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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 #29 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{24}-\frac{ sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{43})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.454,0.891)\text{ and }\overrightarrow{u_2}=(0.063,-0.998)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.05

-0.14

0.42

0.07

Correct answer:

-0.14

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{24}-\frac{ sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{43})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (rcos(\frac{(2\cdot r^{2})}{3}))}{24}-\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{43})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.891}{-0.454})=0.65\pi;\theta_2=arctan(\frac{-0.998}{0.063})=1.52\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.65\pi}^{1.52\pi}\int_{0}^{1.63}(-\frac{ (rcos(\frac{(2\cdot r^{2})}{3}))}{24}-\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{43})drd\theta=(\frac{(3cos(\frac{(2\cdot r^{2})}{3}))}{172}-\frac{ sin(\frac{(2\cdot r^{2})}{3})}{32})d\theta|_{0}^{1.63}\\&\int_{0.65\pi}^{1.52\pi}(-0.05154)d\theta=(-0.05154\cdot \theta)|_{0.65\pi}^{1.52\pi}=-0.14\end{align*}$

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Example Question #30 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{9}+\frac{ 1}{(41\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.22\text{ and }1.05\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.998,0.063)\text{ and }\overrightarrow{u_2}=(0.918,-0.397)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

6.93

1.39

-0.46

-2.77

Correct answer:

1.39

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{9}+\frac{ 1}{(41\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{9}+\frac{ 3}{(82\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{-0.998})=0.98\pi;\theta_2=arctan(\frac{-0.397}{0.918})=1.87\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.98\pi}^{1.87\pi}\int_{0.22}^{1.05}(\frac{(5\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{9}+\frac{ 3}{(82\cdot r)})drd\theta=(\frac{(5e^{(\frac{(2\cdot r^{2})}{3})})}{12}+\frac{ (3ln(r))}{82})d\theta|_{0.22}^{1.05}\\&\int_{0.98\pi}^{1.87\pi}(0.4958)d\theta=(0.4958\cdot \theta)|_{0.98\pi}^{1.87\pi}=1.39\end{align*}$

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Example Question #31 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{3}-\frac{ (11sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{4})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.37\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(0.827,-0.562)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.28

-0.09

0.06

1.68

Correct answer:

-0.28

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{3}-\frac{ (11sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{4})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot re^{(\frac{r^{2}}{2})})}{3}-\frac{ (11\cdot rsin(\frac{r^{2}}{2}))}{4})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.562}{0.827})=1.81\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.79\pi}^{1.81\pi}\int_{0}^{1.37}(\frac{(2\cdot re^{(\frac{r^{2}}{2})})}{3}-\frac{ (11\cdot rsin(\frac{r^{2}}{2}))}{4})drd\theta=(\frac{(11cos(\frac{r^{2}}{2}))}{4}+\frac{ (2e^{(\frac{r^{2}}{2})})}{3})d\theta|_{0}^{1.37}\\&\int_{0.79\pi}^{1.81\pi}(-0.0873)d\theta=(-0.0873\cdot \theta)|_{0.79\pi}^{1.81\pi}=-0.28\end{align*}$

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Example Question #32 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(20sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) +\frac{ 3}{(7\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.41\text{ and }1.57\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.637,0.771)\text{ and }\overrightarrow{u_2}=(0.707,-0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-8.83

211.96

52.99

-264.94

Correct answer:

52.99

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(20sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) +\frac{ 3}{(7\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(20\cdot rsin(\frac{(2\cdot r^{2})}{3}) +\frac{ 2}{(7\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.771}{-0.637})=0.72\pi;\theta_2=arctan(\frac{-0.707}{0.707})=1.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.72\pi}^{1.75\pi}\int_{0.41}^{1.57}(20\cdot rsin(\frac{(2\cdot r^{2})}{3}) +\frac{ 2}{(7\cdot r)})drd\theta=(\frac{(2ln(r))}{7}- 15cos(\frac{(2\cdot r^{2})}{3}))d\theta|_{0.41}^{1.57}\\&\int_{0.72\pi}^{1.75\pi}(16.38)d\theta=(16.38\cdot \theta)|_{0.72\pi}^{1.75\pi}=52.99\end{align*}$

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Example Question #33 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{28}-\frac{ 12}{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.4\text{ and }1.52\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.000)\text{ and }\overrightarrow{u_2}=(0.729,-0.685)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

8.48

50.86

-25.43

-4.24

Correct answer:

-25.43

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{28}-\frac{ 12}{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rcos(\frac{r^{2}}{2}))}{28}-\frac{ 8}{r})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.000}{-1.000})=1\pi;\theta_2=arctan(\frac{-0.685}{0.729})=1.76\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{1\pi}^{1.76\pi}\int_{0.4}^{1.52}(\frac{(rcos(\frac{r^{2}}{2}))}{28}-\frac{ 8}{r})drd\theta=(\frac{sin(\frac{r^{2}}{2})}{28}- 8ln(r))d\theta|_{0.4}^{1.52}\\&\int_{1\pi}^{1.76\pi}(-10.65)d\theta=(-10.65\cdot \theta)|_{1\pi}^{1.76\pi}=-25.43\end{align*}$

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Example Question #34 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{44}-\frac{ e^{(2x^{2} + 2y^{2})}}{8})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.44\text{ and }1.83\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.918,0.397)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-99.2

99.2

-33.07

19.84

Correct answer:

-99.2

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}{44}-\frac{ e^{(2x^{2} + 2y^{2})}}{8})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(\frac{r^{2}}{2}))}{44}-\frac{ (re^{(2\cdot r^{2})})}{8})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.397}{0.918})=0.13\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.13\pi}^{1.38\pi}\int_{0.44}^{1.83}(\frac{(rsin(\frac{r^{2}}{2}))}{44}-\frac{ (re^{(2\cdot r^{2})})}{8})drd\theta=(-\frac{ cos(\frac{r^{2}}{2})}{44}-\frac{ e^{(2\cdot r^{2})}}{32})d\theta|_{0.44}^{1.83}\\&\int_{0.13\pi}^{1.38\pi}(-25.26)d\theta=(-25.26\cdot \theta)|_{0.13\pi}^{1.38\pi}=-99.2\end{align*}$

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Example Question #1401 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{4}- 39e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.125,0.992)\text{ and }\overrightarrow{u_2}=(0.613,-0.790)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-262.23

262.23

-524.46

43.7

Correct answer:

-262.23

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{4}- 39e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(\frac{(2\cdot r^{2})}{3}))}{4}- 39\cdot re^{(\frac{(3\cdot r^{2})}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.992}{0.125})=0.46\pi;\theta_2=arctan(\frac{-0.790}{0.613})=1.71\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.46\pi}^{1.71\pi}\int_{0}^{1.1}(\frac{(rsin(\frac{(2\cdot r^{2})}{3}))}{4}- 39\cdot re^{(\frac{(3\cdot r^{2})}{2})})drd\theta=(-\frac{ (3cos(\frac{(2\cdot r^{2})}{3}))}{16}- 13e^{(\frac{(3\cdot r^{2})}{2})})d\theta|_{0}^{1.1}\\&\int_{0.46\pi}^{1.71\pi}(-66.78)d\theta=(-66.78\cdot \theta)|_{0.46\pi}^{1.71\pi}=-262.23\end{align*}$

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Example Question #36 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(32e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{7})dxdy\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.84\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.861,0.509)\text{ and }\overrightarrow{u_2}=(0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-25.52

5.1

25.52

-2.55

Correct answer:

5.1

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.509}{-0.861})=0.83\pi;\theta_2=arctan(\frac{-0.861}{0.509})=1.67\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.83\pi}^{1.67\pi}\int_{0}^{0.84}(\frac{(32\cdot re^{(\frac{r^{2}}{2})})}{7})drd\theta=(\frac{(32e^{(\frac{r^{2}}{2})})}{7})d\theta|_{0}^{0.84}\\&\int_{0.83\pi}^{1.67\pi}(1.934)d\theta=(1.934\cdot \theta)|_{0.83\pi}^{1.67\pi}=5.1\end{align*}$

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Example Question #37 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{1}{(8\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}-\frac{ (26e^{(- x^{2} - y^{2})})}{3})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.45\text{ and }1.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.588,0.809)\text{ and }\overrightarrow{u_2}=(0.482,-0.876)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

54.48

-1.82

3.03

-9.08

Correct answer:

-9.08

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{1}{(8\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}-\frac{ (26e^{(- x^{2} - y^{2})})}{3})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{3}{(16\cdot r)}-\frac{ (26\cdot re^{(-r^{2})})}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.809}{-0.588})=0.7\pi;\theta_2=arctan(\frac{-0.876}{0.482})=1.66\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.7\pi}^{1.66\pi}\int_{0.45}^{1.65}(\frac{3}{(16\cdot r)}-\frac{ (26\cdot re^{(-r^{2})})}{3})drd\theta=(\frac{(13e^{(-r^{2})})}{3}+\frac{ (3ln(r))}{16})d\theta|_{0.45}^{1.65}\\&\int_{0.7\pi}^{1.66\pi}(-3.011)d\theta=(-3.011\cdot \theta)|_{0.7\pi}^{1.66\pi}=-9.08\end{align*}$

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Example Question #38 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(2x^{2} + 2y^{2}))}{23}- 19e^{(- x^{2} - y^{2})})dxdy\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.26\text{ and }1.39\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.790,0.613)\text{ and }\overrightarrow{u_2}=(0.562,-0.827)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-35.01

5.83

35.01

-8.75

Correct answer:

-35.01

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dxdy=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3cos(2x^{2} + 2y^{2}))}{23}- 19e^{(- x^{2} - y^{2})})dxdy\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rcos(2\cdot r^{2}))}{23}- 19\cdot re^{(-r^{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{0.790})=0.21\pi;\theta_2=arctan(\frac{-0.827}{0.562})=1.69\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.21\pi}^{1.69\pi}\int_{0.26}^{1.39}(\frac{(3\cdot rcos(2\cdot r^{2}))}{23}- 19\cdot re^{(-r^{2})})drd\theta=(\frac{(19e^{(-r^{2})})}{2}+\frac{ (3sin(2\cdot r^{2}))}{92})d\theta|_{0.26}^{1.39}\\&\int_{0.21\pi}^{1.69\pi}(-7.529)d\theta=(-7.529\cdot \theta)|_{0.21\pi}^{1.69\pi}=-35.01\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #29 : Double Integration In Polar Coordinates

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