Originally Posted by

**skeeter** first note that if y = 2x, then x = y/2 .

hope you have made a sketch and know that this problem involves horizontal disks with respect to y.

for rotation about x = 1, your radius of revolution is a horizontal segment from the line x = 1 to the line x = y/2.

since x = 1 is to the right of x = y/2, the length of the radius is 1 - y/2.

so, your volume found by disks is ...

$\displaystyle V = \pi \int_0^2 \left(1 - \frac{y}{2}\right)^2 \, dy$

for rotation about x = 2, note that this will involve the washer method w/r to y.

your inner radius goes from x = 2 to x = 1 ... r = 2-1 = 1

your outer radius goes from x = 2 to x = y/2 ... R = 2 - y/2

$\displaystyle

V = \pi \int_0^2 \left(2 - \frac{y}{2}\right)^2 - (1)^2 \, dy$