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Calculus 3 : Double Integrals

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Example Questions

Example Question #171 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (13\cdot (\frac{2}{7})^{(2x^{2} + 2y^{2})})}{6}-\frac{ (2\cdot 3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.48\text{ and }0.98\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.279,0.960)\text{ and }\overrightarrow{u_2}=(-0.982,-0.187)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.91

-1.82

-10.94

9.11

Correct answer:

-1.82

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (13\cdot (\frac{2}{7})^{(2x^{2} + 2y^{2})})}{6}-\frac{ (2\cdot 3^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})}\cdot r)}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.960}{0.279})=0.41\pi;\theta_2=arctan(\frac{-0.187}{-0.982})=1.06\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.41\pi}^{1.06\pi}\int_{0.48}^{0.98}(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})}\cdot r)}{6})drd\theta=(-\frac{ (2\cdot 3^{(\frac{(3\cdot r^{2})}{2})})}{(9ln(3))}-\frac{ (13\cdot (\frac{2}{7})^{(2\cdot r^{2})})}{(24ln(\frac{2}{7}))})d\theta|_{0.48}^{0.98}\\&\int_{0.41\pi}^{1.06\pi}(-0.8927)d\theta=(-0.8927\theta)|_{0.41\pi}^{1.06\pi}=-1.82\end{align*}$

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Example Question #172 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 5e^{(- x^{2} - y^{2})} -\frac{ (5\cdot (\frac{2}{3})^{(x^{2} + y^{2})})}{17})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.6\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.482,0.876)\text{ and }\overrightarrow{u_2}=(0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-10.22

1.7

40.87

-5.11

Correct answer:

-10.22

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.876}{0.482})=0.34\pi;\theta_2=arctan(\frac{-0.930}{0.368})=1.62\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.34\pi}^{1.62\pi}\int_{0}^{1.6}(-\frac{ (5\cdot (\frac{2}{3})^{(r^{2})}\cdot r)}{17}- 5\cdot re^{(-r^{2})})drd\theta=(\frac{(5e^{(-r^{2})})}{2}-\frac{ (5\cdot (\frac{2}{3})^{(r^{2})})}{(34ln(\frac{2}{3}))})d\theta|_{0}^{1.6}\\&\int_{0.34\pi}^{1.62\pi}(-2.541)d\theta=(-2.541\theta)|_{0.34\pi}^{1.62\pi}=-10.22\end{align*}$

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Example Question #173 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{cos(2x^{2} + 2y^{2})}{4}+\frac{ (2e^{(x^{2} + y^{2})})}{13})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.17\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.56

-0.09

-2.59

0.52

Correct answer:

0.52

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.43\pi}^{1.09\pi}\int_{0}^{1.17}(\frac{(2\cdot re^{(r^{2})})}{13}+\frac{ (rcos(2\cdot r^{2}))}{4})drd\theta=(\frac{e^{(r^{2})}}{13}+\frac{ sin(2\cdot r^{2})}{16})d\theta|_{0}^{1.17}\\&\int_{0.43\pi}^{1.09\pi}(0.25)d\theta=(0.25\theta)|_{0.43\pi}^{1.09\pi}=0.52\end{align*}$

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Example Question #174 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (\frac{29}{4^{(x^{2} + y^{2})}})}{4}-\frac{ 43}{(6\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.13\text{ and }1.83\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-286.14

17.88

-71.54

71.54

Correct answer:

-71.54

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (\frac{29}{4^{(x^{2} + y^{2})}})}{4}-\frac{ 43}{(6\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (29\cdot r)}{(4\cdot 4^{(r^{2})})}-\frac{ 43}{(6\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{-0.031})=0.51\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.51\pi}^{1.57\pi}\int_{0.13}^{1.83}(-\frac{ (29\cdot r)}{(4\cdot 4^{(r^{2})})}-\frac{ 43}{(6\cdot r)})drd\theta=(\frac{29}{(16\cdot 2^{(2\cdot r^{2})}ln(2))}-\frac{ (43ln(r))}{6})d\theta|_{0.13}^{1.83}\\&\int_{0.51\pi}^{1.57\pi}(-21.48)d\theta=(-21.48\theta)|_{0.51\pi}^{1.57\pi}=-71.54\end{align*}$

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Example Question #175 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{46}-\frac{ (29sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{6})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.55\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(-0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.09

-0.28

-1.42

1.42

Correct answer:

-0.28

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (3e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{46}-\frac{ (29sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{6})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{46}-\frac{ (29\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{-0.707})=0.75\pi;\theta_2=arctan(\frac{-0.750}{-0.661})=1.27\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.75\pi}^{1.27\pi}\int_{0}^{0.55}(-\frac{ (3\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{46}-\frac{ (29\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{6})drd\theta=(\frac{(29cos(\frac{(3\cdot r^{2})}{2}))}{18}-\frac{ (9e^{(\frac{(2\cdot r^{2})}{3})})}{184})d\theta|_{0}^{0.55}\\&\int_{0.75\pi}^{1.27\pi}(-0.174)d\theta=(-0.174\theta)|_{0.75\pi}^{1.27\pi}=-0.28\end{align*}$

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Example Question #171 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(- 2x^{2} - 2y^{2})}}{4}-\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{35})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.7\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.941,0.339)\text{ and }\overrightarrow{u_2}=(0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.12

-0.02

-0.35

0.35

Correct answer:

0.12

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.339}{-0.941})=0.89\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.89\pi}^{1.97\pi}\int_{0}^{1.7}(\frac{(re^{(-2\cdot r^{2})})}{4}-\frac{ (rcos(\frac{r^{2}}{2}))}{35})drd\theta=(-\frac{ e^{(-2\cdot r^{2})}}{16}-\frac{ sin(\frac{r^{2}}{2})}{35})d\theta|_{0}^{1.7}\\&\int_{0.89\pi}^{1.97\pi}(0.03396)d\theta=(0.03396\theta)|_{0.89\pi}^{1.97\pi}=0.12\end{align*}$

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Example Question #177 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (42cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{5}-\frac{ (\frac{8}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.21\text{ and }1.35\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.771,0.637)\text{ and }\overrightarrow{u_2}=(-0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

9.13

-4.56

-18.25

91.27

Correct answer:

-18.25

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (42cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{5}-\frac{ (\frac{8}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (8\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}-\frac{ (42\cdot rcos(\frac{r^{2}}{2}))}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.637}{-0.771})=0.78\pi;\theta_2=arctan(\frac{-0.918}{-0.397})=1.37\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.78\pi}^{1.37\pi}\int_{0.21}^{1.35}(-\frac{ (8\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}-\frac{ (42\cdot rcos(\frac{r^{2}}{2}))}{5})drd\theta=(\frac{6}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{7}))}-\frac{ (42sin(\frac{r^{2}}{2}))}{5})d\theta|_{0.21}^{1.35}\\&\int_{0.78\pi}^{1.37\pi}(-9.848)d\theta=(-9.848\theta)|_{0.78\pi}^{1.37\pi}=-18.25\end{align*}$

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Example Question #652 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{35}-\frac{ 41}{(3\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.46\text{ and }1.53\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.249,0.969)\text{ and }\overrightarrow{u_2}=(-0.000,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-55.6

9.27

55.6

-333.58

Correct answer:

-55.6

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (6cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{35}-\frac{ 41}{(3\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (6\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{35}-\frac{ 41}{(3\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.969}{0.249})=0.42\pi;\theta_2=arctan(\frac{-1.000}{-0.000})=1.5\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.42\pi}^{1.5\pi}\int_{0.46}^{1.53}(-\frac{ (6\cdot rcos(\frac{(3\cdot r^{2})}{2}))}{35}-\frac{ 41}{(3\cdot r)})drd\theta=(-\frac{ (2sin(\frac{(3\cdot r^{2})}{2}))}{35}-\frac{ (41ln(r))}{3})d\theta|_{0.46}^{1.53}\\&\int_{0.42\pi}^{1.5\pi}(-16.39)d\theta=(-16.39\theta)|_{0.42\pi}^{1.5\pi}=-55.6\end{align*}$

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Example Question #179 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5}- 32cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.28\text{ and }1.58\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

38.7

-19.35

19.35

-3.23

Correct answer:

19.35

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5}- 32cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}- 32\cdot rcos(\frac{r^{2}}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.77\pi}^{1.33\pi}\int_{0.28}^{1.58}(\frac{(48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}- 32\cdot rcos(\frac{r^{2}}{2}))drd\theta=(- 32sin(\frac{r^{2}}{2}) -\frac{ 36}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{7}))})d\theta|_{0.28}^{1.58}\\&\int_{0.77\pi}^{1.33\pi}(11.0)d\theta=(11.0\theta)|_{0.77\pi}^{1.33\pi}=19.35\end{align*}$

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Example Question #180 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ (2\cdot (\frac{7}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{25})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.34\text{ and }1.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.998,0.063)\text{ and }\overrightarrow{u_2}=(-0.094,0.996)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-255.98

-42.66

127.99

511.96

Correct answer:

127.99

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ (2\cdot (\frac{7}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{25})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{25}+ re^{(\frac{(3\cdot r^{2})}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{0.998})=0.02\pi;\theta_2=arctan(\frac{0.996}{-0.094})=0.53\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.02\pi}^{0.53\pi}\int_{0.34}^{1.91}(\frac{(2\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{25}+ re^{(\frac{(3\cdot r^{2})}{2})})drd\theta=(\frac{e^{(\frac{(3\cdot r^{2})}{2})}}{3}+\frac{ (3\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})})}{(50ln(\frac{7}{2}))})d\theta|_{0.34}^{1.91}\\&\int_{0.02\pi}^{0.53\pi}(79.88)d\theta=(79.88\theta)|_{0.02\pi}^{0.53\pi}=127.99\end{align*}$

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Calculus 3 : Double Integrals

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