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Triple intergrals exer

  1. Sep 13, 2004 #1

    ori

    User Avatar

    how can i calculate this integral value :
    SSSdxdydz/(x^2+y^2+z^2)
    at the area: x^2+y^2+(z-1)^2<=1

    thats what i tried - i got double integral that i dont know how to solve
    i tried to use the ball cordinations
    x=rcos(t)sin(p)
    y=rsin(t)sin(p)
    z=rcos(p)+1
    j=rsin^2(p)
    0<=r<=1
    0<=t<2pi
    0<=p<=pi

    i get:

    SSS drdtdp/[r^2cos^2(t)sin^2(p)+r^2sin^2(t)sin^2(p)+(rcos(p)+1)^2]
    0<=r<=1
    0<=t<2pi
    0<=p<=pi

    SSS drdtdp/[r^2sin^2(p)+r^2cos^2(p)+2rcos(p)+1]

    2pi * SS drdp/[r^2+2rcos(p)+1]

    what's next?

    10x
     
  2. jcsd
  3. Sep 13, 2004 #2
    You have an error :
    the spherical coordinate is:
    x=rcos(t)sin(p)
    y=rsin(t)sin(p)
    z=rcos(p)
    only!!!!!!!
     
  4. Sep 13, 2004 #3

    ori

    User Avatar

    my transformation deals with the area that the integral works on
    urs deals with the integrand , both ways i dont know how to solve the integral
    that we get
     
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