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Calculus 3 : Triple Integrals

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

Example Questions

Example Question #481 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(29\cdot 3^{(x^{2} + y^{2})}sin(z + 1))dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.15\\&\text{and length }0.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.562,-0.827)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

370.46

-24.7

-444.55

74.09

Correct answer:

74.09

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(29\cdot 3^{(x^{2} + y^{2})}sin(z + 1))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(29\cdot 3^{(r^{2})}\cdot rsin(z + 1))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{0.309})=0.4\pi;\theta_2=arctan(\frac{-0.827}{-0.562})=1.31\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.63}\int_{0.4\pi}^{1.31\pi}\int_{0}^{1.15}(29\cdot 3^{(r^{2})}\cdot rsin(z + 1))drd\theta dz=(\frac{(29\cdot 3^{(r^{2})}sin(z + 1))}{(2ln(3))})d\theta dz|_{0}^{1.15}\\&\int_{0}^{0.63}\int_{0.4\pi}^{1.31\pi}(43.23sin(z + 1))d\theta dz=(43.23\theta sin(z + 1))dz|_{0.4\pi}^{1.31\pi}\\&\int_{0}^{0.63}(123.6sin(z + 1))dz=(-123.6cos(z + 1))|_{0}^{0.63}=74.09\end{align*}$

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Example Question #32 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}\cdot (z + 1)^{2})}{41})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.63\\&\text{and length }0.88\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.368,0.930)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.08

0.39

-0.02

0.02

Correct answer:

-0.08

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}\cdot (z + 1)^{2})}{41})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})}\cdot (z + 1)^{2})}{41})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.930}{-0.368})=0.62\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.88}\int_{0.62\pi}^{1.65\pi}\int_{0}^{0.63}(-\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})}\cdot (z + 1)^{2})}{41})drd\theta dz=(\frac{(9e^{(-\frac{(2\cdot r^{2})}{3})}\cdot (z + 1)^{2})}{164})d\theta dz|_{0}^{0.63}\\&\int_{0}^{0.88}\int_{0.62\pi}^{1.65\pi}(-0.01276\cdot (z + 1)^{2})d\theta dz=(-0.01276\theta \cdot (z + 1)^{2})dz|_{0.62\pi}^{1.65\pi}\\&\int_{0}^{0.88}(-0.04128\cdot (z + 1)^{2})dz=(-0.01376\cdot (z + 1)^{3})|_{0}^{0.88}=-0.08\end{align*}$

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Example Question #33 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(cos(2x^{2} + 2y^{2})e^{(2z)})}{8})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.18\\&\text{and length }1.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.905,0.426)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.11

-0.43

0.43

0.07

Correct answer:

-0.43

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(cos(2x^{2} + 2y^{2})e^{(2z)})}{8})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(re^{(2z)}cos(2\cdot r^{2}))}{8})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.426}{-0.905})=0.86\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.63}\int_{0.86\pi}^{1.87\pi}\int_{0}^{1.18}(-\frac{(re^{(2z)}cos(2\cdot r^{2}))}{8})drd\theta dz=(-\frac{(e^{(2z)}sin(2\cdot r^{2}))}{32})d\theta dz|_{0}^{1.18}\\&\int_{0}^{1.63}\int_{0.86\pi}^{1.87\pi}(-0.01091e^{(2z)})d\theta dz=(-0.01091\theta e^{(2z)})dz|_{0.86\pi}^{1.87\pi}\\&\int_{0}^{1.63}(-0.03463e^{(2z)})dz=(-0.01732e^{(2z)})|_{0}^{1.63}=-0.43\end{align*}$

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Example Question #491 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{29}{(2\cdot (\frac{x^{2}}{2}+\frac{ y^{2}}{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.17\text{ and }1.26\\&\text{and length }1.09\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

151.18

907.06

-453.53

-75.59

Correct answer:

151.18

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0}^{1.09}\int_{0.59\pi}^{1.35\pi}\int_{0.17}^{1.26}(\frac{29}{r})drd\theta dz=(29ln(r))d\theta dz|_{0.17}^{1.26}\\&\int_{0}^{1.09}\int_{0.59\pi}^{1.35\pi}(58.09)d\theta dz=(58.09\theta)dz|_{0.59\pi}^{1.35\pi}\\&\int_{0}^{1.09}(138.7)dz=(138.7z)|_{0}^{1.09}=151.18\end{align*}$

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Example Question #492 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{19}{(\frac{7}{2})^{(2x^{2} + 2y^{2})}})}{(3\cdot 2^z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.39\text{ and }1.86\\&\text{and length }0.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.37

1.48

8.86

-1.48

Correct answer:

1.48

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{19}{(\frac{7}{2})^{(2x^{2} + 2y^{2})}})}{(3\cdot 2^z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot r)}{(3\cdot 2^z\cdot (\frac{7}{2})^{(2\cdot r^{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.988}{0.156})=1.55\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{0.99}\int_{0.79\pi}^{1.55\pi}\int_{0.39}^{1.86}(\frac{(19\cdot r)}{(3\cdot 2^z\cdot (\frac{7}{2})^{(2\cdot r^{2})})})drd\theta dz=(-\frac{(19\cdot (\frac{4}{49})^{(r^{2})}\cdot (\frac{1}{2})^z)}{(12ln(\frac{7}{2}))})d\theta dz|_{0.39}^{1.86}\\&\int_{0}^{0.99}\int_{0.79\pi}^{1.55\pi}(\frac{0.8632}{2^{(z)}})d\theta dz=(\frac{(0.8632\theta )}{2^{(z)}})dz|_{0.79\pi}^{1.55\pi}\\&\int_{0}^{0.99}(\frac{2.061}{2^{(z)}})dz=(-\frac{2.973}{2^{(z)}})|_{0}^{0.99}=1.48\end{align*}$

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Example Question #36 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})\cdot (\frac{(3z)}{13}+\frac{ 3}{13}))dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.15\\&\text{and length }1.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.279,0.960)\text{ and }\overrightarrow{u_2}=(0.969,-0.249)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-8.26

-0.41

1.65

4.95

Correct answer:

1.65

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})\cdot (\frac{(3z)}{13}+\frac{ 3}{13}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(rcos(\frac{(2\cdot r^{2})}{3})\cdot (\frac{(3z)}{13}+\frac{ 3}{13}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.249}{0.969})=1.92\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.63}\int_{0.59\pi}^{1.92\pi}\int_{0}^{1.15}(rcos(\frac{(2\cdot r^{2})}{3})\cdot (\frac{(3z)}{13}+\frac{ 3}{13}))drd\theta dz=(sin(\frac{(2\cdot r^{2})}{3})\cdot (\frac{(9z)}{52}+\frac{ 9}{52}))d\theta dz|_{0}^{1.15}\\&\int_{0}^{1.63}\int_{0.59\pi}^{1.92\pi}(0.1336z + 0.1336)d\theta dz=(0.1336\theta \cdot (z + 1))dz|_{0.59\pi}^{1.92\pi}\\&\int_{0}^{1.63}(0.5581z + 0.5581)dz=(0.2791\cdot (z + 1)^{2})|_{0}^{1.63}=1.65\end{align*}$

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Example Question #31 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(3cos(2x^{2} + 2y^{2})e^{(-2z)})}{28})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.49\text{ and }1.53\\&\text{and length }1.95\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.187,0.982)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.02

0.07

-0.15

0.36

Correct answer:

0.07

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(3cos(2x^{2} + 2y^{2})e^{(-2z)})}{28})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(3\cdot re^{(-2z)}cos(2\cdot r^{2}))}{28})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.982}{-0.187})=0.56\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.95}\int_{0.56\pi}^{1.77\pi}\int_{0.49}^{1.53}(-\frac{(3\cdot re^{(-2z)}cos(2\cdot r^{2}))}{28})drd\theta dz=(-\frac{(3e^{(-2z)}sin(2\cdot r^{2}))}{112})d\theta dz|_{0.49}^{1.53}\\&\int_{0}^{1.95}\int_{0.56\pi}^{1.77\pi}(0.03915e^{(-2z)})d\theta dz=(0.03915\theta e^{(-2z)})dz|_{0.56\pi}^{1.77\pi}\\&\int_{0}^{1.95}(0.1488e^{(-2z)})dz=(-0.07441e^{(-2z)})|_{0}^{1.95}=0.07\end{align*}$

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Example Question #31 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(44\cdot 3^{(\frac{z}{2})}e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.82\\&\text{and length }1.21\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.988,0.156)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-5088.37

282.69

1696.12

-424.03

Correct answer:

1696.12

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(44\cdot 3^{(\frac{z}{2})}e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(44\cdot 3^{(\frac{z}{2})}\cdot re^{(\frac{(3\cdot r^{2})}{2})})}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{0.156}{-0.988})=0.95\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.21}\int_{0.2\pi}^{0.95\pi}\int_{0}^{1.82}(\frac{(44\cdot 3^{(\frac{z}{2})}\cdot re^{(\frac{(3\cdot r^{2})}{2})})}{5})drd\theta dz=(\frac{(44\cdot 3^{(\frac{z}{2})}e^{(\frac{(3\cdot r^{2})}{2})})}{15})d\theta dz|_{0}^{1.82}\\&\int_{0}^{1.21}\int_{0.2\pi}^{0.95\pi}(419\cdot 3^{(0.5z)})d\theta dz=(419\cdot 3^{(0.5z)}\theta)dz|_{0.2\pi}^{0.95\pi}\\&\int_{0}^{1.21}(987.1\cdot 3^{(0.5z)})dz=(1797\cdot 3^{(0.5z)})|_{0}^{1.21}=1696.12\end{align*}$

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Example Question #39 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-e^{(2z)}cos(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.32\text{ and }1.19\\&\text{and length }0.6\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.66

5.96

-9.93

-1.99

Correct answer:

-1.99

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.729}{-0.685})=0.74\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.6}\int_{0.74\pi}^{1.97\pi}\int_{0.32}^{1.19}(-rcos(r^{2})e^{(2z)})drd\theta dz=(-\frac{(sin(r^{2})e^{(2z)})}{2})d\theta dz|_{0.32}^{1.19}\\&\int_{0}^{0.6}\int_{0.74\pi}^{1.97\pi}(-0.4429e^{(2z)})d\theta dz=(-0.4429\theta e^{(2z)})dz|_{0.74\pi}^{1.97\pi}\\&\int_{0}^{0.6}(-1.712e^{(2z)})dz=(-0.8558e^{(2z)})|_{0}^{0.6}=-1.99\end{align*}$

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Example Question #40 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(27e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}\cdot (z + 1)^{2})}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.39\\&\text{and length }0.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.992,0.125)\text{ and }\overrightarrow{u_2}=(-0.339,0.941)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

22.07

5.52

-1.1

-33.1

Correct answer:

5.52

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(27e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}\cdot (z + 1)^{2})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(27\cdot re^{(-\frac{(3\cdot r^{2})}{2})}\cdot (z + 1)^{2})}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.125}{0.992})=0.04\pi;\theta_2=arctan(\frac{0.941}{-0.339})=0.61\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.86}\int_{0.04\pi}^{0.61\pi}\int_{0}^{1.39}(\frac{(27\cdot re^{(-\frac{(3\cdot r^{2})}{2})}\cdot (z + 1)^{2})}{5})drd\theta dz=(-\frac{(9e^{(-\frac{(3\cdot r^{2})}{2})}\cdot (z + 1)^{2})}{5})d\theta dz|_{0}^{1.39}\\&\int_{0}^{0.86}\int_{0.04\pi}^{0.61\pi}(1.701\cdot (z + 1)^{2})d\theta dz=(1.701\theta \cdot (z + 1)^{2})dz|_{0.04\pi}^{0.61\pi}\\&\int_{0}^{0.86}(3.046\cdot (z + 1)^{2})dz=(1.015\cdot (z + 1)^{3})|_{0}^{0.86}=5.52\end{align*}$

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If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

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Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

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I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

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This notification is accurate.

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Calculus 3 : Triple Integrals

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