# Disc Method

1. Apr 30, 2008

### frumdogg

1. The problem statement, all variables and given/known data

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

2. Relevant equations
$$\pi$$0$$\int$$2(x^3)dx

3. The attempt at a solution

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

Am I on the right path?

2. Apr 30, 2008

### 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?

3. Apr 30, 2008

### HallsofIvy

Staff Emeritus
Revolved around what axis?

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

4. Apr 30, 2008

### frumdogg

Yes, rotating around the x-axis.

5. Apr 30, 2008

### tiny-tim

Hi frumdogg!

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

6. Apr 30, 2008

### 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.

7. Apr 30, 2008

### tiny-tim

hmm … they might as well have said:

y=x^3
x=1;

the rest is just a cylinder.

8. Apr 30, 2008

### Dick

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.