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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 #88 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(23e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{4}-\frac{ (25e^{(x^{2} + y^{2})})}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.33\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.707,0.707)\text{ and }\overrightarrow{u_2}=(-1.000,0.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-21.78

21.78

-43.56

43.56

Correct answer:

-43.56

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.707}{0.707})=0.25\pi;\theta_2=arctan(\frac{0.000}{-1.000})=1\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.25\pi}^{1\pi}\int_{0}^{1.33}(\frac{(23\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{4}-\frac{ (25\cdot re^{(r^{2})})}{3})drd\theta=(-\frac{ (25e^{(r^{2})})}{6}-\frac{ (23e^{(-\frac{(3\cdot r^{2})}{2})})}{12})d\theta|_{0}^{1.33}\\&\int_{0.25\pi}^{1\pi}(-18.49)d\theta=(-18.49\theta)|_{0.25\pi}^{1\pi}=-43.56\end{align*}$

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Example Question #89 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(15e^{(x^{2} + y^{2})} -\frac{ (43cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{6})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.19\text{ and }1.59\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.613,0.790)\text{ and }\overrightarrow{u_2}=(0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

427.44

-2564.62

2564.62

-106.86

Correct answer:

427.44

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(15e^{(x^{2} + y^{2})} -\frac{ (43cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{6})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(15\cdot re^{(r^{2})} -\frac{ (43\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{0.613})=0.29\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.29\pi}^{1.97\pi}\int_{0.19}^{1.59}(15\cdot re^{(r^{2})} -\frac{ (43\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{6})drd\theta=(\frac{(15e^{(r^{2})})}{2}-\frac{ (43sin(\frac{(2\cdot r^{2})}{3}))}{8})d\theta|_{0.19}^{1.59}\\&\int_{0.29\pi}^{1.97\pi}(80.99)d\theta=(80.99\theta)|_{0.29\pi}^{1.97\pi}=427.44\end{align*}$

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Example Question #91 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(15sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) - 2sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.34\text{ and }1.38\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.960,0.279)\text{ and }\overrightarrow{u_2}=(-0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

20.8

-3.47

6.93

-83.19

Correct answer:

20.8

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(15sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) - 2sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(15\cdot rsin(\frac{(2\cdot r^{2})}{3}) - 2\cdot rsin(\frac{r^{2}}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{0.960})=0.09\pi;\theta_2=arctan(\frac{-0.094}{-0.996})=1.03\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.09\pi}^{1.03\pi}\int_{0.34}^{1.38}(15\cdot rsin(\frac{(2\cdot r^{2})}{3}) - 2\cdot rsin(\frac{r^{2}}{2}))drd\theta=(2cos(\frac{r^{2}}{2}) -\frac{ (45cos(\frac{(2\cdot r^{2})}{3}))}{4})d\theta|_{0.34}^{1.38}\\&\int_{0.09\pi}^{1.03\pi}(7.042)d\theta=(7.042\theta)|_{0.09\pi}^{1.03\pi}=20.8\end{align*}$

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Example Question #92 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{4}-\frac{ (33sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.12\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.218,0.976)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-17.84

-2.97

0.74

8.92

Correct answer:

-2.97

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{4}-\frac{ (33sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(\frac{(3\cdot r^{2})}{2})})}{4}-\frac{ (33\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{-0.218})=0.57\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.57\pi}^{1.38\pi}\int_{0}^{1.12}(\frac{(re^{(\frac{(3\cdot r^{2})}{2})})}{4}-\frac{ (33\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{5})drd\theta=(\frac{(99cos(\frac{(2\cdot r^{2})}{3}))}{20}+\frac{ e^{(\frac{(3\cdot r^{2})}{2})}}{12})d\theta|_{0}^{1.12}\\&\int_{0.57\pi}^{1.38\pi}(-1.169)d\theta=(-1.169\theta)|_{0.57\pi}^{1.38\pi}=-2.97\end{align*}$

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Example Question #91 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(79cos(x^{2} + y^{2}))}{260})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.1\text{ and }1.01\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.06

-1.37

1.37

0.23

Correct answer:

0.23

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(79cos(x^{2} + y^{2}))}{260})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(79\cdot rcos(r^{2}))}{260})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{-0.809})=0.8\pi;\theta_2=arctan(\frac{-0.918}{-0.397})=1.37\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.8\pi}^{1.37\pi}\int_{0.1}^{1.01}(\frac{(79\cdot rcos(r^{2}))}{260})drd\theta=(\frac{(79sin(r^{2}))}{520})d\theta|_{0.1}^{1.01}\\&\int_{0.8\pi}^{1.37\pi}(0.1279)d\theta=(0.1279\theta)|_{0.8\pi}^{1.37\pi}=0.23\end{align*}$

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Example Question #94 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{28}{(\frac{2}{7})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{3})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.15\text{ and }0.76\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.930,0.368)\text{ and }\overrightarrow{u_2}=(-0.918,0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

2.8

-67.32

-2.8

11.22

Correct answer:

11.22

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{28}{(\frac{2}{7})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(28\cdot r)}{(3\cdot (\frac{2}{7})^{(\frac{(3\cdot r^{2})}{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.368}{0.930})=0.12\pi;\theta_2=arctan(\frac{0.397}{-0.918})=0.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.12\pi}^{0.87\pi}\int_{0.15}^{0.76}(\frac{(28\cdot r)}{(3\cdot (\frac{2}{7})^{(\frac{(3\cdot r^{2})}{2})})})drd\theta=(-\frac{(28\cdot 7^{(\frac{(3\cdot r^{2})}{2})})}{(9\cdot 2^{(\frac{(3\cdot r^{2})}{2})}ln(\frac{2}{7}))})d\theta|_{0.15}^{0.76}\\&\int_{0.12\pi}^{0.87\pi}(4.762)d\theta=(4.762\theta)|_{0.12\pi}^{0.87\pi}=11.22\end{align*}$

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Example Question #95 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{45}- 43sin(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.38\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.094,-0.996)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

14.1

-70.48

-11.75

140.96

Correct answer:

-70.48

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{45}- 43sin(x^{2} + y^{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot re^{(\frac{r^{2}}{2})})}{45}- 43\cdot rsin(r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{-0.685})=0.74\pi;\theta_2=arctan(\frac{-0.996}{0.094})=1.53\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.74\pi}^{1.53\pi}\int_{0}^{1.38}(\frac{(4\cdot re^{(\frac{r^{2}}{2})})}{45}- 43\cdot rsin(r^{2}))drd\theta=(\frac{(43cos(r^{2}))}{2}+\frac{ (4e^{(\frac{r^{2}}{2})})}{45})d\theta|_{0}^{1.38}\\&\int_{0.74\pi}^{1.53\pi}(-28.4)d\theta=(-28.4\theta)|_{0.74\pi}^{1.53\pi}=-70.48\end{align*}$

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Example Question #96 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{43}-\frac{ (3sin(2x^{2} + 2y^{2}))}{20})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.562,0.827)\text{ and }\overrightarrow{u_2}=(-0.998,-0.063)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.01

-0.11

0.22

-0.06

Correct answer:

-0.06

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{43}-\frac{ (3sin(2x^{2} + 2y^{2}))}{20})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-\frac{r^{2}}{2})})}{43}-\frac{ (3\cdot rsin(2\cdot r^{2}))}{20})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.827}{0.562})=0.31\pi;\theta_2=arctan(\frac{-0.063}{-0.998})=1.02\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.31\pi}^{1.02\pi}\int_{0}^{0.99}(\frac{(3\cdot re^{(-\frac{r^{2}}{2})})}{43}-\frac{ (3\cdot rsin(2\cdot r^{2}))}{20})drd\theta=(\frac{(3cos(2\cdot r^{2}))}{80}-\frac{ (3e^{(-\frac{r^{2}}{2})})}{43})d\theta|_{0}^{0.99}\\&\int_{0.31\pi}^{1.02\pi}(-0.02471)d\theta=(-0.02471\theta)|_{0.31\pi}^{1.02\pi}=-0.06\end{align*}$

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Example Question #93 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3sin(2x^{2} + 2y^{2}))}{17}-\frac{ 2^{(2x^{2} + 2y^{2})}}{2})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.16\text{ and }1.46\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.426,0.905)\text{ and }\overrightarrow{u_2}=(-0.876,-0.482)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-5.46

1.82

5.46

-1.36

Correct answer:

-5.46

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.905}{-0.426})=0.64\pi;\theta_2=arctan(\frac{-0.482}{-0.876})=1.16\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.64\pi}^{1.16\pi}\int_{0.16}^{1.46}(-\frac{ (2^{(2\cdot r^{2})}\cdot r)}{2}-\frac{ (3\cdot rsin(2\cdot r^{2}))}{17})drd\theta=(\frac{(3cos(r^{2})^{2})}{34}-\frac{ 2^{(2\cdot r^{2})}}{(8ln(2))})d\theta|_{0.16}^{1.46}\\&\int_{0.64\pi}^{1.16\pi}(-3.339)d\theta=(-3.339\theta)|_{0.64\pi}^{1.16\pi}=-5.46\end{align*}$

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Example Question #94 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{10}{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}+\frac{ 1}{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.46\text{ and }1.18\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.613,-0.790)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-97.55

146.32

48.77

-16.26

Correct answer:

48.77

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{10}{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}+\frac{ 1}{(x^{2} + y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{16}{r})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-0.790}{-0.613})=1.29\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.26\pi}^{1.29\pi}\int_{0.46}^{1.18}(\frac{16}{r})drd\theta=(16ln(r))d\theta|_{0.46}^{1.18}\\&\int_{0.26\pi}^{1.29\pi}(15.07)d\theta=(15.07\theta)|_{0.26\pi}^{1.29\pi}=48.77\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #88 : Double Integration In Polar Coordinates

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