How do you find the volume of the solid generated by revolving the region bounded by #y= 2x-1# #y= sqrt(x)# and #x=0# and revolve about the y-axis?

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Answer 1 · 2015-04-27T18:15:40

To avoid doing 2 integral (which the disc method would require), I would use cylindrical shells.

A representative slice parallel to the #y# axis has area #(sqrtx - (2x-1))dx#.

Revolving around the #y# axis, gives us a shell of volume:

#2 pi x (sqrtx - 2x+1))dx#. The volume of the solid is:

#2 pi int_0^1 x (sqrtx - 2x+1)dx = 2 pi int_0^1 (x^(3/2) - 2x^2+x)dx #

That's a straightforward integral to evaluate and finish the problem.

You should get #(7pi)/15#.

Alternative:

You could use discs (washers), but you'l have to #int_-1^0# using the

difference between the lines #x=0# and #x=(y+1)/2# And then

#int_0^1# using the difference between #x=(y+1)/2# and #x=y^2#.