Surface Integral With Divergence Thm

1. Mar 26, 2013

Karnage1993

1. The problem statement, all variables and given/known data
Let $\mathit{F}(x,y,z) = (e^y\cos z, \sqrt{x^3 + 1}\sin z, x^2 + y^2 + 3)$ and let $S$ be the graph of $z = (1-x^2-y^2)e^{(1-x^2-3y^2)}$ for $z \ge 0$, oriented by the upward unit normal. Evaluate $\int_{S} \mathit{F} \ dS$. (Hint: Close up this surface and use the Divergence Theorem)

2. Relevant equations

3. The attempt at a solution
It's clear that $div \ F = 0$, so if I was working with a closed surface, $\int_{S} \mathit{F} \ dS$ would equal 0. For the graph of $S$, I am unsure as to what "closing up" a surface means. How would one close up a surface?

2. Mar 26, 2013

mathskier

Add the surface z=0 to creates closed surface. Then, the divergence theorem gives the surface integral of the sum of the surfaces; subtract the integral over what you added to get the surface integral of the desired surface.

3. Mar 26, 2013

Karnage1993

Ok, so what I think you're saying is I have to do:

$0 - \int_{plane \ z = 0} F \ dS$

which will get me only the surface integral of the graph of z?

4. Mar 26, 2013

Dick

Yes. And you'll need to pick the correct direction for the normal vector for the surface z=0.

5. Mar 26, 2013

Karnage1993

For it to be outward pointing, would the z-component have to be negative, ie, (0,0,-1)?

6. Mar 26, 2013

Exactly.