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Calculus 3 : Triple Integrals

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Example Questions

Example Question #91 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(2z)})}{11})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.44\text{ and }1.06\\&\text{and length }1.11\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.844,0.536)\text{ and }\overrightarrow{u_2}=(-0.685,0.729)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

3.09

-3.09

-0.1

0.62

Correct answer:

0.62

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(2z)})}{11})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(2z)}cos(\frac{(3\cdot r^{2})}{2}))}{11})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{0.844})=0.18\pi;\theta_2=arctan(\frac{0.729}{-0.685})=0.74\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.11}\int_{0.18\pi}^{0.74\pi}\int_{0.44}^{1.06}(\frac{(4\cdot re^{(2z)}cos(\frac{(3\cdot r^{2})}{2}))}{11})drd\theta dz=(\frac{(4e^{(2z)}sin(\frac{(3\cdot r^{2})}{2}))}{33})d\theta dz|_{0.44}^{1.06}\\&\int_{0}^{1.11}\int_{0.18\pi}^{0.74\pi}(0.08571e^{(2z)})d\theta dz=(0.08571\theta e^{(2z)})dz|_{0.18\pi}^{0.74\pi}\\&\int_{0}^{1.11}(0.1508e^{(2z)})dz=(0.07539e^{(2z)})|_{0}^{1.11}=0.62\end{align*}$

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Example Question #92 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(-2z)}sin(x^{2} + y^{2}))}{5})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.04\\&\text{and length }0.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.729,0.685)\text{ and }\overrightarrow{u_2}=(0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.17

-0.03

-0.17

0.69

Correct answer:

0.17

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.685}{-0.729})=0.76\pi;\theta_2=arctan(\frac{-0.771}{0.637})=1.72\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.63}\int_{0.76\pi}^{1.72\pi}\int_{0}^{1.04}(\frac{(3\cdot rsin(r^{2})e^{(-2z)})}{5})drd\theta dz=(-\frac{(3cos(r^{2})e^{(-2z)})}{10})d\theta dz|_{0}^{1.04}\\&\int_{0}^{0.63}\int_{0.76\pi}^{1.72\pi}(0.159e^{(-2z)})d\theta dz=(0.159\theta e^{(-2z)})dz|_{0.76\pi}^{1.72\pi}\\&\int_{0}^{0.63}(0.4796e^{(-2z)})dz=(-0.2398e^{(-2z)})|_{0}^{0.63}=0.17\end{align*}$

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Example Question #91 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(11e^{(z)})}{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.22\text{ and }0.86\\&\text{and length }1.06\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.998,0.063)\text{ and }\overrightarrow{u_2}=(0.426,-0.905)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

575.89

-143.97

-719.86

47.99

Correct answer:

-143.97

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.063}{0.998})=0.02\pi;\theta_2=arctan(\frac{-0.905}{0.426})=1.64\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.06}\int_{0.02\pi}^{1.64\pi}\int_{0.22}^{0.86}(-\frac{(11e^{(z)})}{r})drd\theta dz=(-11e^{(z)}ln(r))d\theta dz|_{0.22}^{0.86}\\&\int_{0}^{1.06}\int_{0.02\pi}^{1.64\pi}(-15e^{(z)})d\theta dz=(-15.0\theta e^{(z)})dz|_{0.02\pi}^{1.64\pi}\\&\int_{0}^{1.06}(-76.32e^{(z)})dz=(-76.32e^{(z)})|_{0}^{1.06}=-143.97\end{align*}$

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Example Question #92 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(45e^{(z)})}{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.25\text{ and }1.12\\&\text{and length }1.01\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.368,0.930)\text{ and }\overrightarrow{u_2}=(0.685,-0.729)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1509.91

2516.51

503.3

-100.66

Correct answer:

503.3

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(45e^{(z)})}{(x^{2} + y^{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(45e^{(z)})}{r})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.930}{0.368})=0.38\pi;\theta_2=arctan(\frac{-0.729}{0.685})=1.74\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.01}\int_{0.38\pi}^{1.74\pi}\int_{0.25}^{1.12}(\frac{(45e^{(z)})}{r})drd\theta dz=(45e^{(z)}ln(r))d\theta dz|_{0.25}^{1.12}\\&\int_{0}^{1.01}\int_{0.38\pi}^{1.74\pi}(67.48e^{(z)})d\theta dz=(67.48\theta e^{(z)})dz|_{0.38\pi}^{1.74\pi}\\&\int_{0}^{1.01}(288.3e^{(z)})dz=(288.3e^{(z)})|_{0}^{1.01}=503.3\end{align*}$

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Example Question #91 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{22}{(\frac{7}{2})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}sin(4z)})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.64\\&\text{and length }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.588,0.809)\text{ and }\overrightarrow{u_2}=(0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.08

-0.12

0.25

-0.99

Correct answer:

0.25

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{22}{(\frac{7}{2})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}sin(4z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(22\cdot rsin(4z))}{(\frac{7}{2})^{(\frac{r^{2}}{2})}})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.809}{-0.588})=0.7\pi;\theta_2=arctan(\frac{-0.279}{0.960})=1.91\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{1.48}\int_{0.7\pi}^{1.91\pi}\int_{0}^{0.64}(\frac{(22\cdot rsin(4z))}{(\frac{7}{2})^{(\frac{r^{2}}{2})}})drd\theta dz=(-\frac{(22\cdot 2^{(\frac{r^{2}}{2})}sin(4z))}{(7^{(\frac{r^{2}}{2})}ln(\frac{7}{2}))})d\theta dz|_{0}^{0.64}\\&\int_{0}^{1.48}\int_{0.7\pi}^{1.91\pi}(3.974sin(4z))d\theta dz=(3.974\theta sin(4z))dz|_{0.7\pi}^{1.91\pi}\\&\int_{0}^{1.48}(15.11sin(4z))dz=(-3.777cos(4z))|_{0}^{1.48}=0.25\end{align*}$

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Example Question #96 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(22e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}e^{(-z)})}{3})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.86\\&\text{and length }1.89\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.536,0.844)\text{ and }\overrightarrow{u_2}=(-0.891,-0.454)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-16.14

-1.79

10.76

5.38

Correct answer:

5.38

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.844}{0.536})=0.32\pi;\theta_2=arctan(\frac{-0.454}{-0.891})=1.15\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.89}\int_{0.32\pi}^{1.15\pi}\int_{0}^{1.86}(\frac{(22\cdot re^{(-z)}e^{(-\frac{(3\cdot r^{2})}{2})})}{3})drd\theta dz=(-\frac{(22e^{(- z -\frac{ (3\cdot r^{2})}{2})})}{9})d\theta dz|_{0}^{1.86}\\&\int_{0}^{1.89}\int_{0.32\pi}^{1.15\pi}(2.431e^{(-z)})d\theta dz=(2.431\theta e^{(-z)})dz|_{0.32\pi}^{1.15\pi}\\&\int_{0}^{1.89}(6.338e^{(-z)})dz=(-6.338e^{(-z)})|_{0}^{1.89}=5.38\end{align*}$

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Example Question #97 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(4e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}e^{(2z)})}{13})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.2\\&\text{and length }0.69\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.509,0.861)\text{ and }\overrightarrow{u_2}=(0.707,-0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.94

-0.47

0.47

3.78

Correct answer:

-0.94

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.861}{0.509})=0.33\pi;\theta_2=arctan(\frac{-0.707}{0.707})=1.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.69}\int_{0.33\pi}^{1.75\pi}\int_{0}^{1.2}(-\frac{(4\cdot re^{(2z)}e^{(-\frac{(2\cdot r^{2})}{3})})}{13})drd\theta dz=(\frac{(3e^{(2z -\frac{ (2\cdot r^{2})}{3})})}{13})d\theta dz|_{0}^{1.2}\\&\int_{0}^{0.69}\int_{0.33\pi}^{1.75\pi}(-0.1424e^{(2z)})d\theta dz=(-0.1424\theta e^{(2z)})dz|_{0.33\pi}^{1.75\pi}\\&\int_{0}^{0.69}(-0.6353e^{(2z)})dz=(-0.3176e^{(2z)})|_{0}^{0.69}=-0.94\end{align*}$

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Example Question #98 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(e^{(2z)}sin(x^{2} + y^{2}))}{3})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.11\\&\text{and length }1.19\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.426,0.905)\text{ and }\overrightarrow{u_2}=(-0.613,-0.790)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.56

-1.11

2.23

0.28

Correct answer:

-1.11

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.905}{-0.426})=0.64\pi;\theta_2=arctan(\frac{-0.790}{-0.613})=1.29\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.19}\int_{0.64\pi}^{1.29\pi}\int_{0}^{1.11}(-\frac{(rsin(r^{2})e^{(2z)})}{3})drd\theta dz=(\frac{(cos(r^{2})e^{(2z)})}{6})d\theta dz|_{0}^{1.11}\\&\int_{0}^{1.19}\int_{0.64\pi}^{1.29\pi}(-0.1113e^{(2z)})d\theta dz=(-0.1113\theta e^{(2z)})dz|_{0.64\pi}^{1.29\pi}\\&\int_{0}^{1.19}(-0.2273e^{(2z)})dz=(-0.1136e^{(2z)})|_{0}^{1.19}=-1.11\end{align*}$

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Example Question #551 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2\cdot 2^{(\frac{z}{2})}e^{(- 2x^{2} - 2y^{2})})}{29})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.58\\&\text{and length }0.95\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.156,0.988)\text{ and }\overrightarrow{u_2}=(0.187,-0.982)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.03

-0.01

0.01

0.13

Correct answer:

-0.03

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.988}{0.156})=0.45\pi;\theta_2=arctan(\frac{-0.982}{0.187})=1.56\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{0.95}\int_{0.45\pi}^{1.56\pi}\int_{0}^{0.58}(-\frac{(2\cdot 2^{(\frac{z}{2})}\cdot re^{(-2\cdot r^{2})})}{29})drd\theta dz=(\frac{(2^{(\frac{z}{2})}e^{(-2\cdot r^{2})})}{58})d\theta dz|_{0}^{0.58}\\&\int_{0}^{0.95}\int_{0.45\pi}^{1.56\pi}(-0.008443\cdot 2^{(0.5z)})d\theta dz=(-0.008443\cdot 2^{(0.5z)}\theta)dz|_{0.45\pi}^{1.56\pi}\\&\int_{0}^{0.95}(-0.02944\cdot 2^{(0.5z)})dz=(-0.08496\cdot 2^{(0.5z)})|_{0}^{0.95}=-0.03\end{align*}$

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Example Question #100 : Triple Integration In Cylindrical Coordinates

2.99

Possible Answers:

-1.5

0.5

-2.99

Correct answer:

-1.5

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(21sin(x^{2} + y^{2}))}{(4\cdot 3^z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(21\cdot rsin(r^{2}))}{(4\cdot 3^z)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.13}\int_{0.47\pi}^{1.57\pi}\int_{0.25}^{1.03}(\frac{(21\cdot rsin(r^{2}))}{(4\cdot 3^z)})drd\theta dz=(-\frac{(21cos(r^{2}))}{(8\cdot 3^z)})d\theta dz|_{0.25}^{1.03}\\&\int_{0}^{1.13}\int_{0.47\pi}^{1.57\pi}(\frac{1.339}{3^{(z)}})d\theta dz=(\frac{(1.339\theta )}{3^{(z)}})dz|_{0.47\pi}^{1.57\pi}\\&\int_{0}^{1.13}(\frac{4.626}{3^{(z)}})dz=(-\frac{4.211}{3^{(z)}})|_{0}^{1.13}=2.99\end{align*}$

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Calculus 3 : Triple Integrals

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Example Question #91 : Triple Integrals

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