# Solids of Revolution - Volume

1. Feb 7, 2012

### Heroesrule99

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

Find the volume of the solid obtained by rotating the region bounded by y=2x2-x3 and y=0 about the y-axis

2. Relevant equations

There no required method (between Disk, Washer, Shell). In my attempt below, I used the Shell method, I believe. 2π(shell radius)(shell height)dx

3. The attempt at a solution

I graphed the equation and ended up with V = ∫20 2π(x)(2x2-x3)dx

V =2π*limit 0 to 2* (2x3-x4)dx

V = 2π *limit 0 to 2* x4/2 - x5/5

V =$$2π (1/2 - 1/5)$$
V= 3π/5

Am I on the right track or did I mess up horribly in the setup? I feel like the area where I might have made a mistake was the limits I set... and still getting used to the functions on here lol sorry

2. Feb 7, 2012

### Dick

The setup looks ok to me but it looks like you are putting x=1 for the upper limit of your integral instead of x=2 when you evaluate it. Why would you do that?

Last edited: Feb 7, 2012
3. Feb 8, 2012

### Heroesrule99

Ah, just a silly error. So when I plug in 2 for x rather than one. I get 8 - 6.4 = 3.2pi. But other than that it looks alright? I was worried about how I choose to set up my limits since I found the x intercept for the upper limit...but kinda sorta just guessed that the other was 0. There has to be a more technical way to go about that. @_@

4. Feb 8, 2012

### Dick

Yes, that's fine. There are two x intercepts. Factor. 2x^2-x^3=x^2*(2-x)=0. So either x=0 or x=2.

5. Feb 8, 2012

### Heroesrule99

Thanks so much!