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Calculus 3 : Calculus 3

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

Example Questions

Example Question #561 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{((\frac{2}{5})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 3^z)}{9})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.1\text{ and }1.89\\&\text{and length }1.12\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.309,0.951)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.14

0.71

-4.23

-0.18

Correct answer:

0.71

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{((\frac{2}{5})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}\cdot 3^z)}{9})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{((\frac{2}{5})^{(\frac{(2\cdot r^{2})}{3})}\cdot 3^z\cdot r)}{9})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{-0.309})=0.6\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.12}\int_{0.6\pi}^{1.87\pi}\int_{0.1}^{1.89}(\frac{((\frac{2}{5})^{(\frac{(2\cdot r^{2})}{3})}\cdot 3^z\cdot r)}{9})drd\theta dz=(\frac{(2^{(\frac{(2\cdot r^{2})}{3})}\cdot 3^z)}{(12\cdot 5^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{5}))})d\theta dz|_{0.1}^{1.89}\\&\int_{0}^{1.12}\int_{0.6\pi}^{1.87\pi}(0.08013\cdot 3^z)d\theta dz=(0.08013\cdot 3^z\theta)dz|_{0.6\pi}^{1.87\pi}\\&\int_{0}^{1.12}(0.3197\cdot 3^z)dz=(0.291\cdot 3^z)|_{0}^{1.12}=0.71\end{align*}$

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Example Question #562 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}}{(10\cdot (z + 1))})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.14\text{ and }1.1\\&\text{and length }0.55\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.960,0.279)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.008

-0.16

0.05

0.11

Correct answer:

0.05

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}}{(10\cdot (z + 1))})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{r^{2}}{2})})}{(10\cdot (z + 1))})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{-0.960})=0.91\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{0.55}\int_{0.91\pi}^{1.77\pi}\int_{0.14}^{1.1}(\frac{(re^{(-\frac{r^{2}}{2})})}{(10\cdot (z + 1))})drd\theta dz=(-\frac{e^{(-\frac{r^{2}}{2})}}{(10\cdot (z + 1))})d\theta dz|_{0.14}^{1.1}\\&\int_{0}^{0.55}\int_{0.91\pi}^{1.77\pi}(\frac{0.04442}{(z + 1)})d\theta dz=(\frac{(0.04442\theta )}{(z + 1)})dz|_{0.91\pi}^{1.77\pi}\\&\int_{0}^{0.55}(\frac{0.12}{(z + 1)})dz=(0.12ln(z + 1))|_{0}^{0.55}=0.05\end{align*}$

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Example Question #102 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})cos(z + 2))}{18})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.16\text{ and }1.56\\&\text{and length }1.43\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.03

-0.13

0.006

-0.008

Correct answer:

0.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})cos(z + 2))}{18})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rcos(z + 2)cos(\frac{(3\cdot r^{2})}{2}))}{18})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1.43}\int_{0.77\pi}^{1.65\pi}\int_{0.16}^{1.56}(\frac{(rcos(z + 2)cos(\frac{(3\cdot r^{2})}{2}))}{18})drd\theta dz=(\frac{(cos(z + 2)sin(\frac{(3\cdot r^{2})}{2}))}{54})d\theta dz|_{0.16}^{1.56}\\&\int_{0}^{1.43}\int_{0.77\pi}^{1.65\pi}(-0.009732cos(z + 2))d\theta dz=(-0.009732\theta cos(z + 2))dz|_{0.77\pi}^{1.65\pi}\\&\int_{0}^{1.43}(-0.02691cos(z + 2))dz=(-0.02691sin(z + 2))|_{0}^{1.43}=0.03\end{align*}$

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Example Question #106 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(\frac{17}{4^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}cos(3z)})}{2})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.58\\&\text{and length }1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.827,0.562)\text{ and }\overrightarrow{u_2}=(-0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.05

-0.21

1.26

-1.26

Correct answer:

-0.21

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.562}{0.827})=0.19\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[cos(az)]=\frac{sin(az)}{a}\\&\int_{0}^{1}\int_{0.19\pi}^{1.35\pi}\int_{0}^{0.58}(-\frac{(17\cdot rcos(3z))}{(2\cdot 4^{(\frac{(2\cdot r^{2})}{3})})})drd\theta dz=(\frac{(51cos(3z))}{(8\cdot 2^{(\frac{(4\cdot r^{2})}{3})}ln(4))})d\theta dz|_{0}^{0.58}\\&\int_{0}^{1}\int_{0.19\pi}^{1.35\pi}(-1.229cos(3z))d\theta dz=(-1.229\theta cos(3z))dz|_{0.19\pi}^{1.35\pi}\\&\int_{0}^{1}(-4.478cos(3z))dz=(-1.493sin(3z))|_{0}^{1}=-0.21\end{align*}$

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Example Question #107 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(2x^{2} + 2y^{2}))}{(25\cdot 3^{(\frac{z}{2})})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.05\\&\text{and length }0.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.976,0.218)\text{ and }\overrightarrow{u_2}=(0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.09

0.18

-0.18

-0.02

Correct answer:

0.09

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.218}{-0.976})=0.93\pi;\theta_2=arctan(\frac{-0.661}{0.750})=1.77\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{0.91}\int_{0.93\pi}^{1.77\pi}\int_{0}^{1.05}(\frac{(3\cdot rsin(2\cdot r^{2}))}{(25\cdot 3^{(\frac{z}{2})})})drd\theta dz=(-\frac{(3cos(r^{2})^{2})}{(50\cdot 3^{(\frac{z}{2})})})d\theta dz|_{0}^{1.05}\\&\int_{0}^{0.91}\int_{0.93\pi}^{1.77\pi}(\frac{0.04778}{3^{(0.5z)}})d\theta dz=(\frac{(0.04778\theta )}{3^{(0.5z)}})dz|_{0.93\pi}^{1.77\pi}\\&\int_{0}^{0.91}(\frac{0.1261}{3^{(0.5z)}})dz=(-\frac{0.2295}{3^{(0.5z)}})|_{0}^{0.91}=0.09\end{align*}$

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Example Question #108 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2sin(2x^{2} + 2y^{2}))}{(3\cdot 3^z)})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.1\text{ and }0.88\\&\text{and length }1.2\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.397,0.918)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.39

2.31

0.06

-2.31

Correct answer:

-0.39

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.918}{-0.397})=0.63\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.2}\int_{0.63\pi}^{1.76\pi}\int_{0.1}^{0.88}(-\frac{(2\cdot rsin(2\cdot r^{2}))}{(3\cdot 3^z)})drd\theta dz=(cos(\frac{r^{2})^{2}}{(3\cdot 3^z)})d\theta dz|_{0.1}^{0.88}\\&\int_{0}^{1.2}\int_{0.63\pi}^{1.76\pi}(-\frac{0.163}{3^{(z)}})d\theta dz=(-\frac{(0.163\theta )}{3^{(z)}})dz|_{0.63\pi}^{1.76\pi}\\&\int_{0}^{1.2}(-\frac{0.5785}{3^{(z)}})dz=(\frac{0.5266}{3^{(z)}})|_{0}^{1.2}=-0.39\end{align*}$

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Example Question #109 : Triple Integration In Cylindrical Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(e^{(- x^{2} - y^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.56\\&\text{and length }1.4\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.187,0.982)\text{ and }\overrightarrow{u_2}=(-0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.48

7.25

1.45

-5.8

Correct answer:

1.45

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(e^{(- x^{2} - y^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(re^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.982}{-0.187})=0.56\pi;\theta_2=arctan(\frac{-0.685}{-0.729})=1.24\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0}^{1.4}\int_{0.56\pi}^{1.24\pi}\int_{0}^{1.56}(re^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))drd\theta dz=(-\frac{(e^{(-r^{2})}\cdot (\frac{(5z)}{8}+\frac{ 5}{8}))}{2})d\theta dz|_{0}^{1.56}\\&\int_{0}^{1.4}\int_{0.56\pi}^{1.24\pi}(0.2851z + 0.2851)d\theta dz=(0.2851\theta \cdot (z + 1))dz|_{0.56\pi}^{1.24\pi}\\&\int_{0}^{1.4}(0.609z + 0.609)dz=(0.3045\cdot (z + 1)^{2})|_{0}^{1.4}=1.45\end{align*}$

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Example Question #561 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(2sin(3z)cos(x^{2} + y^{2}))}{13})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&0.67\\&\text{and length }0.69\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.941,0.339)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.00641

0.06

-0.00641

-0.03

Correct answer:

-0.03

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.339}{-0.941})=0.89\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[sin(az)]=-\frac{cos(az)}{a}\\&\int_{0}^{0.69}\int_{0.89\pi}^{1.51\pi}\int_{0}^{0.67}(-\frac{(2\cdot rcos(r^{2})sin(3z))}{13})drd\theta dz=(-\frac{(sin(r^{2})sin(3z))}{13})d\theta dz|_{0}^{0.67}\\&\int_{0}^{0.69}\int_{0.89\pi}^{1.51\pi}(-0.03338sin(3z))d\theta dz=(-0.03338\theta sin(3z))dz|_{0.89\pi}^{1.51\pi}\\&\int_{0}^{0.69}(-0.06502sin(3z))dz=(0.02167cos(3z))|_{0}^{0.69}=-0.03\end{align*}$

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Example Question #562 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{(3\cdot (\frac{3}{2})^{(x^{2} + y^{2})})}{49})dA\text{, where D is}\\&\text{the region of a cylindrical annulus centered on the origin,}\\&\text{with inner/outer radii }\\&0.49\text{ and }1.16\\&\text{and length }1.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.12

-0.35

1.4

-0.17

Correct answer:

-0.35

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{(3\cdot (\frac{3}{2})^{(x^{2} + y^{2})})}{49})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{(3\cdot (\frac{3}{2})^{(r^{2})}\cdot r)}{49})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{-0.031})=0.51\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0}^{1.86}\int_{0.51\pi}^{1.78\pi}\int_{0.49}^{1.16}(-\frac{(3\cdot (\frac{3}{2})^{(r^{2})}\cdot r)}{49})drd\theta dz=(-\frac{(3\cdot (\frac{3}{2})^{(r^{2})})}{(98ln(\frac{3}{2}))})d\theta dz|_{0.49}^{1.16}\\&\int_{0}^{1.86}\int_{0.51\pi}^{1.78\pi}(-0.04706)d\theta dz=(-0.04706\theta)dz|_{0.51\pi}^{1.78\pi}\\&\int_{0}^{1.86}(-0.1878)dz=(-0.1878z)|_{0}^{1.86}=-0.35\end{align*}$

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Example Question #563 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-7\cdot 3^ze^{(- 2x^{2} - 2y^{2})})dA\text{, where D is}\\&\text{the region of a cylinder centered on the origin with a radius of }\\&1.1\\&\text{and length }1.89\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.339,0.941)\text{ and }\overrightarrow{u_2}=(-0.876,-0.482)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

122.44

12.24

-48.98

-24.49

Correct answer:

-24.49

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to cylindrical form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-7\cdot 3^ze^{(- 2x^{2} - 2y^{2})})dA\rightarrow\int_{z_1}^{z_2}\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-7\cdot 3^z\cdot re^{(-2\cdot r^{2})})drd\theta dz\\&\text{The bounds of our radius and height we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.941}{0.339})=0.39\pi;\theta_2=arctan(\frac{-0.482}{-0.876})=1.16\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[b^{az}]=\frac{b^{az}}{aln(b)}\\&\int_{0}^{1.89}\int_{0.39\pi}^{1.16\pi}\int_{0}^{1.1}(-7\cdot 3^z\cdot re^{(-2\cdot r^{2})})drd\theta dz=(\frac{(7\cdot 3^ze^{(-2\cdot r^{2})})}{4})d\theta dz|_{0}^{1.1}\\&\int_{0}^{1.89}\int_{0.39\pi}^{1.16\pi}(-1.594\cdot 3^z)d\theta dz=(-1.594\cdot 3^z\theta)dz|_{0.39\pi}^{1.16\pi}\\&\int_{0}^{1.89}(-3.857\cdot 3^z)dz=(-3.511\cdot 3^z)|_{0}^{1.89}=-24.49\end{align*}$

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Calculus 3 : Calculus 3

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