Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the quations about the x-axis.
Y=x^2+4
Y=-x^2+2x+8
X=0
X=3

Answer 1

The region is lens-shaped, with tips at (-1,5) and (2,8). Unfortunately, our domain crosses over the right end of the enclosed region, ending at the strip from (3,7) to (13)

Let's say
y1 = -x^2+2x+8 and y2 = x^2+4

Using discs of thickness dx, we have

v = ∫π(R^2-r^2) dx
= ∫[0,2] π((y1)^2-(y2)^2) dx + ∫[2,3] π((y2)^2-(y1)^2) dx

Using shells of thickness dy, we have

v = ∫2πrh dy
where r=y and h is the horizontal distance between the curves x1 and x2, which gets a bit more involved.

Answer 2

followup. Using discs,
v = 272π/3 + 191π/3 = 463π/3

Using shells, it's a bit stranger. I made a typo, and the right boundary goes from y=5 to 13.
v
= ∫[4,8] 2πy√(y-4) dy
+ ∫[8,9] 2πy(1+√(9-y)-(1-√(9-y))) dy
+ ∫[5,8] 2πy(3-(1+√(9-y))) dy
+ ∫[8,13] 2πy(3-√(y-4)) dy
= 1024π/15 + 112π/5 + 94π/5 + 673π/15
= 463π/3

see

http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2%2B4,+y%3D-x%5E2%2B2x%2B8