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Homework Help: Converting triple integral coordinates

  1. Dec 13, 2009 #1
    1. consider the triple integral (x^2 +Y^2) dV where it is bounded by a solid sphere of radius R. Set up the integral using rectangular coordinates


    I tried setting this up with the bounds [ -sqrt(R^2-x^2-Y^2) <= Z <= sqrt(R^2-x^2-Y^2) ,
    -R <= X <= R , -sqrt(R^2-x^2) <= Y <= sqrt(R^2-x^2) ] am I on the right path???
     
    Last edited: Dec 13, 2009
  2. jcsd
  3. Dec 13, 2009 #2

    LCKurtz

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    Yes, but make sure you have them in the correct order when you write down the integrals.
     
  4. Dec 13, 2009 #3
    Fantastic! Yea I know that it goes dz dy dx
    now How about if I rewrite to cylindrical coordinates?
    0<= theta <= 2pi , 0<=r<=R , -sqrt(R^2-r^2)<=z<= sqrt(R^2-r^2) with the integrand being r^3 dz dr dtheta
     
  5. Dec 13, 2009 #4

    LCKurtz

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    Yes, you've got it.
     
  6. Dec 13, 2009 #5
    first, thanks so much for the help

    just to make sure I fully understand all of these triple integrals, to set it up in spherical I should get
    0<= theta<= 2pi, 0<= phi <= pi, 0<= p <= R,
    where the integrand is (p^4) (sin(FI))^3 dp dphI dtheta
     
  7. Dec 13, 2009 #6

    LCKurtz

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    Yes. Very nice. And welcome to the Physics forum.
     
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