Volumes of revolution not around the axis

[SheldonG]

Homework Statement

Find the volume of $y = 2x^2$ y = 0, x = 2 when it is revolved around the line y = 8.

Homework Equations

Integral formulas for volumes by discs, washers and cylinders.

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.