1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spherical coordinate integral

  1. Mar 14, 2005 #1
    My professor said this problem is straightforward in spherical coordinates,

    [tex] \int_{|\vec{r_1}|\leq a } \int_{|\vec{r_2}|\leq a } \frac{1}{|\vec{r_1}-\vec{r_2}|} d^3 r_2d^3r_1[/tex]

    i set up the integral in spherical coordinate and found it really ugly...
     
  2. jcsd
  3. Mar 14, 2005 #2

    StatusX

    User Avatar
    Homework Helper

    Think about what this double integral means. The inner one is the integral over all points inside a sphere of the reciprocal of the distance between each point and a given point, r1. It's clear that this value will depend only on the distance from the origin of r1, so you might as well place it on the [itex]\theta[/itex]=0 axis. Once you get the inner integral into the form of a function of r1 (which will have no [itex]\theta[/itex] or [itex]\phi[/itex] dependance), the outer integral should be pretty easy.
     
  4. Mar 14, 2005 #3
    thank you for your quick reply, i don't think i can do this because the distance is depend on both r1 and r2 (not the origin to r1).

    EDIT:
    the angle between r1 and r2 will make the whole thing very messy
     
  5. Mar 14, 2005 #4

    StatusX

    User Avatar
    Homework Helper

    Yes, but I'm saying the angle of r1 won't matter in the inner integral because you're integrating over every point in the sphere. You'll get a value that only depends on the distance of r1 from the origin.

    Also, you'll still have a rough integral to do over r2. A couple hints: Do the theta integral first, and when you get an expression that involves the square roots of perfect squares, make sure you take the positive root, which may mean you'll have to split up the integral into two ranges.

    EDIT: The answer I get is: 32/15 pi^2 a^5
     
    Last edited: Mar 14, 2005
  6. Mar 14, 2005 #5
    okay, I follow your way and find the first integral is 2 pi a^2, which is independent of r1... surely i did somthing wrong....

    BTW, your answer is right
     
  7. Mar 14, 2005 #6

    StatusX

    User Avatar
    Homework Helper

    Well I can't help you if you don't show me some of your work. How did you set up the first integral? Use the law of cosinesfor the distance.
     
  8. Mar 14, 2005 #7
    this is how my integral look like

    [tex] \int \frac{ r_2^2sin \theta dr_2 d \theta d \phi }{\sqrt{r_1^2+r_2^2-2r_1r_2cos \theta }}[/tex]
     
  9. Mar 14, 2005 #8

    StatusX

    User Avatar
    Homework Helper

    That looks right. Try using the substitution [itex]u=r_1^2+r_2^2-2r_1r_2 cos \theta[/itex] to do the integral over theta first.
     
    Last edited: Mar 15, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Spherical coordinate integral
  1. Spherical coordinates (Replies: 5)

Loading...