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Calculus 3 : Double Integration in Polar Coordinates

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Example Questions

Example Question #161 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(15e^{(2x^{2} + 2y^{2})} -\frac{ (2cos(2x^{2} + 2y^{2}))}{21})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.07\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.951,0.309)\text{ and }\overrightarrow{u_2}=(-0.637,-0.771)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-739.69

123.28

20.55

-61.64

Correct answer:

123.28

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.309}{0.951})=0.1\pi;\theta_2=arctan(\frac{-0.771}{-0.637})=1.28\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.1\pi}^{1.28\pi}\int_{0}^{1.07}(15\cdot re^{(2\cdot r^{2})} -\frac{ (2\cdot rcos(2\cdot r^{2}))}{21})drd\theta=(\frac{(15e^{(2\cdot r^{2})})}{4}-\frac{ sin(2\cdot r^{2})}{42})d\theta|_{0}^{1.07}\\&\int_{0.1\pi}^{1.28\pi}(33.26)d\theta=(33.26\theta)|_{0.1\pi}^{1.28\pi}=123.28\end{align*}$

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Example Question #1531 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{44}-\frac{ 3}{(43\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.39\text{ and }1.66\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(0.905,-0.426)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.1

-0.42

-0.1

0.42

Correct answer:

-0.42

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{44}-\frac{ 3}{(43\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{44}-\frac{ 3}{(43\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.426}{0.905})=1.86\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.77\pi}^{1.86\pi}\int_{0.39}^{1.66}(-\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{44}-\frac{ 3}{(43\cdot r)})drd\theta=(\frac{(3cos(\frac{(2\cdot r^{2})}{3}))}{176}-\frac{ (3ln(r))}{43})d\theta|_{0.39}^{1.66}\\&\int_{0.77\pi}^{1.86\pi}(-0.1225)d\theta=(-0.1225\theta)|_{0.77\pi}^{1.86\pi}=-0.42\end{align*}$

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Example Question #163 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(6e^{(- 2x^{2} - 2y^{2})})}{31}-\frac{ cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{35})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.37\text{ and }1.51\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.930,0.368)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.39

0.06

-0.03

0.19

Correct answer:

0.06

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(6e^{(- 2x^{2} - 2y^{2})})}{31}-\frac{ cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{35})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(6\cdot re^{(-2\cdot r^{2})})}{31}-\frac{ (rcos(\frac{(3\cdot r^{2})}{2}))}{35})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.368}{-0.930})=0.88\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.88\pi}^{1.38\pi}\int_{0.37}^{1.51}(\frac{(6\cdot re^{(-2\cdot r^{2})})}{31}-\frac{ (rcos(\frac{(3\cdot r^{2})}{2}))}{35})drd\theta=(-\frac{ (3e^{(-2\cdot r^{2})})}{62}-\frac{ sin(\frac{(3\cdot r^{2})}{2})}{105})d\theta|_{0.37}^{1.51}\\&\int_{0.88\pi}^{1.38\pi}(0.04085)d\theta=(0.04085\theta)|_{0.88\pi}^{1.38\pi}=0.06\end{align*}$

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Example Question #164 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5\cdot (\frac{2}{5})^{(x^{2} + y^{2})})}{16}- 13cos(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.82\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(1.000,0.031)\text{ and }\overrightarrow{u_2}=(-0.536,0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-41.78

1.39

-8.36

8.36

Correct answer:

-8.36

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5\cdot (\frac{2}{5})^{(x^{2} + y^{2})})}{16}- 13cos(x^{2} + y^{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot (\frac{2}{5})^{(r^{2})}\cdot r)}{16}- 13\cdot rcos(r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.031}{1.000})=0.01\pi;\theta_2=arctan(\frac{0.844}{-0.536})=0.68\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.01\pi}^{0.68\pi}\int_{0}^{0.82}(\frac{(5\cdot (\frac{2}{5})^{(r^{2})}\cdot r)}{16}- 13\cdot rcos(r^{2}))drd\theta=(\frac{(5\cdot (\frac{2}{5})^{(r^{2})})}{(32ln(\frac{2}{5}))}-\frac{ (13sin(r^{2}))}{2})d\theta|_{0}^{0.82}\\&\int_{0.01\pi}^{0.68\pi}(-3.97)d\theta=(-3.97\theta)|_{0.01\pi}^{0.68\pi}=-8.36\end{align*}$

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Example Question #165 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{22}- 3cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.33\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.249,0.969)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-5.07

0.84

30.41

-15.2

Correct answer:

-5.07

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.969}{-0.249})=0.58\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.58\pi}^{1.38\pi}\int_{0}^{1.33}(\frac{(3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{22}- 3\cdot rcos(\frac{(2\cdot r^{2})}{3}))drd\theta=(-\frac{ (9cos(\frac{(2\cdot r^{2})}{3}))}{88}-\frac{ (9sin(\frac{(2\cdot r^{2})}{3}))}{4})d\theta|_{0}^{1.33}\\&\int_{0.58\pi}^{1.38\pi}(-2.016)d\theta=(-2.016\theta)|_{0.58\pi}^{1.38\pi}=-5.07\end{align*}$

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Example Question #166 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(48\cdot (\frac{1}{3})^{(x^{2} + y^{2})})}{5}-\frac{ (47sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{4})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.47\text{ and }1.38\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.790,0.613)\text{ and }\overrightarrow{u_2}=(-0.930,0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

28.88

-57.76

-9.63

1.93

Correct answer:

-9.63

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(48\cdot (\frac{1}{3})^{(x^{2} + y^{2})})}{5}-\frac{ (47sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(48\cdot (\frac{1}{3})^{(r^{2})}\cdot r)}{5}-\frac{ (47\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{4})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{0.790})=0.21\pi;\theta_2=arctan(\frac{0.368}{-0.930})=0.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.21\pi}^{0.88\pi}\int_{0.47}^{1.38}(\frac{(48\cdot (\frac{1}{3})^{(r^{2})}\cdot r)}{5}-\frac{ (47\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{4})drd\theta=(\frac{(47cos(\frac{(3\cdot r^{2})}{2}))}{12}-\frac{ 24}{(5\cdot 3^{(r^{2})}ln(3))})d\theta|_{0.47}^{1.38}\\&\int_{0.21\pi}^{0.88\pi}(-4.574)d\theta=(-4.574\theta)|_{0.21\pi}^{0.88\pi}=-9.63\end{align*}$

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Example Question #167 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{10}+\frac{ (3\cdot (\frac{2}{3})^{(x^{2} + y^{2})})}{2})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.3\text{ and }1.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.562,0.827)\text{ and }\overrightarrow{u_2}=(0.426,-0.905)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1.09

6.53

-13.06

19.59

Correct answer:

6.53

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.827}{0.562})=0.31\pi;\theta_2=arctan(\frac{-0.905}{0.426})=1.64\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.31\pi}^{1.64\pi}\int_{0.3}^{1.77}(\frac{(3\cdot (\frac{2}{3})^{(r^{2})}\cdot r)}{2}+\frac{ (3\cdot rsin(\frac{r^{2}}{2}))}{10})drd\theta=(\frac{(3\cdot (\frac{2}{3})^{(r^{2})})}{(4ln(\frac{2}{3}))}-\frac{ (3cos(\frac{r^{2}}{2}))}{10})d\theta|_{0.3}^{1.77}\\&\int_{0.31\pi}^{1.64\pi}(1.563)d\theta=(1.563\theta)|_{0.31\pi}^{1.64\pi}=6.53\end{align*}$

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Example Question #168 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{28}{(3\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}-\frac{ (5sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{21})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.1\text{ and }1.61\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.482,0.876)\text{ and }\overrightarrow{u_2}=(-0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

82.65

-6.89

41.33

-82.65

Correct answer:

41.33

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{28}{(3\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}-\frac{ (5sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{21})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{56}{(9\cdot r)}-\frac{ (5\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{21})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.876}{-0.482})=0.66\pi;\theta_2=arctan(\frac{-0.976}{-0.218})=1.43\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.66\pi}^{1.43\pi}\int_{0.1}^{1.61}(\frac{56}{(9\cdot r)}-\frac{ (5\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{21})drd\theta=(\frac{(5cos(\frac{(2\cdot r^{2})}{3}))}{28}+\frac{ (56ln(r))}{9})d\theta|_{0.1}^{1.61}\\&\int_{0.66\pi}^{1.43\pi}(17.08)d\theta=(17.08\theta)|_{0.66\pi}^{1.43\pi}=41.33\end{align*}$

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Example Question #169 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(2e^{(2x^{2} + 2y^{2})} - 4e^{(- x^{2} - y^{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1964.7

785.88

-15717.6

3929.4

Correct answer:

3929.4

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.15\pi}^{1.5\pi}\int_{0}^{1.94}(2\cdot re^{(2\cdot r^{2})} - 4\cdot re^{(-r^{2})})drd\theta=(\frac{(e^{(-r^{2})}\cdot (e^{(3\cdot r^{2})} + 4))}{2})d\theta|_{0}^{1.94}\\&\int_{0.15\pi}^{1.5\pi}(926.5)d\theta=(926.5\theta)|_{0.15\pi}^{1.5\pi}=3929.4\end{align*}$

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Example Question #170 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{49}- 2\cdot (\frac{7}{2})^{(2x^{2} + 2y^{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.22\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.454,0.891)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-233.52

-46.7

23.35

46.7

Correct answer:

-46.7

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{49}- 2\cdot (\frac{7}{2})^{(2x^{2} + 2y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{49}- 2\cdot (\frac{7}{2})^{(2\cdot r^{2})}\cdot r)drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.891}{-0.454})=0.65\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.65\pi}^{1.57\pi}\int_{0}^{1.22}(\frac{(5\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{49}- 2\cdot (\frac{7}{2})^{(2\cdot r^{2})}\cdot r)drd\theta=(\frac{(15sin(\frac{(2\cdot r^{2})}{3}))}{196}- (\frac{\frac{7}{2})^{(2\cdot r^{2})}}{(2ln(\frac{7}{2}))})d\theta|_{0}^{1.22}\\&\int_{0.65\pi}^{1.57\pi}(-16.16)d\theta=(-16.16\theta)|_{0.65\pi}^{1.57\pi}=-46.7\end{align*}$

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Calculus 3 : Double Integration in Polar Coordinates

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Example Question #161 : Double Integration In Polar Coordinates

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