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Math Help - [SOLVED] triple integral issue

  1. #1
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    [SOLVED] triple integral issue

    I need some help with a triple integral problem that I keep getting wrong in the last step somehow.

    the problem is:

    Evaluate \int\int\int on E xyz dV where E lies between the spheres \rho=2 and \rho=4 and above the cone \phi= \pi/3.

    There work I've done so far is set up the integral integrated it and got:

    1/4 sin^4 \phi from 0 to \pi * 1/2sin^2 \theta from 0 to 2 \pi * 1/6 \rho^6 from 2 to 4.

    I get that the middle term should be zero, but for some reason my prof gives the answer to be
    (1562 \pi)/15
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  2. #2
    Eater of Worlds
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    That is a large value your professor got. I am not so sure he/she is correct.
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  3. #3
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    That's what I was thinking as I am pretty sure the answer should be zero, but I thought that it wouldn't hurt to check.
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  4. #4
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    Consider symmetry. The region of integration is rotationally symmetric about the z axis, and so if we rotate the function being integrated about the z axis, the integral should be unchanged. Now, let us rotate the integrand by 90 degrees about the z axis: we transform x and y to x' and y' via x'=-y and y'=x. Then the integrand is f(x',y',z)=xyz=(y')(-x')z=-x'y'z.
    Thus, we have \iiint_{E}xyz\,dx\,dy\,dz=\iiint_{E'}-x'y'z\,dx'\,dy'\,dz. But, if we rename the variables x' and y' as x and y, (and as E'=E), we have
    \iiint_{E}xyz\,dx\,dy\,dz=-\iiint_{E}xyz\,dx\,dy\,dz
    which means \iiint_{E}xyz\,dx\,dy\,dz=0.
    Thus, due to the symmetry, the integral is zero.
    [We can also look at this as the fact that xyz's dependence on the azimuthal angle \theta is of the form \sin(2\theta), and the region has no dependence on the azimuthal angle (the rotational symmetry), so the integral over \theta will be 0]

    --Kevin C.
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  5. #5
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    Thanks everyone. I got an e-mail back from my professor and he said that it was a typo. Thanks again!
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