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Rotation Question

  1. Mar 2, 2010 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    The 3 equations
    y= sqrt x
    y = sqrt (2x-1)
    y = 0

    3. 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?
  2. jcsd
  3. Mar 2, 2010 #2


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    Homework Helper

    Welcome to PF!

    Hi ssk13809! Welcome to PF! :smile:

    (have a pi: π and a square-root: √ :wink:)
    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:)
  4. Mar 2, 2010 #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.
  5. Mar 2, 2010 #4


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    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:
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