Question

The region in the first quadrant bounded by the ๐‘ฆ-axis and the curve ๐‘ฅ = 2๐‘ฆ2 โˆ’ ๐‘ฆ3 (graph please) is revolved about the ๐‘ฅ-axis. Find the volume?

Answers

Steve
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

โˆซ[0,2] 2ฯ€y(2y^2-y^3) dy = 16ฯ€/5

Related Questions