Can the Spherical Coordinate Integral be Simplified using a Substitution?

In summary, my professor said this problem is straightforward in spherical coordinates, but it can get messy if you're not careful.
  • #1
vincentchan
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0
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...
 
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  • #2
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.
 
  • #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
 
  • #4
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
 
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  • #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
 
  • #6
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.
 
  • #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]
 
  • #8
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.
 
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1. What is a spherical coordinate integral?

A spherical coordinate integral is a type of integral that is used to calculate the volume or surface area of a three-dimensional shape in spherical coordinates. It involves integrating over the surface or volume of the shape using spherical coordinates, which consist of a radial distance, an azimuthal angle, and a polar angle.

2. How is a spherical coordinate integral different from a regular integral?

A spherical coordinate integral involves integrating over a shape in spherical coordinates, whereas a regular integral involves integrating over a shape in Cartesian coordinates. This means that the limits of integration and the integrand will be expressed in terms of spherical coordinates rather than Cartesian coordinates.

3. What are the advantages of using a spherical coordinate integral?

One advantage of using a spherical coordinate integral is that it is particularly useful for solving problems involving symmetrical three-dimensional shapes, such as spheres, cylinders, and cones. Additionally, it can often simplify the integration process compared to using Cartesian coordinates.

4. Can a spherical coordinate integral be used for non-spherical shapes?

Yes, a spherical coordinate integral can still be used for non-spherical shapes. In these cases, the shape may need to be broken down into smaller, symmetrical parts in order to apply the spherical coordinate system. Alternatively, other coordinate systems such as cylindrical or spherical polar coordinates may be more appropriate.

5. Are there any limitations to using a spherical coordinate integral?

One limitation of using a spherical coordinate integral is that it is not suitable for all shapes. It works best for symmetrical shapes, and can be difficult to apply to irregular or asymmetrical shapes. Additionally, the integrals may become more complicated for more complex shapes, making it challenging to solve analytically.

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