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Math Help - rotation to form a solid... cant figure out error

  1. #1
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    rotation to form a solid... cant figure out error

    The region bounded by y=e^{-x^2}, y=0, x=0, and x=1 is revolved about the y-axis. Find the volume of the resulting solid.

    so it should be the integral (0,1) of e^(-x^2). u=-x^2, du=-2xdx, coefficient=-1/2

    therefore, (-1/2) int(0,1) e^u*du = (-1/2) e^u evaluated at |(0,1)

    so, ((-e^-1)/2) - ((-e^0)/2)= -.1839397206 - (-.5) = .3160602794

    multiply this by pi to get my answer, which was .9929326519, but incorrect.

    what am i doing wrong?
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  2. #2
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    Quote Originally Posted by thedoge View Post
    The region bounded by y=e^{-x^2}, y=0, x=0, and x=1 is revolved about the y-axis. Find the volume of the resulting solid.
    The formula is,
    2\pi \int_0^1 xe^{-x^2} dx
    To find,
    \int xe^{-x^2} dx=-\frac{1}{2}\int e^{-x^2}(-2x) dx
    Let,
    u=-x^2 then u'=-2x,
    Thus,
    -\frac{1}{2}\int e^{u} du=-\frac{1}{2}e^u+C=-\frac{1}{2}e^{-x^2}+C
    Evaluate,
    2\pi\left(-\frac{1}{2}e^{-1^2}+\frac{1}{2}e^0\right)
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  3. #3
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    Hrm, so my error was using 1*pi instead of 2*pi? Thanks again PH, you never cease to amaze me.
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