Question

Can someone please check my set up for a disk?
Integration

Answer 1

y=(1/8)x^2
y=2-x
x=0
revolve around y axis

Answer 2

\[ pi \int\limits_{0}^{48-32\sqrt{2}}\left( \sqrt{8y} \right)^2dy-\int\limits_{0}^{2} pi \left( -y+2 \right)^2 dy\]

Answer 3

I also, tried to do it as a shell, not sure I am correct

Answer 4

you're revolving around the y axis, so the radius for any disk is x(y) not y(x).

Answer 5

\[\int\limits_{0}^{-4+4\sqrt{2}}2 pi x (2-x-(1/8)x^2)dx\]

Answer 6

sorry, I can't see straight anymore. Maybe I should call this a night and look at it tomorrow.