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?

Apr 27, 2015

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 $\left(\sqrt{x} - \left(2 x - 1\right)\right) \mathrm{dx}$.

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

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

$2 \pi {\int}_{0}^{1} x \left(\sqrt{x} - 2 x + 1\right) \mathrm{dx} = 2 \pi {\int}_{0}^{1} \left({x}^{\frac{3}{2}} - 2 {x}^{2} + x\right) \mathrm{dx}$

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

You should get $\frac{7 \pi}{15}$.

Alternative:

You could use discs (washers), but you'l have to ${\int}_{-} {1}^{0}$ using the

difference between the lines $x = 0$ and $x = \frac{y + 1}{2}$ And then

${\int}_{0}^{1}$ using the difference between $x = \frac{y + 1}{2}$ and $x = {y}^{2}$.