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## Homework Statement

Find the Volume of the solid obtained by rotating the region R, that is bounded by the graphs y = sqrt x, and y = sqrt (2x-1) and y = 0, about the x-axis

## Homework Equations

The 3 equations

y= sqrt x

y = sqrt (2x-1)

y = 0

## The Attempt at a Solution

Well, the tricky part of this problem is that the 2 curves, x^(1/2) and (2x-1)^(1/2) intersect at just 1 point, which is 1. So it's not your usual problem. Nonetheless, it is still a closed region.

So here is what I decided to do

find the volume of the region sqrt(x) rotated from 0 to 1 around the x-axis. Then find the volume of the region sqrt(2x-1) rotated from 1/2 (because that is when it is equal y=0) and 1, and then subtract the first and bigger volume from the smaller one.

I ended up with pi/4.

Good strategy?