# Homework Help: Calculate integral using Gauss theorem

1. Nov 29, 2013

### mahler1

The problem statement, all variables and given/known data.

Let
$D =\left\{\left(x,y\right)\ \in\ {\mathbb R}^{2}: x^{2} + y^{2} \leq 1\right\}$ and let ${\rm f}:D \to {\mathbb R}$ defined as
${\rm f}\left(x,y\right) = \left(1 - x^{2} -y^{2}\right)\exp\left(x^{4}y^{10}\right)$.
Consider the surface $S$ given by the graph of ${\rm f}$ restricted to $D$, oriented with the exterior normal vector. Let $G$ be the vector field
$G:{\mathbb R}^{3} \to {\mathbb R}^{3}$ given by

$G(x,y,z) = \left(\,-y,\, x,\, x^{2} + y^{2}\,\right)\,,\qquad\mbox{Calculate}\ \int_{S} G\cdot dS$

The attempt at a solution.

I am having a hard time trying to visualize the surface $S$.

One possibility is to use Gauss theorem, which says that if $W$ is an elementary symmetric region in space where $\partial W$ is a closed oriented surface and $F$ is a function of class $C^1$, then

$\iiint_w (divF).dV=\iint_{\partial W} F.dS$.

I've calculated the divergence of $G$ and it gives $0$. If I could find a region $W$ bounded by $S$, then

$0=\iiint_w (divG).dV=\iint_{\partial W} G.dS=\int_S G.dS$

Another doubt that I have is: how do I know $S$ is a closed surface?

2. Nov 29, 2013

### LCKurtz

It isn't a closed surface. And as such, the "exterior" normal doesn't make sense. Perhaps you mean the upward directed normal. Your surface is defined on the domain $D:~x^2+y^2\le 1$. As a hint, what does your $f(x,y)$ look like on the boundary of $D$?

3. Nov 29, 2013

### mahler1

Yes, I've meant the upward directed normal, sorry, I am translating from spanish to english and sometimes I pick the wrong words. Hmm, $f(x,y)=0$ on the boundary of $D$, and this tells me...? (I need a little more push).

Last edited: Nov 29, 2013
4. Nov 29, 2013

### LCKurtz

It tells you that your open surface and your domain $D$ share the same boundary. So if you add the domain $D$ to your picture, you would have a closed surface. And that buys you ...