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Calculus 3 : Double Integrals

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Example Questions

Example Question #1451 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3cos(x^{2} + y^{2}))}{44}-\frac{ 16}{(\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.27\text{ and }1.66\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(0.482,-0.876)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-346.42

2078.53

57.74

-86.61

Correct answer:

-346.42

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (3cos(x^{2} + y^{2}))}{44}-\frac{ 16}{(\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot rcos(r^{2}))}{44}-\frac{ (16\cdot r)}{(\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.876}{0.482})=1.66\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.79\pi}^{1.66\pi}\int_{0.27}^{1.66}(-\frac{ (3\cdot rcos(r^{2}))}{44}-\frac{ (16\cdot r)}{(\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}})drd\theta=(-\frac{ (3sin(r^{2}))}{88}-\frac{ (16\cdot 2^{(\frac{(3\cdot r^{2})}{2})})}{(3ln(2))})d\theta|_{0.27}^{1.66}\\&\int_{0.79\pi}^{1.66\pi}(-126.7)d\theta=(-126.7\theta)|_{0.79\pi}^{1.66\pi}=-346.42\end{align*}$

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Example Question #81 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{16}{(\frac{3}{2})^{(x^{2} + y^{2})}}-\frac{ cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{6})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.16\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.509,0.861)\text{ and }\overrightarrow{u_2}=(-0.063,-0.998)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

20.99

3.5

-5.25

-41.97

Correct answer:

20.99

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{16}{(\frac{3}{2})^{(x^{2} + y^{2})}}-\frac{ cos(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{6})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(16\cdot r)}{(\frac{3}{2})^{(r^{2})}}-\frac{ (rcos(\frac{(3\cdot r^{2})}{2}))}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.861}{-0.509})=0.67\pi;\theta_2=arctan(\frac{-0.998}{-0.063})=1.48\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.67\pi}^{1.48\pi}\int_{0}^{1.16}(\frac{(16\cdot r)}{(\frac{3}{2})^{(r^{2})}}-\frac{ (rcos(\frac{(3\cdot r^{2})}{2}))}{6})drd\theta=(-\frac{ sin(\frac{(3\cdot r^{2})}{2})}{18}-\frac{ 8}{((\frac{3}{2})^{(r^{2})}ln(\frac{3}{2}))})d\theta|_{0}^{1.16}\\&\int_{0.67\pi}^{1.48\pi}(8.247)d\theta=(8.247\theta)|_{0.67\pi}^{1.48\pi}=20.99\end{align*}$

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Example Question #82 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{7}+\frac{ (24\cdot (\frac{1}{2})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.75\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.998,0.063)\text{ and }\overrightarrow{u_2}=(-0.309,0.951)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.45

4.53

-2.26

2.26

Correct answer:

2.26

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{7}+\frac{ (24\cdot (\frac{1}{2})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(24\cdot (\frac{1}{2})^{(\frac{r^{2}}{2})}\cdot r)}{5}+\frac{ (rsin(\frac{(3\cdot r^{2})}{2}))}{7})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{0.998})=0.02\pi;\theta_2=arctan(\frac{0.951}{-0.309})=0.6\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.02\pi}^{0.6\pi}\int_{0}^{0.75}(\frac{(24\cdot (\frac{1}{2})^{(\frac{r^{2}}{2})}\cdot r)}{5}+\frac{ (rsin(\frac{(3\cdot r^{2})}{2}))}{7})drd\theta=(-\frac{ cos(\frac{(3\cdot r^{2})}{2})}{21}-\frac{ 24}{(5\cdot 2^{(\frac{r^{2}}{2})}ln(2))})d\theta|_{0}^{0.75}\\&\int_{0.02\pi}^{0.6\pi}(1.243)d\theta=(1.243\theta)|_{0.02\pi}^{0.6\pi}=2.26\end{align*}$

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Example Question #83 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (\frac{19}{(\frac{2}{3})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})}{2}-\frac{ 1}{(4\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.21\text{ and }1.93\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.918,0.397)\text{ and }\overrightarrow{u_2}=(-0.809,0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

14.09

-56.35

-338.11

169.05

Correct answer:

-56.35

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (\frac{19}{(\frac{2}{3})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})}{2}-\frac{ 1}{(4\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (19\cdot r)}{(2\cdot (\frac{2}{3})^{(\frac{r^{2}}{2})})}-\frac{ 1}{(4\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.397}{0.918})=0.13\pi;\theta_2=arctan(\frac{0.588}{-0.809})=0.8\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.13\pi}^{0.8\pi}\int_{0.21}^{1.93}(-\frac{ (19\cdot r)}{(2\cdot (\frac{2}{3})^{(\frac{r^{2}}{2})})}-\frac{ 1}{(4\cdot r)})drd\theta=(\frac{(19\cdot (3^{(r^{2})})^{(\frac{1}{2})})}{(2ln(\frac{2}{3})\cdot (2^{(r^{2})})^{(\frac{1}{2})})}-\frac{ ln(r)}{4})d\theta|_{0.21}^{1.93}\\&\int_{0.13\pi}^{0.8\pi}(-26.77)d\theta=(-26.77\theta)|_{0.13\pi}^{0.8\pi}=-56.35\end{align*}$

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Example Question #81 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(38sin(x^{2} + y^{2}))}{3}-\frac{ (8sin(2x^{2} + 2y^{2}))}{3})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.13\text{ and }1.55\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.988,0.156)\text{ and }\overrightarrow{u_2}=(-0.951,-0.309)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.72

-34.33

-11.44

34.33

Correct answer:

34.33

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(38sin(x^{2} + y^{2}))}{3}-\frac{ (8sin(2x^{2} + 2y^{2}))}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(38\cdot rsin(r^{2}))}{3}-\frac{ (8\cdot rsin(2\cdot r^{2}))}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.156}{0.988})=0.05\pi;\theta_2=arctan(\frac{-0.309}{-0.951})=1.1\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.05\pi}^{1.1\pi}\int_{0.13}^{1.55}(\frac{(38\cdot rsin(r^{2}))}{3}-\frac{ (8\cdot rsin(2\cdot r^{2}))}{3})drd\theta=(\frac{(cos(r^{2})\cdot (4cos(r^{2}) - 19))}{3})d\theta|_{0.13}^{1.55}\\&\int_{0.05\pi}^{1.1\pi}(10.41)d\theta=(10.41\theta)|_{0.05\pi}^{1.1\pi}=34.33\end{align*}$

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Example Question #81 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})} -\frac{ sin(x^{2} + y^{2})}{31})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.79\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1.02

-0.26

-4.1

0.51

Correct answer:

1.02

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-1.000}{-0.031})=1.49\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.26\pi}^{1.49\pi}\int_{0}^{0.79}(re^{(-\frac{r^{2}}{2})} -\frac{ (rsin(r^{2}))}{31})drd\theta=(\frac{cos(r^{2})}{62}- e^{(-\frac{r^{2}}{2})})d\theta|_{0}^{0.79}\\&\int_{0.26\pi}^{1.49\pi}(0.265)d\theta=(0.265\theta)|_{0.26\pi}^{1.49\pi}=1.02\end{align*}$

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Example Question #82 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{16}-\frac{ (4cos(x^{2} + y^{2}))}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.97\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.960,0.279)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

4.23

-0.7

0.12

-0.35

Correct answer:

-0.7

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{16}-\frac{ (4cos(x^{2} + y^{2}))}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{16}-\frac{ (4\cdot rcos(r^{2}))}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{-0.960})=0.91\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.91\pi}^{1.78\pi}\int_{0}^{0.97}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{16}-\frac{ (4\cdot rcos(r^{2}))}{5})drd\theta=(-\frac{ (2sin(r^{2}))}{5}-\frac{ (9e^{(-\frac{(2\cdot r^{2})}{3})})}{64})d\theta|_{0}^{0.97}\\&\int_{0.91\pi}^{1.78\pi}(-0.2577)d\theta=(-0.2577\theta)|_{0.91\pi}^{1.78\pi}=-0.7\end{align*}$

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Example Question #87 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3\cdot (\frac{2}{7})^{(x^{2} + y^{2})})}{7}-\frac{ (3sin(x^{2} + y^{2}))}{44})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.47\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.38

0.19

-0.13

-0.76

Correct answer:

0.38

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3\cdot (\frac{2}{7})^{(x^{2} + y^{2})})}{7}-\frac{ (3sin(x^{2} + y^{2}))}{44})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot (\frac{2}{7})^{(r^{2})}\cdot r)}{7}-\frac{ (3\cdot rsin(r^{2}))}{44})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.279}{0.960})=1.91\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.77\pi}^{1.91\pi}\int_{0}^{1.47}(\frac{(3\cdot (\frac{2}{7})^{(r^{2})}\cdot r)}{7}-\frac{ (3\cdot rsin(r^{2}))}{44})drd\theta=(\frac{(3cos(r^{2}))}{88}+\frac{ (3\cdot (\frac{2}{7})^{(r^{2})})}{(14ln(\frac{2}{7}))})d\theta|_{0}^{1.47}\\&\int_{0.77\pi}^{1.91\pi}(0.1066)d\theta=(0.1066\theta)|_{0.77\pi}^{1.91\pi}=0.38\end{align*}$

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Example Question #88 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(23e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{4}-\frac{ (25e^{(x^{2} + y^{2})})}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.33\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.707,0.707)\text{ and }\overrightarrow{u_2}=(-1.000,0.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-21.78

21.78

-43.56

43.56

Correct answer:

-43.56

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(23e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{4}-\frac{ (25e^{(x^{2} + y^{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(23\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{4}-\frac{ (25\cdot re^{(r^{2})})}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{0.707})=0.25\pi;\theta_2=arctan(\frac{0.000}{-1.000})=1\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.25\pi}^{1\pi}\int_{0}^{1.33}(\frac{(23\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{4}-\frac{ (25\cdot re^{(r^{2})})}{3})drd\theta=(-\frac{ (25e^{(r^{2})})}{6}-\frac{ (23e^{(-\frac{(3\cdot r^{2})}{2})})}{12})d\theta|_{0}^{1.33}\\&\int_{0.25\pi}^{1\pi}(-18.49)d\theta=(-18.49\theta)|_{0.25\pi}^{1\pi}=-43.56\end{align*}$

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Example Question #89 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(15e^{(x^{2} + y^{2})} -\frac{ (43cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{6})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.19\text{ and }1.59\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.613,0.790)\text{ and }\overrightarrow{u_2}=(0.996,-0.094)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

427.44

-2564.62

2564.62

-106.86

Correct answer:

427.44

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(15e^{(x^{2} + y^{2})} -\frac{ (43cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{6})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(15\cdot re^{(r^{2})} -\frac{ (43\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{0.613})=0.29\pi;\theta_2=arctan(\frac{-0.094}{0.996})=1.97\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.29\pi}^{1.97\pi}\int_{0.19}^{1.59}(15\cdot re^{(r^{2})} -\frac{ (43\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{6})drd\theta=(\frac{(15e^{(r^{2})})}{2}-\frac{ (43sin(\frac{(2\cdot r^{2})}{3}))}{8})d\theta|_{0.19}^{1.59}\\&\int_{0.29\pi}^{1.97\pi}(80.99)d\theta=(80.99\theta)|_{0.29\pi}^{1.97\pi}=427.44\end{align*}$

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Calculus 3 : Double Integrals

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Example Question #1451 : Calculus 3

Example Question #81 : Double Integration In Polar Coordinates

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Example Question #81 : Double Integrals

Example Question #81 : Double Integrals

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