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Calculus 3 : Double Integration in Polar Coordinates

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Example Questions

Example Question #61 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{6}+ 7\cdot 2^{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.22\text{ and }1.73\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.613,0.790)\text{ and }\overrightarrow{u_2}=(0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

834.35

-556.23

139.06

-46.35

Correct answer:

139.06

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{6}+ 7\cdot 2^{(x^{2} + y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(7\cdot 2^{(r^{2})}\cdot r +\frac{ (rsin(\frac{r^{2}}{2}))}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{-0.613})=0.71\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.71\pi}^{1.97\pi}\int_{0.22}^{1.73}(7\cdot 2^{(r^{2})}\cdot r +\frac{ (rsin(\frac{r^{2}}{2}))}{6})drd\theta=(\frac{(7\cdot 2^{(r^{2})})}{(2ln(2))}-\frac{ cos(\frac{r^{2}}{2})}{6})d\theta|_{0.22}^{1.73}\\&\int_{0.71\pi}^{1.97\pi}(35.13)d\theta=(35.13\cdot \theta)|_{0.71\pi}^{1.97\pi}=139.06\end{align*}$

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Example Question #62 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(881sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{22})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.31\text{ and }1.3\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.982,-0.187)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

7.15

-10.72

42.9

-42.9

Correct answer:

42.9

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(881sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{22})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(881\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{22})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-0.187}{-0.982})=1.06\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.26\pi}^{1.06\pi}\int_{0.31}^{1.3}(\frac{(881\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{22})drd\theta=(-\frac{(2643cos(\frac{(2\cdot r^{2})}{3}))}{88})d\theta|_{0.31}^{1.3}\\&\int_{0.26\pi}^{1.06\pi}(17.07)d\theta=(17.07\cdot \theta)|_{0.26\pi}^{1.06\pi}=42.9\end{align*}$

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Example Question #63 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(12sin(2x^{2} + 2y^{2}) -\frac{ (5e^{(- 2x^{2} - 2y^{2})})}{41})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.9\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.827,0.562)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

9.03

-4.51

-18.05

45.13

Correct answer:

9.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(12sin(2x^{2} + 2y^{2}) -\frac{ (5e^{(- 2x^{2} - 2y^{2})})}{41})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(12\cdot rsin(2\cdot r^{2}) -\frac{ (5\cdot re^{(-2\cdot r^{2})})}{41})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.562}{-0.827})=0.81\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.81\pi}^{1.73\pi}\int_{0}^{0.9}(12\cdot rsin(2\cdot r^{2}) -\frac{ (5\cdot re^{(-2\cdot r^{2})})}{41})drd\theta=(\frac{(5e^{(-2\cdot r^{2})})}{164}- 6cos(r^{2})^{2})d\theta|_{0}^{0.9}\\&\int_{0.81\pi}^{1.73\pi}(3.123)d\theta=(3.123\cdot \theta)|_{0.81\pi}^{1.73\pi}=9.03\end{align*}$

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Example Question #64 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{2}{13}-\frac{ e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}}{12})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.48\text{ and }1.38\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(-0.976,-0.218)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.02

0.25

-1.52

-0.04

Correct answer:

0.25

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{2}{13}-\frac{ e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}}{12})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot r)}{13}-\frac{ (re^{(-\frac{(3\cdot r^{2})}{2})})}{12})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.218}{-0.976})=1.07\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.34\pi}^{1.07\pi}\int_{0.48}^{1.38}(\frac{(2\cdot r)}{13}-\frac{ (re^{(-\frac{(3\cdot r^{2})}{2})})}{12})drd\theta=(\frac{e^{(-\frac{(3\cdot r^{2})}{2})}}{36}+\frac{ r^{2}}{13})d\theta|_{0.48}^{1.38}\\&\int_{0.34\pi}^{1.07\pi}(0.1107)d\theta=(0.1107\cdot \theta)|_{0.34\pi}^{1.07\pi}=0.25\end{align*}$

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Example Question #61 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (29sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{6}- (\frac{\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{17})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.64\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-3.56

-1.19

0.3

4.74

Correct answer:

-1.19

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (29sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{6}- (\frac{\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{17})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ ((\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{17}-\frac{ (29\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.031}{1.000})=1.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.75\pi}^{1.99\pi}\int_{0}^{0.64}(-\frac{ ((\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{17}-\frac{ (29\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{6})drd\theta=(\frac{(29cos(\frac{(3\cdot r^{2})}{2}))}{18}+\frac{ 1}{(51\cdot 2^{(\frac{(3\cdot r^{2})}{2})}ln(2))})d\theta|_{0}^{0.64}\\&\int_{0.75\pi}^{1.99\pi}(-0.3045)d\theta=(-0.3045\cdot \theta)|_{0.75\pi}^{1.99\pi}=-1.19\end{align*}$

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Example Question #62 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- (\frac{\frac{1}{4})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}}{3}- 10\cdot 3^{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.21\text{ and }1.18\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.685,0.729)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

181.92

-6.06

-30.32

10.11

Correct answer:

-30.32

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- (\frac{\frac{1}{4})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}}{3}- 10\cdot 3^{(x^{2} + y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- 10\cdot 3^{(r^{2})}\cdot r -\frac{ ((\frac{1}{4})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{0.729}{-0.685})=0.74\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.15\pi}^{0.74\pi}\int_{0.21}^{1.18}(- 10\cdot 3^{(r^{2})}\cdot r -\frac{ ((\frac{1}{4})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{3})drd\theta=(\frac{1}{(8\cdot 4^{(\frac{(2\cdot r^{2})}{3})}ln(2))}-\frac{ (5\cdot 3^{(r^{2})})}{ln(3)})d\theta|_{0.21}^{1.18}\\&\int_{0.15\pi}^{0.74\pi}(-16.36)d\theta=(-16.36\cdot \theta)|_{0.15\pi}^{0.74\pi}=-30.32\end{align*}$

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Example Question #67 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{19}{(3\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}+\frac{ (2\cdot (\frac{2}{5})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{21})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.2\text{ and }1.96\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.790,0.613)\text{ and }\overrightarrow{u_2}=(-0.771,-0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.14

-30.84

30.84

-15.42

Correct answer:

30.84

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{19}{(3\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}+\frac{ (2\cdot (\frac{2}{5})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{21})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot (\frac{2}{5})^{(\frac{r^{2}}{2})}\cdot r)}{21}+\frac{ 38}{(9\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{0.790})=0.21\pi;\theta_2=arctan(\frac{-0.637}{-0.771})=1.22\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.21\pi}^{1.22\pi}\int_{0.2}^{1.96}(\frac{(2\cdot (\frac{2}{5})^{(\frac{r^{2}}{2})}\cdot r)}{21}+\frac{ 38}{(9\cdot r)})drd\theta=(\frac{(38ln(r))}{9}+\frac{ (2\cdot (\frac{2}{5})^{(\frac{r^{2}}{2})})}{(21ln(\frac{2}{5}))})d\theta|_{0.2}^{1.96}\\&\int_{0.21\pi}^{1.22\pi}(9.721)d\theta=(9.721\cdot \theta)|_{0.21\pi}^{1.22\pi}=30.84\end{align*}$

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Example Question #68 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(23cos(2x^{2} + 2y^{2}) +\frac{ (\frac{26}{3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.960,0.279)\text{ and }\overrightarrow{u_2}=(-0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.99

-35.84

-1.99

11.95

Correct answer:

11.95

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(23cos(2x^{2} + 2y^{2}) +\frac{ (\frac{26}{3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(26\cdot r)}{(5\cdot 3^{(\frac{(3\cdot r^{2})}{2})})}+ 23\cdot rcos(2\cdot r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{-0.960})=0.91\pi;\theta_2=arctan(\frac{-0.976}{-0.218})=1.43\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.91\pi}^{1.43\pi}\int_{0}^{1.99}(\frac{(26\cdot r)}{(5\cdot 3^{(\frac{(3\cdot r^{2})}{2})})}+ 23\cdot rcos(2\cdot r^{2}))drd\theta=(\frac{(23sin(2\cdot r^{2}))}{4}-\frac{ 26}{(15\cdot 3^{(\frac{(3\cdot r^{2})}{2})}ln(3))})d\theta|_{0}^{1.99}\\&\int_{0.91\pi}^{1.43\pi}(7.313)d\theta=(7.313\cdot \theta)|_{0.91\pi}^{1.43\pi}=11.95\end{align*}$

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Example Question #63 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(42sin(2x^{2} + 2y^{2}))}{5}+\frac{ (13e^{(x^{2} + y^{2})})}{6})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.47\text{ and }1.05\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.588,0.809)\text{ and }\overrightarrow{u_2}=(0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

21.12

105.59

-4.22

-63.35

Correct answer:

21.12

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(42sin(2x^{2} + 2y^{2}))}{5}+\frac{ (13e^{(x^{2} + y^{2})})}{6})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(13\cdot re^{(r^{2})})}{6}+\frac{ (42\cdot rsin(2\cdot r^{2}))}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.809}{0.588})=0.3\pi;\theta_2=arctan(\frac{-0.918}{0.397})=1.63\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.3\pi}^{1.63\pi}\int_{0.47}^{1.05}(\frac{(13\cdot re^{(r^{2})})}{6}+\frac{ (42\cdot rsin(2\cdot r^{2}))}{5})drd\theta=(\frac{(13e^{(r^{2})})}{12}-\frac{ (21cos(r^{2})^{2})}{5})d\theta|_{0.47}^{1.05}\\&\int_{0.3\pi}^{1.63\pi}(5.054)d\theta=(5.054\cdot \theta)|_{0.3\pi}^{1.63\pi}=21.12\end{align*}$

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Example Question #64 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(13e^{(x^{2} + y^{2})})}{2}+\frac{ 4}{(\frac{1}{4})^{(x^{2} + y^{2})}})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.49\text{ and }1.57\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.861,0.509)\text{ and }\overrightarrow{u_2}=(0.844,-0.536)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2365.5

-78.85

394.25

78.85

Correct answer:

394.25

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.509}{0.861})=0.17\pi;\theta_2=arctan(\frac{-0.536}{0.844})=1.82\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.17\pi}^{1.82\pi}\int_{0.49}^{1.57}(\frac{(13\cdot re^{(r^{2})})}{2}+\frac{ (4\cdot r)}{(\frac{1}{4})^{(r^{2})}})drd\theta=(\frac{(13e^{(r^{2})})}{4}+\frac{ 2^{(2\cdot r^{2})}}{ln(2)})d\theta|_{0.49}^{1.57}\\&\int_{0.17\pi}^{1.82\pi}(76.06)d\theta=(76.06\cdot \theta)|_{0.17\pi}^{1.82\pi}=394.25\end{align*}$

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Calculus 3 : Double Integration in Polar Coordinates

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Example Question #61 : Double Integration In Polar Coordinates

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Example Question #61 : Double Integration In Polar Coordinates

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