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Solids of revolution volume

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Volume of the region bounded by y = x^2 and x = y^2 about y = 1


    2. Relevant equations
    [itex]\pi r^2[/itex]


    3. The attempt at a solution
    So the functions look something like this:

    ICX4Rud.png

    I decided to use method of washers with respect to x.

    The radius if the center is at y = 1 of the washers is going to be [itex]\sqrt{x} - x^2[/itex] and the inside space is going to have a radius of [itex]1 - \sqrt{x}[/itex]. So we want to subtract these 2 volumes:

    [itex]\pi\int_{0}^{1}(\sqrt{x} - x^2)^2 dx - \pi\int_{0}^{1}(1 - \sqrt{x})^2 dx[/itex]

    So I get:

    [itex]\pi\int_{0}^{1}(\sqrt{x} - x^2)^2 dx - \pi\int_{0}^{1}(1 - \sqrt{x})^2 dx \\
    \pi\int_{0}^{1}(x - 2x^{\frac{5}{2}} + x^4) - (1 - 2x^\frac{1}{2} + x) dx \\
    \pi\int_{0}^{1}(-2x^{\frac{5}{2}} + x^4 - 1 + 2x^\frac{1}{2}) dx \\
    \pi[(\frac{-4}{7}x^{\frac{7}{2}} + \frac{x^5}{5} - x + \frac{4}{3}x^\frac{3}{2})]_{0}^{1} \\
    \pi(-\frac{4}{7} + \frac{1}{5} - 1 + \frac{4}{3}) \\
    \pi\frac{-4}{105}[/itex]

    Not sure how I get a negative number. What did I do wrong?
     
  2. jcsd
  3. Feb 3, 2013 #2

    eumyang

    User Avatar
    Homework Helper

    I don't think the outer radius is correct. We want the distance from the curve y = x2 to the line y = 1. Shouldn't it be 1 - x2?
     
  4. Feb 3, 2013 #3
    Oh, I was using the wrong radius...thanks, I see it now
     
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