Green's first and second identities quick question on greens identities
given that i know $\nabla \cdot (fv) = \nabla f \cdot v + f\nabla \cdot v$ 
then deduce
$\int\int_\Omega \nabla f \cdot v$ $d{A} = \int_{\partial\Omega} fv\cdot n$ $ ds - \int\int_\Omega f\nabla \cdot v$ $ dA$
I understand that i use the identity, and then deduce that $\int_{\partial\Omega} fv \cdot n $ $ds = \int\int_\Omega \nabla \cdot (fv) $ $ dA$
But i dont really understand what $\partial\Omega$ is (except the boundary?) so im not sure how to show that the two are equal
 A: You don't deduce that $\int_{\partial\Omega} fv \cdot n\,ds = \int\int_\Omega \nabla \cdot (fv)\,dA$ you use it to make the desired conclusion. This fact is often known as the divergence theorem.
To spell it all out:

*

*yes, $\partial \Omega$ is the boundary of $\Omega$.


*The divergence theorem states that, if $V$ is a continuously differentiable vector field, then
$$
   \int_{\partial\Omega} V \cdot n\,ds = \iint_\Omega \nabla \cdot V\,dA
$$


*$\int\int_\Omega \nabla f \cdot v\,d{A} = \iint_{\Omega} \nabla \cdot (fv)\, dA - \int\int_\Omega f\nabla \cdot v\,dA$ as you've observed (basically by the 'product' rule for divergence).


*By steps 2 and 3, $\iint_\Omega \nabla f \cdot v\,d{A} = \int_{\partial\Omega} fv \cdot n \, ds - \int\int_\Omega f\nabla \cdot v\,dA$


*Now, if we let $v = \nabla g$ for some function $g$, we have $$\int\int_\Omega \nabla f \cdot \nabla g \,d{A} = \int_{\partial\Omega} f \nabla g \cdot n \, ds - \int\int_\Omega f (\nabla \cdot \nabla g)\,dA$$
i.e. $$\int\int_\Omega \nabla f \cdot \nabla g \,d{A}  + \int\int_\Omega f (\Delta g)\,dA= \int_{\partial\Omega} f \nabla g \cdot n \, ds $$
which is Green's first identity.


*To derive Green's second identity, write Green's first identity again, with the roles of $f$ and $g$ exchanged, and then take the difference of the two equations.
