I'd just plot it. Looks like you only want the loop which goes from 0 to 3 by setting y=0. The top is the same as the bottom, so just multiply the integral by two and use the positive term for y.
The loop 9y^2=x(3-x)^2 is revolved about the y-axis. Find the area of the surface generated this way.
Okay so I solved for y, derived it and squared it. Area = integral of 2(pi)ds And ds is (square root of 1+(dy/dx)^2) dx
Am I doing this right? Also, how would I find the limits of integration?