Compute Volume Integral of $\vec V = xe^{-r}\hat i$: What is 'r'?

[Reshma]

I have this vector function:
\vec V = xe^{-r}\hat i

I have to obtain the volume integral:
I = \int(\vec \nabla \cdot \vec V)d^3x

What is that 'r' and how do I compute the volume integral?

 

[Tide]

r is the distance from the origin: You start by evaluating the divergence of the vector. Offhand, I would recommend using spherical coordinates after evaluating the divergence.

 

[Reshma]

Thanks, Tide!
So r = \sqrt{x^2 + y^2 + z^2}. I have to substitute this in the original vector expression right?

 

[Hootenanny]

I would say so.

 

[Meir Achuz]

I have this vector function:
\vec V = xe^{-r}\hat i

I have to obtain the volume integral:
I = \int(\vec \nabla \cdot \vec V)d^3x

What is that 'r' and how do I compute the volume integral?

I assume the integral is over all space.
If you know the divergence theorem, you can use it to show that the integral=0.

 

[Reshma]

I assume the integral is over all space.
If you know the divergence theorem, you can use it to show that the integral=0.

Sorry, took me a while to get back to this problem.
Yes, the integral is over all space.
The divergence comes out to be:
\vec \nabla \cdot \vec v = \frac{-x^2 e^{-r}}{r} + e^{-r}
I have to integrate this over a sphere of radius R. The volume integration seems too cumbersome. However, if I consider the surface integral, how do I choose the area element d\vec a?

 

[Physics Monkey]

A typical choice for the surface of integration would be a spherical shell of radius R in the limit as R goes to infinity (since the shell must include all space). You should be able to convince yourself that the surface integral is zero without actually calculating it.

 

[Reshma]

I am sorry, I don't get it . Suppose I take the surface integral as S.
S = \int \vec V \cdot d\vec a
As you explained, the area element for the shell would be 4\pi r^2 dr. This means, I have to convert \vec V into spherical coordinates, right?

 

[dextercioby]

Note that your vector field vanishes when evaluated at infinity, so does its flux thrugh the closed surface at infinity...

Daniel.

 

[Reshma]

Ok, so that means, e^-r vanishes with r tends to infinity?

 

[dextercioby]

Of course it does. As Tide said, there's no need to use Gauss's formula if u need to calculate it explicitely. Do the divergence first, then switch to spherical coordinates and then do the integral.

Daniel.