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Calculus 3 : Multiple Integration

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #493 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{1}{(2\cdot (\frac{x^{2}}{2}+\frac{ y^{2}}{2})\cdot (z + 1)^{3})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.35\text{ and }1.23\\&\text{and length }1.71\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.339,0.941)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.19

2.39

-7.16

-0.2

Correct answer:

1.19

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.941}{0.339})=0.39\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0}^{1.71}\int_{0.39\pi}^{1.09\pi}\int_{0.35}^{1.23}(\frac{1}{(r\cdot (z + 1)^{3})})drd\theta dz=(\frac{ln(r)}{(z + 1)^{3}})d\theta dz|_{0.35}^{1.23}\\&\int_{0}^{1.71}\int_{0.39\pi}^{1.09\pi}(\frac{1.257}{(z + 1)^{3}})d\theta dz=(\frac{(1.257\theta )}{(z + 1)^{3}})dz|_{0.39\pi}^{1.09\pi}\\&\int_{0}^{1.71}(\frac{2.764}{(z + 1)^{3}})dz=(-\frac{1.382}{(z + 1)^{2}})|_{0}^{1.71}=1.19\end{align*}$

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Example Question #494 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{e^{(- 2x^{2} - 2y^{2})}}{(8\cdot 2^z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.44\text{ and }1.89\\&\text{and length }1.37\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.125,0.992)\text{ and }\overrightarrow{u_2}=(0.309,-0.951)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.07

-0.34

0.02

0.13

Correct answer:

-0.07

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.992}{0.125})=0.46\pi;\theta_2=arctan(\frac{-0.951}{0.309})=1.6\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.37}\int_{0.46\pi}^{1.6\pi}\int_{0.44}^{1.89}(-\frac{(re^{(-2\cdot r^{2})})}{(8\cdot 2^z)})drd\theta dz=(\frac{e^{(-2\cdot r^{2})}}{(32\cdot 2^z)})d\theta dz|_{0.44}^{1.89}\\&\int_{0}^{1.37}\int_{0.46\pi}^{1.6\pi}(-\frac{0.02119}{2^{(z)}})d\theta dz=(-\frac{(0.02119\theta )}{2^{(z)}})dz|_{0.46\pi}^{1.6\pi}\\&\int_{0}^{1.37}(-\frac{0.0759}{2^{(z)}})dz=(\frac{0.1095}{2^{(z)}})|_{0}^{1.37}=-0.07\end{align*}$

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Example Question #495 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(5e^{(-2z)})}{(42\cdot (\frac{x^{2}}{2}+\frac{ y^{2}}{2}))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.19\text{ and }1.66\\&\text{and length }1.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.960,0.279)\text{ and }\overrightarrow{u_2}=(-0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.73

3.67

0.24

-0.18

Correct answer:

-0.73

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(5e^{(-2z)})}{(42\cdot (\frac{x^{2}}{2}+\frac{ y^{2}}{2}))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5e^{(-2z)})}{(21\cdot r)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{0.960})=0.09\pi;\theta_2=arctan(\frac{-0.125}{-0.992})=1.04\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.54}\int_{0.09\pi}^{1.04\pi}\int_{0.19}^{1.66}(-\frac{(5e^{(-2z)})}{(21\cdot r)})drd\theta dz=(-\frac{(5e^{(-2z)}ln(r))}{21})d\theta dz|_{0.19}^{1.66}\\&\int_{0}^{1.54}\int_{0.09\pi}^{1.04\pi}(-0.5161e^{(-2z)})d\theta dz=(-0.5161\theta e^{(-2z)})dz|_{0.09\pi}^{1.04\pi}\\&\int_{0}^{1.54}(-1.54e^{(-2z)})dz=(0.7701e^{(-2z)})|_{0}^{1.54}=-0.73\end{align*}$

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Example Question #501 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{11}{(2\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})\cdot (z + 1)^{3})})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.12\text{ and }1.67\\&\text{and length }1.4\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.031)\text{ and }\overrightarrow{u_2}=(0.339,-0.941)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.3

-46.62

7.77

-2.59

Correct answer:

7.77

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.031}{-1.000})=0.99\pi;\theta_2=arctan(\frac{-0.941}{0.339})=1.61\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0}^{1.4}\int_{0.99\pi}^{1.61\pi}\int_{0.12}^{1.67}(\frac{11}{(3\cdot r\cdot (z + 1)^{3})})drd\theta dz=(\frac{(11ln(r))}{(3\cdot (z + 1)^{3})})d\theta dz|_{0.12}^{1.67}\\&\int_{0}^{1.4}\int_{0.99\pi}^{1.61\pi}(\frac{9.655}{(z + 1)^{3}})d\theta dz=(\frac{(9.655\theta )}{(z + 1)^{3}})dz|_{0.99\pi}^{1.61\pi}\\&\int_{0}^{1.4}(\frac{18.81}{(z + 1)^{3}})dz=(-\frac{9.403}{(z + 1)^{2}})|_{0}^{1.4}=7.77\end{align*}$

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Example Question #502 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(5\cdot 2^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{(19\cdot 3^z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.24\text{ and }1.53\\&\text{and length }1.7\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.951,0.309)\text{ and }\overrightarrow{u_2}=(0.339,-0.941)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.25

13.48

1.12

-8.99

Correct answer:

-2.25

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(5\cdot 2^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{(19\cdot 3^z)})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(5\cdot 2^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{(19\cdot 3^z)})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.309}{-0.951})=0.9\pi;\theta_2=arctan(\frac{-0.941}{0.339})=1.61\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.7}\int_{0.9\pi}^{1.61\pi}\int_{0.24}^{1.53}(-\frac{(5\cdot 2^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{(19\cdot 3^z)})drd\theta dz=(-\frac{(5\cdot (\frac{1}{3})^z\cdot 2^{(\frac{(3\cdot r^{2})}{2})})}{(57ln(2))})d\theta dz|_{0.24}^{1.53}\\&\int_{0}^{1.7}\int_{0.9\pi}^{1.61\pi}(-\frac{1.309}{3^{(z)}})d\theta dz=(-\frac{(1.309\theta )}{3^{(z)}})dz|_{0.9\pi}^{1.61\pi}\\&\int_{0}^{1.7}(-\frac{2.919}{3^{(z)}})dz=(\frac{2.657}{3^{(z)}})|_{0}^{1.7}=-2.25\end{align*}$

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Example Question #503 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(11sin(2x^{2} + 2y^{2})sin(4z))}{3})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.24\text{ and }0.79\\&\text{and length }0.82\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.876,0.482)\text{ and }\overrightarrow{u_2}=(-0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.28

2.27

-5.67

-1.13

Correct answer:

-1.13

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.482}{0.876})=0.16\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.82}\int_{0.16\pi}^{1.33\pi}\int_{0.24}^{0.79}(-\frac{(11\cdot rsin(4z)sin(2\cdot r^{2}))}{3})drd\theta dz=(\frac{(11cos(r^{2})^{2}sin(4z))}{6})d\theta dz|_{0.24}^{0.79}\\&\int_{0}^{0.82}\int_{0.16\pi}^{1.33\pi}(-0.62sin(4z))d\theta dz=(-0.62\theta sin(4z))dz|_{0.16\pi}^{1.33\pi}\\&\int_{0}^{0.82}(-2.279sin(4z))dz=(0.5697cos(4z))|_{0}^{0.82}=-1.13\end{align*}$

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Example Question #504 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{6}{(\frac{2}{5})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}e^{(-2z)}})}{5})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.33\text{ and }1.83\\&\text{and length }0.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.339,0.941)\text{ and }\overrightarrow{u_2}=(-0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-55.53

-111.06

166.58

9.25

Correct answer:

-55.53

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.941}{0.339})=0.39\pi;\theta_2=arctan(\frac{-0.918}{-0.397})=1.37\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{0.91}\int_{0.39\pi}^{1.37\pi}\int_{0.33}^{1.83}(-\frac{(6\cdot re^{(-2z)})}{(5\cdot (\frac{2}{5})^{(\frac{(3\cdot r^{2})}{2})})})drd\theta dz=(\frac{(2\cdot 5^{(\frac{(3\cdot r^{2})}{2})}e^{(-2z)})}{(5\cdot 2^{(\frac{(3\cdot r^{2})}{2})}ln(\frac{2}{5}))})d\theta dz|_{0.33}^{1.83}\\&\int_{0}^{0.91}\int_{0.39\pi}^{1.37\pi}(-43.05e^{(-2z)})d\theta dz=(-43.05\theta e^{(-2z)})dz|_{0.39\pi}^{1.37\pi}\\&\int_{0}^{0.91}(-132.5e^{(-2z)})dz=(66.26e^{(-2z)})|_{0}^{0.91}=-55.53\end{align*}$

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Example Question #41 : Triple Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})e^{(z)})}{16})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.16\text{ and }0.97\\&\text{and length }1.56\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.9

0.08

2.71

-0.45

Correct answer:

-0.45

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.56}\int_{0.34\pi}^{1.77\pi}\int_{0.16}^{0.97}(-\frac{(3\cdot re^{(z)}sin(\frac{(2\cdot r^{2})}{3}))}{16})drd\theta dz=(\frac{(9e^{(z)}cos(\frac{(2\cdot r^{2})}{3}))}{64})d\theta dz|_{0.16}^{0.97}\\&\int_{0}^{1.56}\int_{0.34\pi}^{1.77\pi}(-0.02675e^{(z)})d\theta dz=(-0.02675\theta e^{(z)})dz|_{0.34\pi}^{1.77\pi}\\&\int_{0}^{1.56}(-0.1202e^{(z)})dz=(-0.1202e^{(z)})|_{0}^{1.56}=-0.45\end{align*}$

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Example Question #506 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(3\cdot 3^zcos(x^{2} + y^{2}))}{2})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.27\\&\text{and length }1.11\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.92

9.2

-4.6

2.3

Correct answer:

-4.6

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.951}{-0.309})=0.6\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.11}\int_{0.6\pi}^{1.5\pi}\int_{0}^{1.27}(-\frac{(3\cdot 3^z\cdot rcos(r^{2}))}{2})drd\theta dz=(-\frac{(3\cdot 3^zsin(r^{2}))}{4})d\theta dz|_{0}^{1.27}\\&\int_{0}^{1.11}\int_{0.6\pi}^{1.5\pi}(-0.7493\cdot 3^z)d\theta dz=(-0.7493\cdot 3^z\theta)dz|_{0.6\pi}^{1.5\pi}\\&\int_{0}^{1.11}(-2.119\cdot 3^z)dz=(-1.929\cdot 3^z)|_{0}^{1.11}=-4.6\end{align*}$

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Example Question #507 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.87\\&\text{and length }1.45\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.982,-0.187)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-7.15

42.92

-42.92

8.58

Correct answer:

-42.92

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(49e^{(2z)}cos(x^{2} + y^{2}))}{6})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.187}{0.982})=1.94\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[e^{az}]=\frac{e^{az}}{a}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}\int_{0}^{1.87}(\frac{(49\cdot rcos(r^{2})e^{(2z)})}{6})drd\theta dz=(\frac{(49sin(r^{2})e^{(2z)})}{12})d\theta dz|_{0}^{1.87}\\&\int_{0}^{1.45}\int_{0.82\pi}^{1.94\pi}(-1.421e^{(2z)})d\theta dz=(-1.421\theta e^{(2z)})dz|_{0.82\pi}^{1.94\pi}\\&\int_{0}^{1.45}(-4.998e^{(2z)})dz=(-2.499e^{(2z)})|_{0}^{1.45}=-42.92\end{align*}$

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