Can volume of a rotated function be calculated using definite integrals?

[Vodkacannon]

Can I calculate the volume of any function rotated once about the y-axis by multiplying the definite integral of that function by 2*pi*r?

For example if we want to generate a solid 3d shape from the function -x^2+1 we multiply the integral of it, (x - x^3 / 3), by 2*pi*1. The reason r is one in this case is because the points where the function crosses the x-axis are 1 unit away from the y axis: (-1,0) and (1,0.)

(I have not taken a calculus class yet, this is just my personal reasoning.)

 

[SteamKing]

Check out the Theorem of Pappus.

 

[Vodkacannon]

Thanks. It also turns out that r in 2*pi*r is just the size of the closed interval in the integral. So in theory this should work for any definite integral.