Call Now to Set Up Tutoring:

(888) 888-0446

Calculus AB : Finding Volume Using Integration

Study concepts, example questions & explanations for Calculus AB

Create An Account

All Calculus AB Resources

45 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Find Cross Sections: Squares & Rectangles

Find the volume of a pyramid whose base is a square with sides of length and whose height is .

Possible Answers:

Correct answer:

Explanation:

First, it is important to consider the shape of this solid. This solid is a pyramid, with one square face and four triangular faces. Through a relationship of similar triangles, we are able to relate the known information (a height of and a base side length of ) to our general variables for side length (x) and height of the pyramid (y) . We can think of this plotted on the coordinate plane, with the width of the pyramid solid being in the direction.

$\frac{x}{3}=\frac{y}{10} \Rightarrow x=\frac{3y}{10}$

Since the side length can be squared to find a general formula for the area of the pyramid’s base (A=x^2) , this can be applied to the volume using cross-sections formula $(V=\int_{a}^{b}A(x) dx)$ as the next step. Note that the equation $x=\frac{3y}{10}$ solved for above is in terms of . Our new volume function, therefore, is also in terms of $y: V=\int_{a}^{b}(\frac{3y}{10})^2 \: dy$

Because the pyramid reaches a maximum height of , and we assume the pyramid’s starting height is at y=0 , the appropriate bounds for the integral expression are: $0\leq y\leq 10$ .

To wrap up this problem, combine the above information into one cohesive expression:

$V=\int_{0}^{10}(\frac{3y}{10})^2 dy =\frac{9}{100} \int_{0}^{10}y^2dy =\frac{9}{100}[\frac{1}{3}y^3] |_0^{10}=\frac{9}{100}*\frac{1}{3}*1000=30$

Report an Error

Example Question #1 : Find Cross Sections: Squares & Rectangles

Find the volume of the solid whose base is bounded by y=x+4 and y=x^2-2 and whose cross-sections are rectangles of height and perpendicular to the axis.

Possible Answers:

50.5

62.5

Correct answer:

62.5

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a rectangle is A=l*w . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) dx)$ , we get the following: $V=\int_{a}^{b}lw \: dx$ .

Next, an expression for must be determined. The problem specifies the length (or height) of the rectangle cross-sections is . This just leaves the value of w to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between y=x+4 and y=x^2-2 , therefore, w=(x+4)-(x^2-2)=6+x-x^2 .

Since the region is bounded by y=x+4 and y=x^2-2 , the base has the following domain: $-2\leq x\leq 3$ .

Putting this all together, we find the following:

$V=\int_{-2}^{3}(3)(6+x-x^2) dx$ $= 3[6x+\frac{1}{2}x^2-\frac{1}{3}x^3] |_{-2}^3=3(18+\frac{9}{2}-9+12-2-\frac{8}{3})=62.5$

Report an Error

Example Question #3 : Finding Volume Using Integration

Identify the correct expression for the volume of a solid whose base is bounded by y=x-3 , y=0 and y=5-x and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$V=\int_{3}^{5}4y^2-32y+64 \: dy$

$V=\int_{3}^{5}x^2 \: dx$

$V=\int_{3}^{5}y^2 \: dy$

$V=\int_{3}^{5}4x^2-32x+64 \: dx$

Correct answer:

$V=\int_{3}^{5}4x^2-32x+64 \: dx$

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a square is A=s^2 , where s is the side length of the square. By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}s^2 \: dx$ .

Next, an expression for s2 must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for s2.

s=(x-3)-(5-x)=2x-8

s^2=(2x-8)^2=4x^2-32x+64

Since the region is bounded by y=x-3 and y=5-x , the base has the following domain: 3x5 .

Putting this all together, we find the following:

$V=\int_{3}^{5}4x^2-32x+64 \: dx$

Report an Error

Example Question #1 : Find Cross Sections: Squares & Rectangles

Identify the correct expression for the volume of a solid whose base is bounded by a disk of radius and whose cross-sections are squares parallel to the axis.

Possible Answers:

$\int_{-R}^{R}R^2 \: dy$

$\int_{-R}^{R}R^2-x^2 \: dx$

$4\int_{-R}^{R}R^2-y^2 \: dy$

$4\int_{-R}^{R}y^2 \: dy$

Correct answer:

$4\int_{-R}^{R}R^2-y^2 \: dy$

Explanation:

The cross-sections are parallel to the axis; in other words, they are perpendicular to the axis. This indicates that the expression should be in terms of .

Because the disk is of radius , the base is defined by the following formula: x^2+y^2=R^2 .

The area of a square is A=s^2 , with s being the side length. By applying this formula to our general volume formula $(V=\int_{a}^{b}A(y) \: dy)$ , we get the following: $V=\int_{a}^{b}s^2 \: dy$ .

The radius defines the bounds as being $-R\leq y\leq R$ . Next, can be found by understanding that differs as the width of the circle changes. The value of is the distance from one side of the circle to the other at any given point along $-R\leq y\leq R$ . The length of one side of the square, therefore, is $s=2\sqrt{R^2-y^2}$ .

Putting it all together, the following is obtained:

$V=\int_{-R}^{R}(2\sqrt{R2-y^2})^2\: dy= 4\int_{-R}^{R}R^2-y^2 dy$

Report an Error

Example Question #5 : Finding Volume Using Integration

Let be the region bounded by y=e^x , x=0 , and x=4 . Find the volume of the solid whose base is region and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$\frac{e^8-1}{2}$

$\frac{e^4-1}{2}$

e^8

e^4

Correct answer:

$\frac{e^8-1}{2}$

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

Next, an expression for s^2 must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for s^2 .

s=e^x-0

$s^2=e^{2x}$

Since the region is bounded by x=0 and x=4 , the base has the following domain: $0\leq x\leq 4$ .

Putting this all together, we find the following:

$V=\int_{0}^{4}e^{2x} \: dx =[\frac{1}{2}e^{2x}] |_0^4=\frac{1}{2}(e^8-e^0)=\frac{e^8-1}{2}$

Report an Error

Example Question #2 : Find Cross Sections: Squares & Rectangles

Let be the region bounded by y=x^2-9 and y=0 . Find the volume of the solid whose base is region and whose cross-sections are rectangles perpendicular to the axis with height .

Possible Answers:

278

360

300

265

Correct answer:

360

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a rectangle is A=l*w . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}lw \: dx$ .

Since the region is bounded by y=x^2-9 and y=0 , the base has the following domain: $-3\leq x\leq 3$ .

Putting this all together, we find the following:

$V=\int_{-3}^{3}10(9-x^2) dx =10[9x-\frac{1}{3}x^3] |_{-3}^3=10(27-9+27-9)=360$

Report an Error

Example Question #712 : Calculus Ab

Let be the region bounded by y=x^2+3 , x=0 and y=12 . Find the volume of the solid whose base is region and whose cross-sections are rectangles perpendicular to the axis with height .

Possible Answers:

108

Correct answer:

108

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a rectangle is A=l*w . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(y) \: dy)$ , we get the following: $V=\int_{a}^{b}lw \: dy$ .

Next, an expression for must be determined. The problem specifies the length (or height) of the rectangle cross-sections is . This just leaves the value of to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between y=x2+3 , x=0 and y=12 . The function y=x^2+3 is rewritten in terms of to become $x=\sqrt{y-3}$ , because the final expression should reflect the fact that the cross sections should be written in terms of . Therefore, $w=\sqrt{y-3} -0 =\sqrt{y-3 }$ .

Since the region is bounded by y=x^2+3 and y=12 , the base has the following domain: $3\leq y\leq 12$ .

Putting this all together, we find the following:

$V=\int_{3}^{12}6(\sqrt{y-3}) dy$ $=6\int_{3}^{12}(y-3)^{\frac{1}{2}} dy =6[\frac{2}{3}(y-3)^{\frac{3}{2}}] |_3^{12} =4(27-0)=108$

Report an Error

Example Question #1 : Finding Volume Using Integration

Identify the correct expression for the volume of a solid whose base is bounded by y=ln(x)+2 and the axis along $1\leq x\leq 4$ , and whose cross-sections are rectangles perpendicular to the axis with a height three times the width.

Possible Answers:

$3\int_{1}^{4}ln(x)^2+4ln(x)+4 \: dx$

$3\int_{1}^{4}ln(x)^2+4 \: dx$

$3\int_{1}^{4}ln(x)^2 \: dx$

$3\int_{1}^{4}ln(x)+2ln(x) \: dx$

Correct answer:

$3\int_{1}^{4}ln(x)^2+4ln(x)+4 \: dx$

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a rectangle is A=l*w . By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^b{}lw \: dx$ .

Next, an expression for must be determined. The problem specifies the length (or height) of the rectangle cross-sections is three times the value of the width, or l=3w . The volume expression can now be modified: $V=\int_{a}^{b}3w^2 \: dx$

This just leaves the value of to be found. The width of the rectangle will vary as the region creating the base of the solid varies. The region is defined by the area enclosed between y=ln(x)+2 and y=0 , along $1\leq x\leq 4$ . Therefore, w=ln(x)+2-0 .

Putting this all together, we find the following:

$V=\int_{1}^{4}3(ln(x)+2)^2 dx=3\int_{1}^{4}ln(x)^2+4ln(x)+4 \: dx$

Report an Error

Example Question #3 : Find Cross Sections: Squares & Rectangles

Identify the correct expression for the volume of a solid whose base is bounded by y=x^5-x^3+4 and y=4 and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$V=\int_{0}^{1}x^6-2x^8+x^{10} \: dx$

$V=\int_{0}^{4}x^6-2x^8+x^{10} \: dx$

$V=\int_{0}^{4}x^5-x^3-4x \: dx$

$V=\int_{0}^{1}x^3- x^5 \: dx$

Correct answer:

$V=\int_{0}^{1}x^6-2x^8+x^{10} \: dx$

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

The area of a square is A=s^2 , where is the side length of the square. By applying this formula to our general volume formula $(V=\int_{a}^{b}A(x) \: dx)$ , we get the following: $V=\int_{a}^{b}s^2\: dx$ .

Next, an expression for s^2 must be determined. Because s should be the width of the solid’s base, the expression for that length can be used to solve for s^2 .

s=(4)-(x^5-x^3+4)=x^3-x^5

$s^2=(x^3-x^5)^2=x^6-2x^8+x^{10}$

Since the region is bounded by y=x^5-x^3+4 and y=4 , the base has the following domain: $0\leq x\leq 1$ .

Putting this all together, we find the following:

$V=\int_{0}^{1}x^6-2x^8+x^{10} \: dx$

Report an Error

Example Question #1 : Finding Volume Using Integration

Identify the correct expression for the volume of a solid whose base is bounded by y=sin(x) and y=0 along and whose cross-sections are squares perpendicular to the axis.

Possible Answers:

$V=\int_{0}^{\pi}sin (x) - \frac{\pi}{2}\: dx$

$V=\int_{\frac{\pi}{2}}^{\pi}sin (x)\: dx$

$V=\int_{\frac{\pi}{2}}^{\pi}sin^2(x)\: dx$

$V=\int_{\frac{\pi}{2}}^{\pi}sin^2(y)\: dy$

Correct answer:

$V=\int_{\frac{\pi}{2}}^{\pi}sin^2(x)\: dx$

Explanation:

The cross-sections are perpendicular to the axis; therefore, the volume expression will be in terms of .

Next, an expression for s^2 must be determined. Because should be the width of the solid’s base, the expression for that length can be used to solve for s^2 .

s=sin(x)-0

s^2=sin^2(x)

Putting this all together, we find the following:

$V=\int_{\frac{\pi}{2}}^{\pi}sin^2(x) \: dx$

Report an Error

View Tutors

Trennon
Certified Tutor

Appalachian State University, Bachelor of Science, Actuarial Science. Appalachian State University, Master of Science, Statis...

View Tutors

Humayun
Certified Tutor

New York University, Master's/Graduate, International Relations.

View Tutors

Kaylee
Certified Tutor

Brigham Young University, Bachelor, MFA - Animation.

All Calculus AB Resources

45 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in San Diego, Statistics Tutors in San Diego, Reading Tutors in Miami, Math Tutors in San Diego, Math Tutors in San Francisco-Bay Area, Physics Tutors in Denver, English Tutors in Houston, Biology Tutors in Phoenix, Chemistry Tutors in Denver, Calculus Tutors in Boston

Popular Courses & Classes

ACT Courses & Classes in Houston, MCAT Courses & Classes in Miami, ISEE Courses & Classes in New York City, GMAT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in New York City, GRE Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Philadelphia, GRE Courses & Classes in Chicago, ISEE Courses & Classes in Phoenix, GRE Courses & Classes in Seattle

Popular Test Prep

ISEE Test Prep in Phoenix, LSAT Test Prep in Los Angeles, GRE Test Prep in San Francisco-Bay Area, GMAT Test Prep in Los Angeles, SAT Test Prep in Seattle, LSAT Test Prep in Miami, GRE Test Prep in Chicago, SSAT Test Prep in Washington DC, LSAT Test Prep in Washington DC, MCAT Test Prep in Miami

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 AB : Finding Volume Using Integration

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Find Cross Sections: Squares & Rectangles

Example Question #1 : Find Cross Sections: Squares & Rectangles

Example Question #3 : Finding Volume Using Integration

Example Question #1 : Find Cross Sections: Squares & Rectangles

Example Question #5 : Finding Volume Using Integration

Example Question #2 : Find Cross Sections: Squares & Rectangles

Example Question #712 : Calculus Ab

Example Question #1 : Finding Volume Using Integration

Example Question #3 : Find Cross Sections: Squares & Rectangles

Example Question #1 : Finding Volume Using Integration

All Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: