Asked by ANON
The region in the first quadrant bounded by the ๐ฆ-axis and the curve ๐ฅ = 2๐ฆ2 โ ๐ฆ3 (graph please) is revolved about the ๐ฅ-axis. Find the volume?
Answers
Answered by
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
โซ[0,2] 2ฯy(2y^2-y^3) dy = 16ฯ/5
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
โซ[0,2] 2ฯy(2y^2-y^3) dy = 16ฯ/5
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