Question

Find the area of the surface of revolution generated by revolving the curve y = 3 sqrt (x), 0 <= x <= 4, about the x-axis.


Okay, so I've set up the integral like this:
2pi ∫[0,4] (3 sqrt (x))(sqrt(1+(1/4x)))dx

Which is coming out to 108.5, but that's not giving me the right answer. Can you help me set up the integral correctly or tell me what I'm doing wrong?

Answer 1

okay. You seem to be using shells, so you need to be integrating along y, since the shells have a thickness of dy.

v = ∫[0,6] 2πrh dy
where r = y and h = 4-x = 4-y^2/9
v = ∫[0,6] 2πy(4-y^2/9) dy = 72π

If you want to integrate along x, you need to use discs of thickness dx:

v = ∫[0,4] πr^2 dx
where r=y
v = ∫[0,4] π(3√x)^2 dx = 72π