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Calculus 3 : Triple Integrals

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Example Questions

Example Question #552 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(6cos(z + 1)e^{(- 2x^{2} - 2y^{2})})}{13})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.97\\&\text{and length }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.309,-0.951)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.05

-0.27

-0.05

0.09

Correct answer:

0.09

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(6cos(z + 1)e^{(- 2x^{2} - 2y^{2})})}{13})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(6\cdot rcos(z + 1)e^{(-2\cdot r^{2})})}{13})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.031})=0.49\pi;\theta_2=arctan(\frac{-0.951}{0.309})=1.6\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.48}\int_{0.49\pi}^{1.6\pi}\int_{0}^{1.97}(-\frac{(6\cdot rcos(z + 1)e^{(-2\cdot r^{2})})}{13})drd\theta dz=(\frac{(3cos(z + 1)e^{(-2\cdot r^{2})})}{26})d\theta dz|_{0}^{1.97}\\&\int_{0}^{1.48}\int_{0.49\pi}^{1.6\pi}(-0.1153cos(z + 1))d\theta dz=(-0.1153\theta cos(z + 1))dz|_{0.49\pi}^{1.6\pi}\\&\int_{0}^{1.48}(-0.4022cos(z + 1))dz=(-0.4022sin(z + 1))|_{0}^{1.48}=0.09\end{align*}$

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Example Question #101 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(32sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2})sin(z + 2))}{5})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.26\text{ and }1.51\\&\text{and length }0.64\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.368,0.930)\text{ and }\overrightarrow{u_2}=(-0.094,-0.996)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.87

-1.47

-35.25

1.96

Correct answer:

5.87

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.930}{0.368})=0.38\pi;\theta_2=arctan(\frac{-0.996}{-0.094})=1.47\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.64}\int_{0.38\pi}^{1.47\pi}\int_{0.26}^{1.51}(\frac{(32\cdot rsin(z + 2)sin(\frac{r^{2}}{2}))}{5})drd\theta dz=(-\frac{(32sin(z + 2)cos(\frac{r^{2}}{2}))}{5})d\theta dz|_{0.26}^{1.51}\\&\int_{0}^{0.64}\int_{0.38\pi}^{1.47\pi}(3.724sin(z + 2))d\theta dz=(3.724\theta sin(z + 2))dz|_{0.38\pi}^{1.47\pi}\\&\int_{0}^{0.64}(12.75sin(z + 2))dz=(-12.75cos(z + 2))|_{0}^{0.64}=5.87\end{align*}$

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Example Question #101 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2^ze^{(x^{2} + y^{2})})}{22})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.3\text{ and }0.86\\&\text{and length }1.67\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.562,0.827)\text{ and }\overrightarrow{u_2}=(-0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.2

-0.6

1.2

0.04

Correct answer:

-0.2

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.827}{0.562})=0.31\pi;\theta_2=arctan(\frac{-0.588}{-0.809})=1.2\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.67}\int_{0.31\pi}^{1.2\pi}\int_{0.3}^{0.86}(-\frac{(2^z\cdot re^{(r^{2})})}{22})drd\theta dz=(-\frac{(2^ze^{(r^{2})})}{44})d\theta dz|_{0.3}^{0.86}\\&\int_{0}^{1.67}\int_{0.31\pi}^{1.2\pi}(-0.02275\cdot 2^z)d\theta dz=(-0.02275\cdot 2^z\theta)dz|_{0.31\pi}^{1.2\pi}\\&\int_{0}^{1.67}(-0.0636\cdot 2^z)dz=(-0.09176\cdot 2^z)|_{0}^{1.67}=-0.2\end{align*}$

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Example Question #561 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{((\frac{2}{5})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 3^z)}{9})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.1\text{ and }1.89\\&\text{and length }1.12\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.309,0.951)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.14

0.71

-4.23

-0.18

Correct answer:

0.71

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{((\frac{2}{5})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 3^z)}{9})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{((\frac{2}{5})^{(\frac{(2\cdot r^{2})}{3})}\cdot 3^z\cdot r)}{9})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{-0.309})=0.6\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.12}\int_{0.6\pi}^{1.87\pi}\int_{0.1}^{1.89}(\frac{((\frac{2}{5})^{(\frac{(2\cdot r^{2})}{3})}\cdot 3^z\cdot r)}{9})drd\theta dz=(\frac{(2^{(\frac{(2\cdot r^{2})}{3})}\cdot 3^z)}{(12\cdot 5^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{5}))})d\theta dz|_{0.1}^{1.89}\\&\int_{0}^{1.12}\int_{0.6\pi}^{1.87\pi}(0.08013\cdot 3^z)d\theta dz=(0.08013\cdot 3^z\theta)dz|_{0.6\pi}^{1.87\pi}\\&\int_{0}^{1.12}(0.3197\cdot 3^z)dz=(0.291\cdot 3^z)|_{0}^{1.12}=0.71\end{align*}$

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Example Question #562 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}}{(10\cdot (z + 1))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.14\text{ and }1.1\\&\text{and length }0.55\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.960,0.279)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.008

-0.16

0.05

0.11

Correct answer:

0.05

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}}{(10\cdot (z + 1))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{r^{2}}{2})})}{(10\cdot (z + 1))})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{-0.960})=0.91\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.55}\int_{0.91\pi}^{1.77\pi}\int_{0.14}^{1.1}(\frac{(re^{(-\frac{r^{2}}{2})})}{(10\cdot (z + 1))})drd\theta dz=(-\frac{e^{(-\frac{r^{2}}{2})}}{(10\cdot (z + 1))})d\theta dz|_{0.14}^{1.1}\\&\int_{0}^{0.55}\int_{0.91\pi}^{1.77\pi}(\frac{0.04442}{(z + 1)})d\theta dz=(\frac{(0.04442\theta )}{(z + 1)})dz|_{0.91\pi}^{1.77\pi}\\&\int_{0}^{0.55}(\frac{0.12}{(z + 1)})dz=(0.12ln(z + 1))|_{0}^{0.55}=0.05\end{align*}$

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Example Question #102 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})cos(z + 2))}{18})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.16\text{ and }1.56\\&\text{and length }1.43\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.03

-0.13

0.006

-0.008

Correct answer:

0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})cos(z + 2))}{18})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rcos(z + 2)cos(\frac{(3\cdot r^{2})}{2}))}{18})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.43}\int_{0.77\pi}^{1.65\pi}\int_{0.16}^{1.56}(\frac{(rcos(z + 2)cos(\frac{(3\cdot r^{2})}{2}))}{18})drd\theta dz=(\frac{(cos(z + 2)sin(\frac{(3\cdot r^{2})}{2}))}{54})d\theta dz|_{0.16}^{1.56}\\&\int_{0}^{1.43}\int_{0.77\pi}^{1.65\pi}(-0.009732cos(z + 2))d\theta dz=(-0.009732\theta cos(z + 2))dz|_{0.77\pi}^{1.65\pi}\\&\int_{0}^{1.43}(-0.02691cos(z + 2))dz=(-0.02691sin(z + 2))|_{0}^{1.43}=0.03\end{align*}$

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Example Question #106 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{17}{4^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}cos(3z)})}{2})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.58\\&\text{and length }1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.827,0.562)\text{ and }\overrightarrow{u_2}=(-0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.05

-0.21

1.26

-1.26

Correct answer:

-0.21

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(\frac{17}{4^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}cos(3z)})}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(17\cdot rcos(3z))}{(2\cdot 4^{(\frac{(2\cdot r^{2})}{3})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.562}{0.827})=0.19\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1}\int_{0.19\pi}^{1.35\pi}\int_{0}^{0.58}(-\frac{(17\cdot rcos(3z))}{(2\cdot 4^{(\frac{(2\cdot r^{2})}{3})})})drd\theta dz=(\frac{(51cos(3z))}{(8\cdot 2^{(\frac{(4\cdot r^{2})}{3})}ln(4))})d\theta dz|_{0}^{0.58}\\&\int_{0}^{1}\int_{0.19\pi}^{1.35\pi}(-1.229cos(3z))d\theta dz=(-1.229\theta cos(3z))dz|_{0.19\pi}^{1.35\pi}\\&\int_{0}^{1}(-4.478cos(3z))dz=(-1.493sin(3z))|_{0}^{1}=-0.21\end{align*}$

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Example Question #107 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(2x^{2} + 2y^{2}))}{(25\cdot 3^{(\frac{z}{2})})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.05\\&\text{and length }0.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.976,0.218)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.09

0.18

-0.18

-0.02

Correct answer:

0.09

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3sin(2x^{2} + 2y^{2}))}{(25\cdot 3^{(\frac{z}{2})})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rsin(2\cdot r^{2}))}{(25\cdot 3^{(\frac{z}{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.218}{-0.976})=0.93\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{0.91}\int_{0.93\pi}^{1.77\pi}\int_{0}^{1.05}(\frac{(3\cdot rsin(2\cdot r^{2}))}{(25\cdot 3^{(\frac{z}{2})})})drd\theta dz=(-\frac{(3cos(r^{2})^{2})}{(50\cdot 3^{(\frac{z}{2})})})d\theta dz|_{0}^{1.05}\\&\int_{0}^{0.91}\int_{0.93\pi}^{1.77\pi}(\frac{0.04778}{3^{(0.5z)}})d\theta dz=(\frac{(0.04778\theta )}{3^{(0.5z)}})dz|_{0.93\pi}^{1.77\pi}\\&\int_{0}^{0.91}(\frac{0.1261}{3^{(0.5z)}})dz=(-\frac{0.2295}{3^{(0.5z)}})|_{0}^{0.91}=0.09\end{align*}$

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Example Question #108 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2sin(2x^{2} + 2y^{2}))}{(3\cdot 3^z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.1\text{ and }0.88\\&\text{and length }1.2\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.397,0.918)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.39

2.31

0.06

-2.31

Correct answer:

-0.39

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.918}{-0.397})=0.63\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.2}\int_{0.63\pi}^{1.76\pi}\int_{0.1}^{0.88}(-\frac{(2\cdot rsin(2\cdot r^{2}))}{(3\cdot 3^z)})drd\theta dz=(cos(\frac{r^{2})^{2}}{(3\cdot 3^z)})d\theta dz|_{0.1}^{0.88}\\&\int_{0}^{1.2}\int_{0.63\pi}^{1.76\pi}(-\frac{0.163}{3^{(z)}})d\theta dz=(-\frac{(0.163\theta )}{3^{(z)}})dz|_{0.63\pi}^{1.76\pi}\\&\int_{0}^{1.2}(-\frac{0.5785}{3^{(z)}})dz=(\frac{0.5266}{3^{(z)}})|_{0}^{1.2}=-0.39\end{align*}$

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Example Question #109 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(e^{(- x^{2} - y^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.56\\&\text{and length }1.4\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.187,0.982)\text{ and }\overrightarrow{u_2}=(-0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.48

7.25

1.45

-5.8

Correct answer:

1.45

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(e^{(- x^{2} - y^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(re^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.982}{-0.187})=0.56\pi;\theta_2=arctan(\frac{-0.685}{-0.729})=1.24\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{1.4}\int_{0.56\pi}^{1.24\pi}\int_{0}^{1.56}(re^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))drd\theta dz=(-\frac{(e^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))}{2})d\theta dz|_{0}^{1.56}\\&\int_{0}^{1.4}\int_{0.56\pi}^{1.24\pi}(0.2851z + 0.2851)d\theta dz=(0.2851\theta \cdot (z + 1))dz|_{0.56\pi}^{1.24\pi}\\&\int_{0}^{1.4}(0.609z + 0.609)dz=(0.3045\cdot (z + 1)^{2})|_{0}^{1.4}=1.45\end{align*}$

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Calculus 3 : Triple Integrals

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