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Calculus 3 : Triple Integrals

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Example Questions

Example Question #508 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}((\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}sin(z + 2))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.41\text{ and }1.72\\&\text{and length }1.09\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.891,0.454)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.55

-2.33

0.78

-0.16

Correct answer:

0.78

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.454}{-0.891})=0.85\pi;\theta_2=arctan(\frac{-0.125}{0.992})=1.96\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.09}\int_{0.85\pi}^{1.96\pi}\int_{0.41}^{1.72}((\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot rsin(z + 2))drd\theta dz=(-\frac{sin(z + 2)}{(3\cdot 2^{(\frac{(3\cdot r^{2})}{2})}ln(2))})d\theta dz|_{0.41}^{1.72}\\&\int_{0}^{1.09}\int_{0.85\pi}^{1.96\pi}(0.3816sin(z + 2))d\theta dz=(0.3816\theta sin(z + 2))dz|_{0.85\pi}^{1.96\pi}\\&\int_{0}^{1.09}(1.331sin(z + 2))dz=(-1.331cos(z + 2))|_{0}^{1.09}=0.78\end{align*}$

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Example Question #509 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(12cos(z + 1)cos(x^{2} + y^{2}))}{5})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.21\text{ and }1.59\\&\text{and length }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.454,0.891)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1.01

1.01

-0.13

0.51

Correct answer:

0.51

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(12cos(z + 1)cos(x^{2} + y^{2}))}{5})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(12\cdot rcos(r^{2})cos(z + 1))}{5})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.891}{-0.454})=0.65\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.48}\int_{0.65\pi}^{1.76\pi}\int_{0.21}^{1.59}(-\frac{(12\cdot rcos(r^{2})cos(z + 1))}{5})drd\theta dz=(-\frac{(6cos(z + 1)sin(r^{2}))}{5})d\theta dz|_{0.21}^{1.59}\\&\int_{0}^{1.48}\int_{0.65\pi}^{1.76\pi}(-0.638cos(z + 1))d\theta dz=(-0.638\theta cos(z + 1))dz|_{0.65\pi}^{1.76\pi}\\&\int_{0}^{1.48}(-2.225cos(z + 1))dz=(-2.225sin(z + 1))|_{0}^{1.48}=0.51\end{align*}$

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Example Question #510 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2cos(z + 2)cos(x^{2} + y^{2}))}{15})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.19\text{ and }1.91\\&\text{and length }1.02\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.876,0.482)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.06

-0.12

0.01

0.23

Correct answer:

-0.06

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.482}{-0.876})=0.84\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.02}\int_{0.84\pi}^{1.51\pi}\int_{0.19}^{1.91}(-\frac{(2\cdot rcos(r^{2})cos(z + 2))}{15})drd\theta dz=(-\frac{(cos(z + 2)sin(r^{2}))}{15})d\theta dz|_{0.19}^{1.91}\\&\int_{0}^{1.02}\int_{0.84\pi}^{1.51\pi}(0.03475cos(z + 2))d\theta dz=(0.03475\theta cos(z + 2))dz|_{0.84\pi}^{1.51\pi}\\&\int_{0}^{1.02}(0.07314cos(z + 2))dz=(0.07314sin(z + 2))|_{0}^{1.02}=-0.06\end{align*}$

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Example Question #51 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(9cos(2x^{2} + 2y^{2})cos(z + 1))}{4})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.35\\&\text{and length }1.2\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.771,0.637)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.01

-0.02

0.13

-0.03

Correct answer:

-0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(9cos(2x^{2} + 2y^{2})cos(z + 1))}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(9\cdot rcos(z + 1)cos(2\cdot r^{2}))}{4})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.637}{-0.771})=0.78\pi;\theta_2=arctan(\frac{-0.125}{0.992})=1.96\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.2}\int_{0.78\pi}^{1.96\pi}\int_{0}^{1.35}(-\frac{(9\cdot rcos(z + 1)cos(2\cdot r^{2}))}{4})drd\theta dz=(-\frac{(9cos(z + 1)sin(2\cdot r^{2}))}{16})d\theta dz|_{0}^{1.35}\\&\int_{0}^{1.2}\int_{0.78\pi}^{1.96\pi}(0.2714cos(z + 1))d\theta dz=(0.2714\theta cos(z + 1))dz|_{0.78\pi}^{1.96\pi}\\&\int_{0}^{1.2}(1.006cos(z + 1))dz=(1.006sin(z + 1))|_{0}^{1.2}=-0.03\end{align*}$

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Example Question #51 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(2cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})sin(z + 1))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.24\text{ and }1.39\\&\text{and length }1.1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.905,0.426)\text{ and }\overrightarrow{u_2}=(0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.88

8.65

-25.96

4.33

Correct answer:

8.65

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.426}{0.905})=0.14\pi;\theta_2=arctan(\frac{-0.588}{0.809})=1.8\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.1}\int_{0.14\pi}^{1.8\pi}\int_{0.24}^{1.39}(2\cdot rsin(z + 1)cos(\frac{r^{2}}{2}))drd\theta dz=(2sin(z + 1)sin(\frac{r^{2}}{2}))d\theta dz|_{0.24}^{1.39}\\&\int_{0}^{1.1}\int_{0.14\pi}^{1.8\pi}(1.588sin(z + 1))d\theta dz=(1.588\theta sin(z + 1))dz|_{0.14\pi}^{1.8\pi}\\&\int_{0}^{1.1}(8.28sin(z + 1))dz=(-8.28cos(z + 1))|_{0}^{1.1}=8.65\end{align*}$

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Example Question #51 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-cos(z + 2)e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.69\\&\text{and length }1.92\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.729,0.685)\text{ and }\overrightarrow{u_2}=(-0.707,-0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.08

0.54

-2.17

-0.22

Correct answer:

1.08

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.685}{0.729})=0.24\pi;\theta_2=arctan(\frac{-0.707}{-0.707})=1.25\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.92}\int_{0.24\pi}^{1.25\pi}\int_{0}^{0.69}(-rcos(z + 2)e^{(-\frac{r^{2}}{2})})drd\theta dz=(cos(z + 2)e^{(-\frac{r^{2}}{2})})d\theta dz|_{0}^{0.69}\\&\int_{0}^{1.92}\int_{0.24\pi}^{1.25\pi}(-0.2118cos(z + 2))d\theta dz=(-0.2118\theta cos(z + 2))dz|_{0.24\pi}^{1.25\pi}\\&\int_{0}^{1.92}(-0.6722cos(z + 2))dz=(-0.6722sin(z + 2))|_{0}^{1.92}=1.08\end{align*}$

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Example Question #52 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(5e^{(-2z)}e^{(x^{2} + y^{2})})}{41})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.13\text{ and }1.91\\&\text{and length }1.67\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.397,0.918)\text{ and }\overrightarrow{u_2}=(0.249,-0.969)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-4.18

16.72

-0.7

0.7

Correct answer:

-4.18

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(5e^{(-2z)}e^{(x^{2} + y^{2})})}{41})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5\cdot re^{(r^{2})}e^{(-2z)})}{41})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.918}{0.397})=0.37\pi;\theta_2=arctan(\frac{-0.969}{0.249})=1.58\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.67}\int_{0.37\pi}^{1.58\pi}\int_{0.13}^{1.91}(-\frac{(5\cdot re^{(r^{2})}e^{(-2z)})}{41})drd\theta dz=(-\frac{(5e^{(r^{2} - 2z)})}{82})d\theta dz|_{0.13}^{1.91}\\&\int_{0}^{1.67}\int_{0.37\pi}^{1.58\pi}(-2.28e^{(-2z)})d\theta dz=(-2.28\theta e^{(-2z)})dz|_{0.37\pi}^{1.58\pi}\\&\int_{0}^{1.67}(-8.665e^{(-2z)})dz=(4.333e^{(-2z)})|_{0}^{1.67}=-4.18\end{align*}$

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Example Question #55 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(8e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}\cdot (z + 1)^{2})}{3})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.21\\&\text{and length }0.84\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

10.88

-43.5

-5.44

32.63

Correct answer:

10.88

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.960}{-0.279})=0.59\pi;\theta_2=arctan(\frac{-0.771}{-0.637})=1.28\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.84}\int_{0.59\pi}^{1.28\pi}\int_{0}^{1.21}(\frac{(8\cdot re^{(\frac{r^{2}}{2})}\cdot (z + 1)^{2})}{3})drd\theta dz=(\frac{(8e^{(\frac{r^{2}}{2})}\cdot (z + 1)^{2})}{3})d\theta dz|_{0}^{1.21}\\&\int_{0}^{0.84}\int_{0.59\pi}^{1.28\pi}(2.878\cdot (z + 1)^{2})d\theta dz=(2.878\theta \cdot (z + 1)^{2})dz|_{0.59\pi}^{1.28\pi}\\&\int_{0}^{0.84}(6.239\cdot (z + 1)^{2})dz=(2.08\cdot (z + 1)^{3})|_{0}^{0.84}=10.88\end{align*}$

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Example Question #56 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-3sin(z + 1)sin(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.19\text{ and }0.84\\&\text{and length }0.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.21

-1.26

-0.32

2.53

Correct answer:

-1.26

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.94}\int_{0.26\pi}^{1.51\pi}\int_{0.19}^{0.84}(-3\cdot rsin(r^{2})sin(z + 1))drd\theta dz=(\frac{(3cos(r^{2})sin(z + 1))}{2})d\theta dz|_{0.19}^{0.84}\\&\int_{0}^{0.94}\int_{0.26\pi}^{1.51\pi}(-0.3572sin(z + 1))d\theta dz=(-0.3572\theta sin(z + 1))dz|_{0.26\pi}^{1.51\pi}\\&\int_{0}^{0.94}(-1.403sin(z + 1))dz=(1.403cos(z + 1))|_{0}^{0.94}=-1.26\end{align*}$

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Example Question #53 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(17cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{(5\cdot (z + 1))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.26\text{ and }1.21\\&\text{and length }1.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

7.77

46.6

-1.29

-38.83

Correct answer:

7.77

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(17cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{(5\cdot (z + 1))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(17\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{(5\cdot (z + 1))})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.031})=0.49\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.99}\int_{0.49\pi}^{1.62\pi}\int_{0.26}^{1.21}(\frac{(17\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{(5\cdot (z + 1))})drd\theta dz=(\frac{(51sin(\frac{(2\cdot r^{2})}{3}))}{(20\cdot (z + 1))})d\theta dz|_{0.26}^{1.21}\\&\int_{0}^{1.99}\int_{0.49\pi}^{1.62\pi}(\frac{1.997}{(z + 1)})d\theta dz=(\frac{(1.997\theta )}{(z + 1)})dz|_{0.49\pi}^{1.62\pi}\\&\int_{0}^{1.99}(\frac{7.09}{(z + 1)})dz=(7.09ln(z + 1))|_{0}^{1.99}=7.77\end{align*}$

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Calculus 3 : Triple Integrals

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