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Integration help

  1. Mar 6, 2008 #1
    I tried to find the volume of a torus revolved around the y-axis using the washer method. The equation for the circle is (x-2)^2 +y^2=1.
    I need to integrate (1-y^2)^1/2 dy from 1 to 0 multiplied by a constant 16pi. The constant is not really important but I can't seem to integrate the integrand. I am aware that there's a formula for finding a vol. of a torus, as well.
    Last edited: Mar 6, 2008
  2. jcsd
  3. Mar 6, 2008 #2


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    Unless you misstated the problem, it is trivial. The integral of 1 is y and the integral of y2 is y3/3. Just put them together.
  4. Mar 6, 2008 #3
    Yes, but in this case, the (1-y^2) are under the square root, you can't separate the two terms.
  5. Mar 6, 2008 #4
    [tex]\int_0^1 \sqrt (1-y^{2}) dy[/tex]
    a trig substitution would work here, like y=sinx, so dy=cos(x)dx, when y=0, x=0, when y=1,[tex] x=\frac{\pi}{2}[/tex]

    [tex]\int_0^\frac{\pi}{2} \sqrt(1-sin^{2}x )cosxdx[/tex] now remember

    [tex]1-sin^{2}(x)=cos^{2}(x)[/tex] so

    [tex]\int_0^\frac{\pi}{2} cos^{2}(x) dx=\int_0^\frac{\pi}{2} \frac{1+cos(2x)}{2} dx[/tex] and you are almost done!!
    Last edited: Mar 6, 2008
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