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

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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.
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ssk13809
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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?
 
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  • #2
Welcome to PF!

Hi ssk13809! Welcome to PF! :smile:

(have a pi: π and a square-root: √ :wink:)
ssk13809 said:
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! :biggrin:

(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? :wink:)
 
  • #3
Thanks for the feedback!

I never learned the shell method or the cylindrical method, so I would be curious to see how that works.
 
  • #4
ok, using two circular cookie-cutters (or napkin-rings, if you're posh :wink:), 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. :smile:
 

1. How do I find the volume of a rotated region using the given equations?

To find the volume of a rotated region, you will need to use the formula V = π∫ (f(x))^2 dx, where f(x) represents the height of the function at a given point and the integral is taken over the limits of the region. In this case, the limits will be determined by the x-values where the two functions intersect, which can be found by setting them equal to each other and solving for x.

2. Why is the region rotated?

The region is rotated because we are trying to find the volume of a three-dimensional shape formed by rotating the given region around the x-axis. This rotation allows us to use the formula for the volume of a solid of revolution.

3. How do I determine the limits of integration for this problem?

The limits of integration can be determined by finding the x-values where the two given functions intersect. To do this, set the two functions equal to each other and solve for x. These x-values will be the limits of the integral.

4. Can I use a different axis of rotation?

Yes, you can use a different axis of rotation if the problem specifies. In this case, the x-axis was specified as the axis of rotation, but the formula for the volume of a solid of revolution can be modified to use a different axis.

5. Is there a shortcut to finding the volume of a rotated region?

Yes, there is a shortcut method known as the "disk method" that can be used to find the volume of a rotated region. This involves slicing the region into thin disks and using the formula V = π∫ (f(x))^2 dx for each disk, then adding up all of the volumes. However, this method can only be used when the region is rotated around the x-axis.

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