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Calculus 3 : Double Integration in Polar Coordinates

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Example Questions

Example Question #151 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (17e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{2}-\frac{ 31}{(5\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.49\text{ and }1.43\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.750,0.661)\text{ and }\overrightarrow{u_2}=(0.969,-0.249)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

15.67

-10.45

-62.68

376.06

Correct answer:

-62.68

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (17e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{2}-\frac{ 31}{(5\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (17\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{2}-\frac{ 93}{(10\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{0.750})=0.23\pi;\theta_2=arctan(\frac{-0.249}{0.969})=1.92\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.23\pi}^{1.92\pi}\int_{0.49}^{1.43}(-\frac{ (17\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{2}-\frac{ 93}{(10\cdot r)})drd\theta=(\frac{(17e^{(-\frac{(3\cdot r^{2})}{2})})}{6}-\frac{ (93ln(r))}{10})d\theta|_{0.49}^{1.43}\\&\int_{0.23\pi}^{1.92\pi}(-11.81)d\theta=(-11.81\theta)|_{0.23\pi}^{1.92\pi}=-62.68\end{align*}$

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Example Question #152 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(x^{2} + y^{2})}{41}-\frac{ (38e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{5})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.2\text{ and }1.32\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

245.08

10.21

-61.27

-20.42

Correct answer:

-61.27

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(x^{2} + y^{2})}{41}-\frac{ (38e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(r^{2}))}{41}-\frac{ (38\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.2\pi}^{1.78\pi}\int_{0.2}^{1.32}(\frac{(rsin(r^{2}))}{41}-\frac{ (38\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{5})drd\theta=(-\frac{ cos(r^{2})}{82}-\frac{ (57e^{(\frac{(2\cdot r^{2})}{3})})}{10})d\theta|_{0.2}^{1.32}\\&\int_{0.2\pi}^{1.78\pi}(-12.34)d\theta=(-12.34\theta)|_{0.2\pi}^{1.78\pi}=-61.27\end{align*}$

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Example Question #153 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{22}{(5\cdot (\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}-\frac{ 16}{(3\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.31\text{ and }0.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.996,0.094)\text{ and }\overrightarrow{u_2}=(0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

9.58

-57.49

47.9

-3.19

Correct answer:

9.58

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{22}{(5\cdot (\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}-\frac{ 16}{(3\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{236}{(45\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.094}{-0.996})=0.97\pi;\theta_2=arctan(\frac{-1.000}{0.031})=1.51\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.97\pi}^{1.51\pi}\int_{0.31}^{0.91}(\frac{236}{(45\cdot r)})drd\theta=(\frac{(236ln(r))}{45})d\theta|_{0.31}^{0.91}\\&\int_{0.97\pi}^{1.51\pi}(5.648)d\theta=(5.648\theta)|_{0.97\pi}^{1.51\pi}=9.58\end{align*}$

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Example Question #154 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4sin(x^{2} + y^{2}))}{21}+\frac{ 47}{(5\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.26\text{ and }1.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.998,0.063)\text{ and }\overrightarrow{u_2}=(0.844,-0.536)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-14.92

5.97

-179.09

29.85

Correct answer:

29.85

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(4sin(x^{2} + y^{2}))}{21}+\frac{ 47}{(5\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(4\cdot rsin(r^{2}))}{21}+\frac{ 94}{(15\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{-0.998})=0.98\pi;\theta_2=arctan(\frac{-0.536}{0.844})=1.82\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.98\pi}^{1.82\pi}\int_{0.26}^{1.54}(\frac{(4\cdot rsin(r^{2}))}{21}+\frac{ 94}{(15\cdot r)})drd\theta=(\frac{(94ln(r))}{15}-\frac{ (2cos(r^{2}))}{21})d\theta|_{0.26}^{1.54}\\&\int_{0.98\pi}^{1.82\pi}(11.31)d\theta=(11.31\theta)|_{0.98\pi}^{1.82\pi}=29.85\end{align*}$

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Example Question #155 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(3cos(2x^{2} + 2y^{2}) +\frac{ (35e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{4})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.2\text{ and }1.96\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.707,-0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.01

-125.26

-6.26

25.05

Correct answer:

25.05

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(3cos(2x^{2} + 2y^{2}) +\frac{ (35e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(3\cdot rcos(2\cdot r^{2}) +\frac{ (35\cdot re^{(-\frac{r^{2}}{2})})}{4})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.707}{0.707})=1.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.75\pi}^{1.75\pi}\int_{0.2}^{1.96}(3\cdot rcos(2\cdot r^{2}) +\frac{ (35\cdot re^{(-\frac{r^{2}}{2})})}{4})drd\theta=(\frac{(3sin(2\cdot r^{2}))}{4}-\frac{ (35e^{(-\frac{r^{2}}{2})})}{4})d\theta|_{0.2}^{1.96}\\&\int_{0.75\pi}^{1.75\pi}(7.974)d\theta=(7.974\theta)|_{0.75\pi}^{1.75\pi}=25.05\end{align*}$

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Example Question #156 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(2e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})} +\frac{ sin(x^{2} + y^{2})}{3})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.12\text{ and }1.26\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.941,0.339)\text{ and }\overrightarrow{u_2}=(0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-19.89

6.63

13.26

-1.1

Correct answer:

6.63

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.339}{0.941})=0.11\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.11\pi}^{1.97\pi}\int_{0.12}^{1.26}(\frac{(rsin(r^{2}))}{3}+ 2\cdot re^{(-\frac{(2\cdot r^{2})}{3})})drd\theta=(-\frac{ cos(r^{2})}{6}-\frac{ (3e^{(-\frac{(2\cdot r^{2})}{3})})}{2})d\theta|_{0.12}^{1.26}\\&\int_{0.11\pi}^{1.97\pi}(1.135)d\theta=(1.135\theta)|_{0.11\pi}^{1.97\pi}=6.63\end{align*}$

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Example Question #157 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{21}- 13cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.992,0.125)\text{ and }\overrightarrow{u_2}=(0.941,-0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-25.11

2.09

-12.56

25.11

Correct answer:

-12.56

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{21}- 13cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{21}- 13\cdot rcos(\frac{(3\cdot r^{2})}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.125}{-0.992})=0.96\pi;\theta_2=arctan(\frac{-0.339}{0.941})=1.89\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.96\pi}^{1.89\pi}\int_{0}^{1}(\frac{(2\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{21}- 13\cdot rcos(\frac{(3\cdot r^{2})}{2}))drd\theta=(-\frac{ (2e^{(-\frac{(3\cdot r^{2})}{2})})}{63}-\frac{ (13sin(\frac{(3\cdot r^{2})}{2}))}{3})d\theta|_{0}^{1}\\&\int_{0.96\pi}^{1.89\pi}(-4.298)d\theta=(-4.298\theta)|_{0.96\pi}^{1.89\pi}=-12.56\end{align*}$

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Example Question #158 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{4}{(41\cdot (x^{2} + y^{2}))}-\frac{ 1}{(5\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.36\text{ and }1.6\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.976,0.218)\text{ and }\overrightarrow{u_2}=(-0.125,-0.992)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.09

-0.09

0.01

-0.02

Correct answer:

-0.09

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{4}{(41\cdot (x^{2} + y^{2}))}-\frac{ 1}{(5\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{22}{(615\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.218}{-0.976})=0.93\pi;\theta_2=arctan(\frac{-0.992}{-0.125})=1.46\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.93\pi}^{1.46\pi}\int_{0.36}^{1.6}(-\frac{22}{(615\cdot r)})drd\theta=(-\frac{(22ln(r))}{615})d\theta|_{0.36}^{1.6}\\&\int_{0.93\pi}^{1.46\pi}(-0.05336)d\theta=(-0.05336\theta)|_{0.93\pi}^{1.46\pi}=-0.09\end{align*}$

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Example Question #151 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{7}+\frac{ (\frac{5}{4^{(x^{2} + y^{2})}})}{13})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.73\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.368,0.930)\text{ and }\overrightarrow{u_2}=(-0.750,-0.661)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

3.07

-1.54

0.77

-0.13

Correct answer:

0.77

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{7}+\frac{ (\frac{5}{4^{(x^{2} + y^{2})}})}{13})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot r)}{(13\cdot 4^{(r^{2})})}+\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{7})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.930}{0.368})=0.38\pi;\theta_2=arctan(\frac{-0.661}{-0.750})=1.23\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.38\pi}^{1.23\pi}\int_{0}^{1.73}(\frac{(5\cdot r)}{(13\cdot 4^{(r^{2})})}+\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{7})drd\theta=(-\frac{ (3cos(\frac{(2\cdot r^{2})}{3}))}{28}-\frac{ 5}{(52\cdot 2^{(2\cdot r^{2})}ln(2))})d\theta|_{0}^{1.73}\\&\int_{0.38\pi}^{1.23\pi}(0.2878)d\theta=(0.2878\theta)|_{0.38\pi}^{1.23\pi}=0.77\end{align*}$

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Example Question #159 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (5cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ e^{(- 2x^{2} - 2y^{2})}}{2})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.6\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.613,0.790)\text{ and }\overrightarrow{u_2}=(-0.094,-0.996)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.95

0.05

0.63

-0.32

Correct answer:

-0.32

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (5cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{41}-\frac{ e^{(- 2x^{2} - 2y^{2})}}{2})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (5\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (re^{(-2\cdot r^{2})})}{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{0.613})=0.29\pi;\theta_2=arctan(\frac{-0.996}{-0.094})=1.47\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.29\pi}^{1.47\pi}\int_{0}^{0.6}(-\frac{ (5\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{41}-\frac{ (re^{(-2\cdot r^{2})})}{2})drd\theta=(\frac{e^{(-2\cdot r^{2})}}{8}-\frac{ (5sin(\frac{(3\cdot r^{2})}{2}))}{123})d\theta|_{0}^{0.6}\\&\int_{0.29\pi}^{1.47\pi}(-0.08506)d\theta=(-0.08506\theta)|_{0.29\pi}^{1.47\pi}=-0.32\end{align*}$

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Calculus 3 : Double Integration in Polar Coordinates

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Example Question #151 : Double Integration In Polar Coordinates

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