Question

The region enclosed by the graph of y = x^2 , the line x = 2, and the x-axis is revolved abut the y-axis. The volume of the solid generated is:

A. 8pi
B. 32pi/5
C. 16pi/3
D. 4pi
5. 8pi/3

I solved for x as √y and set up this integral:

2pi * integral from 0 to 2 of y√y.

But it doesn't seem to give the answer, and I'm not sure what's wrong with it.

Answer 1

Using shells of thickness dx,

v = ∫[0,2] 2πrh dx
where r=x and h=y=x^2
v = ∫[0,2] 2πx^3 dx = 8π

using discs (washers) of thickness dy,

v = ∫[0,4] π(R^2-r^2) dy
where R=2 and r=x=√y
v = ∫[0,4] π(4-y) dy = 8π

You appear to have tried to rotate around the x-axis.

Answer 2

also, if you did that rotation, the height of the shells would be 2-x, not x.