Derivation of a vector calculus identity The following is from a past qualifying exam, and so should have a relatively slick solution.
Let $g: \mathbb{R}^2 \to (0, \infty)$ be a $C^2$ function, and define $\Sigma \in \mathbb{R}^3$ to be the graph of $g$ restricted to the unit disk, i.e. $\{x, y, g(x,y)|x^2 + y^2 \leq 1\}$. Suppose that $\Sigma$ is contained in the ball of radius $R$, and that the ray from the origin to $R$ intersects with $\Sigma$ at most once. If $E$ denotes the set of points on $\mathbb{S}^2(R)$ for which the ray does intersect $\Sigma$ precisely once, then the task is to find an equation relating the area of $E$, $R$ and the following integral: 
$$ \int_\Sigma \nabla \Gamma(x) \cdot N(x) dS$$
where $\Gamma = \frac{1}{|x|},$ $N$ denotes a unit normal as always.
I have tinkered with the idea of the divergence theorem. The idea is something like using the fact that the rays leaving the origin are normal to the sphere, to find some kind of flux integral, but it's just not coming together for me after working through the practice exam.
Apologies if it's really simple!
 A: As $\Sigma$ is contained by that ball, so is $\partial\Sigma$. Consider the rays from the origin to $\partial\Sigma$, they form a surface $T$ that in spherical coordinates, as each ray is defined by some $\theta_0$ and $\phi_0$ fixed, can be parametrised by $\sigma=(r,\theta,\phi)=(u,f(v),g(v))$ (with suitable range of values for $u,v$, being them not relevant now), so is, $\theta$ and $\phi$ don't depend on the value of $u$ into its range. Thus, we have $\dfrac{\partial\sigma}{\partial u}=\hat r$, that is tangent to the surface (in into it, in fact) and being the unitary vector along $r$. Further, the surface $T$ and $\Sigma$ enclose some volume $V_\Sigma$
Now, with  $\nabla \Gamma(x)=\nabla\dfrac{1}{\vert x\vert}=-\dfrac{\hat r}{x^2}$, we calculate the integral 
$$\int_{\Sigma\cup T} \nabla \Gamma(x) \cdot N(x) dS=\int_T -\dfrac{\hat r}{x^2}\cdot N(x)dS+\int_\Sigma \nabla \Gamma(x) \cdot N(x) dS$$
But $\hat r\cdot N(x)=0$ because $\hat r$ tangent to the surface and $N(x)$ it's normal.
$$I_{V_\Sigma}=\int_{\Sigma\cup T} \nabla \Gamma(x) \cdot N(x) dS=\int_\Sigma \nabla \Gamma(x) \cdot N(x) dS$$
We proceed the same way for $E$, considering rays from the origin to $\partial E$ and forming a surface enclosing the volume $V_E$. In the same way, the integral for the whole surface is too equal to the integral for $E$:
$$I_{V_E}=\int_E \nabla \Gamma(x) \cdot N(x) dS$$
But $E$ is simpler than $\Sigma$ and using spherical coordinates, with $\theta(\phi)$ on $\partial\Sigma$:
$$I_{V_E}=\int_0^{2\pi}\int_0^{\theta(\phi)}\int_0^R-\dfrac{\hat r}{R^2}\cdot\hat r r^2\sin\theta dr d\theta d\phi=-\dfrac{1}{R^2}\int_0^{2\pi}\int_0^{\theta(\phi)}\int_0^Rr^2\sin\theta dr d\theta d\phi$$
$$I_{V_E}=-\dfrac{1}{R^2}A(E)$$
With $A(E)$ the area of $E$.
Consider the region $V_E\backslash V_\Sigma$, formed by $\Sigma$, $E$ and the corresponding rays. Then,
$$I_{V_E}=I_{V_\Sigma}+\int_{\partial(V_E\backslash V_\Sigma)} \nabla \Gamma(x) \cdot N(x) dS$$
as the integral for the common surface $\Sigma$ is of different sign for $\partial(V_\Sigma)$ as for $\partial(V_E\backslash V_\Sigma)$
But by the divergence theorem 
$$\int_{\partial(V_E\backslash V_\Sigma)} \nabla \Gamma(x) \cdot N(x) dS=\int_{V_E\backslash V_\Sigma}\nabla\cdot\nabla\Gamma(x)dv=0$$
because $x=0$ in the volume. So,
$$I_{V_\Sigma}=-\dfrac{1}{R^2}A(E)$$
