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Calculus 3 : Multiple Integration

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

Example Questions

Example Question #99 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(10e^{(x^{2} + y^{2})} +\frac{ (\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.16\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.941,-0.339)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

116.6

583

-349.8

-19.43

Correct answer:

116.6

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(10e^{(x^{2} + y^{2})} +\frac{ (\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(10\cdot re^{(r^{2})} +\frac{ (48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.339}{0.941})=1.89\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.47\pi}^{1.89\pi}\int_{0}^{1.16}(10\cdot re^{(r^{2})} +\frac{ (48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})})drd\theta=(5e^{(r^{2})} -\frac{ 36}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{7}))})d\theta|_{0}^{1.16}\\&\int_{0.47\pi}^{1.89\pi}(26.14)d\theta=(26.14\theta)|_{0.47\pi}^{1.89\pi}=116.6\end{align*}$

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Example Question #100 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (17e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{4}- 5sin(2x^{2} + 2y^{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.62\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

209.42

-69.81

34.9

-279.23

Correct answer:

-69.81

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (17e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{4}- 5sin(2x^{2} + 2y^{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (17\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{4}- 5\cdot rsin(2\cdot r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{-0.861}{0.509})=1.67\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.26\pi}^{1.67\pi}\int_{0}^{1.62}(-\frac{ (17\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{4}- 5\cdot rsin(2\cdot r^{2}))drd\theta=(\frac{(5cos(2\cdot r^{2}))}{4}-\frac{ (51e^{(\frac{(2\cdot r^{2})}{3})})}{16})d\theta|_{0}^{1.62}\\&\int_{0.26\pi}^{1.67\pi}(-15.76)d\theta=(-15.76\theta)|_{0.26\pi}^{1.67\pi}=-69.81\end{align*}$

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Example Question #101 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}}{2}- (\frac{\frac{2}{3})^{(x^{2} + y^{2})}}{27})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.86\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.976,0.218)\text{ and }\overrightarrow{u_2}=(0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1.63

1.31

6.53

-26.14

Correct answer:

6.53

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}}{2}- (\frac{\frac{2}{3})^{(x^{2} + y^{2})}}{27})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(\frac{(2\cdot r^{2})}{3})})}{2}-\frac{ ((\frac{2}{3})^{(r^{2})}\cdot r)}{27})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.218}{-0.976})=0.93\pi;\theta_2=arctan(\frac{-0.988}{0.156})=1.55\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.93\pi}^{1.55\pi}\int_{0}^{1.86}(\frac{(re^{(\frac{(2\cdot r^{2})}{3})})}{2}-\frac{ ((\frac{2}{3})^{(r^{2})}\cdot r)}{27})drd\theta=(\frac{(3e^{(\frac{(2\cdot r^{2})}{3})})}{8}- (\frac{\frac{2}{3})^{(r^{2})}}{(54ln(\frac{2}{3}))})d\theta|_{0}^{1.86}\\&\int_{0.93\pi}^{1.55\pi}(3.355)d\theta=(3.355\theta)|_{0.93\pi}^{1.55\pi}=6.53\end{align*}$

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Example Question #102 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{13}+\frac{ (11e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{2})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.26\text{ and }1.59\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.509,0.861)\text{ and }\overrightarrow{u_2}=(0.988,-0.156)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-36.33

72.67

-436.01

436.01

Correct answer:

72.67

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{13}+\frac{ (11e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{2})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{13}+\frac{ (11\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.861}{-0.509})=0.67\pi;\theta_2=arctan(\frac{-0.156}{0.988})=1.95\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.67\pi}^{1.95\pi}\int_{0.26}^{1.59}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{13}+\frac{ (11\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{2})drd\theta=(\frac{(33e^{(\frac{(2\cdot r^{2})}{3})})}{8}-\frac{ (9e^{(-\frac{(2\cdot r^{2})}{3})})}{52})d\theta|_{0.26}^{1.59}\\&\int_{0.67\pi}^{1.95\pi}(18.07)d\theta=(18.07\theta)|_{0.67\pi}^{1.95\pi}=72.67\end{align*}$

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Example Question #103 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(18e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ 11}{(\frac{2}{3})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.95\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.562,-0.827)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-25.49

101.95

-152.92

50.97

Correct answer:

50.97

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(18e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ 11}{(\frac{2}{3})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(11\cdot r)}{(\frac{2}{3})^{(\frac{(2\cdot r^{2})}{3})}}+ 18\cdot re^{(\frac{(3\cdot r^{2})}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{-0.309})=0.6\pi;\theta_2=arctan(\frac{-0.827}{-0.562})=1.31\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.6\pi}^{1.31\pi}\int_{0}^{0.95}(\frac{(11\cdot r)}{(\frac{2}{3})^{(\frac{(2\cdot r^{2})}{3})}}+ 18\cdot re^{(\frac{(3\cdot r^{2})}{2})})drd\theta=(6e^{(\frac{(3\cdot r^{2})}{2})} -\frac{ (33\cdot (\frac{3}{2})^{(\frac{(2\cdot r^{2})}{3})})}{(4ln(\frac{2}{3}))})d\theta|_{0}^{0.95}\\&\int_{0.6\pi}^{1.31\pi}(22.85)d\theta=(22.85\theta)|_{0.6\pi}^{1.31\pi}=50.97\end{align*}$

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Example Question #104 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 36cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}}{20})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.08\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.729,0.685)\text{ and }\overrightarrow{u_2}=(-0.861,-0.509)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

27.69

-11.08

166.15

-55.38

Correct answer:

-55.38

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- 36cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})}}{20})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- 36\cdot rcos(\frac{(2\cdot r^{2})}{3}) -\frac{ (re^{(-\frac{(3\cdot r^{2})}{2})})}{20})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.685}{0.729})=0.24\pi;\theta_2=arctan(\frac{-0.509}{-0.861})=1.17\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.24\pi}^{1.17\pi}\int_{0}^{1.08}(- 36\cdot rcos(\frac{(2\cdot r^{2})}{3}) -\frac{ (re^{(-\frac{(3\cdot r^{2})}{2})})}{20})drd\theta=(\frac{e^{(-\frac{(3\cdot r^{2})}{2})}}{60}- 27sin(\frac{(2\cdot r^{2})}{3}))d\theta|_{0}^{1.08}\\&\int_{0.24\pi}^{1.17\pi}(-18.96)d\theta=(-18.96\theta)|_{0.24\pi}^{1.17\pi}=-55.38\end{align*}$

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Example Question #105 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(39e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{5}+\frac{ (\frac{47}{4^{(x^{2} + y^{2})}})}{2})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.88\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.905,0.426)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

20.73

103.67

-10.37

-82.94

Correct answer:

20.73

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(39e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{5}+\frac{ (\frac{47}{4^{(x^{2} + y^{2})}})}{2})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(47\cdot r)}{(2\cdot 4^{(r^{2})})}+\frac{ (39\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.426}{-0.905})=0.86\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.26\pi}^{0.86\pi}\int_{0}^{1.88}(\frac{(47\cdot r)}{(2\cdot 4^{(r^{2})})}+\frac{ (39\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{5})drd\theta=(-\frac{ (13e^{(-\frac{(3\cdot r^{2})}{2})})}{5}-\frac{ 47}{(8\cdot 4^{(r^{2})}ln(2))})d\theta|_{0}^{1.88}\\&\int_{0.26\pi}^{0.86\pi}(11.0)d\theta=(11.0\theta)|_{0.26\pi}^{0.86\pi}=20.73\end{align*}$

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Example Question #106 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(8cos(2x^{2} + 2y^{2}) +\frac{ (5sin(2x^{2} + 2y^{2}))}{33})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.74\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.844,0.536)\text{ and }\overrightarrow{u_2}=(-0.771,0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.21

-0.85

0.43

2.55

Correct answer:

-0.85

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(8cos(2x^{2} + 2y^{2}) +\frac{ (5sin(2x^{2} + 2y^{2}))}{33})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(8\cdot rcos(2\cdot r^{2}) +\frac{ (5\cdot rsin(2\cdot r^{2}))}{33})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{0.844})=0.18\pi;\theta_2=arctan(\frac{0.637}{-0.771})=0.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.18\pi}^{0.78\pi}\int_{0}^{1.74}(8\cdot rcos(2\cdot r^{2}) +\frac{ (5\cdot rsin(2\cdot r^{2}))}{33})drd\theta=(2sin(2\cdot r^{2}) -\frac{ (5cos(2\cdot r^{2}))}{132})d\theta|_{0}^{1.74}\\&\int_{0.18\pi}^{0.78\pi}(-0.4511)d\theta=(-0.4511\theta)|_{0.18\pi}^{0.78\pi}=-0.85\end{align*}$

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Example Question #107 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(10\cdot 3^{(2x^{2} + 2y^{2})} -\frac{ e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{7})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.69\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.976,0.218)\text{ and }\overrightarrow{u_2}=(-0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-50.87

5.65

16.96

-8.48

Correct answer:

16.96

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(10\cdot 3^{(2x^{2} + 2y^{2})} -\frac{ e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{7})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(10\cdot 3^{(2\cdot r^{2})}\cdot r -\frac{ (re^{(\frac{(3\cdot r^{2})}{2})})}{7})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.218}{0.976})=0.07\pi;\theta_2=arctan(\frac{-0.918}{-0.397})=1.37\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.07\pi}^{1.37\pi}\int_{0}^{0.69}(10\cdot 3^{(2\cdot r^{2})}\cdot r -\frac{ (re^{(\frac{(3\cdot r^{2})}{2})})}{7})drd\theta=(\frac{(5\cdot 3^{(2\cdot r^{2})})}{(2ln(3))}-\frac{ e^{(\frac{(3\cdot r^{2})}{2})}}{21})d\theta|_{0}^{0.69}\\&\int_{0.07\pi}^{1.37\pi}(4.152)d\theta=(4.152\theta)|_{0.07\pi}^{1.37\pi}=16.96\end{align*}$

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Example Question #108 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (2e^{(- x^{2} - y^{2})})}{13}-\frac{ (4e^{(x^{2} + y^{2})})}{25})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.9\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.998,0.063)\text{ and }\overrightarrow{u_2}=(0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-7.6

38.02

-1.9

1.52

Correct answer:

-7.6

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (2e^{(- x^{2} - y^{2})})}{13}-\frac{ (4e^{(x^{2} + y^{2})})}{25})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (4\cdot re^{(r^{2})})}{25}-\frac{ (2\cdot re^{(-r^{2})})}{13})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{-0.998})=0.98\pi;\theta_2=arctan(\frac{-0.588}{0.809})=1.8\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.98\pi}^{1.8\pi}\int_{0}^{1.9}(-\frac{ (4\cdot re^{(r^{2})})}{25}-\frac{ (2\cdot re^{(-r^{2})})}{13})drd\theta=(\frac{e^{(-r^{2})}}{13}-\frac{ (2e^{(r^{2})})}{25})d\theta|_{0}^{1.9}\\&\int_{0.98\pi}^{1.8\pi}(-2.952)d\theta=(-2.952\theta)|_{0.98\pi}^{1.8\pi}=-7.6\end{align*}$

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Calculus 3 : Multiple Integration

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Example Question #99 : Double Integrals

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