Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=x^3/2 about the x-axis over the interval [1,4]

Answer 1

Vol = π∫ y^2 dx from x = 1 to 4
= π ∫ x^3 dx from 1 to 4
= π[(1/4 x^4] from 1 to 4
= π( 4^4/4 - 1/4)
= (255/4)π

Answer 2

or, using shells of thickness dy, you can get the volume with a curved boundaty by
v = ∫[1,8] 2πrh dy
where r=y and h=4-x
v = ∫[1,8] 2πy(4-y^(2/3)) dy = (243/4)π
Now, add to that the small cylinder of radius 1 and height 3, which lies under the line y=1, and has volume 3π, and you get
(243/4 + 12/4)π = (255/4)π