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Math Help - Find the integral over the cone

  1. #1
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    Find the integral over the cone

    Evaluate the integral of 5(x^2+y^2+z^2) over the region D above the cone 3z^2=x^2+y^2 and inside the sphere x^2+y^2+z^2=a^2.
    (D looks like an ice-cream cone)

    obviously, we may use the sphercical polar form to solve this, but I have trouble converting x, y, z into the polar from such as x=ecos\theta sin\varphi, y=esin\theta sin\varphi, z =ecos\varphi
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  2. #2
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    I presume you mean \rho rather than "e"!

    Certainly you should have no trouble putting 5(x^2+ y^2+z^2) and x^2+ y^2+ z^2= a^2 into spherical coordinates ( x^2+ y^2+ z^2= a^2 is a sphere with radius a. So I assume the problem is 3z^2= x^2+ y^2. It's not really hard to put the formulas for x, y, and z, in terms of \rho, \theta, and \phi and reduce but I think I would be inclined to add a z^2 to each side to get 4z^2= x^2+ y^2+ z^2. Now the left side is 4z^2= 4(\rho cos(\phi))^2= 4\rho^2cos^2(\phi) and the right side is, as I said before, very easy.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    I presume you mean \rho rather than "e"!

    Certainly you should have no trouble putting 5(x^2+ y^2+z^2) and x^2+ y^2+ z^2= a^2 into spherical coordinates ( x^2+ y^2+ z^2= a^2 is a sphere with radius a. So I assume the problem is 3z^2= x^2+ y^2. It's not really hard to put the formulas for x, y, and z, in terms of \rho, \theta, and \phi and reduce but I think I would be inclined to add a z^2 to each side to get 4z^2= x^2+ y^2+ z^2. Now the left side is 4z^2= 4(\rho cos(\phi))^2= 4\rho^2cos^2(\phi) and the right side is, as I said before, very easy.
    thanks for the help, so the integral will be \int _\,^\! \int _\,^\! \int _0\,^a\! \rho^4 sin\varphi d\rho d\theta d\varphi

    but im still not sure what the intrand of d\theta d\varphi would be, any hints?

    would d\varphi be from 0 to a/2?
    Last edited by wopashui; January 31st 2011 at 06:51 PM.
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  4. #4
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    Quote Originally Posted by wopashui View Post
    thanks for the help, so the integral will be \int _\,^\! \int _\,^\! \int _0\,^a\! \rho^4 sin\varphi d\rho d\theta d\varphi

    but im still not sure what the intrand of d\theta d\varphi would be, any hints?

    would d\varphi be from 0 to a/2?

    actually, it would be \int _\,^\! \int _\,^\! \int _0\,^a\! 5 \rho^4 sin\varphi d\rho d\theta d\varphi
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