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Triple integral in spherical form

  1. Jul 10, 2011 #1
    consider this following triple integral
    1/(x^2+y^2+z^2)dxdydz

    bounded above by sphere z=(9-x^2-y^2)^1/2 and below by the cone z=(x^2+y^2)^1/2

    what i have done:

    z=Pcospi

    P^2=x^2+y^2+z^2

    9=x^2+y^2+z^2

    P=0 to 3

    pi=0 to pi/4

    theta=0 to 2pi

    is this the correct range?
     
  2. jcsd
  3. Jul 10, 2011 #2

    hunt_mat

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    I would first equate the z's to find that [itex]x^{2}+y^{2}=9/2[/itex], then this means that the radius is [itex]3/\sqrt{2}[/itex] and i think that you limits are: [itex]3/\sqrt{2}\leqslant r\leqslant 3[/itex] and all your limits are okay.
     
  4. Jul 10, 2011 #3

    LCKurtz

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    I think his P is supposed to be the spherical coordinate ρ, so his limits are correct. naspek, don't forget the spherical dV element.
     
  5. Jul 10, 2011 #4

    hunt_mat

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    I think you're right coming to think of it, the point being that the intersection (which is a point on the sphere) is a circle of radius [itex]3/\sqrt{2}[/itex].
     
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