We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Volume

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3932 : Calculus

Find the volume of the function $f(x)=\sqrt{cos \,x}$ revolved around the -axis on the interval $\left[0, \frac{\pi}{2}\right]$ .

Possible Answers:

$\pi^2$ cubic units

$\pi$ cubic units

$\frac{\pi}{2}$ cubic units

$3.14\pi$ cubic units

Correct answer:

$\pi$ cubic units

Explanation:

The formula for the volume of the region revolved around the x-axis on the interval [a,b] is given as

$V=\pi \, \int_{a}^{b} r^2 \, dx$

where r=f(x) .

As such,

$V=\pi \, \int_{0}^{\pi/2} (\sqrt{cos\,x})^2 \, dx$

$=\pi \, \int_{0}^{\pi/2} cos\,x \, dx$

$=\pi(sin\,x)|_{0}^{\pi/2}$ .

And by the corollary of the First Fundamental Theorem of Calculus

$\pi(sin\,x)|_{0}^{\pi/2}=\pi[sin(\frac{\pi}{2})-sin(0)]=\pi$ .

As such, the volume is

$\pi$ cubic units.

Report an Error

Example Question #21 : Regions

Find the volume generated by rotating the function $f(x)=3xe^{x^3}$ around the -axis over the interval x=[0,2] .

Possible Answers:

$\frac{3}{2}\pi(e^{16}-1)$

$\frac{3}{2}\pi(e^{8}-1)$

$\frac{3}{2}(e^{16}-1)$

$\frac{3}{2}\pi(e^{16})$

$\frac{3}{2}\pi(e^{8})$

Correct answer:

$\frac{3}{2}\pi(e^{16}-1)$

Explanation:

To approach this problem, treat the function $f(x)=3xe^{x^3}$ as the radius of a disc, with infintesimal thickness dx (which has the x-axis going through its center).

This infinitesismal volume of this disc is:

$dV=\pi(3xe^{x^3})^2dx$

$dV=9\pi x^2e^{2x^3}dx$

Now the volume of the solid rotated around the x-axis is the sum of these discs. In other word, it is the integral:

$V=9\pi \int_0^2 x^2e^{2x^3}dx$

Note that the derivative of $e^{2x^3}$ is $6x^2e^{2x^3}$ , so from this we can infer that the integral of $x^2e^{2x^3}$ is $\frac{1}{6}e^{2x^3}$ :

$V=\frac{9}{6}\pi(e^{2x^3})|_0^2$

$V=\frac{9}{6}\pi(e^{16}-e^0)$

$V=\frac{3}{2}\pi(e^{16}-1)$

Report an Error

Example Question #3931 : Calculus

Find the volume of the solid generated by rotating the parabola f(x)=25-x^2 around the -axis, over the region where the function is greater than or equal to zero.

Possible Answers:

$\frac{5000}{3}\pi$

$\frac{10000}{3}\pi$

$5000\pi$

$10000\pi$

Correct answer:

$\frac{10000}{3}\pi$

Explanation:

In stating the bounds, the question is asking that we use those values of x where the function intercepts the x-axis:

25-x^2=0

$x=\pm 5$

Now, treat the function f(x)=25-x^2 as the radius of a disc with infinitesimal thickness . The volume of this disc is thus

dV=Adx

$dV=\pi f(x)^2 dx$

$dV=\pi (x^4-50x^2+625)dx$

And the volume can found via integration:

$V= \int_{-5}^5\pi(x^4-50x^2+625)dx$

$V=\pi (\frac{x^5}{5}-\frac{50x^3}{3}+625x)|_{-5}^5$

$V=\pi (\frac{3125}{5}-\frac{6250}{3}+3125)-\pi(-\frac{3125}{5}+\frac{6250}{3}-3125)$

$V=\frac{10000}{3}\pi$

Report an Error

Example Question #3934 : Calculus

For this problem assume that volume is given by:

$V=\int S(x)dz$ , where S(x) is the surface area function.

Given the formula for surface area of a structure is given by S(x)=32sec^2(x) .

Determine the volume function of this structure.

For this problem assume that is not a function of . given below is a constant.

Possible Answers:

V=32sec^2(x)*z +C

V=32tan(x)(z+C)

V=32sec^2(x)(z+C)

V=z(32tan(x)+C)

Correct answer:

V=32sec^2(x)(z+C)

Explanation:

We're given that

$V=\int S(x)dz$

Since the indefinite integral is with respect to , we can move the surface area function in front of the integral since we are told that is not a function of .

$V=S(x)\int 1dz$

Using the power rule,

$\int 1dz=z+C$ , where is a constant.

V=S(x)(z+C)

In our problem S(x)=32sec^2(x)

Therefore, volume can be given as

V=S(x)(z+C)=32sec^2(x)(z+C)

Report an Error

Example Question #2908 : Functions

Using the method of cylindrical disks, find the volume of the region of the graph of the function

f(x)=x-5

on the interval [5,10] .

Possible Answers:

$150\pi$ units cubed

$250\pi$ units cubed

$300\pi$ units cubed

$\frac{125}{3}\pi$ units cubed

Correct answer:

$\frac{125}{3}\pi$ units cubed

Explanation:

The formula for volume is given as

$V=\pi \int_{a}^{b} r^2\,dx$

where r=f(x) .

As such,

$V=\pi \int_{5}^{10} (x-5)^2\,dx =\pi \int_{5}^{10}x^2-10x+25\,dx$ .

When taking the integral, we will use the inverse power rule which states

$\int x^n = \frac{x^{n+1}}{n+1}$ .

Applying this rule term by term we get

$\pi \int_{5}^{10}x^2-10x+25\,dx =\pi \left ( \frac{x^3}{3} - \frac{10x^2}{2} + 25x\right) |_{5}^{10}$

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi \left ( \frac{x^3}{3} - \frac{10x^2}{2} + 25x\right |_{5}^{10}=\pi \left ( [\frac{10^3}{3}-\frac{10(10)^2}{2}+25(10)]-[\frac{5^3}{3}-\frac{10(5)^2}{2}+25(5)]\right )$

$=\frac{125}{3}\pi$

As such, the volume is

$\frac{125}{3}\pi$ units cubed.

Report an Error

Example Question #2909 : Functions

Using the method of cylindrical disks, find the volume of the region of the graph of the function

$f(x)=\sqrt{x-\pi}$

on the interval $[\pi,9+\pi]$ .

Possible Answers:

$50\pi$ units cubed

$55\pi$ units cubed

$35\pi$ units cubed

$40.5}\pi$ units cubed

Correct answer:

$40.5}\pi$ units cubed

Explanation:

The formula for volume is given as

$V=\pi \int_{a}^{b} r^2\,dx$

where r=f(x) .

As such,

$V=\pi \int_{\pi}^{9+\pi} (\sqrt{x-\pi})^2\,dx =\pi \int_{\pi}^{9+\pi} x-\pi\,dx$ .

When taking the integral, we will use the inverse power rule which states

$\int x^n = \frac{x^{n+1}}{n+1}$ .

Applying this rule term by term we get

$\pi \int_{\pi}^{9+\pi} x-\pi\,dx=\pi \left ( \frac{x^2}{2}-\pi x\ \right) |_{\pi}^{9+\pi}$ .

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi \left ( \frac{x^2}{2}-\pi x\ \right )|_{\pi}^{9+\pi}=\pi\left ( [\frac{(9+\pi)^2}{2}-\pi (9+\pi)]-[\frac{\pi^2}{2}-\pi^2] \right )$

$=40.5\pi$ .

As such, the volume is

$40.5}\pi$ units cubed.

Report an Error

Example Question #2910 : Functions

Set up the integral to find the volume of the solid if the region is bounded by y=x^2 and y=4x-3 , revolved about x=5 .

Possible Answers:

$\int_{1}^{3} 2\pi(5-x)(4x-3-x^2)\:dx$

$\int_{1}^{5} \pi[(4x-3)^2-x^2]^2 \:dx$

$\int_{1}^{3} 2\pi(x-5)(4x-3-x^2)\:dx$

$\int_{1}^{5} \pi(4x-3-x^2)^2\:dx$

$\int_{1}^{3} \pi(5-x)^2(4x-3-x^2)\:dx$

Correct answer:

$\int_{1}^{3} 2\pi(5-x)(4x-3-x^2)\:dx$

Explanation:

For this problem, we will use the shell method. Write the formula for the shell method.

$V=\int_{a}^{b} 2\pi rh \:dr$

The thickness of the shell is .

The radius of the shell is the distance from the y-axis to the vertical rectangles, 5-x .

The height of the shell is the top curve subtract the bottom curve. The top curve is y=4x-3 and the bottom curve is y=x^2 . The height is: 4x-3-x^2 .

Determine the bounds by setting the top and bottom equations equal to each other.

x^2=4x-3

Solve for roots.

x^2-4x+3=0

(x-3)(x-1)=0

The roots are x = 1,3 and will be the bounds to the integral. Write the integral.

$V=\int_{a}^{b} 2\pi rh \:dr= \int_{1}^{3} 2\pi(5-x)(4x-3-x^2)\:dx$

Report an Error

Example Question #3935 : Calculus

Sphere segment

A sphere with a radius of 5 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 4 units to the left of the origin and the second cut is 2 units to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{49}{3}\pi$

$\frac{128}{3}\pi$

$\frac{94}{3}\pi$

$\frac{251}{3}\pi$

$\frac{23}{3}\pi$

Correct answer:

$\frac{94}{3}\pi$

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{a}^{b}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{a}^{b}\pi(r^2-x^2)dx$

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(5^2x-\frac{x^3}{3})|_{-4}^{-2}$

$V=\pi(5^2(-2)-\frac{(-2)^3}{3})-\pi(5^2(-4)-\frac{(-4)^3}{3})$

$V=\frac{94}{3}\pi$

Report an Error

Example Question #22 : How To Find Volume Of A Region

Sphere segment

A sphere with a radius of 10 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 4 units to the left of the origin and the second cut is 3 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$\frac{802}{3}\pi$

$\frac{3112}{3}\pi$

$\frac{4129}{3}\pi$

$\frac{2009}{3}\pi$

$\frac{1984}{3}\pi$

Correct answer:

$\frac{2009}{3}\pi$

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{a}^{b}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{a}^{b}\pi(r^2-x^2)dx$

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(10^2x-\frac{x^3}{3})|_{-4}^{3}$

$V=\pi(10^2(3)-\frac{(3)^3}{3})-\pi(10^2(-4)-\frac{(-4)^3}{3})$

$V=\frac{2009}{3}\pi$

Report an Error

Example Question #21 : Regions

Sphere segment

A sphere with a radius of 3 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 2 units to the left of the origin and the second cut is 1 unit to the right of the origin, what is the volume of the segment?

Possible Answers:

$24\pi$

$\frac{40}{3}\pi$

$30\pi$

$\frac{20}{3}\pi$

$12\pi$

Correct answer:

$24\pi$

Explanation:

To approach this problem, begin by picturing how the sphere might be projected onto the two-dimensional Cartesian plane:

Sphere segment outlined

In its two-dimensional projection, we notice the outline of a circle. The formula for a circle of radius in the Cartesian coordinate system is given as:

x^2+y^2=r^2

Which can be rewritten in terms of as $y=\sqrt{r^2-x^2}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{a}^{b}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our f(x) , this integral can be written as:

$V=\int_{a}^{b}\pi(r^2-x^2)dx$

$V=\pi(r^2x-\frac{x^3}{3})|_{a}^{b}$

Consider the points where circle contacts the x-axis, the greatest and smallest possible values of on the circle:

$V=\pi(r^2x-\frac{x^3}{3})|_{-r}^{r}$

$V=\pi(r^2r-\frac{r^3}{3})-\pi(r^2(-r)-\frac{(-r)^3}{3})$

$V=\pi(\frac{2}{3}r^3)-\pi(-\frac{2}{3}r^3)$

$V=\frac{4}{3}\pi r^3$

This is how the volume of a sphere can be derived, and similary, this is how we can derive the area of our segment:

$V=\pi(3^2x-\frac{x^3}{3})|_{-2}^{1}$

$V=\pi(3^2(1)-\frac{(1)^3}{3})-\pi(3^2(-2)-\frac{(-2)^3}{3})$

$V=24\pi$

Report an Error

View Calculus Tutors

Michelle
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Mathematics.

View Calculus Tutors

Leonard
Certified Tutor

Florida State University, Bachelor of Science, Environmental Science.

View Calculus Tutors

Mohammad
Certified Tutor

Rutgers University-New Brunswick, Bachelor of Science, Mathematics. Rutgers University-New Brunswick, Master of Science, Stat...

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in New York City, SSAT Tutors in Philadelphia, Chemistry Tutors in Philadelphia, ISEE Tutors in Phoenix, LSAT Tutors in Washington DC, Computer Science Tutors in Denver, Reading Tutors in San Diego, SAT Tutors in Boston, Calculus Tutors in Boston, ACT Tutors in Miami

Popular Courses & Classes

ISEE Courses & Classes in Washington DC, GRE Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in Washington DC, Spanish Courses & Classes in Houston, Spanish Courses & Classes in Phoenix, GRE Courses & Classes in Philadelphia, ISEE Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Seattle, ISEE Courses & Classes in Miami, LSAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

SSAT Test Prep in Houston, ISEE Test Prep in Chicago, SSAT Test Prep in Washington DC, MCAT Test Prep in Miami, GMAT Test Prep in Philadelphia, SAT Test Prep in San Francisco-Bay Area, ACT Test Prep in New York City, SAT Test Prep in Dallas Fort Worth, MCAT Test Prep in New York City, ISEE Test Prep in Denver

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 1 : Volume

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3932 : Calculus

Example Question #21 : Regions

Example Question #3931 : Calculus

Example Question #3934 : Calculus

Example Question #2908 : Functions

Example Question #2909 : Functions

Example Question #2910 : Functions

Example Question #3935 : Calculus

Example Question #22 : How To Find Volume Of A Region

Example Question #21 : Regions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: