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Calculus 3 : Double Integrals

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Example Questions

Example Question #121 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(23e^{(- 2x^{2} - 2y^{2})} +\frac{ (3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{17})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

12.02

-60.11

-6.01

Correct answer:

12.02

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(23e^{(- 2x^{2} - 2y^{2})} +\frac{ (3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{17})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(23\cdot re^{(-2\cdot r^{2})} +\frac{ (3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{17})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{0.309})=0.4\pi;\theta_2=arctan(\frac{-0.125}{-0.992})=1.04\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.4\pi}^{1.04\pi}\int_{0}^{1.91}(23\cdot re^{(-2\cdot r^{2})} +\frac{ (3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{17})drd\theta=(-\frac{ (9cos(\frac{(2\cdot r^{2})}{3}))}{68}-\frac{ (23e^{(-2\cdot r^{2})})}{4})d\theta|_{0}^{1.91}\\&\int_{0.4\pi}^{1.04\pi}(5.979)d\theta=(5.979\theta)|_{0.4\pi}^{1.04\pi}=12.02\end{align*}$

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Example Question #1492 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 4e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})} - 3\cdot 2^{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.2\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.000)\text{ and }\overrightarrow{u_2}=(0.426,-0.905)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-44.71

-11.18

11.18

5.59

Correct answer:

-11.18

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.000}{-1.000})=1\pi;\theta_2=arctan(\frac{-0.905}{0.426})=1.64\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{1\pi}^{1.64\pi}\int_{0}^{1.2}(- 3\cdot 2^{(r^{2})}\cdot r - 4\cdot re^{(-\frac{(2\cdot r^{2})}{3})})drd\theta=(3e^{(-\frac{(2\cdot r^{2})}{3})} -\frac{ (3\cdot 2^{(r^{2})})}{(2ln(2))})d\theta|_{0}^{1.2}\\&\int_{1\pi}^{1.64\pi}(-5.559)d\theta=(-5.559\theta)|_{1\pi}^{1.64\pi}=-11.18\end{align*}$

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Example Question #1493 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{2}-\frac{ cos(2x^{2} + 2y^{2})}{8})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.25\text{ and }1.93\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(-0.941,-0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.17

-0.52

-1.04

4.14

Correct answer:

-1.04

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{2}-\frac{ cos(2x^{2} + 2y^{2})}{8})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rcos(r^{2}))}{2}-\frac{ (rcos(2\cdot r^{2}))}{8})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.339}{-0.941})=1.11\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.43\pi}^{1.11\pi}\int_{0.25}^{1.93}(\frac{(3\cdot rcos(r^{2}))}{2}-\frac{ (rcos(2\cdot r^{2}))}{8})drd\theta=(-\frac{(sin(r^{2})\cdot (cos(r^{2}) - 12))}{16})d\theta|_{0.25}^{1.93}\\&\int_{0.43\pi}^{1.11\pi}(-0.4848)d\theta=(-0.4848\theta)|_{0.43\pi}^{1.11\pi}=-1.04\end{align*}$

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Example Question #1494 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(4e^{(- x^{2} - y^{2})})}{3}-\frac{ e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}}{2})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.38\text{ and }1.96\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.588,0.809)\text{ and }\overrightarrow{u_2}=(-0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.45

-1.81

0.15

0.9

Correct answer:

0.9

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.809}{-0.588})=0.7\pi;\theta_2=arctan(\frac{-0.918}{-0.397})=1.37\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.7\pi}^{1.37\pi}\int_{0.38}^{1.96}(\frac{(4\cdot re^{(-r^{2})})}{3}-\frac{ (re^{(-\frac{(3\cdot r^{2})}{2})})}{2})drd\theta=(\frac{e^{(-\frac{(3\cdot r^{2})}{2})}}{6}-\frac{ (2e^{(-r^{2})})}{3})d\theta|_{0.38}^{1.96}\\&\int_{0.7\pi}^{1.37\pi}(0.429)d\theta=(0.429\theta)|_{0.7\pi}^{1.37\pi}=0.9\end{align*}$

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Example Question #1495 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(7e^{(x^{2} + y^{2})} -\frac{ (\frac{46}{2^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})}{5})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.43\text{ and }1.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.905,-0.426)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

38.13

-19.07

76.26

-381.31

Correct answer:

76.26

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(7e^{(x^{2} + y^{2})} -\frac{ (\frac{46}{2^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(7\cdot re^{(r^{2})} -\frac{ (46\cdot r)}{(5\cdot 2^{(\frac{r^{2}}{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{0.750})=0.23\pi;\theta_2=arctan(\frac{-0.426}{-0.905})=1.14\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.23\pi}^{1.14\pi}\int_{0.43}^{1.54}(7\cdot re^{(r^{2})} -\frac{ (46\cdot r)}{(5\cdot 2^{(\frac{r^{2}}{2})})})drd\theta=(\frac{(7e^{(r^{2})})}{2}+\frac{ 46}{(5\cdot 2^{(\frac{r^{2}}{2})}ln(2))})d\theta|_{0.43}^{1.54}\\&\int_{0.23\pi}^{1.14\pi}(26.68)d\theta=(26.68\theta)|_{0.23\pi}^{1.14\pi}=76.26\end{align*}$

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Example Question #126 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}) -\frac{ 17}{(6\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.46\text{ and }1.52\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.976,0.218)\text{ and }\overrightarrow{u_2}=(0.613,-0.790)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-38.98

4.87

48.73

-9.75

Correct answer:

-9.75

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}) -\frac{ 17}{(6\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- rsin(\frac{r^{2}}{2}) -\frac{ 17}{(6\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.218}{-0.976})=0.93\pi;\theta_2=arctan(\frac{-0.790}{0.613})=1.71\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.93\pi}^{1.71\pi}\int_{0.46}^{1.52}(- rsin(\frac{r^{2}}{2}) -\frac{ 17}{(6\cdot r)})drd\theta=(cos(\frac{r^{2}}{2}) -\frac{ (17ln(r))}{6})d\theta|_{0.46}^{1.52}\\&\int_{0.93\pi}^{1.71\pi}(-3.977)d\theta=(-3.977\theta)|_{0.93\pi}^{1.71\pi}=-9.75\end{align*}$

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Example Question #121 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(- 2x^{2} - 2y^{2})}}{3}+ (\frac{\frac{2}{3})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}}{4})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.37\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(0.861,-0.509)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.26

-0.26

-5.18

1.29

Correct answer:

1.29

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.509}{0.861})=1.83\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.34\pi}^{1.83\pi}\int_{0}^{1.37}(\frac{((\frac{2}{3})^{(\frac{r^{2}}{2})}\cdot r)}{4}+\frac{ (re^{(-2\cdot r^{2})})}{3})drd\theta=((\frac{\frac{2}{3})^{(\frac{r^{2}}{2})}}{(4ln(\frac{2}{3}))}-\frac{ e^{(-2\cdot r^{2})}}{12})d\theta|_{0}^{1.37}\\&\int_{0.34\pi}^{1.83\pi}(0.2765)d\theta=(0.2765\theta)|_{0.34\pi}^{1.83\pi}=1.29\end{align*}$

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Example Question #604 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5e^{(- x^{2} - y^{2})})}{14}- 49cos(2x^{2} + 2y^{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.72\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.536,0.844)\text{ and }\overrightarrow{u_2}=(0.309,-0.951)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

14.04

-10.53

-42.11

84.22

Correct answer:

-42.11

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5e^{(- x^{2} - y^{2})})}{14}- 49cos(2x^{2} + 2y^{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot re^{(-r^{2})})}{14}- 49\cdot rcos(2\cdot r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{0.536})=0.32\pi;\theta_2=arctan(\frac{-0.951}{0.309})=1.6\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.32\pi}^{1.6\pi}\int_{0}^{0.72}(\frac{(5\cdot re^{(-r^{2})})}{14}- 49\cdot rcos(2\cdot r^{2}))drd\theta=(-\frac{ (5e^{(-r^{2})})}{28}-\frac{ (49sin(2\cdot r^{2}))}{4})d\theta|_{0}^{0.72}\\&\int_{0.32\pi}^{1.6\pi}(-10.47)d\theta=(-10.47\theta)|_{0.32\pi}^{1.6\pi}=-42.11\end{align*}$

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Example Question #121 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{32}+\frac{ (3\cdot (\frac{1}{4})^{(2x^{2} + 2y^{2})})}{13})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.95\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.8

0.4

0.8

-0.07

Correct answer:

0.4

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{32}+\frac{ (3\cdot (\frac{1}{4})^{(2x^{2} + 2y^{2})})}{13})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot (\frac{1}{4})^{(2\cdot r^{2})}\cdot r)}{13}+\frac{ (5\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{32})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{0.707})=0.25\pi;\theta_2=arctan(\frac{-0.279}{0.960})=1.91\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.25\pi}^{1.91\pi}\int_{0}^{0.95}(\frac{(3\cdot (\frac{1}{4})^{(2\cdot r^{2})}\cdot r)}{13}+\frac{ (5\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{32})drd\theta=(-\frac{ (5e^{(-\frac{(3\cdot r^{2})}{2})})}{96}-\frac{ 3}{(104\cdot 4^{(2\cdot r^{2})}ln(2))})d\theta|_{0}^{0.95}\\&\int_{0.25\pi}^{1.91\pi}(0.07684)d\theta=(0.07684\theta)|_{0.25\pi}^{1.91\pi}=0.4\end{align*}$

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Example Question #122 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(14cos(2x^{2} + 2y^{2}))}{3}+\frac{ (15cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.426,-0.905)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.91

104.68

17.45

-69.79

Correct answer:

17.45

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(14cos(2x^{2} + 2y^{2}))}{3}+\frac{ (15cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(14\cdot rcos(2\cdot r^{2}))}{3}+\frac{ (15\cdot rcos(\frac{r^{2}}{2}))}{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{0.309})=0.4\pi;\theta_2=arctan(\frac{-0.905}{-0.426})=1.36\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.4\pi}^{1.36\pi}\int_{0}^{1.54}(\frac{(14\cdot rcos(2\cdot r^{2}))}{3}+\frac{ (15\cdot rcos(\frac{r^{2}}{2}))}{2})drd\theta=(\frac{(7sin(2\cdot r^{2}))}{6}+\frac{ (15sin(\frac{r^{2}}{2}))}{2})d\theta|_{0}^{1.54}\\&\int_{0.4\pi}^{1.36\pi}(5.785)d\theta=(5.785\theta)|_{0.4\pi}^{1.36\pi}=17.45\end{align*}$

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Calculus 3 : Double Integrals

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Example Question #121 : Double Integration In Polar Coordinates

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