# Divergence Theorem

1. Feb 25, 2006

This problem is either really easy, or I'm really dumb, and since there are no answers to check my work I figured someone here might want to help :)

Q: $w=(x,y,z)$ what is the flux $\int \int w \cdot n\,\, dS$ out of a unit cube and a unit sphere? Compute both sides in the divergence theorem?

A: ?

$$\vec w = (x,y,z)$$
$$\int \int \vec w \cdot n \,\,dS = \int\int\int div \,\vec w \,\,dV$$

$$grad \cdot \vec w = 3$$

$$3\int dV = 3 \times vol_{sphere} = 4\pi$$

Last edited: Feb 25, 2006
2. Feb 26, 2006

### siddharth

Over what volume are you supposed to integrate? For a unit sphere, what you did is right. What about the cube?
Check the definition of the divergence theorem again.

3. Feb 26, 2006

I just wanted to make sure I wasn't misinterpretting the definition of the divergence theorem. It's been awhile since I've had practice using it.

To complicate things for my own good. (maybe this should be another thread... but I guess it still does pertain to this homework problem). Lets say $\vec w = (x^2,y^2,z^2)$.

And lets say I'm asked to find the flux through a unit sphere.

Would this be a logical way to arrive at the answer.

First compute the divergence: $div\, \vec w = 2x+2y+2z$

Now the integral would be of the form: $2\int \int \int (x+y+z) \,\,dV$

I could convert $(x+y+z)$ to spherical coordinates. Setup the limits of integration in spherical coordinates, and that would be the flux through the sphere?

I'm thinking this right. But perhaps I'm not visualizing it properly. The way I'm visualizing it right now is that I have this cube in space full of numbers, I can get to each number by feeding coordinates to it $x,y,z$. Then in this visualization exercise the volume of integration is actually all the numbers I want to feed to this cube and pull those numbers. So I'm actually feeding every coordinate within the sphere to this cube. I hope that actually makes sense and am applying this thought properly. Thanks for the help.