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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 #101 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(40e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})} -\frac{ (4sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{9})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.21\text{ and }1.65\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.771,0.637)\text{ and }\overrightarrow{u_2}=(0.844,-0.536)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

35.84

143.37

-716.84

-28.67

Correct answer:

143.37

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(40e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})} -\frac{ (4sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{9})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(40\cdot re^{(-\frac{r^{2}}{2})} -\frac{ (4\cdot rsin(\frac{r^{2}}{2}))}{9})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.637}{0.771})=0.22\pi;\theta_2=arctan(\frac{-0.536}{0.844})=1.82\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.22\pi}^{1.82\pi}\int_{0.21}^{1.65}(40\cdot re^{(-\frac{r^{2}}{2})} -\frac{ (4\cdot rsin(\frac{r^{2}}{2}))}{9})drd\theta=(\frac{(4cos(\frac{r^{2}}{2}))}{9}- 40e^{(-\frac{r^{2}}{2})})d\theta|_{0.21}^{1.65}\\&\int_{0.22\pi}^{1.82\pi}(28.52)d\theta=(28.52\theta)|_{0.22\pi}^{1.82\pi}=143.37\end{align*}$

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Example Question #102 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5\cdot (\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{32}- 12e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.68\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.809,0.588)\text{ and }\overrightarrow{u_2}=(0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

2.06

8.26

-8.26

-2.75

Correct answer:

-8.26

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5\cdot (\frac{1}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{32}- 12e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot (\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{32}- 12\cdot re^{(\frac{r^{2}}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{-0.809})=0.8\pi;\theta_2=arctan(\frac{-0.891}{0.454})=1.65\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.8\pi}^{1.65\pi}\int_{0}^{0.68}(\frac{(5\cdot (\frac{1}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{32}- 12\cdot re^{(\frac{r^{2}}{2})})drd\theta=(- 12e^{(\frac{r^{2}}{2})} -\frac{ 5}{(96\cdot 2^{(\frac{(3\cdot r^{2})}{2})}ln(2))})d\theta|_{0}^{0.68}\\&\int_{0.8\pi}^{1.65\pi}(-3.093)d\theta=(-3.093\theta)|_{0.8\pi}^{1.65\pi}=-8.26\end{align*}$

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Example Question #111 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{2}{(\frac{7}{2})^{(x^{2} + y^{2})}})}{13}+ 46\cdot (\frac{1}{2})^{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.51\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.729,0.685)\text{ and }\overrightarrow{u_2}=(0.279,-0.960)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

224

112

Correct answer:

112

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{2}{(\frac{7}{2})^{(x^{2} + y^{2})}})}{13}+ 46\cdot (\frac{1}{2})^{(x^{2} + y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(46\cdot (\frac{1}{2})^{(r^{2})}\cdot r +\frac{ (2\cdot r)}{(13\cdot (\frac{7}{2})^{(r^{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.685}{0.729})=0.24\pi;\theta_2=arctan(\frac{-0.960}{0.279})=1.59\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.24\pi}^{1.59\pi}\int_{0}^{1.51}(46\cdot (\frac{1}{2})^{(r^{2})}\cdot r +\frac{ (2\cdot r)}{(13\cdot (\frac{7}{2})^{(r^{2})})})drd\theta=(-\frac{ 23}{(2^{(r^{2})}ln(2))}-\frac{ 1}{(13\cdot (\frac{7}{2})^{(r^{2})}ln(\frac{7}{2}))})d\theta|_{0}^{1.51}\\&\int_{0.24\pi}^{1.59\pi}(26.41)d\theta=(26.41\theta)|_{0.24\pi}^{1.59\pi}=112\end{align*}$

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Example Question #112 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (13cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{6}-\frac{ (35cos(x^{2} + y^{2}))}{4})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.41\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.094,0.996)\text{ and }\overrightarrow{u_2}=(-0.930,-0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

2.38

59.38

-11.88

-71.26

Correct answer:

-11.88

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (13cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{6}-\frac{ (35cos(x^{2} + y^{2}))}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (35\cdot rcos(r^{2}))}{4}-\frac{ (13\cdot rcos(\frac{r^{2}}{2}))}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{0.094})=0.47\pi;\theta_2=arctan(\frac{-0.368}{-0.930})=1.12\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.47\pi}^{1.12\pi}\int_{0}^{1.41}(-\frac{ (35\cdot rcos(r^{2}))}{4}-\frac{ (13\cdot rcos(\frac{r^{2}}{2}))}{6})drd\theta=(-\frac{ (35sin(r^{2}))}{8}-\frac{ (13sin(\frac{r^{2}}{2}))}{6})d\theta|_{0}^{1.41}\\&\int_{0.47\pi}^{1.12\pi}(-5.816)d\theta=(-5.816\theta)|_{0.47\pi}^{1.12\pi}=-11.88\end{align*}$

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Example Question #113 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 7e^{(- x^{2} - y^{2})} -\frac{ (\frac{9}{(\frac{1}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{2})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.28\text{ and }1.34\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.930,0.368)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

224

11.2

Correct answer:

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(- 7e^{(- x^{2} - y^{2})} -\frac{ (\frac{9}{(\frac{1}{3})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})}})}{2})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (9\cdot r)}{(2\cdot (\frac{1}{3})^{(\frac{(3\cdot r^{2})}{2})})}- 7\cdot re^{(-r^{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{0.750})=0.23\pi;\theta_2=arctan(\frac{0.368}{-0.930})=0.88\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.23\pi}^{0.88\pi}\int_{0.28}^{1.34}(-\frac{ (9\cdot r)}{(2\cdot (\frac{1}{3})^{(\frac{(3\cdot r^{2})}{2})})}- 7\cdot re^{(-r^{2})})drd\theta=(\frac{(7e^{(-r^{2})})}{2}-\frac{ (3\cdot 3^{(\frac{(3\cdot r^{2})}{2})})}{(2ln(3))})d\theta|_{0.28}^{1.34}\\&\int_{0.23\pi}^{0.88\pi}(-27.42)d\theta=(-27.42\theta)|_{0.23\pi}^{0.88\pi}=-56\end{align*}$

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Example Question #114 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(7cos(x^{2} + y^{2}) -\frac{ 1}{(10\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.18\text{ and }0.82\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.637,0.771)\text{ and }\overrightarrow{u_2}=(0.397,-0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.47

-27.37

-0.91

21.9

Correct answer:

5.47

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(7cos(x^{2} + y^{2}) -\frac{ 1}{(10\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(7\cdot rcos(r^{2}) -\frac{ 1}{(10\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.771}{-0.637})=0.72\pi;\theta_2=arctan(\frac{-0.918}{0.397})=1.63\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.72\pi}^{1.63\pi}\int_{0.18}^{0.82}(7\cdot rcos(r^{2}) -\frac{ 1}{(10\cdot r)})drd\theta=(\frac{(7sin(r^{2}))}{2}-\frac{ ln(r)}{10})d\theta|_{0.18}^{0.82}\\&\int_{0.72\pi}^{1.63\pi}(1.915)d\theta=(1.915\theta)|_{0.72\pi}^{1.63\pi}=5.47\end{align*}$

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Example Question #115 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{5}+ 7\cdot 2^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.72\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.685,0.729)\text{ and }\overrightarrow{u_2}=(0.426,-0.905)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-18.85

1.57

6.28

-1.05

Correct answer:

6.28

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.729}{-0.685})=0.74\pi;\theta_2=arctan(\frac{-0.905}{0.426})=1.64\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.74\pi}^{1.64\pi}\int_{0}^{0.72}(7\cdot 2^{(\frac{r^{2}}{2})}\cdot r +\frac{ (3\cdot re^{(\frac{(3\cdot r^{2})}{2})})}{5})drd\theta=(\frac{e^{(\frac{(3\cdot r^{2})}{2})}}{5}+\frac{ (7\cdot 2^{(\frac{r^{2}}{2})})}{ln(2)})d\theta|_{0}^{0.72}\\&\int_{0.74\pi}^{1.64\pi}(2.223)d\theta=(2.223\theta)|_{0.74\pi}^{1.64\pi}=6.28\end{align*}$

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Example Question #116 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (2cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{25}-\frac{ sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{24})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.4\text{ and }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(0.187,-0.982)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.18

-0.04

0.04

0.9

Correct answer:

-0.18

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (2cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{25}-\frac{ sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{24})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (2\cdot rcos(\frac{r^{2}}{2}))}{25}-\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{24})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.982}{0.187})=1.56\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.77\pi}^{1.56\pi}\int_{0.4}^{1.31}(-\frac{ (2\cdot rcos(\frac{r^{2}}{2}))}{25}-\frac{ (rsin(\frac{(2\cdot r^{2})}{3}))}{24})drd\theta=(\frac{cos(\frac{(2\cdot r^{2})}{3})}{32}-\frac{ (2sin(\frac{r^{2}}{2}))}{25})d\theta|_{0.4}^{1.31}\\&\int_{0.77\pi}^{1.56\pi}(-0.07227)d\theta=(-0.07227\theta)|_{0.77\pi}^{1.56\pi}=-0.18\end{align*}$

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Example Question #117 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (25cos(2x^{2} + 2y^{2}))}{2}-\frac{ (4e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{35})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.66\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.536,0.844)\text{ and }\overrightarrow{u_2}=(-0.156,-0.988)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

34.21

-1.71

1.43

-8.55

Correct answer:

-8.55

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (25cos(2x^{2} + 2y^{2}))}{2}-\frac{ (4e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})})}{35})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (25\cdot rcos(2\cdot r^{2}))}{2}-\frac{ (4\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{35})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.844}{0.536})=0.32\pi;\theta_2=arctan(\frac{-0.988}{-0.156})=1.45\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.32\pi}^{1.45\pi}\int_{0}^{0.66}(-\frac{ (25\cdot rcos(2\cdot r^{2}))}{2}-\frac{ (4\cdot re^{(-\frac{(3\cdot r^{2})}{2})})}{35})drd\theta=(\frac{(4e^{(-\frac{(3\cdot r^{2})}{2})})}{105}-\frac{ (25sin(2\cdot r^{2}))}{8})d\theta|_{0}^{0.66}\\&\int_{0.32\pi}^{1.45\pi}(-2.409)d\theta=(-2.409\theta)|_{0.32\pi}^{1.45\pi}=-8.55\end{align*}$

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Example Question #118 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (5e^{(x^{2} + y^{2})})}{43}-\frac{ (\frac{6}{3^{(x^{2} + y^{2})}})}{37})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.25\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.063,0.998)\text{ and }\overrightarrow{u_2}=(0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.58

-6.96

-1.16

6.96

Correct answer:

-1.16

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (5e^{(x^{2} + y^{2})})}{43}-\frac{ (\frac{6}{3^{(x^{2} + y^{2})}})}{37})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (5\cdot re^{(r^{2})})}{43}-\frac{ (6\cdot r)}{(37\cdot 3^{(r^{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.998}{0.063})=0.48\pi;\theta_2=arctan(\frac{-0.588}{0.809})=1.8\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.48\pi}^{1.8\pi}\int_{0}^{1.25}(-\frac{ (5\cdot re^{(r^{2})})}{43}-\frac{ (6\cdot r)}{(37\cdot 3^{(r^{2})})})drd\theta=(\frac{3}{(37\cdot 3^{(r^{2})}ln(3))}-\frac{ (5e^{(r^{2})})}{86})d\theta|_{0}^{1.25}\\&\int_{0.48\pi}^{1.8\pi}(-0.2798)d\theta=(-0.2798\theta)|_{0.48\pi}^{1.8\pi}=-1.16\end{align*}$

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Calculus 3 : Calculus 3

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Example Question #101 : Double Integrals

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