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Calculus 3 : Multiple Integration

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{2}{(\frac{5}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}e^{(-z)}})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.3\text{ and }1.38\\&\text{and length }1.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.827,0.562)\text{ and }\overrightarrow{u_2}=(-0.661,0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.74

1.49

-1.49

7.43

Correct answer:

1.49

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{2}{(\frac{5}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}e^{(-z)}})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot re^{(-z)})}{(\frac{5}{2})^{(\frac{(2\cdot r^{2})}{3})}})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.562}{0.827})=0.19\pi;\theta_2=arctan(\frac{0.750}{-0.661})=0.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.86}\int_{0.19\pi}^{0.73\pi}\int_{0.3}^{1.38}(\frac{(2\cdot re^{(-z)})}{(\frac{5}{2})^{(\frac{(2\cdot r^{2})}{3})}})drd\theta dz=(-\frac{(3\cdot 2^{(\frac{(2\cdot r^{2})}{3})}e^{(-z)})}{(2\cdot 5^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{5}{2}))})d\theta dz|_{0.3}^{1.38}\\&\int_{0}^{1.86}\int_{0.19\pi}^{0.73\pi}(1.038e^{(-z)})d\theta dz=(1.038\theta e^{(-z)})dz|_{0.19\pi}^{0.73\pi}\\&\int_{0}^{1.86}(1.761e^{(-z)})dz=(-1.761e^{(-z)})|_{0}^{1.86}=1.49\end{align*}$

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Example Question #62 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{42}{2^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 2^z})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.43\text{ and }1.71\\&\text{and length }0.72\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.063,0.998)\text{ and }\overrightarrow{u_2}=(-0.941,-0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

12.96

-12.96

-207.37

51.84

Correct answer:

51.84

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{42}{2^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 2^z})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(42\cdot 2^z\cdot r)}{2^{(\frac{(2\cdot r^{2})}{3})}})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.998}{-0.063})=0.52\pi;\theta_2=arctan(\frac{-0.339}{-0.941})=1.11\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{0.72}\int_{0.52\pi}^{1.11\pi}\int_{0.43}^{1.71}(\frac{(42\cdot 2^z\cdot r)}{2^{(\frac{(2\cdot r^{2})}{3})}})drd\theta dz=(-\frac{(63\cdot 2^{(z -\frac{ (2\cdot r^{2})}{3})})}{(2ln(2))})d\theta dz|_{0.43}^{1.71}\\&\int_{0}^{0.72}\int_{0.52\pi}^{1.11\pi}(29.96\cdot 2^z)d\theta dz=(29.96\cdot 2^z\theta)dz|_{0.52\pi}^{1.11\pi}\\&\int_{0}^{0.72}(55.53\cdot 2^z)dz=(80.11\cdot 2^z)|_{0}^{0.72}=51.84\end{align*}$

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Example Question #519 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(10e^{(z)})}{(3\cdot (2x^{2} + 2y^{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.37\text{ and }1.07\\&\text{and length }0.69\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.509,0.861)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-5.25

1.75

15.75

-10.5

Correct answer:

-5.25

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(10e^{(z)})}{(3\cdot (2x^{2} + 2y^{2}))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5e^{(z)})}{(3\cdot r)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.861}{-0.509})=0.67\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.69}\int_{0.67\pi}^{1.62\pi}\int_{0.37}^{1.07}(-\frac{(5e^{(z)})}{(3\cdot r)})drd\theta dz=(-\frac{(5e^{(z)}ln(r))}{3})d\theta dz|_{0.37}^{1.07}\\&\int_{0}^{0.69}\int_{0.67\pi}^{1.62\pi}(-1.77e^{(z)})d\theta dz=(-1.77\theta e^{(z)})dz|_{0.67\pi}^{1.62\pi}\\&\int_{0}^{0.69}(-5.282e^{(z)})dz=(-5.282e^{(z)})|_{0}^{0.69}=-5.25\end{align*}$

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Example Question #521 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(9\cdot 3^{(\frac{z}{2})}sin(x^{2} + y^{2}))}{2})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.08\\&\text{and length }1.38\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.930,0.368)\text{ and }\overrightarrow{u_2}=(-0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

23.19

2.9

-11.6

-23.19

Correct answer:

-11.6

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(9\cdot 3^{(\frac{z}{2})}sin(x^{2} + y^{2}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(9\cdot 3^{(\frac{z}{2})}\cdot rsin(r^{2}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.368}{0.930})=0.12\pi;\theta_2=arctan(\frac{-0.976}{-0.218})=1.43\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.38}\int_{0.12\pi}^{1.43\pi}\int_{0}^{1.08}(-\frac{(9\cdot 3^{(\frac{z}{2})}\cdot rsin(r^{2}))}{2})drd\theta dz=(\frac{(9\cdot 3^{(\frac{z}{2})}cos(r^{2}))}{4})d\theta dz|_{0}^{1.08}\\&\int_{0}^{1.38}\int_{0.12\pi}^{1.43\pi}(-1.365\cdot 3^{(0.5z)})d\theta dz=(-1.365\cdot 3^{(0.5z)}\theta)dz|_{0.12\pi}^{1.43\pi}\\&\int_{0}^{1.38}(-5.616\cdot 3^{(0.5z)})dz=(-10.22\cdot 3^{(0.5z)})|_{0}^{1.38}=-11.6\end{align*}$

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Example Question #522 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{(4\cdot (z + 1)^{3})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.64\\&\text{and length }1.4\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(-0.951,0.309)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.01

0.13

0.04

-0.17

Correct answer:

0.04

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{(4\cdot (z + 1)^{3})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{(4\cdot (z + 1)^{3})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{0.309}{-0.951})=0.9\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{1.4}\int_{0.16\pi}^{0.9\pi}\int_{0}^{0.64}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{(4\cdot (z + 1)^{3})})drd\theta dz=(-\frac{(3e^{(-\frac{(2\cdot r^{2})}{3})})}{(16\cdot (z + 1)^{3})})d\theta dz|_{0}^{0.64}\\&\int_{0}^{1.4}\int_{0.16\pi}^{0.9\pi}(\frac{0.0448}{(z + 1)^{3}})d\theta dz=(\frac{(0.0448\theta )}{(z + 1)^{3}})dz|_{0.16\pi}^{0.9\pi}\\&\int_{0}^{1.4}(\frac{0.1042}{(z + 1)^{3}})dz=(-\frac{0.05208}{(z + 1)^{2}})|_{0}^{1.4}=0.04\end{align*}$

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Example Question #523 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(e^{(z)}sin(x^{2} + y^{2}))}{12})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.9\\&\text{and length }0.73\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.771,0.637)\text{ and }\overrightarrow{u_2}=(0.249,-0.969)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.07

-0.03

0.01

0.07

Correct answer:

-0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(e^{(z)}sin(x^{2} + y^{2}))}{12})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(rsin(r^{2})e^{(z)})}{12})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.637}{-0.771})=0.78\pi;\theta_2=arctan(\frac{-0.969}{0.249})=1.58\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.73}\int_{0.78\pi}^{1.58\pi}\int_{0}^{0.9}(-\frac{(rsin(r^{2})e^{(z)})}{12})drd\theta dz=(\frac{(cos(r^{2})e^{(z)})}{24})d\theta dz|_{0}^{0.9}\\&\int_{0}^{0.73}\int_{0.78\pi}^{1.58\pi}(-0.01294e^{(z)})d\theta dz=(-0.01294\theta e^{(z)})dz|_{0.78\pi}^{1.58\pi}\\&\int_{0}^{0.73}(-0.03252e^{(z)})dz=(-0.03252e^{(z)})|_{0}^{0.73}=-0.03\end{align*}$

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Example Question #524 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{(22\cdot (z + 1)^{3})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.67\\&\text{and length }1.51\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.509,0.861)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.04

0.19

-0.19

0.04

Correct answer:

0.04

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{(22\cdot (z + 1)^{3})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rcos(r^{2}))}{(22\cdot (z + 1)^{3})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.861}{-0.509})=0.67\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.51}\int_{0.67\pi}^{1.87\pi}\int_{0}^{1.67}(\frac{(3\cdot rcos(r^{2}))}{(22\cdot (z + 1)^{3})})drd\theta dz=(\frac{(3sin(r^{2}))}{(44\cdot (z + 1)^{3})})d\theta dz|_{0}^{1.67}\\&\int_{0}^{1.51}\int_{0.67\pi}^{1.87\pi}(\frac{0.02355}{(z + 1)^{3}})d\theta dz=(\frac{(0.02355\theta )}{(z + 1)^{3}})dz|_{0.67\pi}^{1.87\pi}\\&\int_{0}^{1.51}(\frac{0.08879}{(z + 1)^{3}})dz=(-\frac{0.04439}{(z + 1)^{2}})|_{0}^{1.51}=0.04\end{align*}$

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Example Question #525 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5sin(z + 2)e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{29})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.61\\&\text{and length }0.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-28.44

-1.14

34.13

5.69

Correct answer:

5.69

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5sin(z + 2)e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{29})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot rsin(z + 2)e^{(\frac{(3\cdot r^{2})}{2})})}{29})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.988}{0.156})=1.55\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.86}\int_{0.34\pi}^{1.55\pi}\int_{0}^{1.61}(\frac{(5\cdot rsin(z + 2)e^{(\frac{(3\cdot r^{2})}{2})})}{29})drd\theta dz=(\frac{(5sin(z + 2)e^{(\frac{(3\cdot r^{2})}{2})})}{87})d\theta dz|_{0}^{1.61}\\&\int_{0}^{0.86}\int_{0.34\pi}^{1.55\pi}(2.748sin(z + 2))d\theta dz=(2.748\theta sin(z + 2))dz|_{0.34\pi}^{1.55\pi}\\&\int_{0}^{0.86}(10.45sin(z + 2))dz=(-10.45cos(z + 2))|_{0}^{0.86}=5.69\end{align*}$

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Example Question #63 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4\cdot 3^{(\frac{z}{2})}\cdot 4^{(x^{2} + y^{2})})}{31})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.28\text{ and }1.34\\&\text{and length }1.49\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.930,0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-9.59

0.48

2.4

-0.6

Correct answer:

2.4

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4\cdot 3^{(\frac{z}{2})}\cdot 4^{(x^{2} + y^{2})})}{31})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot 3^{(\frac{z}{2})}\cdot 4^{(r^{2})}\cdot r)}{31})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{0.750})=0.23\pi;\theta_2=arctan(\frac{0.368}{-0.930})=0.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.49}\int_{0.23\pi}^{0.88\pi}\int_{0.28}^{1.34}(\frac{(4\cdot 3^{(\frac{z}{2})}\cdot 4^{(r^{2})}\cdot r)}{31})drd\theta dz=(\frac{(2\cdot 2^{(2\cdot r^{2})}\cdot 3^{(\frac{z}{2})})}{(31ln(4))})d\theta dz|_{0.28}^{1.34}\\&\int_{0}^{1.49}\int_{0.23\pi}^{0.88\pi}(0.509\cdot 3^{(0.5z)})d\theta dz=(0.509\cdot 3^{(0.5z)}\theta)dz|_{0.23\pi}^{0.88\pi}\\&\int_{0}^{1.49}(1.039\cdot 3^{(0.5z)})dz=(1.892\cdot 3^{(0.5z)})|_{0}^{1.49}=2.4\end{align*}$

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Example Question #64 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(5cos(z + 2)cos(x^{2} + y^{2}))}{32})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.5\text{ and }1.22\\&\text{and length }0.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.125,0.992)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.1

0.02

-0.57

-0.05

Correct answer:

0.1

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(5cos(z + 2)cos(x^{2} + y^{2}))}{32})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5\cdot rcos(r^{2})cos(z + 2))}{32})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.992}{-0.125})=0.54\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{0.65}\int_{0.54\pi}^{1.73\pi}\int_{0.5}^{1.22}(-\frac{(5\cdot rcos(r^{2})cos(z + 2))}{32})drd\theta dz=(-\frac{(5cos(z + 2)sin(r^{2}))}{64})d\theta dz|_{0.5}^{1.22}\\&\int_{0}^{0.65}\int_{0.54\pi}^{1.73\pi}(-0.05853cos(z + 2))d\theta dz=(-0.05853\theta cos(z + 2))dz|_{0.54\pi}^{1.73\pi}\\&\int_{0}^{0.65}(-0.2188cos(z + 2))dz=(-0.2188sin(z + 2))|_{0}^{0.65}=0.1\end{align*}$

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