Volumes of revolution not around the axis

[SheldonG]

1. The problem statement, all variables and given/known data
Find the volume of y = 2x^2 y = 0, x = 2 when it is revolved around the line y = 8.


2. Relevant equations
Integral formulas for volumes by discs, washers and cylinders.


3. The attempt at a solution
Translate the curve so that axis of revolution is along the X axis. Is this the right idea? This gives y = 2x^2 - 8 . I would integrate this and subtract from the volume of the cylinder with radius 8 and height 2:

\pi(8^2)(2) - \int_0^2 \pi(2x^2 - 8)^2\,dx

Is this the right approach?

Thanks,
Sheldon

[Dick]

This looks correct to me.

[SheldonG]

Thanks Dick, I really appreciate it.

[orb]

\pi\int_0^2 [ (8)^2 - (8 - 2x^2)^2 ] \,dx

I think that works, because if you use the washer method, the outer radius is just the part that has a y-length of 8, and the inner radius is the part above the function and under y=8, so using pi (R^2 - r^2) integrated, that's what I get. Hope that helps :)

[Dick]

\pi\int_0^2 [ (8)^2 - (8 - 2x^2)^2 ] \,dx

I think that works, because if you use the washer method, the outer radius is just the part that has a y-length of 8, and the inner radius is the part above the function and under y=8, so using pi (R^2 - r^2) integrated, that's what I get. Hope that helps :)

It's the same thing he already wrote.