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Calculus 3 : Triple Integrals

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Example Questions

Example Question #81 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{46}{3^{(x^{2} + y^{2})}e^{(-z)}})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.35\text{ and }1.64\\&\text{and length }1.46\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

48.56

-24.28

9.71

-97.13

Correct answer:

48.56

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{46}{3^{(x^{2} + y^{2})}e^{(-z)}})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(46\cdot re^{(-z)})}{3^{(r^{2})}})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.46}\int_{0.15\pi}^{1.32\pi}\int_{0.35}^{1.64}(\frac{(46\cdot re^{(-z)})}{3^{(r^{2})}})drd\theta dz=(-\frac{(23e^{(-z)})}{(3^{(r^{2})}ln(3))})d\theta dz|_{0.35}^{1.64}\\&\int_{0}^{1.46}\int_{0.15\pi}^{1.32\pi}(17.21e^{(-z)})d\theta dz=(17.21\theta e^{(-z)})dz|_{0.15\pi}^{1.32\pi}\\&\int_{0}^{1.46}(63.25e^{(-z)})dz=(-63.25e^{(-z)})|_{0}^{1.46}=48.56\end{align*}$

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Example Question #81 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(22cos(2x^{2} + 2y^{2})sin(z + 2))}{3})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.8\\&\text{and length }1.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.844,0.536)\text{ and }\overrightarrow{u_2}=(-0.951,0.309)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

6.71

-1.12

0.56

-2.24

Correct answer:

-1.12

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.536}{0.844})=0.18\pi;\theta_2=arctan(\frac{0.309}{-0.951})=0.9\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.94}\int_{0.18\pi}^{0.9\pi}\int_{0}^{0.8}(-\frac{(22\cdot rsin(z + 2)cos(2\cdot r^{2}))}{3})drd\theta dz=(-\frac{(11sin(z + 2)sin(2\cdot r^{2}))}{6})d\theta dz|_{0}^{0.8}\\&\int_{0}^{1.94}\int_{0.18\pi}^{0.9\pi}(-1.756sin(z + 2))d\theta dz=(-1.756\theta sin(z + 2))dz|_{0.18\pi}^{0.9\pi}\\&\int_{0}^{1.94}(-3.973sin(z + 2))dz=(3.973cos(z + 2))|_{0}^{1.94}=-1.12\end{align*}$

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Example Question #83 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{(2z)}{47}+\frac{ 2}{47})}{(2x^{2} + 2y^{2})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.45\text{ and }1.91\\&\text{and length }1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.000)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.03

-0.97

0.16

0.04

Correct answer:

0.16

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.000}{-1.000})=1\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0}^{1.77}\int_{1\pi}^{1.5\pi}\int_{0.45}^{1.91}(\frac{(\frac{(2z)}{47}+\frac{ 2}{47})}{(2\cdot r)})drd\theta dz=(ln(r)\cdot (\frac{z}{47}+\frac{ 1}{47}))d\theta dz|_{0.45}^{1.91}\\&\int_{0}^{1.77}\int_{1\pi}^{1.5\pi}(0.03076z + 0.03076)d\theta dz=(0.03076\theta \cdot (z + 1))dz|_{1\pi}^{1.5\pi}\\&\int_{0}^{1.77}(0.04831z + 0.04831)dz=(0.02416\cdot (z + 1)^{2})|_{0}^{1.77}=0.16\end{align*}$

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Example Question #81 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(cos(x^{2} + y^{2})\cdot (z + 1)^{2})}{33})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.43\text{ and }1.85\\&\text{and length }1.1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.861,0.509)\text{ and }\overrightarrow{u_2}=(-1.000,0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.01

0.05

-0.1

-0.01

Correct answer:

0.05

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(cos(x^{2} + y^{2})\cdot (z + 1)^{2})}{33})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(rcos(r^{2})\cdot (z + 1)^{2})}{33})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.509}{0.861})=0.17\pi;\theta_2=arctan(\frac{0.031}{-1.000})=0.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.1}\int_{0.17\pi}^{0.99\pi}\int_{0.43}^{1.85}(-\frac{(rcos(r^{2})\cdot (z + 1)^{2})}{33})drd\theta dz=(-\frac{(sin(r^{2})\cdot (z + 1)^{2})}{66})d\theta dz|_{0.43}^{1.85}\\&\int_{0}^{1.1}\int_{0.17\pi}^{0.99\pi}(0.006986\cdot (z + 1)^{2})d\theta dz=(0.006986\theta \cdot (z + 1)^{2})dz|_{0.17\pi}^{0.99\pi}\\&\int_{0}^{1.1}(0.018\cdot (z + 1)^{2})dz=(0.005999\cdot (z + 1)^{3})|_{0}^{1.1}=0.05\end{align*}$

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Example Question #82 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{1}{3^{(x^{2} + y^{2})}\cdot 3^z})}{3})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.22\\&\text{and length }2\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.790,0.613)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.54

3.27

1.63

-19.61

Correct answer:

3.27

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.613}{0.790})=0.21\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{2}\int_{0.21\pi}^{1.38\pi}\int_{0}^{1.22}(\frac{(3^z\cdot r)}{(3\cdot 3^{(r^{2})})})drd\theta dz=(-\frac{3^z}{(6\cdot 3^{(r^{2})}ln(3))})d\theta dz|_{0}^{1.22}\\&\int_{0}^{2}\int_{0.21\pi}^{1.38\pi}(0.1221\cdot 3^z)d\theta dz=(0.1221\cdot 3^z\theta)dz|_{0.21\pi}^{1.38\pi}\\&\int_{0}^{2}(0.4489\cdot 3^z)dz=(0.4086\cdot 3^z)|_{0}^{2}=3.27\end{align*}$

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Example Question #86 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(11cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})sin(3z))dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.31\\&\text{and length }0.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(1.000,0.031)\text{ and }\overrightarrow{u_2}=(-0.976,-0.218)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.75

-0.6

-14.99

Correct answer:

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(11cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})sin(3z))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(11\cdot rsin(3z)cos(\frac{(3\cdot r^{2})}{2}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.031}{1.000})=0.01\pi;\theta_2=arctan(\frac{-0.218}{-0.976})=1.07\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.65}\int_{0.01\pi}^{1.07\pi}\int_{0}^{1.31}(11\cdot rsin(3z)cos(\frac{(3\cdot r^{2})}{2}))drd\theta dz=(\frac{(11sin(3z)sin(\frac{(3\cdot r^{2})}{2}))}{3})d\theta dz|_{0}^{1.31}\\&\int_{0}^{0.65}\int_{0.01\pi}^{1.07\pi}(1.971sin(3z))d\theta dz=(1.971\theta sin(3z))dz|_{0.01\pi}^{1.07\pi}\\&\int_{0}^{0.65}(6.563sin(3z))dz=(-2.188cos(3z))|_{0}^{0.65}=3\end{align*}$

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Example Question #541 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{39}{(5\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})\cdot (z + 1))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.1\text{ and }1.76\\&\text{and length }1.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(1.000,0.000)\text{ and }\overrightarrow{u_2}=(0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

57.53

-345.18

-28.77

172.59

Correct answer:

172.59

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{39}{(5\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})\cdot (z + 1))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{117}{(10\cdot r\cdot (z + 1))})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.000}{1.000})=0\pi;\theta_2=arctan(\frac{-0.844}{0.536})=1.68\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0}^{1.65}\int_{0\pi}^{1.68\pi}\int_{0.1}^{1.76}(\frac{117}{(10\cdot r\cdot (z + 1))})drd\theta dz=(\frac{(117ln(r))}{(10\cdot (z + 1))})d\theta dz|_{0.1}^{1.76}\\&\int_{0}^{1.65}\int_{0\pi}^{1.68\pi}(\frac{33.55}{(z + 1)})d\theta dz=(\frac{(33.55\theta )}{(z + 1)})dz|_{0\pi}^{1.68\pi}\\&\int_{0}^{1.65}(\frac{177.1}{(z + 1)})dz=(177.1ln(z + 1))|_{0}^{1.65}=172.59\end{align*}$

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Example Question #541 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}((\frac{7}{2})^{(x^{2} + y^{2})}\cdot (\frac{z}{42}+\frac{ 1}{42}))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.33\text{ and }1.27\\&\text{and length }0.82\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.729,0.685)\text{ and }\overrightarrow{u_2}=(-0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.02

0.46

0.11

-0.46

Correct answer:

0.11

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}((\frac{7}{2})^{(x^{2} + y^{2})}\cdot (\frac{z}{42}+\frac{ 1}{42}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}((\frac{7}{2})^{(r^{2})}\cdot r\cdot (\frac{z}{42}+\frac{ 1}{42}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.685}{-0.729})=0.76\pi;\theta_2=arctan(\frac{-0.771}{-0.637})=1.28\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0}^{0.82}\int_{0.76\pi}^{1.28\pi}\int_{0.33}^{1.27}((\frac{7}{2})^{(r^{2})}\cdot r\cdot (\frac{z}{42}+\frac{ 1}{42}))drd\theta dz=(\frac{((\frac{7}{2})^{(r^{2})}\cdot (\frac{z}{42}+\frac{ 1}{42}))}{(2ln(\frac{7}{2}))})d\theta dz|_{0.33}^{1.27}\\&\int_{0}^{0.82}\int_{0.76\pi}^{1.28\pi}(0.06079z + 0.06079)d\theta dz=(0.06079\theta \cdot (z + 1))dz|_{0.76\pi}^{1.28\pi}\\&\int_{0}^{0.82}(0.0993z + 0.0993)dz=(0.04965\cdot (z + 1)^{2})|_{0}^{0.82}=0.11\end{align*}$

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Example Question #83 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(e^{(- 2x^{2} - 2y^{2})}e^{(-2z)})}{6})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.34\\&\text{and length }1.45\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(-1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.07

0.03

-0.01

-0.03

Correct answer:

0.03

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.031}{-1.000})=1.01\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.45}\int_{0.43\pi}^{1.01\pi}\int_{0}^{1.34}(\frac{(re^{(-2z)}e^{(-2\cdot r^{2})})}{6})drd\theta dz=(-\frac{e^{(- 2z - 2\cdot r^{2})}}{24})d\theta dz|_{0}^{1.34}\\&\int_{0}^{1.45}\int_{0.43\pi}^{1.01\pi}(0.04052e^{(-2z)})d\theta dz=(0.04052\theta e^{(-2z)})dz|_{0.43\pi}^{1.01\pi}\\&\int_{0}^{1.45}(0.07383e^{(-2z)})dz=(-0.03691e^{(-2z)})|_{0}^{1.45}=0.03\end{align*}$

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Example Question #90 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(e^{(- x^{2} - y^{2})}e^{(2z)})}{32})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.84\\&\text{and length }1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.000,1.000)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.1

-0.1

0.03

-0.05

Correct answer:

-0.1

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.000})=0.5\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1}\int_{0.5\pi}^{1.77\pi}\int_{0}^{0.84}(-\frac{(re^{(2z)}e^{(-r^{2})})}{32})drd\theta dz=(\frac{e^{(2z - r^{2})}}{64})d\theta dz|_{0}^{0.84}\\&\int_{0}^{1}\int_{0.5\pi}^{1.77\pi}(-0.007909e^{(2z)})d\theta dz=(-0.007909\theta e^{(2z)})dz|_{0.5\pi}^{1.77\pi}\\&\int_{0}^{1}(-0.03156e^{(2z)})dz=(-0.01578e^{(2z)})|_{0}^{1}=-0.1\end{align*}$

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Calculus 3 : Triple Integrals

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Example Question #81 : Triple Integration In Cylindrical Coordinates

Example Question #81 : Triple Integration In Cylindrical Coordinates

Example Question #83 : Multiple Integration

Example Question #81 : Triple Integration In Cylindrical Coordinates

Example Question #82 : Triple Integration In Cylindrical Coordinates

Example Question #86 : Multiple Integration

Example Question #541 : Calculus 3

Example Question #541 : Calculus 3

Example Question #83 : Triple Integration In Cylindrical Coordinates

Example Question #90 : Multiple Integration

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