# Find Vol of Rotated Region R: y=sqrt x, y=sqrt(2x-1), y=0

• ssk13809
In summary, the student attempts to find the volume of a region that is bounded by the graphs y = sqrt x, and y = sqrt (2x-1) and y = 0. Using calculus, he finds that the volume of the region is pi/4.f

## 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

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?

Welcome to PF!

Hi ssk13809! Welcome to PF!

(have a pi: π and a square-root: √ )
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

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?

Perfect!

(though i haven't checked the answer)

(an alternative, if you want to have just one integral, but with both limits variable, would be to slice it into horizontal cylindrical shells of thickness dy …

do you want to to see whether that gives the same result? )

Thanks for the feedback!

I never learned the shell method or the cylindrical method, so I would be curious to see how that works.

ok, using two circular cookie-cutters (or napkin-rings, if you're posh ), cut a slice of thickness dx … that will be a cylindrical shell.

Its volume will be 2π times its radius times its length times dx.