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Integration volume

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the volume of solid generated when the shaded area is rotated through pi radians about the y-axis . The graphs are y=x^2 and y=2-x^2 .

    2. Relevant equations

    3. The attempt at a solution

    i did everything and found the answer to be 16/15 pi but the answer given is pi .

    [tex]V=\pi \int^{1}_{0}x^4 dx+\pi \int^{2}_{1}(2-x^2)^2 dx [/tex]

    any mistakes in my set up ?
  2. jcsd
  3. May 12, 2010 #2


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    Science Advisor

    When I set it up, I got:

    [tex]V=\int_0^1\pi x\int_{x^2}^{2-x^2} dy dx = \int_0^1\pi x(2-2x^2)dx = 2\pi[\frac{x^2}{2}-\frac{x^4}{4}]_0^1 = \frac{\pi}{2}[/tex]

    However, this gives pi/2, not pi. Are you sure pi is the right answer? Anyone else?
  4. May 12, 2010 #3
    ok , i think i see my mistake , its with respect to y .
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