We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Multiple Integration

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 #661 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5}- 32cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.28\text{ and }1.58\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.750,0.661)\text{ and }\overrightarrow{u_2}=(-0.509,-0.861)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

38.7

-3.23

19.35

-19.35

Correct answer:

19.35

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{48}{(\frac{2}{7})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}})}{5}- 32cos(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}- 32\cdot rcos(\frac{r^{2}}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.661}{-0.750})=0.77\pi;\theta_2=arctan(\frac{-0.861}{-0.509})=1.33\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.77\pi}^{1.33\pi}\int_{0.28}^{1.58}(\frac{(48\cdot r)}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})})}- 32\cdot rcos(\frac{r^{2}}{2}))drd\theta=(- 32sin(\frac{r^{2}}{2}) -\frac{ 36}{(5\cdot (\frac{2}{7})^{(\frac{(2\cdot r^{2})}{3})}ln(\frac{2}{7}))})d\theta|_{0.28}^{1.58}\\&\int_{0.77\pi}^{1.33\pi}(11.0)d\theta=(11.0\theta)|_{0.77\pi}^{1.33\pi}=19.35\end{align*}$

Report an Error

Example Question #661 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ (2\cdot (\frac{7}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{25})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.34\text{ and }1.91\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.998,0.063)\text{ and }\overrightarrow{u_2}=(-0.094,0.996)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

127.99

-255.98

-42.66

511.96

Correct answer:

127.99

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(e^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})} +\frac{ (2\cdot (\frac{7}{2})^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{25})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(2\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{25}+ re^{(\frac{(3\cdot r^{2})}{2})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{0.998})=0.02\pi;\theta_2=arctan(\frac{0.996}{-0.094})=0.53\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.02\pi}^{0.53\pi}\int_{0.34}^{1.91}(\frac{(2\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{25}+ re^{(\frac{(3\cdot r^{2})}{2})})drd\theta=(\frac{e^{(\frac{(3\cdot r^{2})}{2})}}{3}+\frac{ (3\cdot (\frac{7}{2})^{(\frac{(2\cdot r^{2})}{3})})}{(50ln(\frac{7}{2}))})d\theta|_{0.34}^{1.91}\\&\int_{0.02\pi}^{0.53\pi}(79.88)d\theta=(79.88\theta)|_{0.02\pi}^{0.53\pi}=127.99\end{align*}$

Report an Error

Example Question #663 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{50}{(3\cdot (x^{2} + y^{2}))})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.22\text{ and }1.31\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.613,0.790)\text{ and }\overrightarrow{u_2}=(-0.454,-0.891)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

29.89

119.58

-59.79

-358.73

Correct answer:

-59.79

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{50}{(3\cdot (x^{2} + y^{2}))})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{50}{(3\cdot r)})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.790}{-0.613})=0.71\pi;\theta_2=arctan(\frac{-0.891}{-0.454})=1.35\pi\\&\text{Now, utilizing integral rules:}\\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\\&\int_{0.71\pi}^{1.35\pi}\int_{0.22}^{1.31}(-\frac{50}{(3\cdot r)})drd\theta=(-\frac{(50ln(r))}{3})d\theta|_{0.22}^{1.31}\\&\int_{0.71\pi}^{1.35\pi}(-29.74)d\theta=(-29.74\theta)|_{0.71\pi}^{1.35\pi}=-59.79\end{align*}$

Report an Error

Example Question #661 : Multiple Integration

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.41\text{ and }1.71\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.790,0.613)\text{ and }\overrightarrow{u_2}=(1.000,-0.031)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

58.46

-292.31

-58.46

1753.86

Correct answer:

-292.31

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(19sin(\frac{x^{2}}{2}+\frac{ y^{2}}{2}))}{2}-\frac{ (\frac{47}{(\frac{2}{5})^{(x^{2} + y^{2})}})}{4})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{-0.790})=0.79\pi;\theta_2=arctan(\frac{-0.031}{1.000})=1.99\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.79\pi}^{1.99\pi}\int_{0.41}^{1.71}(\frac{(19\cdot rsin(\frac{r^{2}}{2}))}{2}-\frac{ (47\cdot r)}{(4\cdot (\frac{2}{5})^{(r^{2})})})drd\theta=(\frac{47}{(8\cdot (\frac{2}{5})^{(r^{2})}ln(\frac{2}{5}))}-\frac{ (19cos(\frac{r^{2}}{2}))}{2})d\theta|_{0.41}^{1.71}\\&\int_{0.79\pi}^{1.99\pi}(-77.54)d\theta=(-77.54\theta)|_{0.79\pi}^{1.99\pi}=-292.31\end{align*}$

Report an Error

Example Question #1 : Triple Integration In Spherical Coordinates

Evaluate $\int \int \int_F 20z \ dV$ , where is the upper half of the sphere x^2+y^2+z^2=1 .

Possible Answers:

$5\pi$

$4\pi$

$\pi$

$2\pi$

Correct answer:

$5\pi$

Explanation:

Since we are only dealing with the upper half of a sphere, we can determine the boundaries easily, and remember to convert to spherical coordinates.

$0 \leq \rho \leq 1$

$0 \leq \theta \leq2\pi$

$0 \leq\phi\leq \frac{\pi}{2}$

$\int \int \int_F 20z \ dV=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} \rho^2\sin(\phi)(20\rho\cos(\phi)) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} \int_{0}^{1} 20\rho^3\sin(\phi)\cos(\phi) d\rho \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\rho^4\sin(\phi)\cos(\phi) \Big|_{0}^{1} \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5(1)^4\sin(\phi)\cos(\phi)-0 \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} \int_{0}^{2\pi} 5\sin(\phi)\cos(\phi) \ d\theta\ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 5\theta \sin(\phi)\cos(\phi) \Big|_{0}^{2\pi} \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi)-0 \ d\phi$

$=\int_{0}^{\frac{\pi}{2}} 10\pi \sin(\phi)\cos(\phi) \ d\phi$

$= -5\pi \cos^2(\phi)\Big|_{0}^{\frac{\pi}{2}}$

$= -5\pi \cos^2(\frac{\pi}{2})+5\pi \cos^2(0)=5\pi$

Report an Error

Example Question #1 : Double Integration Over General Regions

Calculate the following Integral.

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx$

Possible Answers:

$-\frac{1}{120}$

$\frac{1}{120}$

120

$-\frac{3}{120}$

Correct answer:

$-\frac{3}{120}$

Explanation:

$\int_{0}^{1}\int_{x}^{x^2} xy^2 dydx=\int_{0}^{1}\Big(\int_{x}^{x^2} xy^2 dy\Big)dx$

Lets deal with the inner integral first.

$\int_{x}^{x^2}xy^2 dy$

$\int_{x}^{x^2}xy^2 dy=\frac{1}{3}y^3x\Big|_{x}^{x^2}$

$=(\frac{1}{3}(x^2)^3\cdot x)-(\frac{1}{3}x^3\cdot x)$

$=\frac{1}{3}x^7-\frac{1}{3}x^4$

Now we evaluate this expression in the outer integral.

$\int_{0}^{1} \frac{1}{3}x^7-\frac{1}{3}x^4dx$

$=\frac{1}{3}\int_{0}^{1} x^7-x^4dx$

$=\frac{1}{3}\Big( \frac{1}{8}x^8-\frac{1}{5}x^5\Big)\Big|_{0}^{1}$

$=\frac{1}{3}\Big(\frac{1}{8}1^8-\frac{1}{5}1^5\Big)-\frac{1}{3}\Big(\frac{1}{8}0^8-\frac{1}{5}0^5\Big)$

$=\frac{1}{3}\Big(\frac{1}{8}-\frac{1}{5}\Big)-\frac{1}{3}\Big(0-0\Big)$

$=\frac{1}{3}\Big(\frac{5}{40}-\frac{8}{40}\Big)$

$=\frac{1}{3}\Big(-\frac{3}{40}\Big)=-\frac{3}{120}$

Report an Error

Example Question #1 : Double Integration Over General Regions

Calculate the definite integral of the function f(x,y) , given below as

$f(x,y)= 3x^3\sqrt{y}+ \sin(2x)\cos(4y)$

Possible Answers:

Cannot be solved.

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)+C$

$\iint f(x,y)dxdy=x^4y^{3/2}-\cos(2x)\sin(4y)+C$

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)$

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}+\frac{1}{8}\cos(2x)\sin(4y)+C$

Correct answer:

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{8}\cos(2x)\sin(4y)+C$

Explanation:

Because there are no nested terms containing both and , we can rewrite the integral as

$\iint f(x,y)dxdy=\int3x^3dx \int \sqrt{y}dy+\int\sin(2x)dx \int \cos(4y)dy$

This enables us to evaluate the double integral and the product of two independent single integrals. From the integration rules from single-variable calculus, we should arrive at the result

$\iint f(x,y)dxdy=\frac{1}{2}x^4y^{3/2}-\frac{1}{12}\cos(2x)\sin(4y)+C$ .

Report an Error

Example Question #1 : Double Integration Over General Regions

Evaluate the following integral on the region specified:

$\int\int_R(x^3y)\:dxdy$

Where R is the region defined by the conditions:

$0\leq x \leq 2, \; 1\leq y \leq2$

Possible Answers:

$4\sqrt{2}$

$2\sqrt{2}$

7.5

Correct answer:

Explanation:

$\int^2_1\int^2_0(x^3y)\:dxdy=\int^2_1(\frac{x^4y}{4}|^2_0)dy\int^2_14y\:dy =2y^2|^2_1=2(4-1)=2(3)=6$

Report an Error

Example Question #2 : Double Integration Over General Regions

Evaluate:

$\int \int x^2dxdy$

Possible Answers:

$\frac{x^2y}{3}+C$

$\frac{x^3y}{3}+C$

$\frac{x^3}{3}+y+C$

$\frac{x^3}{3}+C$

Correct answer:

$\frac{x^3y}{3}+C$

Explanation:

Because the x and y terms in the integrand are independent of one another, we can move them to their respective integrals:

$\int dy \cdot \int x^2dx= \frac{x^3y}{3}+C$

We used the following rules for integration:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$ , $\int dx=x+C$

Report an Error

Example Question #3 : Double Integration Over General Regions

Evaluate the following integral.

$\int_{0}^{1}\int_{2}^{3}\left (5x^4y \right )dydx$

Possible Answers:

$\frac{5}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

First, you must evaluate the integral with respect to y (because of the notation dydx ).

Using the rules of integration, this gets us

$5x^4\frac{y^2}{2}dx$ .

Evaluated from y=2 to y=3, we get

$5x^4(\frac{9}{2}-\frac{4}{2})dx=\frac{25}{2}x^4dx$ .

Integrating this with respect to x gets us $\frac{5}{2}x^5$ , and evaluating from x=0 to x=1, you get $\frac{5}{2}$ .

Report an Error

View Calculus 3 Tutors

Nicholas
Certified Tutor

Bryant University, Bachelor of Science, Actuarial Science.

View Calculus 3 Tutors

Ryanto
Certified Tutor

Excelsior College, Albany NY, Bachelor, Business with Math Minor. University of Houston, Master's/Graduate, Accounting.

View Calculus 3 Tutors

Mohammed
Certified Tutor

University of Maryland-College Park, Bachelor of Science, Electrical Engineering. University of Maryland-College Park, Master...

All Calculus 3 Resources

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

Popular Subjects

ISEE Tutors in Atlanta, Chemistry Tutors in Chicago, GMAT Tutors in Miami, ACT Tutors in San Francisco-Bay Area, Math Tutors in Atlanta, Calculus Tutors in Dallas Fort Worth, GRE Tutors in Miami, SAT Tutors in Boston, Chemistry Tutors in Philadelphia, French Tutors in Phoenix

Popular Courses & Classes

MCAT Courses & Classes in Atlanta, ACT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Los Angeles, Spanish Courses & Classes in Chicago, LSAT Courses & Classes in Boston, SSAT Courses & Classes in New York City, SSAT Courses & Classes in Phoenix, SAT Courses & Classes in Atlanta, SSAT Courses & Classes in San Diego

Popular Test Prep

MCAT Test Prep in Philadelphia, ISEE Test Prep in Phoenix, SSAT Test Prep in Miami, SAT Test Prep in Washington DC, LSAT Test Prep in Philadelphia, LSAT Test Prep in Washington DC, GRE Test Prep in Seattle, SAT Test Prep in Chicago, SSAT Test Prep in San Diego, SAT Test Prep in Philadelphia

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 : Multiple Integration

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #661 : Multiple Integration

Example Question #661 : Multiple Integration

Example Question #663 : Multiple Integration

Example Question #661 : Multiple Integration

Example Question #1 : Triple Integration In Spherical Coordinates

Example Question #1 : Double Integration Over General Regions

Example Question #1 : Double Integration Over General Regions

Example Question #1 : Double Integration Over General Regions

Example Question #2 : Double Integration Over General Regions

Example Question #3 : Double Integration Over General Regions

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: