# Calculus Volume Integration

1. Sep 20, 2009

### kira137

I've been trying this equation for few hours now and I can't seem to get the right answer..
answer is suppose to be 640(pi)/3

The problem statement, all variables and given/known data

Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^(2)+1 and y = 9-x^2 around the x-axis

The attempt at a solution
2
Pi S (9-x^2)^2 - (x^2+1)^2 dx
-2

2
Pi S (81-2x^2+x^4-x^4-2x^2-1) dx
-2

Pi[80x - (4x^3)/3] (x=-2 to 2)

= (448(pi)/3) - (-448(pi)/3) = 896(pi)/3

that's what I keep getting..
and if that isn't the correct answer for rotating around x=0, isn't it suppose to be the answer for when rotating around y=1?

thank you in advance

2. Sep 20, 2009

### sylas

Check again your algebra when you expand the squares inside the integration.

Cheers -- sylas