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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 #1421 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(7e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{3}-\frac{ (2\cdot 4^{(x^{2} + y^{2})})}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.22\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.982,0.187)\text{ and }\overrightarrow{u_2}=(0.976,-0.218)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-1.33

-3.99

0.33

6.65

Correct answer:

-1.33

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(7e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{3}-\frac{ (2\cdot 4^{(x^{2} + y^{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(7\cdot re^{(-\frac{r^{2}}{2})})}{3}-\frac{ (2\cdot 4^{(r^{2})}\cdot r)}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.187}{-0.982})=0.94\pi;\theta_2=arctan(\frac{-0.218}{0.976})=1.93\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.94\pi}^{1.93\pi}\int_{0}^{1.22}(\frac{(7\cdot re^{(-\frac{r^{2}}{2})})}{3}-\frac{ (2\cdot 4^{(r^{2})}\cdot r)}{3})drd\theta=(-\frac{ (7e^{(-\frac{r^{2}}{2})})}{3}-\frac{ 4^{(r^{2})}}{(3ln(4))})d\theta|_{0}^{1.22}\\&\int_{0.94\pi}^{1.93\pi}(-0.4277)d\theta=(-0.4277\cdot \theta)|_{0.94\pi}^{1.93\pi}=-1.33\end{align*}$

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Example Question #1422 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (15cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (3e^{(2x^{2} + 2y^{2})})}{8})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.83\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.918,0.397)\text{ and }\overrightarrow{u_2}=(-0.930,-0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1037.08

-259.27

-518.54

43.21

Correct answer:

-259.27

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (15cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (3e^{(2x^{2} + 2y^{2})})}{8})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (15\cdot rcos(\frac{r^{2}}{2}))}{2}-\frac{ (3\cdot re^{(2\cdot r^{2})})}{8})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.397}{0.918})=0.13\pi;\theta_2=arctan(\frac{-0.368}{-0.930})=1.12\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.13\pi}^{1.12\pi}\int_{0}^{1.83}(-\frac{ (15\cdot rcos(\frac{r^{2}}{2}))}{2}-\frac{ (3\cdot re^{(2\cdot r^{2})})}{8})drd\theta=(-\frac{ (3e^{(2\cdot r^{2})})}{32}-\frac{ (15sin(\frac{r^{2}}{2}))}{2})d\theta|_{0}^{1.83}\\&\int_{0.13\pi}^{1.12\pi}(-83.36)d\theta=(-83.36\cdot \theta)|_{0.13\pi}^{1.12\pi}=-259.27\end{align*}$

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Example Question #1421 : Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 2e^{(- x^{2} - y^{2})} -\frac{ 40}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.54\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.339,0.941)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

100.84

-8.4

12.6

-50.42

Correct answer:

-50.42

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- 2e^{(- x^{2} - y^{2})} -\frac{ 40}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (40\cdot r)}{3}- 2\cdot re^{(-r^{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.941}{-0.339})=0.61\pi;\theta_2=arctan(\frac{-0.976}{0.218})=1.57\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.61\pi}^{1.57\pi}\int_{0}^{1.54}(-\frac{ (40\cdot r)}{3}- 2\cdot re^{(-r^{2})})drd\theta=(e^{(-r^{2})} -\frac{ (20\cdot r^{2})}{3})d\theta|_{0}^{1.54}\\&\int_{0.61\pi}^{1.57\pi}(-16.72)d\theta=(-16.72\cdot \theta)|_{0.61\pi}^{1.57\pi}=-50.42\end{align*}$

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Example Question #52 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (49sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{4}-\frac{ (5sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{18})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.78\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.031,1.000)\text{ and }\overrightarrow{u_2}=(0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-50.67

202.68

-152.01

10.13

Correct answer:

-50.67

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (49sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{4}-\frac{ (5sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{18})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (49\cdot rsin(\frac{r^{2}}{2}))}{4}-\frac{ (5\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{18})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{-0.031})=0.51\pi;\theta_2=arctan(\frac{-0.588}{0.809})=1.8\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.51\pi}^{1.8\pi}\int_{0}^{1.78}(-\frac{ (49\cdot rsin(\frac{r^{2}}{2}))}{4}-\frac{ (5\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{18})drd\theta=(\frac{(49cos(\frac{r^{2}}{2}))}{4}+\frac{ (5cos(\frac{(3\cdot r^{2})}{2}))}{54})d\theta|_{0}^{1.78}\\&\int_{0.51\pi}^{1.8\pi}(-12.5)d\theta=(-12.5\cdot \theta)|_{0.51\pi}^{1.8\pi}=-50.67\end{align*}$

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Example Question #53 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (26cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{3}-\frac{ (45sin(x^{2} + y^{2}))}{4})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.46\text{ and }1.16\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.918,0.397)\text{ and }\overrightarrow{u_2}=(-0.031,-1.000)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

51.01

2.83

-51.01

Correct answer:

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (26cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{3}-\frac{ (45sin(x^{2} + y^{2}))}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (45\cdot rsin(r^{2}))}{4}-\frac{ (26\cdot rcos(\frac{r^{2}}{2}))}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.397}{-0.918})=0.87\pi;\theta_2=arctan(\frac{-1.000}{-0.031})=1.49\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.87\pi}^{1.49\pi}\int_{0.46}^{1.16}(-\frac{ (45\cdot rsin(r^{2}))}{4}-\frac{ (26\cdot rcos(\frac{r^{2}}{2}))}{3})drd\theta=(-\frac{ (26sin(\frac{r^{2}}{2}))}{3}-\frac{ (45sin(\frac{r^{2}}{2})^{2})}{4})d\theta|_{0.46}^{1.16}\\&\int_{0.87\pi}^{1.49\pi}(-8.729)d\theta=(-8.729\cdot \theta)|_{0.87\pi}^{1.49\pi}=-17\end{align*}$

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Example Question #54 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (5e^{(2x^{2} + 2y^{2})})}{47}-\frac{ (2sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{3})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.23\text{ and }1.15\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.771,0.637)\text{ and }\overrightarrow{u_2}=(-0.309,-0.951)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.21

6.39

-1.28

0.43

Correct answer:

-1.28

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.637}{-0.771})=0.78\pi;\theta_2=arctan(\frac{-0.951}{-0.309})=1.4\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.78\pi}^{1.4\pi}\int_{0.23}^{1.15}(-\frac{ (5\cdot re^{(2\cdot r^{2})})}{47}-\frac{ (2\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{3})drd\theta=(\frac{(2cos(\frac{(3\cdot r^{2})}{2}))}{9}-\frac{ (5e^{(2\cdot r^{2})})}{188})d\theta|_{0.23}^{1.15}\\&\int_{0.78\pi}^{1.4\pi}(-0.6557)d\theta=(-0.6557\cdot \theta)|_{0.78\pi}^{1.4\pi}=-1.28\end{align*}$

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Example Question #55 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (5cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{4}-\frac{ (5e^{(- x^{2} - y^{2})})}{42})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.53\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.809,0.588)\text{ and }\overrightarrow{u_2}=(0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.58

10.34

0.52

-0.65

Correct answer:

-2.58

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (5cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{4}-\frac{ (5e^{(- x^{2} - y^{2})})}{42})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (5\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{4}-\frac{ (5\cdot re^{(-r^{2})})}{42})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{-0.809})=0.8\pi;\theta_2=arctan(\frac{-0.918}{0.397})=1.63\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.8\pi}^{1.63\pi}\int_{0}^{1.53}(-\frac{ (5\cdot rcos(\frac{(2\cdot r^{2})}{3}))}{4}-\frac{ (5\cdot re^{(-r^{2})})}{42})drd\theta=(\frac{(5e^{(-r^{2})})}{84}-\frac{ (15sin(\frac{(2\cdot r^{2})}{3}))}{16})d\theta|_{0}^{1.53}\\&\int_{0.8\pi}^{1.63\pi}(-0.9912)d\theta=(-0.9912\cdot \theta)|_{0.8\pi}^{1.63\pi}=-2.58\end{align*}$

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Example Question #56 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{sin(2x^{2} + 2y^{2})}{9}-\frac{ 49}{(2\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.42\text{ and }1.43\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.988,0.156)\text{ and }\overrightarrow{u_2}=(0.976,-0.218)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

692.46

46.16

-415.47

-138.49

Correct answer:

-138.49

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{sin(2x^{2} + 2y^{2})}{9}-\frac{ 49}{(2\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(rsin(2\cdot r^{2}))}{9}-\frac{ 147}{(4\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.156}{-0.988})=0.95\pi;\theta_2=arctan(\frac{-0.218}{0.976})=1.93\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.95\pi}^{1.93\pi}\int_{0.42}^{1.43}(\frac{(rsin(2\cdot r^{2}))}{9}-\frac{ 147}{(4\cdot r)})drd\theta=(-\frac{ (147ln(r))}{4}- cos(\frac{r^{2})^{2}}{18})d\theta|_{0.42}^{1.43}\\&\int_{0.95\pi}^{1.93\pi}(-44.98)d\theta=(-44.98\cdot \theta)|_{0.95\pi}^{1.93\pi}=-138.49\end{align*}$

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Example Question #57 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(32e^{(x^{2} + y^{2})})}{3}+\frac{ 17}{(4\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.48\text{ and }1.68\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.000)\text{ and }\overrightarrow{u_2}=(0.861,-0.509)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

230.25

-115.12

-920.98

920.98

Correct answer:

230.25

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(32e^{(x^{2} + y^{2})})}{3}+\frac{ 17}{(4\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(32\cdot re^{(r^{2})})}{3}+\frac{ 17}{(4\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.000}{-1.000})=1\pi;\theta_2=arctan(\frac{-0.509}{0.861})=1.83\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{1\pi}^{1.83\pi}\int_{0.48}^{1.68}(\frac{(32\cdot re^{(r^{2})})}{3}+\frac{ 17}{(4\cdot r)})drd\theta=(\frac{(16e^{(r^{2})})}{3}+\frac{ (17ln(r))}{4})d\theta|_{0.48}^{1.68}\\&\int_{1\pi}^{1.83\pi}(88.3)d\theta=(88.3\cdot \theta)|_{1\pi}^{1.83\pi}=230.25\end{align*}$

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Example Question #58 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5cos(x^{2} + y^{2}))}{38}-\frac{ (34e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.38\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.125,0.992)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-241.14

241.14

-964.57

80.38

Correct answer:

-241.14

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5cos(x^{2} + y^{2}))}{38}-\frac{ (34e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot rcos(r^{2}))}{38}-\frac{ (34\cdot re^{(\frac{(3\cdot r^{2})}{2})})}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.992}{-0.125})=0.54\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.54\pi}^{1.78\pi}\int_{0}^{1.38}(\frac{(5\cdot rcos(r^{2}))}{38}-\frac{ (34\cdot re^{(\frac{(3\cdot r^{2})}{2})})}{3})drd\theta=(\frac{(5sin(r^{2}))}{76}-\frac{ (34e^{(\frac{(3\cdot r^{2})}{2})})}{9})d\theta|_{0}^{1.38}\\&\int_{0.54\pi}^{1.78\pi}(-61.9)d\theta=(-61.9\cdot \theta)|_{0.54\pi}^{1.78\pi}=-241.14\end{align*}$

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Calculus 3 : Calculus 3

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