Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Regions

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 #3961 : Calculus

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 3 and a major axes (the horizontal length) of 8, 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 at the origin, what is the volume of the segment?

Possible Answers:

$3\pi$

$28\pi$

$8\pi$

$48\pi$

$24\pi$

Correct answer:

$24\pi$

Explanation:

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

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

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

$V=\int_{m}^{n}\pi f^2(x)dx$

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:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

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

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

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

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

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

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius b=3 . and the length of the spheroid 2a=8 , or a=4 :

$V=\pi (3)^2(x-\frac{x^3}{3(4)^2})|_{-4}^{0}$

$V=\pi (3)^2(0-\frac{0^3}{3(4)^2})-\pi (3)^2(-4-\frac{(-4)^3}{3(4)^2})$

$V=24\pi$

Report an Error

Example Question #3962 : Calculus

Ellipse segment

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 6 and a major axes (the horizontal length) of 14, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 7 units to the left of the origin and the second cut is 7 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$168\pi$

$462\pi$

$336\pi$

$231\pi$

$198\pi$

Correct answer:

$336\pi$

Explanation:

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

Ellipse segment outlined

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

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

$V=\int_{m}^{n}\pi f^2(x)dx$

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:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

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

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

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

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

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

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius b=6 . and the length of the spheroid 2a=14 , or a=7 :

$V=\pi (6)^2(x-\frac{x^3}{3(7)^2})|_{-7}^{7}$

$V=\pi (6)^2(7-\frac{7^3}{3(7)^2})-\pi (6)^2(-7-\frac{(-7)^3}{3(7)^2})$

$V=336\pi$

In the case of how this problem was defined, it turns out we were asked to find the area of the full spheroid.

Report an Error

Example Question #53 : Volume

Sphere segment

A sphere with a radius of 13 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 6 units to the left of the origin and the second cut is 4 units to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{808}{3}\pi$

$\frac{431}{3}\pi$

$\frac{862}{3}\pi$

$\frac{202}{3}\pi$

$\frac{404}{3}\pi$

Correct answer:

$\frac{862}{3}\pi$

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:

$V=\pi(13^2x-\frac{13^3}{3})|_{-6}^{-4}$

$V=\pi(13^2(-4)-\frac{(-4)^3}{3})-\pi(13^2(-6)-\frac{(-6)^3}{3})=\frac{862}{3}\pi$

Report an Error

Example Question #54 : Volume

Sphere segment

A sphere with a radius of 100 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 99 units to the left of the origin and the second cut is 98 units to the left of the origin, what is the volume of the segment?

Possible Answers:

$\frac{905}{3}\pi$

$\frac{287}{3}\pi$

$\frac{893}{3}\pi$

$\frac{514}{3}\pi$

$\frac{781}{3}\pi$

Correct answer:

$\frac{893}{3}\pi$

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:

$V=\pi(100^2x-\frac{x^3}{3})|_{-99}^{-98}$

$V=\pi(100^2((-98)-(-99))-\frac{(-98)^3-(-99)^3}{3})=\frac{893}{3}\pi$

Report an Error

Example Question #55 : Volume

Sphere segment

A sphere with a radius of 3.2 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 1.6 units to the right of the origin and the second cut is 2.4 units to the right of the origin, what is the volume of the segment?

Possible Answers:

22.3

14.1

15.5

6.6

8.9

Correct answer:

15.5

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:

$V=\pi(3.2^2x-\frac{x^3}{3})|_{1.6}^{2.4}$

$V=\pi(3.2^2((2.4)-(1.6))-\frac{(2.4)^3-(1.6)^3}{3})=15.5$

Report an Error

Example Question #3963 : Calculus

Sphere segment

A sphere with a radius of 9.8 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 1.2 units to the right of the origin and the second cut is 2.8 units to the right of the origin, what is the volume of the segment?

Possible Answers:

461.6

732.5

98.3

606.1

309.4

Correct answer:

461.6

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:

$V=\pi(9.8^2x-\frac{x^3}{3})|_{1.2}^{2.8}$

$V=\pi(9.8^2((2.8)-(1.2))-\frac{(2.8)^3-(1.2)^3}{3})=461.6$

Report an Error

Example Question #3964 : Calculus

Sphere segment

A sphere with a radius of 85 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 10 units to the right of the origin and the second cut is 13 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$24510\pi$

$12768\pi$

$18904\pi$

$30816\pi$

$21276\pi$

Correct answer:

$21276\pi$

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:

$V=\pi(85^2x-\frac{x^3}{3})|_{10}^{13}$

$V=\pi(85^2((13)-(10))-\frac{(13)^3-(10)^3}{3})=21276\pi$

Report an Error

Example Question #3965 : Calculus

Sphere segment

A sphere with a radius of 14 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 12 units to the right of the origin and the second cut is 14 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$90\pi$

$\frac{160}{3}\pi$

$50\pi$

$\frac{160}{3}\pi$

$\frac{140}{3}\pi$

Correct answer:

$\frac{160}{3}\pi$

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:

$V=\pi(14^2x-\frac{x^3}{3})|_{12}^{14}$

$V=\pi(14^2((14)-(12))-\frac{(14)^3-(12)^3}{3})=\frac{160}{3}\pi$

Report an Error

Example Question #51 : How To Find Volume Of A Region

Sphere segment

A sphere with a radius of 20 has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 4 units to the right of the origin and the second cut is 18 units to the right of the origin, what is the volume of the segment?

Possible Answers:

$\frac{7892}{3}\pi$

$\frac{14137}{3}\pi$

$\frac{10903}{3}\pi$

$\frac{11032}{3}\pi$

$\frac{5411}{3}\pi$

Correct answer:

$\frac{11032}{3}\pi$

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:

$V=\pi(20^2x-\frac{x^3}{3})|_{4}^{18}$

$V=\pi(20^2((18)-(4))-\frac{(18)^3-(4)^3}{3})=\frac{11032}{3}\pi$

Report an Error

Example Question #52 : 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 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:

$\frac{907}{3}\pi$

$297\pi$

$103\pi$

$\frac{772}{3}\pi$

$\frac{293}{3}\pi$

Correct answer:

$297\pi$

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:

$V=\pi(10^2x-\frac{x^3}{3})|_{-2}^{1}$

$V=\pi(10^2((1)-(-2))-\frac{(1)^3-(-2)^3}{3})=297\pi$

Report an Error

View Calculus Tutors

Michael
Certified Tutor

Kennesaw State University, Bachelor of Science, Mathematics.

View Calculus Tutors

Siddharth
Certified Tutor

University of Petroleum and Energy Studies and, Masters in Business Administration, Aviation Management.

View Calculus Tutors

Emmanuel
Certified Tutor

Baylor University, Bachelor of Science, Biology, General.

All Calculus 1 Resources

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

Popular Subjects

Statistics Tutors in Atlanta, Algebra Tutors in Miami, Calculus Tutors in San Francisco-Bay Area, Calculus Tutors in Miami, LSAT Tutors in Phoenix, Physics Tutors in Phoenix, Math Tutors in New York City, SSAT Tutors in San Francisco-Bay Area, ACT Tutors in Boston, GMAT Tutors in Chicago

Popular Courses & Classes

SSAT Courses & Classes in Denver, MCAT Courses & Classes in Boston, Spanish Courses & Classes in Chicago, GRE Courses & Classes in Phoenix, ISEE Courses & Classes in Washington DC, SSAT Courses & Classes in Phoenix, SSAT Courses & Classes in Dallas Fort Worth, GMAT Courses & Classes in Los Angeles, ACT Courses & Classes in Philadelphia, Spanish Courses & Classes in San Diego

Popular Test Prep

SSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Seattle, ACT Test Prep in Seattle, SAT Test Prep in Atlanta, ISEE Test Prep in Washington DC, SAT Test Prep in Houston, GRE Test Prep in Phoenix, SSAT Test Prep in Miami, MCAT Test Prep in Los Angeles, ISEE Test Prep in Los Angeles

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 : Regions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3961 : Calculus

Example Question #3962 : Calculus

Example Question #53 : Volume

Example Question #54 : Volume

Example Question #55 : Volume

Example Question #3963 : Calculus

Example Question #3964 : Calculus

Example Question #3965 : Calculus

Example Question #51 : How To Find Volume Of A Region

Example Question #52 : How To Find Volume Of A Region

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: