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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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Example Question #71 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(49cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{5}+\frac{ e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})}}{2})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.73\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.063,0.998)\text{ and }\overrightarrow{u_2}=(-0.861,-0.509)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

62.25

-20.75

-6.92

20.75

Correct answer:

20.75

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.998}{-0.063})=0.52\pi;\theta_2=arctan(\frac{-0.509}{-0.861})=1.17\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.52\pi}^{1.17\pi}\int_{0}^{1.73}(\frac{(49\cdot rcos(\frac{r^{2}}{2}))}{5}+\frac{ (re^{(-\frac{r^{2}}{2})})}{2})drd\theta=(\frac{(49sin(\frac{r^{2}}{2}))}{5}-\frac{ e^{(-\frac{r^{2}}{2})}}{2})d\theta|_{0}^{1.73}\\&\int_{0.52\pi}^{1.17\pi}(10.16)d\theta=(10.16\cdot \theta)|_{0.52\pi}^{1.17\pi}=20.75\end{align*}$

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Example Question #71 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{1}{(x^{2} + y^{2})}-\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{33})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.31\text{ and }1.9\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.156,0.988)\text{ and }\overrightarrow{u_2}=(-0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

4.21

-1.4

1.4

-21.03

Correct answer:

4.21

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{1}{(x^{2} + y^{2})}-\frac{ cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{33})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{1}{r}-\frac{ (rcos(\frac{r^{2}}{2}))}{33})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.988}{0.156})=0.45\pi;\theta_2=arctan(\frac{-0.588}{-0.809})=1.2\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.45\pi}^{1.2\pi}\int_{0.31}^{1.9}(\frac{1}{r}-\frac{ (rcos(\frac{r^{2}}{2}))}{33})drd\theta=(ln(r) -\frac{ sin(\frac{r^{2}}{2})}{33})d\theta|_{0.31}^{1.9}\\&\int_{0.45\pi}^{1.2\pi}(1.785)d\theta=(1.785\cdot \theta)|_{0.45\pi}^{1.2\pi}=4.21\end{align*}$

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Example Question #72 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(11cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ (3\cdot (\frac{7}{2})^{(x^{2} + y^{2})})}{29})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.5\text{ and }1.64\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.156,0.988)\text{ and }\overrightarrow{u_2}=(0.249,-0.969)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-3.58

-71.64

17.91

35.82

Correct answer:

17.91

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(11cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ (3\cdot (\frac{7}{2})^{(x^{2} + y^{2})})}{29})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(11\cdot rcos(\frac{(2\cdot r^{2})}{3}) -\frac{ (3\cdot (\frac{7}{2})^{(r^{2})}\cdot r)}{29})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.988}{-0.156})=0.55\pi;\theta_2=arctan(\frac{-0.969}{0.249})=1.58\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.55\pi}^{1.58\pi}\int_{0.5}^{1.64}(11\cdot rcos(\frac{(2\cdot r^{2})}{3}) -\frac{ (3\cdot (\frac{7}{2})^{(r^{2})}\cdot r)}{29})drd\theta=(\frac{(33sin(\frac{(2\cdot r^{2})}{3}))}{4}-\frac{ (3\cdot (\frac{7}{2})^{(r^{2})})}{(58ln(\frac{7}{2}))})d\theta|_{0.5}^{1.64}\\&\int_{0.55\pi}^{1.58\pi}(5.535)d\theta=(5.535\cdot \theta)|_{0.55\pi}^{1.58\pi}=17.91\end{align*}$

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Example Question #73 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}}{24}+\frac{ sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.67\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.397,0.918)\text{ and }\overrightarrow{u_2}=(-0.279,-0.960)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.74

2.22

-3.7

-0.15

Correct answer:

0.74

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.918}{0.397})=0.37\pi;\theta_2=arctan(\frac{-0.960}{-0.279})=1.41\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.37\pi}^{1.41\pi}\int_{0}^{1.67}(\frac{(re^{(\frac{r^{2}}{2})})}{24}+\frac{ (rsin(\frac{(3\cdot r^{2})}{2}))}{5})drd\theta=(\frac{e^{(\frac{r^{2}}{2})}}{24}-\frac{ cos(\frac{(3\cdot r^{2})}{2})}{15})d\theta|_{0}^{1.67}\\&\int_{0.37\pi}^{1.41\pi}(0.2267)d\theta=(0.2267\cdot \theta)|_{0.37\pi}^{1.41\pi}=0.74\end{align*}$

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Example Question #75 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{23}- 32e^{(- 2x^{2} - 2y^{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.26\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.969,0.249)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-15.79

5.26

63.16

-31.58

Correct answer:

-15.79

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{23}- 32e^{(- 2x^{2} - 2y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{23}- 32\cdot re^{(-2\cdot r^{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.249}{-0.969})=0.92\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.26\pi}^{0.92\pi}\int_{0}^{1.26}(\frac{(3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{23}- 32\cdot re^{(-2\cdot r^{2})})drd\theta=(8e^{(-2\cdot r^{2})} -\frac{ (9cos(\frac{(2\cdot r^{2})}{3}))}{92})d\theta|_{0}^{1.26}\\&\int_{0.26\pi}^{0.92\pi}(-7.616)d\theta=(-7.616\theta)|_{0.26\pi}^{0.92\pi}=-15.79\end{align*}$

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Example Question #76 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2cos(x^{2} + y^{2}))}{15}-\frac{ 2}{(23\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.32\text{ and }1.45\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.996,0.094)\text{ and }\overrightarrow{u_2}=(0.876,-0.482)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.22

0.07

-0.11

0.22

Correct answer:

-0.22

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2cos(x^{2} + y^{2}))}{15}-\frac{ 2}{(23\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot rcos(r^{2}))}{15}-\frac{ 2}{(23\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.094}{-0.996})=0.97\pi;\theta_2=arctan(\frac{-0.482}{0.876})=1.84\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.97\pi}^{1.84\pi}\int_{0.32}^{1.45}(\frac{(2\cdot rcos(r^{2}))}{15}-\frac{ 2}{(23\cdot r)})drd\theta=(\frac{sin(r^{2})}{15}-\frac{ (2ln(r))}{23})d\theta|_{0.32}^{1.45}\\&\int_{0.97\pi}^{1.84\pi}(-0.08074)d\theta=(-0.08074\theta)|_{0.97\pi}^{1.84\pi}=-0.22\end{align*}$

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Example Question #77 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (5sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{23}-\frac{ 18}{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.14\text{ and }1.64\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(1.000,0.031)\text{ and }\overrightarrow{u_2}=(-0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-189.97

569.92

37.99

-47.49

Correct answer:

-189.97

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (5sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{23}-\frac{ 18}{(x^{2} + y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (5\cdot rsin(\frac{r^{2}}{2}))}{23}-\frac{ 18}{r})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.031}{1.000})=0.01\pi;\theta_2=arctan(\frac{-0.918}{-0.397})=1.37\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.01\pi}^{1.37\pi}\int_{0.14}^{1.64}(-\frac{ (5\cdot rsin(\frac{r^{2}}{2}))}{23}-\frac{ 18}{r})drd\theta=(\frac{(5cos(\frac{r^{2}}{2}))}{23}- 18ln(r))d\theta|_{0.14}^{1.64}\\&\int_{0.01\pi}^{1.37\pi}(-44.46)d\theta=(-44.46\theta)|_{0.01\pi}^{1.37\pi}=-189.97\end{align*}$

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Example Question #71 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(9sin(2x^{2} + 2y^{2}) - 16sin(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.37\text{ and }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

9.2

-110.4

36.8

-18.4

Correct answer:

-18.4

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(9sin(2x^{2} + 2y^{2}) - 16sin(x^{2} + y^{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(9\cdot rsin(2\cdot r^{2}) - 16\cdot rsin(r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.15\pi}^{1.38\pi}\int_{0.37}^{1.31}(9\cdot rsin(2\cdot r^{2}) - 16\cdot rsin(r^{2}))drd\theta=(-\frac{(cos(r^{2})\cdot (9cos(r^{2}) - 16))}{2})d\theta|_{0.37}^{1.31}\\&\int_{0.15\pi}^{1.38\pi}(-4.762)d\theta=(-4.762\theta)|_{0.15\pi}^{1.38\pi}=-18.4\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #63 : Double Integration In Polar Coordinates

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