Vector Calculus 4th Edition Chapter 5 Multiple Integration
Page 313 Problem 1 Answer
Given: The expression is
∫0 ∫1 y sin x dy dx.
Π 2
To find: The iterated integrals of the given expression.Apply the properties of integrals∫sinx=−cosx.
Let us consider the given expression:
⇒ \(\int_0^\pi \int_1^2 y \sin x d y d x\)
Integrate with respect to y,
⇒ \(\int_0^\pi\left[\frac{y^2}{2} \sin x\right]_1^2 d x\)
Evaluate, apply The Fundamental theorem of Calculus,
⇒ \(\int_0^\pi\left[\frac{(2)^2}{2} \sin x-\frac{(1)^2}{2} \sin x\right] d x\)
= \(\int_0^\pi\left(\frac{3}{2} \sin x\right) dx\)
= \(\frac{3}{2}\int_0^\pi sin x dx\)
Integrate and evaluate.
⇒ \(\frac{3}{2} \int_0^\pi \sin x d x=-\frac{3}{2}[\cos x]_0^\pi\)
= \(-\frac{3}{2}\left[\cos x\right]_0^\pi=-\frac{3}{2}[\cos \pi-\cos 0]\)
= \(-\frac{3}{2}[\cos x]_0^\pi=-\frac{3}{2}[-1-1]\)
= \(-\frac{3}{2}[\cos x]_0^\pi=3\)
Evaluate to obtain the final answer.
The answer of given expression after evaluating is 3.
Page 313 Problem 2 Answer
Given: The expression is
∫−2∫0 xey/dydx.
4 1
To find: The iterated integrals of the given expression.
Apply the properties of integrals∫ex/dx=ex.
Evaluate to obtain the final answer.
Let us consider the given expression;
⇒ \(\int_{-2}^4 \int_0^1 x e^y d y d x\)
Integrate with respect to y,
= \(\int_{-2}^4\left[x e^y\right]_0^1 d x\)
= \(\int_{-2}^4 x\left[e^y\right]_0^1 d x\)
Evaluate.
=\(\int_{-2}^4 x\left[e-e^0\right] d x\)
= \(\int_{-2}^4 x(e-1) d x\)
= \(\int_{-2}^{4^2}(e-1) x d x\)
Integrate and evaluate.
⇒ \(\int_{-2}^4(e-1) x d x=(e-1)\left[\frac{x^2}{2}\right]_{-2}^4\)
= \((e-1)\left[\frac{x^2}{2}\right]_{-2}^4=(e-1)\left[\frac{(4)^2}{2}-\frac{(-2)^2}{2}\right]\)
= \((e-1)\left[\frac{x^2}{2}\right]_{-2}^4=(e-1)(6)\)
= \((e-1)\left[\frac{x^2}{2}\right]_{-2}^4=6 e-6\)
The required result is 6e-6.
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Page 313 Problem 3 Answer
Given: The expression is
∫1∫0(ex+y+x2+lny) dx.
2 1
To find: The iterated integrals of the given expression. Apply the properties of integrals∫ex/dx=ex.
Evaluate to obtain the final answer.
Let us consider the given expression;
⇒ \(\int_1^2 \int_0^1\left(e^{z+y}+x^2+\ln y\right) d x d y\)
Integrate with respect to x,
⇒ \(\int_1^2\left[e^{x+y}+\frac{x^3}{3}+x \ln y\right]_0^1 d y\)
Estimate,
= \(\int_1^2\left[\left(e^{1+y}+\frac{(1)^3}{3}+(1) \ln y\right)-\left(e^{0+y}+\frac{(0)^3}{3}+(0) \ln y\right)\right] d y\)
= \(\int_1^2\left(e^{1+y}+\frac{1}{3}+\ln y-e^y\right) d y\)
Integrate and evaluate,
= \(\left[e^{1+y}+\frac{1}{3} y+y \ln y-y-e^y\right]^{2}\)
= \(\left[e^{1+y}-\frac{1}{3} y+y \ln y-e^y\right]^\frac{1}{2}\)
= \(\left[e^{1+2}-\frac{1}{3}(2)+2 \ln (2)-e^2\right]-\left[e^{1+1}-\frac{1}{3}(1)+2 \ln (1)-e^1\right]^1\)
= \(e^3-\frac{2}{3}+2 \ln (2)-e^2-e^2+\frac{1}{3}+e\)
= \(e^3-2 e^2+e+2 \ln 2-\frac{1}{3}\)
The required result is e3−2e2+e+2ln2−1/3.
Page 313 Problem 4 Answer
Given: The expression is
∫1 ∫1ln√x/xydxdy.
9 e
To find: The iterated integrals of the given expression.
Apply the properties of integrals ∫2lnx/xdx=(lnx)2.
Evaluate to obtain the final answer.
Let us consider the given expression;
∴ \(\int_1^9 \int_1^e \frac{\ln \sqrt{x}}{x y} d x d y=\int_1^9\left(\int_1^e \frac{\ln x^2}{x y} d x\right) d y\)
= \(\int_1^9\left(\int_1^e \frac{\ln x}{2 x y} d x\right) d y\)
= \(\int_1^9\left(\left.\frac{(\ln x)^2}{4 y}\right|_1 ^e\right) d y\)
= \(\int_1^9\left(\frac{(\ln e)^2}{4 y}-\frac{(\ln 1)^2}{4 y}\right) d y\)
= \(\int_1^9\left(\frac{1^2}{4 y}-\frac{0^2}{4 y}\right) d y\)
Integrate and evaluate.
= \(\int_1^9 \frac{1}{4 y} d y\)
= \(\left.\frac{\ln y}{4}\right|_1 ^9\)
= \(\frac{\ln 9}{4}-\frac{\ln 1}{4}\)
= \(\frac{\ln 9}{4}-\frac{0}{4}\)
= \(\frac{1}{4} \ln 9\)
= In \(9^\frac{1}{4}\)
= In √3.
The required result is ln√3.
Page 313 Problem 5 Answer
Given: The paraboloid z=x2+y2+2 and over the rectangle R={(x,y)∣−1≤x≤2,0≤y≤2}.
To find: The volume of the given region.
Apply Cavalieri’s principle.Evaluate to obtain the final answer.
Let us consider the given expression;
V = \(\int_a^b A(x) d x\)
A (x) = \(\int^d f(x, y) d y\)
V = \(\int_{a}^{b} \int_c^d f(x, y) d y d x\)
V = \(\int_{-1}^2 \int_0^2 x^2+y^2+2 d y d x\)
V = \(\int_{-1}^2 x^2 y+\frac{y^3}{3}+\left.2 y\right|_0 ^2 d x\)
V = \(\frac{x^3}{3}+\left.\frac{20}{3} x\right|_{-1} ^2\)
= \(\frac{2}{3} \cdot 2^3+\frac{20}{3} \cdot 2+\frac{2}{3} \cdot 1+\frac{20}{3} \cdot 1\)
= 26
The volume of the given region is 26.
Page 313 Problem 6 Answer
Given: The paraboloid z=x2+y2+2 and over the rectangle R={(x,y)∣−1≤x≤2,0≤y≤2}.
To find: The volume of the given region.Apply Cavalieri’s principle.Evaluate to obtain the final answer.
Let us consider the given expression;
V = \(\int_a^b A(y) d y\)
A (y) = \(\int_c^d f(x, y) d x\)
V = \(\int_c^d \int_a^b f(x, y) d x d y\)
V = \(\int_0^2 \int_{-1}^2 x^2+y^2+2 d x d y\)
V = \(\int_0^2 \frac{x^3}{3}+y^2 x+\left.2 x\right|_{-1} ^2 d y\)
V = \(9 y+\left.y^3\right|_0 ^2\)
= 26.
The volume of the given region is 26.
Page 313 Problem 7 Answer
Given: The plane is z=x+3y+1.
To find: The volume of the given region.
Apply the formula of volume V=∫a∫cf(x,y) dxdy.
B d
Evaluate to obtain the final answer.
Let us consider the given expression
V = \(\int_a^b \int_c^d f(x, y) d x d y\)
V = \(\int_1^2 \int_0^3 x+3 y+1 d x d y\)
V = \(\int_1^2 \frac{x^2}{2}+3 y x+\left.x\right|_0 ^3 d y\)
Integrate and evaluate.
V = \(\int_1^2 \frac{9}{2}+9 y+3 d y\)
V = \(\int_1^2 \frac{15}{2}+9 y d y\)
V = \(\frac{15}{2} y+\left.\frac{9 y^2}{2}\right|_1 ^2\)
= \(\frac{15}{2} \cdot 2+\frac{9}{2} \cdot 2^2-\frac{15}{2} \cdot 1-\frac{9}{2} \cdot 1^2\)
= 21
The volume of the given region is 21.
Page 313 Problem 8 Answer
Given: The function is f(x,y)=2x2+y4sinπx.
To find: The volume of the given region.
Apply the formula of volume V=∫a∫c f(x,y) dxdy.
B d
Evaluate to obtain the final answer.
Let us consider the given expression;
f(x,y) = 2x2 + y4 sin πx,
V = \(\int_a^b \int_c^d f(x, y) d x d y\)
V = \(\int_{-1}^2 \int_0^1 2 x^2+y^4 \sin \pi x d x d y\)
V = \(\int_{-1}^2 \frac{2}{3} x^3-\left.\frac{1}{\pi} y^4 \cos \pi x\right|_0 ^1 d y\)
Integrate and evaluate.
V = \(\int_{-1}^2 \frac{2}{3}(1)^3-\frac{1}{\pi} y^4 \cos \pi(1)-\frac{1}{\pi} y^4 \cos (1) \pi d y\)
V = \(\int_{-1}^2 \frac{2}{3}-\frac{2}{\pi} y^4 d y\)
V = \(\frac{2}{3} y-\left.\frac{2}{5 \pi} y^5\right|_{-1} ^2\)
= \(\frac{2}{3} \cdot 2+\frac{3}{5 \pi} \cdot 2^5+\frac{2}{3} \cdot 1+\frac{2}{5 \pi} \cdot 2^5\)
≈ 6.2.
The volume of the given region is 6.2.
Page 314 Problem 9 Answer
Given: The expression is
∫0∫12dxdy.
2 3
To find: The volume of the given region.
Evaluate to get the final answer.
This is the area of the region whose points are determined by (x,y,z) ∈ [1,3] x [0,2] x [0,2]
⇒ \(\int_0^2 \int_1^3 2 d x d y=2 \int_0^2(3-1) d y\)
= 4(2-0)
= 8.
The volume of the given region is 8.
Page 314 Problem 10 Answer
Given: The expression is
∫1∫−2(16−x2−y2) dydx.
3 2
To find: The volume of the given region.
Evaluate to obtain the final answer.
Let us consider the given expression;
⇒ \(\int_1^3 \int_{-2}^2\left(16-x^2-y^2\right) d y d x\)
This is the volume of the region whose bounded by 1 ≤ x ≤ 3 and -2 ≤ y ≤ 2.
⇒ \(\int_1^3 \int_{-2}^2\left(16-x^2-y^2\right) d y d x=\int_1^3\left[\left(16-x^2\right) y-\frac{y^3}{3}\right]_{y=-2}^{y=2} d x\)
= \(\int_1^3 4\left(16-x^2\right)-\frac{16}{3} d x\)
= \(\left[64 x-\frac{4 x^3}{3}-\frac{16}{3} x\right]_1^3\)
= \(\frac{248}{3}\).
The volume of the given region is 248/3.
Page 314 Problem 11 Answer
Given: The expression is ∫y=−π/2
y=π/2
∫x=0
x=π
sin(x)cos(y)dxdy.
To find: The volume of the given region.
Evaluate to obtain the final answer.
Let us consider the given expression;
⇒ \(\int_{y=-\pi / 2}^{y=\pi / 2} \int_{z=0}^{x=\pi} \sin (x) \cos (y) d x d y .\)
This is the volume of the region whose bounded by z = sin (x) cos (y), 0 ≤ x ≤ π, and -π/2 ≤ y ≤ π/2.
⇒ \(\int_{y=-\pi / 2}^{y=\pi / 2} \int_{x=0}^{x=\pi} \sin (x) \cos (y) d x d y=\int_{y=-\pi / 2}^{y=\pi / 2} \cos (y)\left[\int_{x=0}^{x=\pi} \sin (x) d x\right] d y\)
= \(2 \int_{y=-\pi / 2}^{y=\pi / 2} \cos (y) d y\)
= 4.
The volume of the given region is V=4.
Page 314 Problem 12 Answer
Given: The expression is∫y=0
y=5
∫x=−2
x=2
(4−x2) dxdy.
To find: The volume of the given region.
Evaluate to obtain the final answer.
Let us consider the given expression;
⇒ \(\int_{y=0}^{y=5} \int_{x=-2}^{x=2}\left(4-x^2\right) d x d y,\)
This is the volume of the region whose bounded by z = 4-x2, -2 ≤ x ≤ 2,0 ≤ y ≤ 5.
⇒ \(\int_{y=0}^{y=5} \int_{x=-2}^{x=2}\left(4-x^2\right) d x d y=\int_{y=0}^{y=5}\left[\int_{x=-2}^{x-2}\left(4-x^2\right) d x\right] d y\)
= \(\frac{32}{3} \int_{y=0}^{y=5} d y\)
= \(\frac{160}{3}\)
The volume of the given region is 160/3.
Page 314 Problem 13 Answer
Given: The expression is∫x=−2 x=3
∫y=0y=1∣x∣sin(πy)dydx.
To find: The volume of the given region.
Evaluate to obtain the final answer.
Let us consider the given expression;
⇒ \(\int_{x=-2}^{x=3} \int_{y=0}^{y=1}|x| \sin (\pi y) d y d x\)
This is the volume of the region whose bounded by \(z=|x| \sin (\pi y),-2 \leq x \leq 3,0 \leq y \leq 1\)
⇒ \(\int_{x=-2}^{x-3} \int_{y=0}^{y=1}|x| \sin (\pi y) d y d x=\int_{x=-2}^{x=3}|x|\left[\int_{y=0}^{y=1} \sin (\pi y) d y\right] d x\)
= \(\frac{2}{\pi} \int_{x=-2}^{x-3}|x| d x\)
= \(\frac{13}{\pi}\)
The volume of the given region is 13/π.
Page 314 Problem 14 Answer
Given: The expression is ∫y=−5 y=5
∫x=−1x=2 (5−∣y∣) dxdy.
To find: The volume of the given region.
Evaluate to obtain the final answer.
Let us consider the given expression;
⇒ \(\int_{y=-5}^{y=5} \int_{x=-1}^{x=2}(5-|y|) d x d y\)
This is the volume of the region whose bounded by \(z=5-|y|,-1 \leq x \leq 2,-5 \leq y \leq 5\)
⇒ \(\int_{y=-5}^{y=5} \int_{x=-1}^{z=2}(5-|y|) d x d y=\int_{y=-5}^{y=5}(5-|y|)\left[\int_{x=-1}^{z-2} d x\right] d y\)
= \(3 \int_{y=-5}^{y=5}(5-|y|) d y\)
= 75.
The volume of the given region is 75.
Page 314 Problem 15 Answer
Given: The f(x,y) is a non negative-valued, continuous function defined on R={(x,y)∣a≤x≤b,c≤y≤d}.
To find: The Volume under the given region.
Apply the formula of volumeV=∬R f(x,y)dxdy.
Evaluate to obtain the final answer.
The volume under the surface is given by;
V = \(\iint_R f(x, y) d x d y\)
Yet f(x,y) ⩾ M,
Where M is some constant,
V = \(\int_a^b \int_c^d f(x, y) d y d x\)
= \(\int_a^b \int_c^d M d y d x\)
Integrate and evaluate.
= \(\left.M \int_a^b y\right|_c ^d d x\)
= \(\int_a^b(d-c) d x\)
= \(\left.M(d-c) x\right|_a ^b\)
= M (d-c)(b-a)
= \(V ⩾ M(d-c)(b-a)\)
The volume under the given region is V⩾M(d−c)(b−a).