Surface Integral With Divergence Thm

[Karnage1993]

Homework Statement

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)

Homework Equations

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?

 

[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.

 

[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?

 

[Dick]

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?

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

 

[Karnage1993]

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

 

[Dick]

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

Exactly.