I'm in the middle of a big problem and to get to the next part I need to prove the following:
$\int \int_{\partial \Omega} f\nabla g- g\nabla f dA$ = 0
When $f(x)=g(x)=0$ For any $x$ belonging to $\partial\Omega$
$f$ and $ g$ Are both functions from $R^3$ to $R$, and x is a vector.
How exactly would I be able to prove this. Any help would be greatly appreciated, thanks :)
$$\iint_{\partial \Omega} f\nabla g- g\nabla f\ dA = \iint_{\partial \Omega} f\nabla g\ dA - \iint_{\partial \Omega}g\nabla f\ dA$$
But $f(x) = g(x) = 0$ if $x \in \partial\Omega $.
Hence for each integral one of the functions, $f$ on the first and $g $ on the second, is always 0. This means each integral evaluates to 0.
$$\iint_{\partial \Omega} f\nabla g\ dA = \iint_{\partial \Omega} 0\nabla g\ dA = 0$$
And similarly for the second one.
One didn't even have to split the integrals. You could immediately have
$$\iint_{\partial \Omega} f\nabla g- g\nabla f\ dA = \iint_{\partial \Omega} 0\nabla g- 0\nabla f\ dA = 0$$