Calculating Volume of Solid Formed by Revolving Region

[frumdogg]

Homework Statement

Find the volume of a solid formed by revolving the region bounded by graphs of:
y=x^3
y=1
and
x=2

Homework Equations

\pi0\int2(x^3)dx

The Attempt at a Solution

x^7/7 with boundaries of [0,2]

Am I on the right path?

 

[Dick]

That's ok if you ignore the y=1 boundary, but I don't think you should. Draw a picture. I presume you are rotating around the x-axis?

 

[HallsofIvy]

Homework Statement

Find the volume of a solid formed by revolving the region bounded by graphs of:
y=x^3
y=1
and
x=2

Revolved around what axis?

Homework Equations

\pi0\int2(x^3)dx

If you are the x-axis and are using the disk method, you would be integrating \pi\int_0^1 y^2 dx+ \pi\int_1^2 1 dx

The Attempt at a Solution

x^7/7 with boundaries of [0,2]

Am I on the right path?

 

[frumdogg]

Yes, rotating around the x-axis.

 

[tiny-tim]

Find the volume of a solid formed by revolving the region bounded by graphs of:
y=x^3
y=1
and
x=2

Hi frumdogg!

That doesn't look solid … do you mean y = 2 ?

 

[frumdogg]

The problem says
y=x^3
y=1
x=2

When graphing it, it's a small area with x^3 on the left, y=1 on top, x=2 on right, and x-axis on bottom, at least what I am coming up with.

 

[tiny-tim]

hmm … they might as well have said:

y=x^3
x=1;

the rest is just a cylinder.

 

[Dick]

The problem says
y=x^3
y=1
x=2

When graphing it, it's a small area with x^3 on the left, y=1 on top, x=2 on right, and x-axis on bottom, at least what I am coming up with.

It's a region with y=0 at the bottom and y=x^3 at the top from x equal 0 to 1 and y=1 at the top and y=0 at the bottom from x equal 1 to 2. Halls already set it up for you. tiny-tim is saying the same thing.