The region in the first quadrant bounded by the 𝑦-axis and the curve 𝑥 = 2𝑦2 − 𝑦3 (graph please) is revolved about the 𝑥-axis. Find the volume?

Answer 1

x = y^2(2-y)

So the region in the 1st quadrant is a hump from (0,0) to (0,2)

To revolve that around the x-axis it is best to use shells of radius y, thickness dy and height x. So, the total volume is

∫[0,2] 2πy(2y^2-y^3) dy = 16π/5