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Calculus 3 : Triple Integration in Cylindrical Coordinates

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Example Questions

Example Question #71 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(23cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})cos(4z))}{5})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.4\text{ and }1.47\\&\text{and length }1.02\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.063,0.998)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.7

-8.42

-2.1

4.21

Correct answer:

-2.1

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(23cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})cos(4z))}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(23\cdot rcos(4z)cos(\frac{r^{2}}{2}))}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.998}{0.063})=0.48\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.02}\int_{0.48\pi}^{1.38\pi}\int_{0.4}^{1.47}(\frac{(23\cdot rcos(4z)cos(\frac{r^{2}}{2}))}{5})drd\theta dz=(\frac{(23cos(4z)sin(\frac{r^{2}}{2}))}{5})d\theta dz|_{0.4}^{1.47}\\&\int_{0}^{1.02}\int_{0.48\pi}^{1.38\pi}(3.69cos(4z))d\theta dz=(3.69\theta cos(4z))dz|_{0.48\pi}^{1.38\pi}\\&\int_{0}^{1.02}(10.43cos(4z))dz=(2.609sin(4z))|_{0}^{1.02}=-2.1\end{align*}$

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Example Question #72 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(3\cdot 3^z)}{(34\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.26\text{ and }1.72\\&\text{and length }1.09\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.339,-0.941)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

2.61

-0.87

-1.74

0.22

Correct answer:

-0.87

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(3\cdot 3^z)}{(34\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(3\cdot 3^z)}{(34\cdot r)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.941}{0.339})=1.61\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.09}\int_{0.82\pi}^{1.61\pi}\int_{0.26}^{1.72}(-\frac{(3\cdot 3^z)}{(34\cdot r)})drd\theta dz=(-\frac{(3\cdot 3^zln(r))}{34})d\theta dz|_{0.26}^{1.72}\\&\int_{0}^{1.09}\int_{0.82\pi}^{1.61\pi}(-0.1667\cdot 3^z)d\theta dz=(-0.1667\cdot 3^z\theta)dz|_{0.82\pi}^{1.61\pi}\\&\int_{0}^{1.09}(-0.4138\cdot 3^z)dz=(-0.3766\cdot 3^z)|_{0}^{1.09}=-0.87\end{align*}$

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Example Question #73 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(13\cdot 3^{(\frac{z}{2})}sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{4})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.31\text{ and }1.38\\&\text{and length }1.71\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.918,0.397)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

3.39

13.56

-13.56

-2.26

Correct answer:

-13.56

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(13\cdot 3^{(\frac{z}{2})}sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(13\cdot 3^{(\frac{z}{2})}\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.397}{-0.918})=0.87\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.71}\int_{0.87\pi}^{1.76\pi}\int_{0.31}^{1.38}(-\frac{(13\cdot 3^{(\frac{z}{2})}\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{4})drd\theta dz=(\frac{(39\cdot 3^{(\frac{z}{2})}cos(\frac{(2\cdot r^{2})}{3}))}{16})d\theta dz|_{0.31}^{1.38}\\&\int_{0}^{1.71}\int_{0.87\pi}^{1.76\pi}(-1.709\cdot 3^{(0.5z)})d\theta dz=(-1.709\cdot 3^{(0.5z)}\theta)dz|_{0.87\pi}^{1.76\pi}\\&\int_{0}^{1.71}(-4.779\cdot 3^{(0.5z)})dz=(-8.701\cdot 3^{(0.5z)})|_{0}^{1.71}=-13.56\end{align*}$

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Example Question #71 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(6e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-z)})}{29})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.25\text{ and }1.23\\&\text{and length }1.8\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.891,0.454)\text{ and }\overrightarrow{u_2}=(0.930,-0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.12

-0.04

0.5

-0.25

Correct answer:

-0.25

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(6e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-z)})}{29})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(6\cdot re^{(-z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{29})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.454}{-0.891})=0.85\pi;\theta_2=arctan(\frac{-0.368}{0.930})=1.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}\int_{0.25}^{1.23}(-\frac{(6\cdot re^{(-z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{29})drd\theta dz=(\frac{(9e^{(- z -\frac{ (2\cdot r^{2})}{3})})}{58})d\theta dz|_{0.25}^{1.23}\\&\int_{0}^{1.8}\int_{0.85\pi}^{1.88\pi}(-0.09224e^{(-z)})d\theta dz=(-0.09224\theta e^{(-z)})dz|_{0.85\pi}^{1.88\pi}\\&\int_{0}^{1.8}(-0.2985e^{(-z)})dz=(0.2985e^{(-z)})|_{0}^{1.8}=-0.25\end{align*}$

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Example Question #72 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(32e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}e^{(-2z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.33\text{ and }1.49\\&\text{and length }0.76\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.905,0.426)\text{ and }\overrightarrow{u_2}=(0.988,-0.156)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

4.41

-105.73

26.43

-13.22

Correct answer:

26.43

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.426}{-0.905})=0.86\pi;\theta_2=arctan(\frac{-0.156}{0.988})=1.95\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.76}\int_{0.86\pi}^{1.95\pi}\int_{0.33}^{1.49}(32\cdot re^{(-2z)}e^{(-\frac{r^{2}}{2})})drd\theta dz=(-32e^{(- 2z -\frac{ r^{2}}{2})})d\theta dz|_{0.33}^{1.49}\\&\int_{0}^{0.76}\int_{0.86\pi}^{1.95\pi}(19.76e^{(-2z)})d\theta dz=(19.76\theta e^{(-2z)})dz|_{0.86\pi}^{1.95\pi}\\&\int_{0}^{0.76}(67.66e^{(-2z)})dz=(-33.83e^{(-2z)})|_{0}^{0.76}=26.43\end{align*}$

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Example Question #73 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(4\cdot 2^z)}{(3\cdot (2x^{2} + 2y^{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.31\text{ and }0.99\\&\text{and length }1.07\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.536,0.844)\text{ and }\overrightarrow{u_2}=(0.790,-0.613)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-4.28

-12.84

8.56

0.71

Correct answer:

-4.28

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.844}{-0.536})=0.68\pi;\theta_2=arctan(\frac{-0.613}{0.790})=1.79\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.07}\int_{0.68\pi}^{1.79\pi}\int_{0.31}^{0.99}(-\frac{(2\cdot 2^z)}{(3\cdot r)})drd\theta dz=(-\frac{(2\cdot 2^zln(r))}{3})d\theta dz|_{0.31}^{0.99}\\&\int_{0}^{1.07}\int_{0.68\pi}^{1.79\pi}(-0.7741\cdot 2^z)d\theta dz=(-0.7741\cdot 2^z\theta)dz|_{0.68\pi}^{1.79\pi}\\&\int_{0}^{1.07}(-2.699\cdot 2^z)dz=(-3.894\cdot 2^z)|_{0}^{1.07}=-4.28\end{align*}$

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Example Question #74 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-2z)})}{47})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.17\\&\text{and length }0.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.613,0.790)\text{ and }\overrightarrow{u_2}=(-1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.11

-0.01

0.04

0.02

Correct answer:

0.04

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(-2z)})}{47})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(-2z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{47})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{0.613})=0.29\pi;\theta_2=arctan(\frac{-0.031}{-1.000})=1.01\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.86}\int_{0.29\pi}^{1.01\pi}\int_{0}^{1.17}(\frac{(4\cdot re^{(-2z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{47})drd\theta dz=(-\frac{(3e^{(- 2z -\frac{ (2\cdot r^{2})}{3})})}{47})d\theta dz|_{0}^{1.17}\\&\int_{0}^{0.86}\int_{0.29\pi}^{1.01\pi}(0.0382e^{(-2z)})d\theta dz=(0.0382\theta e^{(-2z)})dz|_{0.29\pi}^{1.01\pi}\\&\int_{0}^{0.86}(0.08641e^{(-2z)})dz=(-0.04321e^{(-2z)})|_{0}^{0.86}=0.04\end{align*}$

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Example Question #75 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{6}{(\frac{2}{5})^{(x^{2} + y^{2})}e^{(2z)}})}{31})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.84\\&\text{and length }1.56\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.969,0.249)\text{ and }\overrightarrow{u_2}=(-0.930,0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.87

-5.22

15.67

-2.61

Correct answer:

-2.61

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(\frac{6}{(\frac{2}{5})^{(x^{2} + y^{2})}e^{(2z)}})}{31})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(6\cdot re^{(2z)})}{(31\cdot (\frac{2}{5})^{(r^{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.249}{0.969})=0.08\pi;\theta_2=arctan(\frac{0.368}{-0.930})=0.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.56}\int_{0.08\pi}^{0.88\pi}\int_{0}^{0.84}(-\frac{(6\cdot re^{(2z)})}{(31\cdot (\frac{2}{5})^{(r^{2})})})drd\theta dz=(\frac{(3\cdot (\frac{5}{2})^{(r^{2})}e^{(2z)})}{(31ln(\frac{2}{5}))})d\theta dz|_{0}^{0.84}\\&\int_{0}^{1.56}\int_{0.08\pi}^{0.88\pi}(-0.096e^{(2z)})d\theta dz=(-0.096\theta e^{(2z)})dz|_{0.08\pi}^{0.88\pi}\\&\int_{0}^{1.56}(-0.2413e^{(2z)})dz=(-0.1206e^{(2z)})|_{0}^{1.56}=-2.61\end{align*}$

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Example Question #76 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(18\cdot (\frac{7}{2})^{(x^{2} + y^{2})}cos(z + 2))}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.66\\&\text{and length }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.187,0.982)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

804.96

-643.97

-160.99

80.5

Correct answer:

-160.99

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(18\cdot (\frac{7}{2})^{(x^{2} + y^{2})}cos(z + 2))}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(18\cdot (\frac{7}{2})^{(r^{2})}\cdot rcos(z + 2))}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.982}{-0.187})=0.56\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.48}\int_{0.56\pi}^{1.5\pi}\int_{0}^{1.66}(\frac{(18\cdot (\frac{7}{2})^{(r^{2})}\cdot rcos(z + 2))}{5})drd\theta dz=(\frac{(9\cdot (\frac{7}{2})^{(r^{2})}cos(z + 2))}{(5ln(\frac{7}{2}))})d\theta dz|_{0}^{1.66}\\&\int_{0}^{1.48}\int_{0.56\pi}^{1.5\pi}(43.92cos(z + 2))d\theta dz=(43.92\theta cos(z + 2))dz|_{0.56\pi}^{1.5\pi}\\&\int_{0}^{1.48}(129.7cos(z + 2))dz=(129.7sin(z + 2))|_{0}^{1.48}=-160.99\end{align*}$

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Example Question #77 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(4sin(x^{2} + y^{2}))}{(29\cdot (z + 1)^{2})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.54\\&\text{and length }0.82\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.844,0.536)\text{ and }\overrightarrow{u_2}=(-0.941,0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.02

0.36

-0.6

-0.12

Correct answer:

-0.12

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(4sin(x^{2} + y^{2}))}{(29\cdot (z + 1)^{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(4\cdot rsin(r^{2}))}{(29\cdot (z + 1)^{2})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{0.844})=0.18\pi;\theta_2=arctan(\frac{0.339}{-0.941})=0.89\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0}^{0.82}\int_{0.18\pi}^{0.89\pi}\int_{0}^{1.54}(-\frac{(4\cdot rsin(r^{2}))}{(29\cdot (z + 1)^{2})})drd\theta dz=(\frac{(2cos(r^{2}))}{(58z + 29z^{2} + 29)})d\theta dz|_{0}^{1.54}\\&\int_{0}^{0.82}\int_{0.18\pi}^{0.89\pi}(-\frac{0.1185}{(z + 1)^{2}})d\theta dz=(-\frac{(0.1185\theta )}{(z + 1)^{2}})dz|_{0.18\pi}^{0.89\pi}\\&\int_{0}^{0.82}(-\frac{0.2643}{(z + 1)^{2}})dz=(\frac{0.2643}{(z + 1)})|_{0}^{0.82}=-0.12\end{align*}$

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Calculus 3 : Triple Integration in Cylindrical Coordinates

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