## set up an integral for the volume cut out.

A hole of radius r is bored through the middle of a cylinder of radius R>r at right angles to the axis of the cylinder. Set up, but do not evaluate, an integral for the volume cut out.

My solution: The V of the bigger cylinder would be: PiR^(2)h
V of smaller cylinder is: Pir^(2)h
V cut out: PiR^(2)h-Pir^(2)h=Pih(R^2-r^2)
And after doing the integral, I kind of got integral from 0 to r of (R^2-y^2)(r^2-y^2)

But the ans. was 8*integral from 0 to r of sqrt. of (R^2-y^2) sqrt. of (r^2-y^2). Where did the 8 and sqrt. come from???