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Volume of a revolved solid

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

    y=x2
    x=y2
    Rotated about y=1

    2. Relevant equations
    Area of cross-section (in this case, a disk) = A(x) = π(outer radius)2 - π(inner radius)2
    Volume = V = ∫A(x) dx

    3. The attempt at a solution
    calc_prob1_zpsqqcr6ens.png

    yellow line => y=x2
    red line => x=y2

    I converted x=y2 to y=√x

    The outer-radius is y=x2
    The inner-radius is x=√x
    The intersection points of the two graphs is (0,0) and (1,1)

    So A(x) = π(x2)2 - π(√x)2 = π(x4-x)

    So V = ∫A(x)dx = π ∫ [ (1/5)x5 - (1/2)x2 ]dx integrated at (x=0 to x=1) = -3π/10 ...

    However the volume can't be negative and the correct answer is 11π/30

    Any help? The book shows the same graph and same intersection points, but I am getting the wrong answer.
     
  2. jcsd
  3. Jan 30, 2016 #2

    SteamKing

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    You want to use the shell method here:

    https://en.wikipedia.org/wiki/Shell_integration
     
  4. Jan 31, 2016 #3
    Thank you for you're reference, I read it and sort of get how it works. However this chapter I am in does not show me that method yet, and only shows the disk/ washer method that I attempted, so it must be solvable using that method. In fact, there is an almost identical problem using y=x2 and y=x revolved around the x-axis (y=0) that uses this method and it works. Any help trying to solve it using this method?
     
  5. Jan 31, 2016 #4
    Turns out I found out what I was doing wrong. I had to subtract 1 (y=1) from each of the outer and inner radius. So the inner radius is π(1-x2)2 and the outer radius is π(1-√x)2. When I integrated the difference of these two at 1 and 0 I got 11π/30.
     
  6. Jan 31, 2016 #5

    SammyS

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    I think you have something close to the right idea.

    Actually, the radii are: ##\ 1-x^2 \ ## and ##\ 1-\sqrt{x\,}\ ##. However, y = x2 is the lower curve (farther away from y = 1), so ##\ 1-x^2 \ ## is the outer radius.
     
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