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Calculus 3 : Triple Integrals

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Example Questions

Example Question #121 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{1}{(\frac{2}{5})^{(x^{2} + y^{2})}\cdot 3^{(\frac{z}{2})}})}{10})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.21\text{ and }2\\&\text{and length }1.75\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.397,0.918)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

3.99

119.78

-7.99

-23.96

Correct answer:

-23.96

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.918}{0.397})=0.37\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.75}\int_{0.37\pi}^{1.62\pi}\int_{0.21}^{2}(-\frac{(3^{(\frac{z}{2})}\cdot r)}{(10\cdot (\frac{2}{5})^{(r^{2})})})drd\theta dz=(\frac{((\frac{5}{2})^{(r^{2})}\cdot 3^{(\frac{z}{2})})}{(20ln(\frac{2}{5}))})d\theta dz|_{0.21}^{2}\\&\int_{0}^{1.75}\int_{0.37\pi}^{1.62\pi}(-2.075\cdot 3^{(0.5z)})d\theta dz=(-2.075\cdot 3^{(0.5z)}\theta)dz|_{0.37\pi}^{1.62\pi}\\&\int_{0}^{1.75}(-8.147\cdot 3^{(0.5z)})dz=(-14.83\cdot 3^{(0.5z)})|_{0}^{1.75}=-23.96\end{align*}$

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Example Question #122 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-10cos(z + 1)sin(2x^{2} + 2y^{2}))dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.22\\&\text{and length }1.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.536,0.844)\text{ and }\overrightarrow{u_2}=(0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-39.72

-2.48

19.86

9.93

Correct answer:

9.93

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.844}{-0.536})=0.68\pi;\theta_2=arctan(\frac{-0.771}{0.637})=1.72\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.91}\int_{0.68\pi}^{1.72\pi}\int_{0}^{1.22}(-10\cdot rcos(z + 1)sin(2\cdot r^{2}))drd\theta dz=(\frac{(5cos(z + 1)cos(2\cdot r^{2}))}{2})d\theta dz|_{0}^{1.22}\\&\int_{0}^{1.91}\int_{0.68\pi}^{1.72\pi}(-4.966cos(z + 1))d\theta dz=(-4.966\theta cos(z + 1))dz|_{0.68\pi}^{1.72\pi}\\&\int_{0}^{1.91}(-16.23cos(z + 1))dz=(-16.23sin(z + 1))|_{0}^{1.91}=9.93\end{align*}$

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Example Question #123 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2sin(3z)e^{(- x^{2} - y^{2})})}{19})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.86\\&\text{and length }1.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.876,0.482)\text{ and }\overrightarrow{u_2}=(0.707,-0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.22

-0.02

0.01

0.04

Correct answer:

0.04

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.482}{-0.876})=0.84\pi;\theta_2=arctan(\frac{-0.707}{0.707})=1.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.65}\int_{0.84\pi}^{1.75\pi}\int_{0}^{1.86}(\frac{(2\cdot rsin(3z)e^{(-r^{2})})}{19})drd\theta dz=(-\frac{(sin(3z)e^{(-r^{2})})}{19})d\theta dz|_{0}^{1.86}\\&\int_{0}^{1.65}\int_{0.84\pi}^{1.75\pi}(0.05098sin(3z))d\theta dz=(0.05098\theta sin(3z))dz|_{0.84\pi}^{1.75\pi}\\&\int_{0}^{1.65}(0.1457sin(3z))dz=(-0.04858cos(3z))|_{0}^{1.65}=0.04\end{align*}$

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Example Question #124 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5cos(z + 1)sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{36})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.98\\&\text{and length }1.68\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.982,0.187)\text{ and }\overrightarrow{u_2}=(1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.05

0.01

-0.21

0.26

Correct answer:

-0.05

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5cos(z + 1)sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{36})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot rcos(z + 1)sin(\frac{(3\cdot r^{2})}{2}))}{36})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.187}{-0.982})=0.94\pi;\theta_2=arctan(\frac{-0.031}{1.000})=1.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.68}\int_{0.94\pi}^{1.99\pi}\int_{0}^{0.98}(\frac{(5\cdot rcos(z + 1)sin(\frac{(3\cdot r^{2})}{2}))}{36})drd\theta dz=(-\frac{(5cos(z + 1)cos(\frac{(3\cdot r^{2})}{2}))}{108})d\theta dz|_{0}^{0.98}\\&\int_{0}^{1.68}\int_{0.94\pi}^{1.99\pi}(0.04029cos(z + 1))d\theta dz=(0.04029\theta cos(z + 1))dz|_{0.94\pi}^{1.99\pi}\\&\int_{0}^{1.68}(0.1329cos(z + 1))dz=(0.1329sin(z + 1))|_{0}^{1.68}=-0.05\end{align*}$

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Example Question #125 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}e^{(2z)})}{15})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.77\\&\text{and length }1.66\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.661,0.750)\text{ and }\overrightarrow{u_2}=(-0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

4.21

-0.7

0.23

-1.4

Correct answer:

-1.4

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(2e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}e^{(2z)})}{15})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(2\cdot re^{(2z)}e^{(-\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.750}{-0.661})=0.73\pi;\theta_2=arctan(\frac{-1.000}{-0.031})=1.49\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.66}\int_{0.73\pi}^{1.49\pi}\int_{0}^{1.77}(-\frac{(2\cdot re^{(2z)}e^{(-\frac{(3\cdot r^{2})}{2})})}{15})drd\theta dz=(\frac{(2e^{(2z -\frac{ (3\cdot r^{2})}{2})})}{45})d\theta dz|_{0}^{1.77}\\&\int_{0}^{1.66}\int_{0.73\pi}^{1.49\pi}(-0.04404e^{(2z)})d\theta dz=(-0.04404\theta e^{(2z)})dz|_{0.73\pi}^{1.49\pi}\\&\int_{0}^{1.66}(-0.1052e^{(2z)})dz=(-0.05258e^{(2z)})|_{0}^{1.66}=-1.4\end{align*}$

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Example Question #126 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(16sin(4z)e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{3})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.35\text{ and }1.42\\&\text{and length }0.59\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

3.99

-7.98

-1.6

7.98

Correct answer:

-7.98

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(16sin(4z)e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{3})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(16\cdot rsin(4z)e^{(\frac{(2\cdot r^{2})}{3})})}{3})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.397}{-0.918})=1.13\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.59}\int_{0.59\pi}^{1.13\pi}\int_{0.35}^{1.42}(-\frac{(16\cdot rsin(4z)e^{(\frac{(2\cdot r^{2})}{3})})}{3})drd\theta dz=(-4sin(4z)e^{(\frac{(2\cdot r^{2})}{3})})d\theta dz|_{0.35}^{1.42}\\&\int_{0}^{0.59}\int_{0.59\pi}^{1.13\pi}(-11sin(4z))d\theta dz=(-11.0\theta sin(4z))dz|_{0.59\pi}^{1.13\pi}\\&\int_{0}^{0.59}(-18.66sin(4z))dz=(4.666cos(4z))|_{0}^{0.59}=-7.98\end{align*}$

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Example Question #581 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(2z)})}{35})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.5\\&\text{and length }1.81\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(-0.996,0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.58

-0.19

-2.32

2.9

Correct answer:

0.58

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(2z)})}{35})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(2z)}sin(\frac{(3\cdot r^{2})}{2}))}{35})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{0.094}{-0.996})=0.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.81}\int_{0.43\pi}^{0.97\pi}\int_{0}^{1.5}(\frac{(re^{(2z)}sin(\frac{(3\cdot r^{2})}{2}))}{35})drd\theta dz=(-\frac{(e^{(2z)}cos(\frac{(3\cdot r^{2})}{2}))}{105})d\theta dz|_{0}^{1.5}\\&\int_{0}^{1.81}\int_{0.43\pi}^{0.97\pi}(0.01879e^{(2z)})d\theta dz=(0.01879\theta e^{(2z)})dz|_{0.43\pi}^{0.97\pi}\\&\int_{0}^{1.81}(0.03188e^{(2z)})dz=(0.01594e^{(2z)})|_{0}^{1.81}=0.58\end{align*}$

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Example Question #128 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(42cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-2z)})}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.76\\&\text{and length }1.59\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.992,0.125)\text{ and }\overrightarrow{u_2}=(0.790,-0.613)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-20.99

10.49

-2.1

31.48

Correct answer:

10.49

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(42cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-2z)})}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(42\cdot re^{(-2z)}cos(\frac{r^{2}}{2}))}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.125}{-0.992})=0.96\pi;\theta_2=arctan(\frac{-0.613}{0.790})=1.79\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.59}\int_{0.96\pi}^{1.79\pi}\int_{0}^{1.76}(\frac{(42\cdot re^{(-2z)}cos(\frac{r^{2}}{2}))}{5})drd\theta dz=(\frac{(42e^{(-2z)}sin(\frac{r^{2}}{2}))}{5})d\theta dz|_{0}^{1.76}\\&\int_{0}^{1.59}\int_{0.96\pi}^{1.79\pi}(8.398e^{(-2z)})d\theta dz=(8.398\theta e^{(-2z)})dz|_{0.96\pi}^{1.79\pi}\\&\int_{0}^{1.59}(21.9e^{(-2z)})dz=(-10.95e^{(-2z)})|_{0}^{1.59}=10.49\end{align*}$

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Example Question #129 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(35cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(z)})}{4})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.31\text{ and }1.04\\&\text{and length }1.41\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.661,0.750)\text{ and }\overrightarrow{u_2}=(0.905,-0.426)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-164.45

27.41

-9.14

82.23

Correct answer:

27.41

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(35cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(z)})}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(35\cdot re^{(z)}cos(\frac{(3\cdot r^{2})}{2}))}{4})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.750}{-0.661})=0.73\pi;\theta_2=arctan(\frac{-0.426}{0.905})=1.86\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.41}\int_{0.73\pi}^{1.86\pi}\int_{0.31}^{1.04}(\frac{(35\cdot re^{(z)}cos(\frac{(3\cdot r^{2})}{2}))}{4})drd\theta dz=(\frac{(35e^{(z)}sin(\frac{(3\cdot r^{2})}{2}))}{12})d\theta dz|_{0.31}^{1.04}\\&\int_{0}^{1.41}\int_{0.73\pi}^{1.86\pi}(2.494e^{(z)})d\theta dz=(2.494\theta e^{(z)})dz|_{0.73\pi}^{1.86\pi}\\&\int_{0}^{1.41}(8.853e^{(z)})dz=(8.853e^{(z)})|_{0}^{1.41}=27.41\end{align*}$

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Example Question #130 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(5cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{(8\cdot (z + 1))})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.19\\&\text{and length }1.93\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.000,1.000)\text{ and }\overrightarrow{u_2}=(0.827,-0.562)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

6.72

-1.68

-3.36

0.56

Correct answer:

-1.68

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(5cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{(8\cdot (z + 1))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{(8\cdot (z + 1))})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.000})=0.5\pi;\theta_2=arctan(\frac{-0.562}{0.827})=1.81\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.93}\int_{0.5\pi}^{1.81\pi}\int_{0}^{1.19}(-\frac{(5\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{(8\cdot (z + 1))})drd\theta dz=(-\frac{(15sin(\frac{(2\cdot r^{2})}{3}))}{(4\cdot (8z + 8))})d\theta dz|_{0}^{1.19}\\&\int_{0}^{1.93}\int_{0.5\pi}^{1.81\pi}(-\frac{0.3797}{(z + 1)})d\theta dz=(-\frac{(0.3797\theta )}{(z + 1)})dz|_{0.5\pi}^{1.81\pi}\\&\int_{0}^{1.93}(-\frac{1.563}{(z + 1)})dz=(-1.563ln(z + 1))|_{0}^{1.93}=-1.68\end{align*}$

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Calculus 3 : Triple Integrals

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