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Math Help - Problem with spherical coordinates

  1. #1
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    Problem with spherical coordinates

    First off, sorry I had to write all the equations in "text", but I hope it's not too impossible to read them.

    I want the volume above the xy-plane, under the paraboloid z = 1 - x^2 - y^2 and in the wedge cut out by -x\< y \< sqrt(3) x . With cylindrical coordinates, I get the correct answer (=7pi/48), but with spherical coordinates I do not. Obviously I'm making some elementary error in my logic or calculations - could someone please check it out ?!

    My spherical coordinates:
    x = rsin(phi)cos(theta)
    y = rsin(phi)sin(theta)
    z = rcos(phi)

    dxdydz = r^2 sin(phi) dr dphi dtheta

    My tripple integral in spherical coordinates:

    int(-pi/4 -> pi/3) dtheta
    int(0 -> pi/2) sin(phi) dphi
    int(0 -> 1) r^2 dr

    According to this, I ought to get 7pi/12 * 1 * 1/3 = 7pi/36
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  2. #2
    MHF Contributor

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    Your problem is that \rho does NOT go from 0 to 1. (And that is why cylindrical coordinates is a much better choice than spherical coordinates). In spherical coordinates, \rho is the straight line distance from (0,0,0) to (x,y,z) which is, of course, [tex]\sqrt{x^2+ y^2+ z^2}[tex]. For any point on the parabola z= 1- x^2-y^2, x^2+ y^2+ z^2= x^2+ y^2+ (1- x^2- y^2)^2 which is, if my algebra is correct, x^2+ y^2+ (1- 2x^2- 2y^2- 2x^2y^2+ x^4+ y^4) = 1- x^2-y^2+ (x^2- y^2)^2. In spherical coordinates, x= \rho cos(\theta)sin(\phi) and y= \rho sin(\theta)sin(\phi) so that becomes \rho^2= 1- \rho^2 cos^2(\theta)sin^2(\phi)- \rho^2 sin^2(\theta)sin^2(\phi)+ \rho^4 cos(2\theta)sin^2(\phi).

    Solve that equation (quadratic in \rho^2; since \rho is never negative, you can discard one root) for \rho and use that as the upper limit of the integral.

    I said cylindrical coordinates was a better choice! You have plenty of circular symmetry but no spherical symmetry.
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  3. #3
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    I guessed it must have something to do with r, but it actually took me a while to understand why r does not go from 0 to 1 (I was thinking of the graph as a sphere rather than a parabola...).

    Thanks for the help !
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