Question

The area enclosed by the loop of the curve x²=y(y-3)² is revolved about the y-axis. Find the volume generated.

Answer 1

The loop closes on itself at (0,3). The y-axis is the axis of symmetry, so you can generate the volume just by revolving one half of the loop. It's easier to express x as a function of y, so use discs of thickness dy.
v = ∫[0,3] πr^2 dy
where r=x
v = ∫[0,3] π(y(3-y)^2) dy = 27π/4