Find the volume of the solid obtained by rotating the region under the graph of f(x)= 9-x^2 for 0<= x<=3 about the vertical axis x= -2.

Question

My answer: 81 pi/2

Answer 1

using washers,

v = ∫[0,9] π(R^2-r^2) dy
where R=2+√(9-y) and r=2
v = ∫[0,9] π((2+√(9-y))^2-4) dy
v = 225π/2

or, using shells

v = ∫[0,3] 2πrh dx
where r = x+2 and h = 9-x^2
v = ∫[0,3] 2π(x+2)(9-x^2) dx
= 225π/2