Calculate the volume integral of divergence over a sphere

[Vitani11]

Homework Statement

For the vector field F(r) = Ar³e^-ar2rˆ+Br^-3θ^ calculate the volume integral of the divergence over a sphere of radius R, centered at the origin.

Homework Equations

Volume of sphere V= ∫∫∫dV = ∫∫∫r²sinθdrdθdφ
Force F(r) = Ar³e^-ar2rˆ+Br^-3θ^ where ^ denote basis (unit vectors), F is a vector and is a function of r which is also a vector.
D(V) divergence over volume
θ = zenith angle (limits 0 to π)
Φ = azimuthal angle (limits 0 to 2π)
R = radius, r = variable of integration (limits 0 to R)

The Attempt at a Solution

The divergence of this thing is A(3r²e^-ar2-2ar⁴e^-ar2)
Here is the integral I set up:

D(V) = A∫∫∫(3r²e^-ar2-2ar⁴e^-ar2)r²sinθdrdθdΦ and I got down to the following: ... = 12Aπ∫r⁴e^-ar2dr - 8aAπ∫r⁶e^-ar2dr. My problem is integrating this. I mean is this even possible? If this were complex I am sure I could do it... but the integral of e^-ar2 alone is already difficult to integrate.e^-r2 I know has the solution of √π when limits are from -∞ to ∞ but these limits are also different.

 

[Vitani11]

Can I do this using a Fourier transform or expansion you think?

 

[TSny]

The divergence of this thing is A(3r²e^-ar2-2ar⁴e^-ar2)

Check the general formula for divergence in spherical coordinates
https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates#Del_formula

 

[Vitani11]

Oh... thank you.

 

[Vitani11]

Okay now even fixing that I am having the same issue. I still need to integrate ∫r⁴e^-ar2dr and ∫r⁶e^-ar2dr. Divergence obtained is A(5r²e^-ar2-2ar⁴e^-ar2)+Bcosθ/r⁴sinθ. When doing the integral the limits are from 0 to π and therefore the term with the B coefficient goes away since the integrals limits become 0 to 0, also after integrating the Φ component to 2π (since it doesn't depend on anything in the integral) I have this: 2πA[∫∫5r⁴e^-ar2sinθdrdθ-2πA∫∫2ar⁶e^-ar2sinθdrdθ. Then after integrating sinθ I get 20πA∫r⁴e^-ar2dr-8πAa∫r⁶e^-ar2dr

 

[TSny]

See if you can modify section 1 on page 1 of these notes:

https://web.williams.edu/Mathematics/lg5/Feynman.pdf

 

[Chestermiller]

Are you familiar with the divergence theorem?

 

[TSny]

Following Chestermiller: If you are allowed to use the divergence theorem then that's definitely the way to go!

 

[Vitani11]

Okay awesome, thank you. I'll learn it

 

[Vitani11]

So when I used the divergence theorem I got a value which was not 0, but when I integrated using grit over the volume instead of Gausses theorem I get something which is zero. I had to use the gamma function and what not for the second approach and there is more room for error when solving the problem, but this broke down simply and equaled 0 even after all the algebra which makes it more convincing than the first approach. Would it make sense for this this to be 0? What does this question even mean? I know the maths and what a divergence is... am I just summing up all the bits of outward force over the volume of the sphere? Using gausses theorem and the other method tells me that the divergence in the θ hat direction =0. Is this because of symmetry for θ in terms of the divergence throughout the spheres volume?

 

[TSny]

So when I used the divergence theorem I got a value which was not 0, but when I integrated using grit over the volume instead of Gausses theorem I get something which is zero. I had to use the gamma function and what not for the second approach and there is more room for error when solving the problem, but this broke down simply and equaled 0 even after all the algebra which makes it more convincing than the first approach.

I let Mathematica do the volume integral of the divergence. It gives a nonzero result which agrees with the result using the divergence theorem.

Using gausses theorem and the other method tells me that the divergence in the θ hat direction =0. Is this because of symmetry for θ in terms of the divergence throughout the spheres volume?

I'm not sure what "divergence in the θ hat direction" means. Can you clarify that?

 

[Vitani11]

I meant that the divergence throughout the volume of the sphere is symmetric when integrating it throughout the zenith angle from 0 to pi i.e. it just cancels because the sum of the vectors represented by θ hat all sum to 0. Does this make sense? Also what did Mathematica tell you? I don't have software like that to check my work.

 

[TSny]

I meant that the divergence throughout the volume of the sphere is symmetric when integrating it throughout the zenith angle from 0 to pi i.e. it just cancels because the sum of the vectors represented by θ hat all sum to 0. Does this make sense?

We are integrating the divergence of the vector field F throughout the volume of the sphere. The divergence at each point is just a number. So, the integral is a sum of numbers rather than a sum of vectors.

However, F does have a $\hat{\theta}$ component. The contribution of this component of F to the divergence of F is Bcosθ/r⁴. As you noted in post #5, the integral of this expression over the volume of the sphere is zero. The only contribution to the volume integral of ∇ ⋅ F over the sphere come from the $\hat{r}$ component of F. This makes sense when considering the divergence theorem, because only the $\hat{r}$ component of F gives any flux through the surface of the sphere.

Also what did Mathematica tell you? I don't have software like that to check my work.

I let Mathematica do the integral ∫(5r²e^-ar2-2ar⁴e^-ar2)r²dr from r = 0 to r = R and it gave me an answer of R⁵e^-aR2.

 

[Vitani11]

Okay, thank you.