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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

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

My attempt at a solution:

I've edited what I wrote before since I think I was taking the wrong approach,

I think that the idea of this exercise is to use Gauss theorem.

The divergence of $G$ gives $0$, so if I could justify that $S$ is a closed surface and I could find $W$ such that $W$ is a region bounded by $S$, then

$0=\iiint_W (divG)dV=\iint_\partial W G.dS=\int_S G.dS$

The problem is I don't know which region $W$ I could define.

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$S$ is a surface consisting of the unit circle at the bottom and a dome on top of it defined by $f$. So if you integrate $G$ against the circle you get the negative of the integral you are looking for. –  Maesumi Nov 30 '13 at 4:26

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