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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 #131 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 16e^{(- x^{2} - y^{2})} - 3cos(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.94\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.998,0.063)\text{ and }\overrightarrow{u_2}=(0.891,-0.454)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

3.2

-47.99

Correct answer:

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.063}{-0.998})=0.98\pi;\theta_2=arctan(\frac{-0.454}{0.891})=1.85\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.98\pi}^{1.85\pi}\int_{0}^{0.94}(- 3\cdot rcos(r^{2}) - 16\cdot re^{(-r^{2})})drd\theta=(8e^{(-r^{2})} -\frac{ (3sin(r^{2}))}{2})d\theta|_{0}^{0.94}\\&\int_{0.98\pi}^{1.85\pi}(-5.853)d\theta=(-5.853\theta)|_{0.98\pi}^{1.85\pi}=-16\end{align*}$

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Example Question #132 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5\cdot (\frac{2}{3})^{(x^{2} + y^{2})})}{27}- 28e^{(- x^{2} - y^{2})})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.72\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.509,0.861)\text{ and }\overrightarrow{u_2}=(0.426,-0.905)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

4.28

-51.38

85.63

-17.13

Correct answer:

-17.13

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5\cdot (\frac{2}{3})^{(x^{2} + y^{2})})}{27}- 28e^{(- x^{2} - y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot (\frac{2}{3})^{(r^{2})}\cdot r)}{27}- 28\cdot re^{(-r^{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.861}{-0.509})=0.67\pi;\theta_2=arctan(\frac{-0.905}{0.426})=1.64\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.67\pi}^{1.64\pi}\int_{0}^{0.72}(\frac{(5\cdot (\frac{2}{3})^{(r^{2})}\cdot r)}{27}- 28\cdot re^{(-r^{2})})drd\theta=(14e^{(-r^{2})} +\frac{ (5\cdot (\frac{2}{3})^{(r^{2})})}{(54ln(\frac{2}{3}))})d\theta|_{0}^{0.72}\\&\int_{0.67\pi}^{1.64\pi}(-5.62)d\theta=(-5.62\theta)|_{0.67\pi}^{1.64\pi}=-17.13\end{align*}$

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Example Question #141 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{3}{(37\cdot (2x^{2} + 2y^{2}))}-\frac{ 22}{(5\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.18\text{ and }1.99\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.000)\text{ and }\overrightarrow{u_2}=(0.279,-0.960)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

175.29

14.61

-7.3

-29.22

Correct answer:

-29.22

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{3}{(37\cdot (2x^{2} + 2y^{2}))}-\frac{ 22}{(5\cdot (\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{2427}{(370\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.000}{-1.000})=1\pi;\theta_2=arctan(\frac{-0.960}{0.279})=1.59\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{1\pi}^{1.59\pi}\int_{0.18}^{1.99}(-\frac{2427}{(370\cdot r)})drd\theta=(-\frac{(2427ln(r))}{370})d\theta|_{0.18}^{1.99}\\&\int_{1\pi}^{1.59\pi}(-15.76)d\theta=(-15.76\theta)|_{1\pi}^{1.59\pi}=-29.22\end{align*}$

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Example Question #142 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}- 6e^{(- x^{2} - y^{2})})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.26\text{ and }1.92\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.905,0.426)\text{ and }\overrightarrow{u_2}=(0.998,-0.063)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

45.14

-9.03

-2.26

2.26

Correct answer:

-9.03

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})}}{4}- 6e^{(- x^{2} - y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}- 6\cdot re^{(-r^{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.426}{-0.905})=0.86\pi;\theta_2=arctan(\frac{-0.063}{0.998})=1.98\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.86\pi}^{1.98\pi}\int_{0.26}^{1.92}(\frac{(re^{(-\frac{(2\cdot r^{2})}{3})})}{4}- 6\cdot re^{(-r^{2})})drd\theta=(-\frac{(3e^{(-r^{2})}\cdot (e^{(\frac{r^{2}}{3})} - 16))}{16})d\theta|_{0.26}^{1.92}\\&\int_{0.86\pi}^{1.98\pi}(-2.566)d\theta=(-2.566\theta)|_{0.86\pi}^{1.98\pi}=-9.03\end{align*}$

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Example Question #143 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{21}+ 17sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.22\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(1.000,0.031)\text{ and }\overrightarrow{u_2}=(0.613,-0.790)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

24.65

4.11

-8.22

-73.95

Correct answer:

24.65

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{21}+ 17sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{21}+ 17\cdot rsin(\frac{r^{2}}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.031}{1.000})=0.01\pi;\theta_2=arctan(\frac{-0.790}{0.613})=1.71\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.01\pi}^{1.71\pi}\int_{0}^{1.22}(\frac{(2\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{21}+ 17\cdot rsin(\frac{r^{2}}{2}))drd\theta=(\frac{e^{(\frac{(2\cdot r^{2})}{3})}}{14}- 17cos(\frac{r^{2}}{2}))d\theta|_{0}^{1.22}\\&\int_{0.01\pi}^{1.71\pi}(4.616)d\theta=(4.616\theta)|_{0.01\pi}^{1.71\pi}=24.65\end{align*}$

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Example Question #144 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(10sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ e^{(- x^{2} - y^{2})}}{12})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.976,-0.218)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-2.75

5.51

-16.52

11.02

Correct answer:

5.51

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(10sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ e^{(- x^{2} - y^{2})}}{12})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(10\cdot rsin(\frac{(2\cdot r^{2})}{3}) -\frac{ (re^{(-r^{2})})}{12})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.218}{0.976})=1.93\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.82\pi}^{1.93\pi}\int_{0}^{1}(10\cdot rsin(\frac{(2\cdot r^{2})}{3}) -\frac{ (re^{(-r^{2})})}{12})drd\theta=(\frac{e^{(-r^{2})}}{24}-\frac{ (15cos(\frac{(2\cdot r^{2})}{3}))}{2})d\theta|_{0}^{1}\\&\int_{0.82\pi}^{1.93\pi}(1.58)d\theta=(1.58\theta)|_{0.82\pi}^{1.93\pi}=5.51\end{align*}$

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Example Question #145 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(3e^{(- 2x^{2} - 2y^{2})} +\frac{ 25}{(2\cdot (\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.24\text{ and }1.73\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.930,0.368)\text{ and }\overrightarrow{u_2}=(-0.397,0.918)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

5.49

-27.44

-6.86

27.44

Correct answer:

27.44

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.368}{0.930})=0.12\pi;\theta_2=arctan(\frac{0.918}{-0.397})=0.63\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.12\pi}^{0.63\pi}\int_{0.24}^{1.73}(3\cdot re^{(-2\cdot r^{2})} +\frac{ 25}{(3\cdot r)})drd\theta=(\frac{(25ln(r))}{3}-\frac{ (3e^{(-2\cdot r^{2})})}{4})d\theta|_{0.24}^{1.73}\\&\int_{0.12\pi}^{0.63\pi}(17.13)d\theta=(17.13\theta)|_{0.12\pi}^{0.63\pi}=27.44\end{align*}$

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Example Question #146 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{2}- 8sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.83\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.637,0.771)\text{ and }\overrightarrow{u_2}=(0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-84.09

9.34

168.17

-28.03

Correct answer:

-28.03

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.771}{-0.637})=0.72\pi;\theta_2=arctan(\frac{-0.861}{0.509})=1.67\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.72\pi}^{1.67\pi}\int_{0}^{1.83}(\frac{(rcos(\frac{(2\cdot r^{2})}{3}))}{2}- 8\cdot rsin(\frac{(2\cdot r^{2})}{3}))drd\theta=(6cos(\frac{(2\cdot r^{2})}{3}) +\frac{ (3sin(\frac{(2\cdot r^{2})}{3}))}{8})d\theta|_{0}^{1.83}\\&\int_{0.72\pi}^{1.67\pi}(-9.391)d\theta=(-9.391\theta)|_{0.72\pi}^{1.67\pi}=-28.03\end{align*}$

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Example Question #147 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{5}{(x^{2} + y^{2})}+ 2\cdot (\frac{2}{5})^{(x^{2} + y^{2})})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.48\text{ and }1.84\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.661,-0.750)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-53.15

-13.29

26.58

53.15

Correct answer:

26.58

Explanation:

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

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{-0.750}{-0.661})=1.27\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.15\pi}^{1.27\pi}\int_{0.48}^{1.84}(2\cdot (\frac{2}{5})^{(r^{2})}\cdot r +\frac{ 5}{r})drd\theta=(5ln(r) + (\frac{\frac{2}{5})^{(r^{2})}}{ln(\frac{2}{5})})d\theta|_{0.48}^{1.84}\\&\int_{0.15\pi}^{1.27\pi}(7.553)d\theta=(7.553\theta)|_{0.15\pi}^{1.27\pi}=26.58\end{align*}$

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Example Question #148 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(4sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ 4}{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.38\text{ and }1.81\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.187,-0.982)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-9.79

9.79

-3.26

1.63

Correct answer:

-9.79

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(4sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}) -\frac{ 4}{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(4\cdot rsin(\frac{(2\cdot r^{2})}{3}) -\frac{ 6}{r})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.982}{-0.187})=1.44\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.77\pi}^{1.44\pi}\int_{0.38}^{1.81}(4\cdot rsin(\frac{(2\cdot r^{2})}{3}) -\frac{ 6}{r})drd\theta=(- 3cos(\frac{(2\cdot r^{2})}{3}) - 6ln(r))d\theta|_{0.38}^{1.81}\\&\int_{0.77\pi}^{1.44\pi}(-4.653)d\theta=(-4.653\theta)|_{0.77\pi}^{1.44\pi}=-9.79\end{align*}$

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Calculus 3 : Calculus 3

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Example Questions

Example Question #131 : Double Integrals

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Example Question #141 : Double Integration In Polar Coordinates

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