.dr along the line joining (0, 0, 0) to (2,1,4) where c F = yz i + (xz + 1) j + xy k .

Solution:

Equation to the line joining (1,0,0) to (2,1,4) is \(\frac{x-1}{1}=\frac{y}{1}=\frac{z}{4}=t\)

∴ \(x=t+1, y=t, z=4 t \Rightarrow d x=d t, d y=d t, d z=4 d t\)

t varies from 0 to 1

⇒ \(\mathbf{F} \cdot d \mathbf{r}=(y z \mathbf{i}+(x z+1) \mathbf{j}+x y \mathbf{k}) \cdot(\mathbf{i} d x+\mathbf{j} d y+\mathbf{k} d z)\)

= \(y z d x+(x z+1) d y+x y d z\)

∴ \(\int_C \mathbf{F} . d \mathbf{r}=\int_0^1 t(4 t) d t+[(t+1) 4 t+1] d t+(t+1) t(4 d t)\)

+ \(\int_0^1\left(12 t^2+8 t+1\right) d t=\left[4 t^3+4 t^2+t\right]=9\)

9. If F = (x- 3y)i + (y- 2x) j find\(\int_{\mathbf{C}}\)F.dr where C is the closed curve in the c xy plane, x = 2 cos t, y = 3 sin t from t = 0 to t = 2π .

Solution:

x = \(2 \cos t \Rightarrow d x=-2 \sin t . d t, \quad y=3 \sin t d t \Rightarrow d y=3 \cos t d t\)

F. dr = (x-3 y) d x+(y-2 x) d y

(2 cos t-9 sin t)(-2 sin t) d t+(3 sin t-4 cos t)(3 cos t) d t

= \(\left(\frac{5}{2} \sin 2 t+18 \sin ^2 t-12 \cos ^2 t\right) d t\)

= \(\left(\frac{5}{2} \sin 2 t-12 \cos 2 t+6 \sin ^2 t\right) d t\)

∴ \(\int_C \mathbf{F} . d \mathbf{r}=\int_0^{2 \pi}\left(6 \sin ^2 t+\frac{5}{2} \sin 2 t-12 \cos 2 t\right) d t\)

= \(6 \cdot \int_0^{2 \pi} \sin ^2 t+\int_0^{2 \pi}\left(+\frac{5}{2} \sin 2 t-12 \cos 2 t\right) d t\)

= \(6.4 \int_0^{\pi / 2} \sin ^2 t+0=6.4 \cdot \frac{1}{2} \frac{\pi}{2}=6 \pi\)

Integration Of Vectors Surface Integrals

A surface r = f (w, v) is called a smooth surface if f (u, V) is continuous and possesses partial derivatives. Let F (r) be a continuous vector point function,
defined over the smooth surface r = f (w, v) Let S be a region of the surface. Divide the region into m sub-regions of areas δS₁ δS₂, …, δS_i… δS_m. Let P_{i be point of} S_i and N_j be the unit normal to δS_i at  P_i.

Let δ  A_i be the vector area of δS_i, then  A_i = N_i δ S₁

Now from the sum I_m\(=\sum_{i=1}^m \mathbf{F}\left(\mathbf{r}_{\mathrm{i}}\right) \cdot \delta \mathbf{A}_{\mathrm{i}} \quad \text { or } \Sigma \mathbf{F}\left(\mathbf{r}_{\mathrm{i}}\right) \mathbf{N}_{\mathrm{i}} \delta \mathbf{S}_{\mathrm{i}}\)

Let m tend to infinity in such a way that each 8 Sj shrinks to a point. The limit of I_m
if it exists, is called the normal surface integral of F (r) over the region S of the surface :

⇒ \(\bar{r}\)=\(\bar{f}(u, v)\)and is denoted by \(\int_S \mathrm{~F}(\bar{r}) \cdot d \mathrm{~A} \text { or } \int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}\)

Note. The other types of surface integrals are \(\int_S \mathbf{F} \times d \mathbf{A}, \quad \int \phi d \mathbf{A}\)

where F is a continuous vector and  Φ  a continuous scalar point function.

Integration Of Vectors Flux

 Let S be a closed surface. Then the normal surface integral \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S} \text { or } \int_S \mathbf{F} \cdot d \mathbf{A}\) called the flux of F over S.

Cartesian form

Let F(r) =F₁i+ F₂j+F₃k where F₁, F₂, and F₃ are continuous and differentiable functions of x, y, and z.

Let cosα, cos β, cos γ be the directions cosines of the unit normal N.

N = i cos α + j cos β + k cos γ and F . N = F₁ cos α + F₂ cos β + F₃ cos γ

∴  \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}\)=\(\int_S\left(F_1 \cos \alpha+\mathrm{F}_2 \cos \beta+\mathrm{F}_3 \cos \gamma\right) d \mathbf{S}\)

But dScos α, dScosβ, and dScos γ  are the projections of dS on yz, zx and xy planes. If dx, dy, dz are the differentials along the axes then

dS cos α= dy dz, dS cos β= dz dx, dS cos ϒ = dx dy

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_S\left(\mathrm{~F}_1 d y d z+\mathrm{F}_2 d z d x+\mathrm{F}_3 d x d y\right)\)

Note.  Let R1 be the projection of S on xy plane. then

⇒ \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_{R_1} \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{\cos \gamma}=\iint_{R_1} \mathrm{~F} \cdot \mathrm{N} \frac{d x d y}{|\mathrm{~N} \cdot \overline{\mathrm{K}}|}\)

Thus the surface integral S can be evaluated with the help of the double integral integrated over in R₁ xy plane.

Similarly , \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}\)=\(\iint_{R_2} \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}\)=\(\iint_{R_3} \mathbf{F} \cdot \mathbf{N} \frac{d z d x}{|\mathbf{N} \cdot \mathbf{j}|}\)

Integration Of Vectors Solved problems

Example. 1. Evaluate \(\int_S\)F.Nds, where \(\int_S\)F=zi+xj-3y²zk and is the surface x²+y²=16 included in the first octant between z=0 and z=5.

Solution:

Given

\(\int_S\)F.Nds

Let \(\phi=x^2+y^2-16\)

∴ The normal to the surface \(\mathbf{S}\) is grad \(\phi\) (i.e. \(\nabla \phi\))

∴ normal = \(\nabla \phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=2 x \mathbf{i}+2 y \mathbf{j}\)

∴ Unit normal \(\mathbf{N}=\frac{2 x \mathbf{i}+2 y \mathbf{j}}{\sqrt{\left(4 x^2+4 y^2\right)}}=\frac{x \mathbf{i}+y \mathbf{j}}{\sqrt{\left(x^2+y^2\right)}}=\frac{x \mathbf{i}+y \mathbf{j}}{4}\) because \(x^2+y^2=16\) on S.

Let R be the projection of S on Y Z plane, then \(\iint_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_S \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}\)

Given \(\mathbf{F}=z \mathbf{i}+x \mathbf{j}-3 y^2 z \mathbf{k}\)

∴ \(\mathbf{F} \cdot \mathbf{N}=\left(z \mathbf{i}+x \mathbf{j}-3 y^2 z \mathbf{k}\right) \cdot \frac{(x \mathbf{i}+y \mathbf{j})}{4}=\frac{1}{4}(x z+x y)\) and \(\mathbf{N} \cdot \mathbf{i}=\frac{1}{4}(x \mathbf{i}+y \mathbf{j}) \cdot \mathbf{i}=\frac{x}{4}\)

For the surface \(x^2+y^2=16\) in yz plane x=0 ⇒ y=4

Hence in first octant y varies from 0 to 4 z varies from 0 to 5

Then the surface integral \(\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}=\iint_R\left(\frac{x z+x \dot{y}}{4}\right)\left(\frac{4}{x}\right) d y d z=\int_{y=0}^4 \int_{z=0}^5(y+z) d y d z\)

= \(\int_{y=0}^4 \int_0^5(y d y)^1 d z+\int_{y=0}^4 \int_{z=0}^5 d y(z d z)=\int_0^4 y d y \int_0^5 d z+\int_0^4 d y \int_0^5 z d z\)

= \({\left[\frac{y^2}{2}\right]_0^4[z]_0^5+[y]_0^4\left[\frac{z^2}{2}\right]_0^5=8 \cdot 5+4 \cdot \frac{25}{2}=90 }\)

Vector Integration Problems In Calculus And Physics

Example.2. Evaluate \(\int_S\)F.Nds where F= 18zi-12j+3yk and Sis the part of the plane 2x+3y+6z=12 located in the first octant

Solution:

Let \(\phi=2 x+3 y+6 z-12\)

∴ Normal to the plane \(\phi\) is \(\nabla \phi\)

⇒ \(\nabla \phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z} \quad=2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}\)

∴ unit normal, \(\mathbf{N}=\frac{2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k}}{\sqrt{(4+9+36)}}=\frac{1}{7}(2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{h}\) ।

Let R be the projection of S on X Y plane, then \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|(\mathbf{N} \cdot \mathbf{k})|}\)

Given \(\mathbf{F}=18 z \mathbf{i}-12 \mathbf{j}+3 y \mathbf{k}\)

∴  \(\mathbf{F} \cdot \mathbf{N}=(8 z \mathbf{i}-12 \mathbf{j}+3 y \mathbf{k}) \cdot \frac{1}{7}(2 \mathbf{i}+3 \mathbf{j}+6 \mathbf{k})=\frac{1}{7}(36 z+36+18 y)=\frac{6}{7}(6 z-6+3 y)\)

⇒ \(\mathrm{N} \cdot \overline{\mathbf{k}} \cdot=\frac{1}{7}(2 \mathrm{i}+3 \mathrm{j}+6 \mathrm{k}) \cdot \mathbf{k}=\frac{6}{7}\)

Given surface S is 2 x+3 y+6 z=12 ∴ R of x y plane is 2 x+3 y=12

⇒ \(y=\frac{12-2 x}{3}\)

∴ y=0 ⇒ x=6

∴ x varies from 0 to 6 and y varies from 0 to \(\frac{12-2 x}{3}\)

The surface integral \(\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}=\iint_R^6 \frac{7}{7}(6 z-6+3 y) \cdot \frac{7}{6} d x d y\)

= \(\iint_R(12-2 x-3 y-6+3 y) d x d y\) because 6 z=12-2 x-3 y

= \(\iint_R(6-2 x) d x d y=2 \int_0^6(3-x)\left[\int_0^{\frac{1}{3}(12-2 x)} d y d x\right.\)

= \(2 \int_0^6(3-x)[y]_0^{\frac{1}{3}(12-2 x)} d x=2 \int_0^6(3-x) \frac{1}{3}(12-2 x) d x\)

= \(\frac{4}{3} \int_0^6\left(18-9 x+x^2\right) d x=\frac{4}{3}\left[18 x-\frac{9}{2} x^2+\frac{1}{3} x^3\right]_0^6=24\)

Example.3.If f=4xzi-y²j+yzk, evaluate 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 surrounded by the following faces.

(1) For the face PQRS, i is the outward normal

∴ \(\mathbf{N}=\mathrm{i}, x=a, d \mathbf{S}=d y d z\)

∴ \(\int_0 \mathrm{~F} \cdot \mathrm{N} d \mathrm{~S}=\int_{y=0}^a \int_{z=0}^a\left(4 x z \mathrm{i}-y^2 \mathrm{j}+y z \mathrm{k}\right) \cdot \overline{\mathrm{i}} d y d z \)

= \(\int_{y=0}^a \int_{z=0}^a 4 x z d y d z=4 a \int_{y=0}^a d y \cdot \int_0^a z d z\)

because x = a on this face = \(4 a[y]_0^a\left[\frac{z^2}{2}\right]_0^a=2 a^4\)

(2) For the face OABC, \(-\overline{\mathrm{i}}\) is the outward normal

∴ \(\mathbf{N}=-\mathrm{i}, x=0\), and \(d \mathbf{S}=d y d z\)

∴ \(\int_{R_2} \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_{R_2}\left(4 x z \mathbf{i}-y^2 \mathbf{j}+y z \mathbf{k}\right) \cdot(-\mathbf{i}) d y d z=-\iint_{R_2} 4 x z d y d z=0\)

because x=0 on this face

(3) For the face BPQC, \(\bar{j}\) is the outward normal

∴ \(\mathrm{N}=\bar{j}, y=a\), and \(d \mathbf{S}=d x d z\)

∴ \(\int_{R_3} \mathrm{~F} \cdot \mathrm{N} d \mathrm{~S}=\iint_{R_3}\left(4 x z \mathrm{i}-y^2 \mathrm{j}+y z \mathrm{k}\right) \cdot \overline{\mathrm{j}} d x d z=-\iint_{R_3} y^2 d x d z\)

= \(-a^2 \int_{x=0}^a \int_{z=0}^a d y d z\) because y=a, on this face \(=-a^2[x]_0^a[z]_0^a=-a^4\)

(4) For the face ORSA, \(-\overline{\mathrm{j}}\) is the outward normal

∴ \(\mathrm{N}=-\overline{\mathrm{j}}, y=0\) and \(d \mathbf{S}=d x d z\)

F. \(\mathrm{N}=\left(4 x z \mathrm{i}-y^2 \mathrm{j}+y z \mathrm{k}\right) \cdot(-\mathrm{j})=y^2\)

∴ \(\int_{R_4} \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_{R_4} y^2 d x d z=0\)

because y=0, on this face

(5) For the face ABPS; \(\overline{\mathrm{k}}\) is the outward normal

∴ \(\mathrm{N}=\overrightarrow{\mathrm{k}}, z=a\), and \(d \mathbf{S}=d x d y\)

F. \(\mathrm{N}=\left(4 x z \mathrm{i}-y^2 \mathrm{j}+y z \mathrm{k}\right) \cdot \overline{\mathrm{k}}=y z\)

∴ \(\int_{R_5} \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_{R_5} y z d x d y=\int_{z=0}^a \int_{y=0}^a a y d x d y=a[x]_0^a \cdot\left[\frac{y^2}{2}\right]_0^a=\frac{a^4}{4}\)

(6) For the face OCQR,\(-\overline{\mathrm{k}}\) is the outward normal

∴ \(\mathrm{N}=-\overline{\mathrm{k}}, z=0\) and \(d \mathrm{~S}=d x d y\)

F . \(\mathrm{N}=\left(4 x z \mathrm{i}-y^2 \mathrm{j}+y z \mathrm{k}\right) \cdot(-\overline{\mathrm{k}})=-y z\)

∴ \(\int_{R_6} \mathbf{F} . \mathbf{N} d \mathbf{S}=-\iint_{R_6} y z d x d y=0\)

because z=0, on this face

Hence adding all we get \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=2 a^4+0-a^4+0+\frac{1}{2} a^4+0=\frac{3}{2} a^4\).

Example.4. If F=(x+y²)i-2xj+2yzk evaluate  \(\int_S\)F.Nds where S is the surface of the plane  2x+y+2z=6 in the first octant.

Solution:

Let \(\phi=2 x+y+2 z-6\)

∴ The vector normal to surface \(\mathbf{S}\) is \(\nabla \phi\)

i.e. \(\nabla \phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=2 \mathbf{i}+\mathbf{j}+2 \mathbf{k}\)

∴ Unit normal, \(\mathbf{N}=\frac{2 \mathbf{i}+\mathbf{j}+2 \mathbf{k}}{\sqrt{(4+1+4)}}=\frac{1}{3}(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})\)

Let R be the projection of S over XY – plane. Now R is bounded by x-axis, y-axis, and the line 2 x+y=6, z=0.

we have \(\mathbf{F} \cdot \mathbf{N}=\left[\left(x+y^2\right) \mathbf{i}-2 x \mathbf{j}+2 y z \mathbf{k}\right] \cdot \frac{1}{3}(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})\)

= \(\frac{1}{3}\left[2 x+2 y^2-2 x+4 y z\right]=\frac{2}{3}\left(y^2+2 y z\right)\)

N \(\cdot \overline{\mathbf{k}}=\frac{1}{3}(2 \mathrm{i}+\mathrm{j}+2 \mathrm{k}) \cdot \mathbf{k}=\frac{2}{3}\)

∴ \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d x d y}{|\mathbf{N} \cdot \mathbf{k}|}\)

= \(\iint_R^2 \frac{2}{3}\left(y^2+2 y z\right)\left(\frac{3}{2}\right) d x d y=\iint_R\left[y^2+2 y\left(\frac{6-2 x-y}{2}\right)\right] d x d y\)

= \(2 \iint_R y(3-x) d x d y\) because \(z=\frac{6-2 x-y}{2}\) on S

To evaluate the double integral keep x fixed and integrate with respect to y from y=0 to y=6-2x, then integrate y function from x=0 to x=3.

∴ The surface integral \(=2 \int_{x=0}^3 \int_{y=0}^{6-2 x} y(3-x) d x d y=\int_{x=0}^3\left[\int_{y=0}^{6-2 x} y d y\right](3-x) d x\)

= \(2 \int_0^3\left[\frac{y^2}{2}\right]_0^5(3-x) d x=\int_0^3(6-2 x)^2(3-x) d x=4 \int_0^3(3-x)^3 d x=4\left[\frac{(3-x)^4}{4}\right]_0^3=81\)

Integration Of Vectors Volume Integrals

Let V be a volume bounded by a surface r = f (w, v).F (r) be a vector point function defined over V .

Divide V into m sub-regions of volumes (δV₁,δV₂,….δV_i,…δV,) let P_i (r_i) be a point δV_i  then form the sum I_m =\(\sum_{i=1}^m \mathbf{F}\left(\mathbf{r}_{\mathrm{i}}\right) \delta \mathbf{V}_{\mathrm{i}}\)

Let m→ ∝ in such a way that δV_i. Shrinks to a point. The limit of I_m  if it exists, is called the volume integrals of F (r) in the region V and is denoted by\(\int_V \mathbf{F}(\mathbf{r}) d \mathbf{V} \text { or } \int_V \mathbf{F} d \mathbf{V}\)

Cartesian form

Let F (r) = F₁i + F₂ j + F₃k where F₁, F₂ , F₃ are functions of x,y,z

Also, we have dV= dx dy dz The volume integral is then given by

The volume integral is then given by\(\int_V \mathbf{F} d \mathbf{V}\)=\(\iiint\left(\mathbf{F}_1 \mathbf{i}+\mathbf{F}_2 \mathbf{j}+\mathbf{F}_3 \mathbf{k}\right) d x d y d z\)

= \(\mathbf{i} \iiint \mathbf{F}_1 d x d y d z+\mathbf{j} \iiint \mathbf{F}_2 d x d y d z+\mathbf{k} \iiint \mathbf{F}_3 d x d y d z\)

Integration Of Vectors Solved problems

Example.1. 2xzi-xj+y²k evaluate \(\int_V\)F.dv where V is the region bounded by the surfaces x=0,x=2,y=0,y=6,z=x²,z=4

Solution:

Given \(\mathbf{F}=2 x z \mathbf{i}-x \mathbf{j}+y^2 \mathbf{k}\)

∴ The volume integral is \(\int_V \mathrm{~F} . d \mathrm{~V}=\iiint_V\left(2 x z \mathrm{i}-x \mathrm{j}+y^2 \mathrm{k}\right) d x d y d z\)

= \(\mathbf{i} \int_{x=0}^2 \int_{y=0}^6 \int_{z=x^2}^4 2 x z d x d y d z-\mathbf{j} \iiint x d x d y d z+\mathbf{k} \iiint y^2 d x d y d z\)

= \(\mathbf{i} \int_{x=0}^2 \int_{y=0}^6 2 x\left[\frac{z^2}{2}\right]_{x^2}^4 d x d y-\mathbf{j} \int_{x=0}^2 \int_{y=0}^6 x\left[z \int_{x^2}^4 d x d y+\mathbf{k} \int_{x=0}^2 \int_{y=0}^6 y^2[z]_{x^2}^4 d x d y\right.\)

= \(\mathbf{i} \int_0^2 x\left(16-x^4\right) d x \int_0^6 d y-\mathbf{j} \int_0^2 x\left(4-x^2\right) d x \int_0^6 d y+\mathbf{k} \int_{x=0}^2\left(4-x^2\right) d x \int_0^6 y^2 d y\)

= \(\mathbf{i}\left[8 x^2-\frac{x^6}{6}\right]_0^2[y]_0^6-\mathbf{j}\left[2 x^2-\frac{x^4}{4}\right]_0^2[y]_0^6+\mathbf{k}\left[4 x-\frac{x^3}{3}\right]_0^2\left[\frac{y^3}{3}\right]_0^6=128 \mathbf{i}-24 \mathbf{j}+384 \mathbf{k}\)

Example.2.  If F = (2x² – 3z)i- 2xyj- 4x k evaluate (1) \(\int_V\)∇F.dv and v (2)\(\int_V\)∇×F.dv where V is the closed region bounded by  x = 0, y = 0, z = 0,2x + 2y + z = 4.

Solution: (1) ∇. F =\(\left(\mathbf{i} \cdot \frac{\partial \mathbf{F}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{F}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{F}}{\partial z}\right)\) 4x-2x=2x

The limits are z=0 to z=4-2x-2y

y = 0 to \(\frac{4-2 x}{2}\) i.e. (2-x) x=0 to \(\frac{4}{2}\) i.e. 2

∴ \(\int_V \nabla . \overline{\mathrm{F}} d \mathrm{~V}=\int_{x=0}^2 \int_{y=0}^{2-x} \int_{z=0}^{4-2 x-2 y} 2 x d x d y d z\)

= \(\int_{x=0}^2 \int_{y=0}^{2-x} 2 x[z]_0^{4-2 x-2 y} d x d y\)

= \(\int_{x=0}^2 \int_{y=0}^{2-x} 2 x(4-2 x-2 y) d x d y=4 \int_{x=0}^2 \int_{y=0}^{2-x}\left(2 x-x^2-x y\right) d x d y\)

= \(4 \int_0^2\left[\left(2 x-x^2\right) y-x \frac{y^2}{2}\right]_0^{2-x} d x=4 \int_0^2\left[\left(2 x-x^2\right)(2-x)-\frac{x}{2}(2-x)^2\right] d x\)

= \(\int_0^2\left(2 x^3-8 x^2+8 x\right) d x=\left[\frac{x^4}{2}-\frac{8 x^3}{3}+4 x^2\right]_0^2=\frac{8}{3}\)

(2) \(\nabla \times \mathbf{F}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2 x^2-3 z & -2 x y & -4 x\end{array}\right|=\mathbf{j}-2 y \mathbf{k}\)

∴ \(\int_V \nabla \times \mathbf{F} d \mathbf{V}=\iiint_V(\mathbf{j}-2 y \mathbf{k}) d x d y d z=\int_{x=0}^2 \int_{y=0}^{2-x} \int(\mathbf{j}-2 y \mathbf{k})[z]_0^{4-2 x-2 y} d x d y\)

= \(\int_{x=0}^2 \int_{y=0}^{2-x}(\mathbf{j}-2 y \mathbf{k})(4-2 x-2 y) d x d y\)

= \(\int_{x=0}^2 \mathbf{j}\left[(4-2 x) y-y^2\right]_0^{2-x} d x-\mathbf{k} \int_{x=0}^2\left[(4-2 x) y^2-\frac{4 y^3}{3}\right] d x\)

= \(\mathbf{j} \int_0^2(2-x)^2 d x-\mathbf{k} \int_0^2 \frac{2}{3}(2-x)^3 d x=\mathbf{j}\left[\frac{(2-x)^3}{3}\right]_0^2 \frac{-2 \mathbf{k}}{3}\left[\frac{(2-x)^4}{4}\right]_0^2=\frac{8}{3}(\mathbf{j}-\mathbf{k})\)

Integration Of Vectors Exercise 6(c)

1. Evaluate \(\int_{\mathbf{S}}\)F.NdS, where F = y i + 2xj-z k and S is the surface of the plane 2x + y = 6 in the first octant, cut off by the plane z = 4 .  [Take R the projection of S on yz plane].
Solution:

Let \(\phi=2 x+y-6\)

∴ The vector normal to the surface is \(\nabla \phi=2 \mathbf{i}+\mathbf{j}\)

∴ Unit normal N = \(\frac{2 \mathbf{i}+\mathbf{j}}{\sqrt{5}}\)

Let R be the projection of \(\mathbf{S}\) over the y z-plane. Here the surface S is \(\perp r\) to y – plane and hence cannot be projected on x y – plane.

∴ Now \(\mathbf{F} \cdot \mathbf{N}=(y \mathbf{i}+2 x \mathbf{j}-z \mathbf{k}) \cdot \frac{1}{\sqrt{5}}(2 \mathbf{i}+\mathbf{j})=\frac{2}{\sqrt{5}}(x+y)\)

∴ We have \(\oint_S \mathbf{F} . \mathbf{N} d \mathbf{S}=\int \oint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} . i|}=\int \oint_R \frac{2}{\sqrt{5}}(x+y) \frac{\sqrt{5}}{2} d y d z\)

= \(\int \oint_R\left(\frac{6-y}{2}+y\right) d y d z\)…….. (because x=\(\frac{6-y}{2}\))

= \(\frac{1}{2} \int_{y=0}^6 \int_{y=0}^4(6+y) d y d z=\frac{1}{2} \int_{y=0}^6(6+y) d y \int_{y=0}^4 d z\)

= \(\frac{1}{2}\left[6 y+\frac{y^3}{2}\right]_0^6[z]_0^4=108\)

2. If F = 2yi- 3 j + x²k and S is the surface y² = 8x in the first octant bounded by the planes y= 4 and z = 6, evaluate\(\int_{\mathbf{S}}\)F.NdS. [Take R by projecting S on yz plane]
Solution:

Let \(\phi=y^2-8 x\) be the surface S. Normal to S = \(\nabla \phi=-8 \mathbf{i}+2 y \mathbf{j}\)

∴ \(\mathbf{N}=\frac{-4 \mathbf{i}+y \mathbf{j}}{\sqrt{\left(16+y^2\right)}}\)

∴ \(\mathbf{N} . \mathbf{i}=\frac{-4}{\sqrt{16+y^2}}\)

∴ \(\mathrm{F} . \mathbf{N}=\frac{-8 y-3 y}{\sqrt{\left(16+y^2\right)}}=\frac{-11 y}{\sqrt{16+y^2}}\)

Let R be the projection of S over the y z-plane. Then

= \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S} \iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}=\frac{11}{4} \iint_R y d y d z\)

= \(\frac{11}{4} \int_{y=0}^4 y d y \int_{z=0}^6 d z=132\)

3. Find \(\int_{\mathbf{S}}\)F.NdS over the entire surface of the region bounded by x² + z² = 9, x = 0, y = 0, z = 0 and y = 8 if F = 6zi + (2x + y)- xk .
Solution:

Let \(\varphi=x^2+z^2-9[/late] be the surface S.

Normal vector to S = [latex]\nabla \phi=2 x \mathbf{i}+2 z \mathbf{k}\)

∴ Unit normal \(\mathbf{N}=\frac{x \mathbf{i}+z \mathbf{k}}{\sqrt{x^3+z^3}}=\frac{1}{3}(x \mathbf{i}+z \mathbf{k})\)

Now \(\mathbf{N}=\frac{1}{3}(6 z x-z x)=\frac{5}{3} z x\)

Let R be the projection of S over the y z-plane. Then

= \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S} \iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}=\frac{5}{3} \iint_R z x\left(\frac{3}{x}\right) d y d z\)

= \(5 \int_0^3 z d z \int_0^8 d y=180\)

4. If F = yz i + zx j + xy k evaluate \(\int_{\mathbf{S}}\)F.NdS over the surface of x² + y² + z²= 1 in the first octant.
Solution:

Let \(\phi=x^2+y^2+z^2-1\) be the surface S.

Normal vector to \(\mathrm{S}, \nabla \phi=2(x \mathbf{i}+y \mathbf{j}+\mathbf{z k})\)

∴ Unit normal \(\mathbf{N}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\sqrt{x^3+y^3+z^3}}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{1}\)

∴ \(\mathbf{F} . \mathbf{N}=3 x y z, \mathbf{N} \cdot \mathbf{i}=x\)

Let R be the projection of S over the y z-plane. Then \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}=\iint_R(3 x y z) \frac{d y d z}{x}\)

Now in the y z-plane x=0, hence the equation of the surface becomes \(x^2+y^2+z^2=1\)

∴ \(\int \mathbf{F} . \mathbf{N} d \mathbf{S}=3 \int_{y=0}^1 y \int_{z=0}^{\sqrt{1-y^2}} z d z d y\)

= \(\frac{3}{2} \int_0^1 y\left(1-y^2\right) d y=\frac{3}{2}\left[\frac{y^2}{2}-\frac{y^4}{4}\right]_0^1=\frac{3}{8}\)

5. Find ∫F.NdS Where F = 4xi- 2y² j+ z² k over the region bounded by x² + y² = 4, z = 0 and z = 3
Solution:

Let \(\phi=x^2+y^2-4, \mathbf{F}=4 x \mathbf{i}-2 y^2 \mathbf{j}+z^2 \mathbf{k}\)

Normal = \(\nabla \phi=2 x \mathbf{i}+2 y \mathbf{j}\)

∴ Unit normal \(\mathbf{N}=\frac{2 x \mathbf{i}+2 y \mathbf{j}}{\sqrt{4 x^2+4 y^2}}=\frac{1}{2}(x \mathbf{i}+y \mathbf{j})\)

Let R be the projection of S over the y z-plane. Then \(\int_S \mathbf{F} \cdot \mathbf{N} d \mathbf{S}=\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}\)

Now \(\mathbf{F} \cdot \mathbf{N}=\left(4 x \mathbf{i}-2 y^2 \mathbf{j}+z^2 \mathbf{k}\right) \cdot \frac{1}{2}(x \mathbf{i}+y \mathbf{j})=2 x^2-y^3\) and \(\mathbf{N} \cdot \mathbf{i}=\frac{1}{2}(x \mathbf{i}+y \mathbf{j}) \cdot \mathbf{i}=\frac{x}{2}\)

For the surface \(x^2+y^2=4\) in y z-plane. x=0 ⇒ \(y= \pm 2\)

Hence y varies from -2 to 2.

z varies from 0 to 3

∴ The surface integral = \(\iint_R \mathbf{F} \cdot \mathbf{N} \frac{d y d z}{|\mathbf{N} \cdot \mathbf{i}|}=\iint_R\left(2 x^2-y^3\right)\left(\frac{2}{x}\right) d y d z\)

= \(\int_{y=-2}^2\left(4 x-\frac{2 y^3}{x}\right) d y \cdot \int_0^3 d z=3 \int_{y=-2}^2\left(4 \sqrt{4-y^2}-\frac{2 y^3}{\sqrt{4-y^2}}\right) d y\)

= \(12 \int_2^2 \sqrt{4-y^2} d y-6 \int_2^2 \frac{y^3}{\sqrt{4-y^2}} d y=24 \int_0^2 \sqrt{4-y^2} d y-0\)

Put \(y=2 \sin \theta, d y=2 \cos \theta, d \theta\)

6. If Φ  = 45x²,y evaluate \(\begin{equation}
\iiint_V \phi d \mathbf{V}\end{equation}\)where V is the closed region bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.
Solution:

The limits are

z=0 to z=8-4 x-2 y

y=0 to y=4-2 x

x=0 to x=2

∴ \(\iiint_V \phi d \mathrm{~V}=\int_{x=0}^2 \int_{y=0}^{4-2 x} \int_{z=0}^{8-4 x-2 y} 45 x^2 y d z d y d x\)

= \(45 \int_{x=0}^2 \int_{y=0}^{4-2 x} x^2 y[z]_0^{8-4 x-2 y} d y \cdot d x\)

= \(45 \int_{x=0}^2 \int_{y=0}^{4-2 x} x^2 y(8-4 x-2 y) d x d y\)

= \(45 \int_0^2 x^2\left[\frac{8 y^2}{2}-\frac{4 x y^2}{2}-\frac{2 y^3}{3}\right]_0^{4-2 x} d x\)