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
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?
1 answer
1 answer