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Calculus 3 : Triple Integration in Cylindrical Coordinates

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Example Questions

Example Question #21 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{14}-\frac{ (cos(3z)sin(x^{2} + y^{2}))}{2})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.19\text{ and }1.27\\&\text{and length }1.53\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.827,0.562)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.29

-0.04

0.26

-0.52

Correct answer:

0.26

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{14}-\frac{ (cos(3z)sin(x^{2} + y^{2}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(\frac{(3\cdot r^{2})}{2}))}{14}-\frac{ (rcos(3z)sin(r^{2}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.562}{-0.827})=0.81\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.53}\int_{0.26\pi}^{0.81\pi}\int_{0.19}^{1.27}(\frac{(rsin(\frac{(3\cdot r^{2})}{2}))}{14}-\frac{ (rcos(3z)sin(r^{2}))}{2})drd\theta dz=(\frac{(cos(r^{2})cos(3z))}{4}-\frac{ cos(\frac{(3\cdot r^{2})}{2})}{42})d\theta dz|_{0.19}^{1.27}\\&\int_{0}^{1.53}\int_{0.26\pi}^{0.81\pi}(0.04164 - 0.2604cos(3z))d\theta dz=(-1.0\theta\cdot (0.2604cos(3z) - 0.04164))dz|_{0.26\pi}^{0.81\pi}\\&\int_{0}^{1.53}(0.07195 - 0.4499cos(3z))dz=(0.07195z - 0.2999cos(1.5z)sin(1.5z))|_{0}^{1.53}=0.26\end{align*}$

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Example Question #22 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5cos(x^{2} + y^{2}))}{38}-\frac{ (45e^{(- 2x^{2} - 2y^{2})}e^{(2z)})}{4})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.51\\&\text{and length }1.16\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-47.12

-23.56

23.56

141.36

Correct answer:

-47.12

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5cos(x^{2} + y^{2}))}{38}-\frac{ (45e^{(- 2x^{2} - 2y^{2})}e^{(2z)})}{4})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot rcos(r^{2}))}{38}-\frac{ (45\cdot re^{(2z)}e^{(-2\cdot r^{2})})}{4})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.16}\int_{0.79\pi}^{1.97\pi}\int_{0}^{1.51}(\frac{(5\cdot rcos(r^{2}))}{38}-\frac{ (45\cdot re^{(2z)}e^{(-2\cdot r^{2})})}{4})drd\theta dz=(\frac{(5sin(r^{2}))}{76}+\frac{ (45e^{(2z - 2\cdot r^{2})})}{16})d\theta dz|_{0}^{1.51}\\&\int_{0}^{1.16}\int_{0.79\pi}^{1.97\pi}(0.04992 - 2.783e^{(2z)})d\theta dz=(-\theta\cdot (2.783e^{(2z)} - 0.04992))dz|_{0.79\pi}^{1.97\pi}\\&\int_{0}^{1.16}(0.1851 - 10.32e^{(2z)})dz=(0.1851z - 493.2e^{(2z - 4.56)})|_{0}^{1.16}=-47.12\end{align*}$

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Example Question #23 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (24e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{5}- 6\cdot 3^{(\frac{z}{2})}e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.58\\&\text{and length }1.78\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.827,0.562)\text{ and }\overrightarrow{u_2}=(-0.482,-0.876)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-3.14

-15.68

2.61

31.36

Correct answer:

-15.68

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (24e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{5}- 6\cdot 3^{(\frac{z}{2})}e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (24\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{5}- 6\cdot 3^{(\frac{z}{2})}\cdot re^{(-\frac{(2\cdot r^{2})}{3})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.562}{0.827})=0.19\pi;\theta_2=arctan(\frac{-0.876}{-0.482})=1.34\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.78}\int_{0.19\pi}^{1.34\pi}\int_{0}^{0.58}(-\frac{ (24\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{5}- 6\cdot 3^{(\frac{z}{2})}\cdot re^{(-\frac{(2\cdot r^{2})}{3})})drd\theta dz=(\frac{(9\cdot 3^{(\frac{z}{2})}e^{(-\frac{(2\cdot r^{2})}{3})})}{2}-\frac{ (18e^{(\frac{(2\cdot r^{2})}{3})})}{5})d\theta dz|_{0}^{0.58}\\&\int_{0}^{1.78}\int_{0.19\pi}^{1.34\pi}(- 0.904\cdot 3^{(0.5z)} - 0.9051)d\theta dz=(-0.1808\theta\cdot (5\cdot 3^{(0.5z)} + 5.006))dz|_{0.19\pi}^{1.34\pi}\\&\int_{0}^{1.78}(- 3.266\cdot 3^{(0.5z)} - 3.27)dz=(- 3.27z - 5.946\cdot 3^{(0.5z)})|_{0}^{1.78}=-15.68\end{align*}$

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Example Question #24 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (7e^{(-2z)})}{(6\cdot (x^{2} + y^{2}))}-\frac{ 7}{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.37\text{ and }1.7\\&\text{and length }1.84\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.509,0.861)\text{ and }\overrightarrow{u_2}=(-0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

73.45

-36.72

9.18

-73.45

Correct answer:

-36.72

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (7e^{(-2z)})}{(6\cdot (x^{2} + y^{2}))}-\frac{ 7}{(x^{2} + y^{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (7e^{(-2z)})}{(6\cdot r)}-\frac{ 7}{r})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.861}{-0.509})=0.67\pi;\theta_2=arctan(\frac{-0.685}{-0.729})=1.24\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.84}\int_{0.67\pi}^{1.24\pi}\int_{0.37}^{1.7}(-\frac{ (7e^{(-2z)})}{(6\cdot r)}-\frac{ 7}{r})drd\theta dz=(-ln(r)\cdot (\frac{(7e^{(-2z)})}{6}+ 7))d\theta dz|_{0.37}^{1.7}\\&\int_{0}^{1.84}\int_{0.67\pi}^{1.24\pi}(- 1.779e^{(-2z)} - 10.67)d\theta dz=(-1.779\theta\cdot (e^{(-2z)} + 6))dz|_{0.67\pi}^{1.24\pi}\\&\int_{0}^{1.84}(- 3.186e^{(-2z)} - 19.11)dz=(1.593e^{(-2z)} - 19.11z)|_{0}^{1.84}=-36.72\end{align*}$

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Example Question #25 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(11sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}) +\frac{ (43e^{(- x^{2} - y^{2})})}{(6\cdot 2^{(\frac{z}{2})})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.45\\&\text{and length }0.92\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.905,0.426)\text{ and }\overrightarrow{u_2}=(-0.941,0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

17.84

-8.92

-71.38

8.92

Correct answer:

17.84

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(11sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}) +\frac{ (43e^{(- x^{2} - y^{2})})}{(6\cdot 2^{(\frac{z}{2})})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(11\cdot rsin(\frac{r^{2}}{2}) +\frac{ (43\cdot re^{(-r^{2})})}{(6\cdot 2^{(\frac{z}{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.426}{0.905})=0.14\pi;\theta_2=arctan(\frac{0.339}{-0.941})=0.89\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{0.92}\int_{0.14\pi}^{0.89\pi}\int_{0}^{1.45}(11\cdot rsin(\frac{r^{2}}{2}) +\frac{ (43\cdot re^{(-r^{2})})}{(6\cdot 2^{(\frac{z}{2})})})drd\theta dz=(- 11cos(\frac{r^{2}}{2}) -\frac{ (43e^{(-r^{2})})}{(12\cdot 2^{(\frac{z}{2})})})d\theta dz|_{0}^{1.45}\\&\int_{0}^{0.92}\int_{0.14\pi}^{0.89\pi}(\frac{3.146}{2^{(0.5z)}}+ 5.539)d\theta dz=(1.0\theta\cdot (\frac{3.146}{2^{(0.5z)}}+ 5.539))dz|_{0.14\pi}^{0.89\pi}\\&\int_{0}^{0.92}(\frac{7.412}{2^{(0.5z)}}+ 13.05)dz=(13.05z -\frac{ 21.39}{2^{(0.5z)}})|_{0}^{0.92}=17.84\end{align*}$

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Example Question #26 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{cos(2x^{2} + 2y^{2})}{16}- 14cos(z + 2)e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.7\\&\text{and length }1.4\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.905,0.426)\text{ and }\overrightarrow{u_2}=(-1.000,0.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-19.25

1.92

-1.92

9.62

Correct answer:

9.62

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{cos(2x^{2} + 2y^{2})}{16}- 14cos(z + 2)e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rcos(2\cdot r^{2}))}{16}- 14\cdot rcos(z + 2)e^{(-\frac{r^{2}}{2})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.426}{0.905})=0.14\pi;\theta_2=arctan(\frac{0.000}{-1.000})=1\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.4}\int_{0.14\pi}^{1\pi}\int_{0}^{0.7}(\frac{(rcos(2\cdot r^{2}))}{16}- 14\cdot rcos(z + 2)e^{(-\frac{r^{2}}{2})})drd\theta dz=(\frac{sin(2\cdot r^{2})}{64}+ 14cos(z + 2)e^{(-\frac{r^{2}}{2})})d\theta dz|_{0}^{0.7}\\&\int_{0}^{1.4}\int_{0.14\pi}^{1\pi}(0.01298 - 3.042cos(z + 2))d\theta dz=(-\theta\cdot (3.042cos(z + 2) - 0.01298))dz|_{0.14\pi}^{1\pi}\\&\int_{0}^{1.4}(0.03506 - 8.219cos(z + 2))dz=(0.03506z - 8.219sin(z + 2))|_{0}^{1.4}=9.62\end{align*}$

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Example Question #27 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(17e^{(2z)}e^{(x^{2} + y^{2})})}{6})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.84\\&\text{and length }1.79\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.771,0.637)\text{ and }\overrightarrow{u_2}=(-0.187,-0.982)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-4384.68

2923.12

730.78

-1461.56

Correct answer:

-1461.56

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(17e^{(2z)}e^{(x^{2} + y^{2})})}{6})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(17\cdot re^{(r^{2})}e^{(2z)})}{6})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.637}{-0.771})=0.78\pi;\theta_2=arctan(\frac{-0.982}{-0.187})=1.44\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.79}\int_{0.78\pi}^{1.44\pi}\int_{0}^{1.84}(-\frac{(17\cdot re^{(r^{2})}e^{(2z)})}{6})drd\theta dz=(-\frac{(17e^{(2z + r^{2})})}{12})d\theta dz|_{0}^{1.84}\\&\int_{0}^{1.79}\int_{0.78\pi}^{1.44\pi}(-40.43e^{(2z)})d\theta dz=(-40.43\theta e^{(2z)})dz|_{0.78\pi}^{1.44\pi}\\&\int_{0}^{1.79}(-83.82e^{(2z)})dz=(-41.91e^{(2z)})|_{0}^{1.79}=-1461.56\end{align*}$

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Example Question #28 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(37sin(2x^{2} + 2y^{2})e^{(z)})}{2})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.33\\&\text{and length }0.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-45.61

91.22

-182.44

9.12

Correct answer:

-45.61

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(37sin(2x^{2} + 2y^{2})e^{(z)})}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(37\cdot re^{(z)}sin(2\cdot r^{2}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{0.309})=0.4\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.91}\int_{0.4\pi}^{1.5\pi}\int_{0}^{1.33}(-\frac{(37\cdot re^{(z)}sin(2\cdot r^{2}))}{2})drd\theta dz=(\frac{(37cos(r^{2})^{2}e^{(z)})}{4})d\theta dz|_{0}^{1.33}\\&\int_{0}^{0.91}\int_{0.4\pi}^{1.5\pi}(-8.892e^{(z)})d\theta dz=(-8.892\theta e^{(z)})dz|_{0.4\pi}^{1.5\pi}\\&\int_{0}^{0.91}(-30.73e^{(z)})dz=(-30.73e^{(z)})|_{0}^{0.91}=-45.61\end{align*}$

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Example Question #29 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{34})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.33\text{ and }1.71\\&\text{and length }1.47\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.000,1.000)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.03

0.16

0.03

Correct answer:

0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{34})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rcos(r^{2}))}{34})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.000})=0.5\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0}^{1.47}\int_{0.5\pi}^{1.76\pi}\int_{0.33}^{1.71}(\frac{(3\cdot rcos(r^{2}))}{34})drd\theta dz=(\frac{(3sin(r^{2}))}{68})d\theta dz|_{0.33}^{1.71}\\&\int_{0}^{1.47}\int_{0.5\pi}^{1.76\pi}(0.004725)d\theta dz=(0.004725\theta)dz|_{0.5\pi}^{1.76\pi}\\&\int_{0}^{1.47}(0.0187)dz=(0.0187z)|_{0}^{1.47}=0.03\end{align*}$

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Example Question #30 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{1}{(\frac{1}{2})^{(x^{2} + y^{2})}cos(z + 2)})}{6})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.12\text{ and }1.24\\&\text{and length }1.37\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.509,0.861)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

2.92

-0.97

0.97

-0.16

Correct answer:

0.97

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.861}{-0.509})=0.67\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.37}\int_{0.67\pi}^{1.87\pi}\int_{0.12}^{1.24}(-\frac{(rcos(z + 2))}{(6\cdot (\frac{1}{2})^{(r^{2})})})drd\theta dz=(-\frac{(2^{(r^{2})}cos(z + 2))}{(12ln(2))})d\theta dz|_{0.12}^{1.24}\\&\int_{0}^{1.37}\int_{0.67\pi}^{1.87\pi}(-0.2276cos(z + 2))d\theta dz=(-0.2276\theta cos(z + 2))dz|_{0.67\pi}^{1.87\pi}\\&\int_{0}^{1.37}(-0.858cos(z + 2))dz=(-0.858sin(z + 2))|_{0}^{1.37}=0.97\end{align*}$

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Calculus 3 : Triple Integration in Cylindrical Coordinates

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