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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 #121 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{32}+\frac{ (3\cdot (\frac{1}{4})^{(2x^{2} + 2y^{2})})}{13})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.95\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.07

0.4

0.8

-0.8

Correct answer:

0.4

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{32}+\frac{ (3\cdot (\frac{1}{4})^{(2x^{2} + 2y^{2})})}{13})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot (\frac{1}{4})^{(2\cdot r^{2})}\cdot r)}{13}+\frac{ (5\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{32})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.707}{0.707})=0.25\pi;\theta_2=arctan(\frac{-0.279}{0.960})=1.91\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.25\pi}^{1.91\pi}\int_{0}^{0.95}(\frac{(3\cdot (\frac{1}{4})^{(2\cdot r^{2})}\cdot r)}{13}+\frac{ (5\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{32})drd\theta=(-\frac{ (5e^{(-\frac{(3\cdot r^{2})}{2})})}{96}-\frac{ 3}{(104\cdot 4^{(2\cdot r^{2})}ln(2))})d\theta|_{0}^{0.95}\\&\int_{0.25\pi}^{1.91\pi}(0.07684)d\theta=(0.07684\theta)|_{0.25\pi}^{1.91\pi}=0.4\end{align*}$

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Example Question #122 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(14cos(2x^{2} + 2y^{2}))}{3}+\frac{ (15cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.426,-0.905)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.91

104.68

-69.79

17.45

Correct answer:

17.45

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(14cos(2x^{2} + 2y^{2}))}{3}+\frac{ (15cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(14\cdot rcos(2\cdot r^{2}))}{3}+\frac{ (15\cdot rcos(\frac{r^{2}}{2}))}{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{0.309})=0.4\pi;\theta_2=arctan(\frac{-0.905}{-0.426})=1.36\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.4\pi}^{1.36\pi}\int_{0}^{1.54}(\frac{(14\cdot rcos(2\cdot r^{2}))}{3}+\frac{ (15\cdot rcos(\frac{r^{2}}{2}))}{2})drd\theta=(\frac{(7sin(2\cdot r^{2}))}{6}+\frac{ (15sin(\frac{r^{2}}{2}))}{2})d\theta|_{0}^{1.54}\\&\int_{0.4\pi}^{1.36\pi}(5.785)d\theta=(5.785\theta)|_{0.4\pi}^{1.36\pi}=17.45\end{align*}$

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Example Question #131 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 5e^{(2x^{2} + 2y^{2})} -\frac{ (3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{26})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.34\text{ and }1.09\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.309,0.951)\text{ and }\overrightarrow{u_2}=(-0.960,-0.279)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-12.9

103.23

-25.81

12.9

Correct answer:

-25.81

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- 5e^{(2x^{2} + 2y^{2})} -\frac{ (3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{26})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(- 5\cdot re^{(2\cdot r^{2})} -\frac{ (3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{26})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.951}{0.309})=0.4\pi;\theta_2=arctan(\frac{-0.279}{-0.960})=1.09\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.4\pi}^{1.09\pi}\int_{0.34}^{1.09}(- 5\cdot re^{(2\cdot r^{2})} -\frac{ (3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{26})drd\theta=(\frac{(9cos(\frac{(2\cdot r^{2})}{3}))}{104}-\frac{ (5e^{(2\cdot r^{2})})}{4})d\theta|_{0.34}^{1.09}\\&\int_{0.4\pi}^{1.09\pi}(-11.91)d\theta=(-11.91\theta)|_{0.4\pi}^{1.09\pi}=-25.81\end{align*}$

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Example Question #131 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (5cos(x^{2} + y^{2}))}{3}- (\frac{\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{20})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.32\text{ and }1.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(1.000,0.031)\text{ and }\overrightarrow{u_2}=(0.891,-0.454)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-37.58

-9.4

3.13

18.79

Correct answer:

-9.4

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (5cos(x^{2} + y^{2}))}{3}- (\frac{\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}}{20})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (5\cdot rcos(r^{2}))}{3}-\frac{ ((\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{20})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.031}{1.000})=0.01\pi;\theta_2=arctan(\frac{-0.454}{0.891})=1.85\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.01\pi}^{1.85\pi}\int_{0.32}^{1.54}(-\frac{ (5\cdot rcos(r^{2}))}{3}-\frac{ ((\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{20})drd\theta=(-\frac{ (5sin(r^{2}))}{6}- (\frac{\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})}}{(60ln(\frac{7}{2}))})d\theta|_{0.32}^{1.54}\\&\int_{0.01\pi}^{1.85\pi}(-1.625)d\theta=(-1.625\theta)|_{0.01\pi}^{1.85\pi}=-9.4\end{align*}$

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Example Question #132 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{2}{(\frac{5}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{17}-\frac{ (4e^{(2x^{2} + 2y^{2})})}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.96\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.562,0.827)\text{ and }\overrightarrow{u_2}=(0.930,-0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-8.59

17.18

2.86

-51.54

Correct answer:

-8.59

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.827}{0.562})=0.31\pi;\theta_2=arctan(\frac{-0.368}{0.930})=1.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.31\pi}^{1.88\pi}\int_{0}^{0.96}(\frac{(2\cdot r)}{(17\cdot (\frac{5}{2})^{(\frac{(3\cdot r^{2})}{2})})}-\frac{ (4\cdot re^{(2\cdot r^{2})})}{3})drd\theta=(-\frac{ e^{(2\cdot r^{2})}}{3}-\frac{ 2}{(51\cdot (\frac{5}{2})^{(\frac{(3\cdot r^{2})}{2})}ln(\frac{5}{2}))})d\theta|_{0}^{0.96}\\&\int_{0.31\pi}^{1.88\pi}(-1.742)d\theta=(-1.742\theta)|_{0.31\pi}^{1.88\pi}=-8.59\end{align*}$

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Example Question #133 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (19e^{(- x^{2} - y^{2})})}{2}-\frac{ cos(x^{2} + y^{2})}{34})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.77\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

4.19

-8.38

-4.19

41.88

Correct answer:

-8.38

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (19e^{(- x^{2} - y^{2})})}{2}-\frac{ cos(x^{2} + y^{2})}{34})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (rcos(r^{2}))}{34}-\frac{ (19\cdot re^{(-r^{2})})}{2})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.844}{0.536})=1.68\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.43\pi}^{1.68\pi}\int_{0}^{0.77}(-\frac{ (rcos(r^{2}))}{34}-\frac{ (19\cdot re^{(-r^{2})})}{2})drd\theta=(\frac{(19e^{(-r^{2})})}{4}-\frac{ sin(r^{2})}{68})d\theta|_{0}^{0.77}\\&\int_{0.43\pi}^{1.68\pi}(-2.133)d\theta=(-2.133\theta)|_{0.43\pi}^{1.68\pi}=-8.38\end{align*}$

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Example Question #135 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(2x^{2} + 2y^{2}))}{34}+\frac{ (46sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.63\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.930,0.368)\text{ and }\overrightarrow{u_2}=(-0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-7.84

-0.78

1.57

0.39

Correct answer:

1.57

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3cos(2x^{2} + 2y^{2}))}{34}+\frac{ (46sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rcos(2\cdot r^{2}))}{34}+\frac{ (46\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.368}{0.930})=0.12\pi;\theta_2=arctan(\frac{-0.844}{-0.536})=1.32\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.12\pi}^{1.32\pi}\int_{0}^{0.63}(\frac{(3\cdot rcos(2\cdot r^{2}))}{34}+\frac{ (46\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{3})drd\theta=(\frac{(3sin(2\cdot r^{2}))}{136}-\frac{ (23cos(\frac{(2\cdot r^{2})}{3}))}{2})d\theta|_{0}^{0.63}\\&\int_{0.12\pi}^{1.32\pi}(0.416)d\theta=(0.416\theta)|_{0.12\pi}^{1.32\pi}=1.57\end{align*}$

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Example Question #611 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{49}- 10sin(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.25\text{ and }1.48\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.637,0.771)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

24.94

-12.47

6.23

-24.94

Correct answer:

-24.94

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{49}- 10sin(x^{2} + y^{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot rcos(\frac{r^{2}}{2}))}{49}- 10\cdot rsin(r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.771}{-0.637})=0.72\pi;\theta_2=arctan(\frac{-0.750}{0.661})=1.73\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.72\pi}^{1.73\pi}\int_{0.25}^{1.48}(\frac{(2\cdot rcos(\frac{r^{2}}{2}))}{49}- 10\cdot rsin(r^{2}))drd\theta=(\frac{(2sin(\frac{r^{2}}{2}))}{49}- 10sin(\frac{r^{2}}{2})^{2})d\theta|_{0.25}^{1.48}\\&\int_{0.72\pi}^{1.73\pi}(-7.859)d\theta=(-7.859\theta)|_{0.72\pi}^{1.73\pi}=-24.94\end{align*}$

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Example Question #134 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(- x^{2} - y^{2})}}{19}+\frac{ (25e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{4})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.17\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.218,0.976)\text{ and }\overrightarrow{u_2}=(0.827,-0.562)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

3.67

-66.03

-5.5

11.01

Correct answer:

11.01

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(- x^{2} - y^{2})}}{19}+\frac{ (25e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-r^{2})})}{19}+\frac{ (25\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{4})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{-0.218})=0.57\pi;\theta_2=arctan(\frac{-0.562}{0.827})=1.81\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.57\pi}^{1.81\pi}\int_{0}^{1.17}(\frac{(re^{(-r^{2})})}{19}+\frac{ (25\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{4})drd\theta=(-\frac{(e^{(-r^{2})}\cdot (1425e^{(\frac{r^{2}}{3})} + 8))}{304})d\theta|_{0}^{1.17}\\&\int_{0.57\pi}^{1.81\pi}(2.825)d\theta=(2.825\theta)|_{0.57\pi}^{1.81\pi}=11.01\end{align*}$

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Example Question #138 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(50e^{(- 2x^{2} - 2y^{2})} -\frac{ e^{(x^{2} + y^{2})}}{7})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.36\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.588,0.809)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

40.06

80.13

-20.03

-160.26

Correct answer:

40.06

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(50e^{(- 2x^{2} - 2y^{2})} -\frac{ e^{(x^{2} + y^{2})}}{7})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(50\cdot re^{(-2\cdot r^{2})} -\frac{ (re^{(r^{2})})}{7})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.809}{-0.588})=0.7\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.7\pi}^{1.78\pi}\int_{0}^{1.36}(50\cdot re^{(-2\cdot r^{2})} -\frac{ (re^{(r^{2})})}{7})drd\theta=(-\frac{ e^{(r^{2})}}{14}-\frac{ (25e^{(-2\cdot r^{2})})}{2})d\theta|_{0}^{1.36}\\&\int_{0.7\pi}^{1.78\pi}(11.81)d\theta=(11.81\theta)|_{0.7\pi}^{1.78\pi}=40.06\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #121 : Double Integrals

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