Simple divergence theorem questions

[bmrick]

So I understand the divergence theorem for the most part. This is the proof that I'm working with http://www.math.ncku.edu.tw/~rchen/Advanced Calculus/divergence theorem.pdf

For right now I'm just looking at the rectangular model. My understanding is that should we find a proof for this model we can validate all transformations of the models from a proof regarding the integral values of transformed regions. (there's also probably an easier method where we just don't assume the object to be square too, right?)

What bothers me is that the area integral of F(x,y,z) dot d(a) is equal to the double integral of F(x,y,z)dydz. What happened to the cos@ value that existed when we were dotting across the area?

the other thing that is bothering me is that when we get rid of this dot product, one the integrals becomes negative. What gives?

 

[Matterwave]

In the proof the dot product gives you simply the first component of F, which is why you see $F_1$ instead of just $F$ in that integral after the dot product has been taken. Perhaps it would have been more clear if they had used $F_x$ instead of $F_1$. Of course $F_1 = F\cos(\theta)$ where $\theta$ is the angle between $\vec{F}$ and $\hat{x}$. Of course this is so since $F_1$ is simply the projection of $\vec{F}$ onto the x-axis.

For the other side (opposite face), the area element is pointing in the negative x-direction, so you get the negative of the first component of $F$ in that integral when you take the dot product.

 

[bmrick]

Ahhhhh i see. Thank you very much. You are both a scholar and a gentleman

 

[Matterwave]

Haha, no problem. :)

 

[bmrick]

Dies the divergence theorem only apply to surfaces that have a net flux?

 

[Matterwave]

No, the divergence theorem applies to any closed volume on which a (differentiable/smooth) vector field is defined, whether the flux is net in, net out, or 0. In fact, in physics, the divergence theorem is often used for when the net flux is 0, e.g. in Gauss's law to show that there is no net charge enclosed in a given volume.