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Calculus 3 : Calculus 3

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Example Questions

Example Question #14 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(3e^{(- 2x^{2} - 2y^{2})} -\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.84\\&\text{and length }1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

8.07

-5.38

-0.54

2.69

Correct answer:

2.69

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}\int_{0}^{0.84}(3\cdot re^{(-2\cdot r^{2})} -\frac{ (rcos(\frac{r^{2}}{2}))}{(11\cdot 2^{(\frac{z}{2})})})drd\theta dz=(-\frac{ (3e^{(-2\cdot r^{2})})}{4}-\frac{ sin(\frac{r^{2}}{2})}{(11\cdot 2^{(\frac{z}{2})})})d\theta dz|_{0}^{0.84}\\&\int_{0}^{1.77}\int_{0.2\pi}^{1.09\pi}(0.5671 -\frac{ 0.03141}{2^{(0.5z)}})d\theta dz=(-1.0\theta\cdot (\frac{0.03141}{2^{(0.5z)}}- 0.5671))dz|_{0.2\pi}^{1.09\pi}\\&\int_{0}^{1.77}(1.586 -\frac{ 0.08783}{2^{(0.5z)}})dz=(1.586z +\frac{ 0.2534}{2^{(0.5z)}})|_{0}^{1.77}=2.69\end{align*}$

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Example Question #15 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(x^{2} + y^{2})})}{25}+\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-2z)})}{23})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.13\text{ and }0.99\\&\text{and length }1.7\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.536,0.844)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.83

-0.42

-4.16

0.14

Correct answer:

0.83

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4e^{(x^{2} + y^{2})})}{25}+\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})e^{(-2z)})}{23})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(r^{2})})}{25}+\frac{ (6\cdot re^{(-2z)}cos(\frac{(3\cdot r^{2})}{2}))}{23})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{0.536})=0.32\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.7}\int_{0.32\pi}^{1.32\pi}\int_{0.13}^{0.99}(\frac{(4\cdot re^{(r^{2})})}{25}+\frac{ (6\cdot re^{(-2z)}cos(\frac{(3\cdot r^{2})}{2}))}{23})drd\theta dz=(\frac{(2e^{(r^{2})})}{25}+\frac{ (2e^{(-2z)}sin(\frac{(3\cdot r^{2})}{2}))}{23})d\theta dz|_{0.13}^{0.99}\\&\int_{0}^{1.7}\int_{0.32\pi}^{1.32\pi}(0.08431e^{(-2z)} + 0.1318)d\theta dz=(\theta\cdot (0.08431e^{(-2z)} + 0.1318))dz|_{0.32\pi}^{1.32\pi}\\&\int_{0}^{1.7}(0.2649e^{(-2z)} + 0.4141)dz=(0.4141z - 0.1324e^{(-2z)})|_{0}^{1.7}=0.83\end{align*}$

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Example Question #16 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(19zsin(2x^{2} + 2y^{2}))}{2}-\frac{ sin(x^{2} + y^{2})}{47})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.27\text{ and }1.58\\&\text{and length }1.44\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.982,0.187)\text{ and }\overrightarrow{u_2}=(-0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.81

-11.61

-0.97

1.45

Correct answer:

5.81

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(19zsin(2x^{2} + 2y^{2}))}{2}-\frac{ sin(x^{2} + y^{2})}{47})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot rzsin(2\cdot r^{2}))}{2}-\frac{ (rsin(r^{2}))}{47})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.187}{0.982})=0.06\pi;\theta_2=arctan(\frac{-0.397}{-0.918})=1.13\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0}^{1.44}\int_{0.06\pi}^{1.13\pi}\int_{0.27}^{1.58}(\frac{(19\cdot rzsin(2\cdot r^{2}))}{2}-\frac{ (rsin(r^{2}))}{47})drd\theta dz=(-\frac{(cos(r^{2})\cdot (893zcos(r^{2}) - 2))}{188})d\theta dz|_{0.27}^{1.58}\\&\int_{0}^{1.44}\int_{0.06\pi}^{1.13\pi}(1.693z - 0.01911)d\theta dz=(\theta\cdot (1.693z - 0.01911))dz|_{0.06\pi}^{1.13\pi}\\&\int_{0}^{1.44}(5.689z - 0.06424)dz=(0.993\cdot (1.693z - 0.01911)^{2})|_{0}^{1.44}=5.81\end{align*}$

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Example Question #17 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{4}- 17cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.38\text{ and }1.63\\&\text{and length }0.76\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.588,0.809)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-31.3

7.82

-15.65

62.59

Correct answer:

-31.3

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (3e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{4}- 17cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})e^{(-z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot re^{(\frac{r^{2}}{2})})}{4}- 17\cdot re^{(-z)}cos(\frac{r^{2}}{2}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.809}{-0.588})=0.7\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.76}\int_{0.7\pi}^{1.73\pi}\int_{0.38}^{1.63}(-\frac{ (3\cdot re^{(\frac{r^{2}}{2})})}{4}- 17\cdot re^{(-z)}cos(\frac{r^{2}}{2}))drd\theta dz=(-\frac{ (3e^{(\frac{r^{2}}{2})})}{4}- 17e^{(-z)}sin(\frac{r^{2}}{2}))d\theta dz|_{0.38}^{1.63}\\&\int_{0}^{0.76}\int_{0.7\pi}^{1.73\pi}(- 15.28e^{(-z)} - 2.025)d\theta dz=(-\theta\cdot (15.28e^{(-z)} + 2.025))dz|_{0.7\pi}^{1.73\pi}\\&\int_{0}^{0.76}(- 49.43e^{(-z)} - 6.553)dz=(49.43e^{(-z)} - 6.553z)|_{0}^{0.76}=-31.3\end{align*}$

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Example Question #18 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(2z)})}{2}-\frac{ (\frac{2}{(\frac{1}{2})^{(x^{2} + y^{2})}})}{11})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.73\\&\text{and length }1.01\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.861,0.509)\text{ and }\overrightarrow{u_2}=(0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-22.09

-198.82

33.14

66.27

Correct answer:

66.27

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(2z)})}{2}-\frac{ (\frac{2}{(\frac{1}{2})^{(x^{2} + y^{2})}})}{11})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}\cdot re^{(2z)})}{2}-\frac{ (2\cdot r)}{(11\cdot (\frac{1}{2})^{(r^{2})})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.509}{0.861})=0.17\pi;\theta_2=arctan(\frac{-0.918}{0.397})=1.63\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.01}\int_{0.17\pi}^{1.63\pi}\int_{0}^{0.73}(\frac{(39\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}\cdot re^{(2z)})}{2}-\frac{ (2\cdot r)}{(11\cdot (\frac{1}{2})^{(r^{2})})})drd\theta dz=(\frac{(13\cdot (\frac{2}{3})^{(\frac{(3\cdot r^{2})}{2})}e^{(2z)})}{(2ln(\frac{2}{3}))}-\frac{ 2^{(r^{2})}}{(11ln(2))})d\theta dz|_{0}^{0.73}\\&\int_{0}^{1.01}\int_{0.17\pi}^{1.63\pi}(4.438e^{(2z)} - 0.0586)d\theta dz=(1.0\theta\cdot (4.438e^{(2z)} - 0.0586))dz|_{0.17\pi}^{1.63\pi}\\&\int_{0}^{1.01}(20.36e^{(2z)} - 0.2688)dz=(10.18e^{(2z)} - 0.2688z)|_{0}^{1.01}=66.27\end{align*}$

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Example Question #19 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(22e^{(-2z)})}{(2x^{2} + 2y^{2})}-\frac{ cos(x^{2} + y^{2})}{11})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin with radii }\\&0.24\text{ and }1.58\\&\text{and length }1.12\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.125,0.992)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-69.04

8.63

-6.9

34.52

Correct answer:

34.52

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(22e^{(-2z)})}{(2x^{2} + 2y^{2})}-\frac{ cos(x^{2} + y^{2})}{11})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(11e^{(-2z)})}{r}-\frac{ (rcos(r^{2}))}{11})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.992}{0.125})=0.46\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.12}\int_{0.46\pi}^{1.65\pi}\int_{0.24}^{1.58}(\frac{(11e^{(-2z)})}{r}-\frac{ (rcos(r^{2}))}{11})drd\theta dz=(11e^{(-2z)}ln(r) -\frac{ sin(r^{2})}{22})d\theta dz|_{0.24}^{1.58}\\&\int_{0}^{1.12}\int_{0.46\pi}^{1.65\pi}(20.73e^{(-2z)} - 0.02472)d\theta dz=(\theta\cdot (20.73e^{(-2z)} - 0.02472))dz|_{0.46\pi}^{1.65\pi}\\&\int_{0}^{1.12}(77.5e^{(-2z)} - 0.09241)dz=(- 0.09241z - 38.75e^{(-2z)})|_{0}^{1.12}=34.52\end{align*}$

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Example Question #20 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(e^{(z)}sin(x^{2} + y^{2}))}{10}-\frac{ (9cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{2})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.11\\&\text{and length }1.17\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.951,0.309)\text{ and }\overrightarrow{u_2}=(-0.707,0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-5.75

1.44

-17.26

23.01

Correct answer:

-5.75

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(e^{(z)}sin(x^{2} + y^{2}))}{10}-\frac{ (9cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{2})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(r^{2})e^{(z)})}{10}-\frac{ (9\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{2})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.309}{0.951})=0.1\pi;\theta_2=arctan(\frac{0.707}{-0.707})=0.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.17}\int_{0.1\pi}^{0.75\pi}\int_{0}^{1.11}(\frac{(rsin(r^{2})e^{(z)})}{10}-\frac{ (9\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{2})drd\theta dz=(-\frac{ (27sin(\frac{(2\cdot r^{2})}{3}))}{8}-\frac{ (cos(r^{2})e^{(z)})}{20})d\theta dz|_{0}^{1.11}\\&\int_{0}^{1.17}\int_{0.1\pi}^{0.75\pi}(0.03339e^{(z)} - 2.471)d\theta dz=(1.0\theta\cdot (0.03339e^{(z)} - 2.471))dz|_{0.1\pi}^{0.75\pi}\\&\int_{0}^{1.17}(0.06818e^{(z)} - 5.046)dz=(0.06818e^{(z)} - 5.046z)|_{0}^{1.17}=-5.75\end{align*}$

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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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