We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Double Integration in Polar Coordinates

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

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

Example Questions

Example Question #531 : Multiple Integration

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

Possible Answers:

100.84

-50.42

12.6

-8.4

Correct answer:

-50.42

Explanation:

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

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

Report an Error

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:

202.68

-152.01

-50.67

10.13

Correct answer:

-50.67

Explanation:

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

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

Report an Error

Example Question #51 : Double Integrals

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

Possible Answers:

51.01

2.83

-51.01

Correct answer:

Explanation:

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

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

Report an Error

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:

6.39

0.43

-0.21

-1.28

Correct answer:

-1.28

Explanation:

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

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

Report an Error

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:

10.34

-0.65

-2.58

0.52

Correct answer:

-2.58

Explanation:

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

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

Report an Error

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:

-415.47

-138.49

46.16

692.46

Correct answer:

-138.49

Explanation:

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

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

Report an Error

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:

920.98

-920.98

230.25

-115.12

Correct answer:

230.25

Explanation:

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

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

Report an Error

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:

80.38

-964.57

-241.14

241.14

Correct answer:

-241.14

Explanation:

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

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

Report an Error

Example Question #51 : Double Integrals

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{5}{(38\cdot (x^{2} + y^{2}))}-\frac{ (3sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{8})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.45\text{ and }1.03\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.339,0.941)\text{ and }\overrightarrow{u_2}=(0.809,-0.588)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.29

-0.06

0.01

-0.01

Correct answer:

-0.06

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{5}{(38\cdot (x^{2} + y^{2}))}-\frac{ (3sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{8})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{5}{(38\cdot r)}-\frac{ (3\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{8})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.941}{0.339})=0.39\pi;\theta_2=arctan(\frac{-0.588}{0.809})=1.8\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.39\pi}^{1.8\pi}\int_{0.45}^{1.03}(\frac{5}{(38\cdot r)}-\frac{ (3\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{8})drd\theta=(\frac{cos(\frac{(3\cdot r^{2})}{2})}{8}+\frac{ (5ln(r))}{38})d\theta|_{0.45}^{1.03}\\&\int_{0.39\pi}^{1.8\pi}(-0.01289)d\theta=(-0.01289\cdot \theta)|_{0.39\pi}^{1.8\pi}=-0.06\end{align*}$

Report an Error

Example Question #51 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(2sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{35}- 8cos(x^{2} + y^{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.094,0.996)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-3.52

2.94

-17.62

52.87

Correct answer:

-17.62

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(2sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))}{35}- 8cos(x^{2} + y^{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{35}- 8\cdot rcos(r^{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.996}{-0.094})=0.53\pi;\theta_2=arctan(\frac{-0.125}{0.992})=1.96\pi\\&\text{Now, utilizing integral rules:}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int_{0.53\pi}^{1.96\pi}\int_{0}^{1.31}(\frac{(2\cdot rsin(\frac{(3\cdot r^{2})}{2}))}{35}- 8\cdot rcos(r^{2}))drd\theta=(- 4sin(r^{2}) -\frac{ (2cos(\frac{(3\cdot r^{2})}{2}))}{105})d\theta|_{0}^{1.31}\\&\int_{0.53\pi}^{1.96\pi}(-3.923)d\theta=(-3.923\cdot \theta)|_{0.53\pi}^{1.96\pi}=-17.62\end{align*}$

Report an Error

View Calculus 3 Tutors

Ezam
Certified Tutor

University of the West Indies, Bachelor of Science, Mathematics and Computer Science. University of Phoenix-Central Valley Ca...

View Calculus 3 Tutors

Alan
Certified Tutor

Clemson University, Bachelor of Science, Mathematics. Clemson University, Master of Science, Mathematics.

View Calculus 3 Tutors

John
Certified Tutor

The Richard Stockton College of New Jersey, Bachelor of Science, Mathematics. University of Texas Rio Grande Valley, Master o...

All Calculus 3 Resources

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

Popular Subjects

Chemistry Tutors in Miami, Reading Tutors in Houston, English Tutors in Philadelphia, Biology Tutors in Houston, GRE Tutors in Los Angeles, French Tutors in Chicago, SAT Tutors in Atlanta, ISEE Tutors in Washington DC, Biology Tutors in Boston, Algebra Tutors in New York City

Popular Courses & Classes

Spanish Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Chicago, LSAT Courses & Classes in Washington DC, GRE Courses & Classes in Boston, SAT Courses & Classes in Denver, LSAT Courses & Classes in Los Angeles, GRE Courses & Classes in Washington DC, Spanish Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in Atlanta, LSAT Courses & Classes in Miami

Popular Test Prep

MCAT Test Prep in Atlanta, SSAT Test Prep in Chicago, GMAT Test Prep in Denver, ISEE Test Prep in Washington DC, GRE Test Prep in Phoenix, SAT Test Prep in Miami, ISEE Test Prep in Houston, SSAT Test Prep in Philadelphia, SSAT Test Prep in Seattle, ISEE Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Double Integration in Polar Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #531 : Multiple Integration

Example Question #52 : Double Integration In Polar Coordinates

Example Question #51 : Double Integrals

Example Question #54 : Double Integration In Polar Coordinates

Example Question #55 : Double Integration In Polar Coordinates

Example Question #56 : Double Integration In Polar Coordinates

Example Question #57 : Double Integration In Polar Coordinates

Example Question #58 : Double Integration In Polar Coordinates

Example Question #51 : Double Integrals

Example Question #51 : Double Integration In Polar Coordinates

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: