The divergence theorem in three dimensions is
$$\iiint _{\mathcal{V}}\nabla^2 \cdot \vec{Q}\ \mathrm dV= \iint_{\mathcal{A}}\vec{Q} \cdot \widehat n\ \mathrm dA,$$
where $\mathcal{V}$ represents the volume of integration, $\mathcal{A}$ represents the surface area of the volume, $\vec{Q}$ is a vector in three-dimensional space, and $\widehat n$ is the outward unit normal to the surface.
Part (a): Work out the analog of Green's first and second identities in three dimensions.
Parb (b): Given $\nabla^2 u=0$ inside of a sphere or radius $R$ centered at the point $(\xi, \eta, \zeta)$ derive the result
$$u(\xi, \eta, \zeta)=\frac{1}{A(S_R)}\iint_{S_R}u\Big|_{r=R}\ \mathrm dA,$$
where $S_R$ represents the surface of the sphere and $A(S_R)$ represents the surface area of the sphere. That is, the value of a harmonic function at the center of a sphere is equal to its average on the surface of the sphere.
Part (c): Identify the Free Space Green's function for the Laplacian operator in three dimensions and show that it is consistent with
$$\nabla^2\Phi=\delta(x-\xi)\delta(y-\eta)\delta(z-\zeta),$$
and
$$1=\iint_{S(R)}\frac{\partial \Phi}{\partial n}\Big|_{r=R}\ \mathrm dA,$$
where $S(R)$ is the sphere of radius $R$ centered at $(\xi, \eta, \zeta)$ and $\frac{\partial \Phi}{\partial n}$ is the outward normal derivative of $\Phi$.
Attempt:
For part (a), I know that you should think of the two-dimensional case where $$\iint_{\mathcal{D}} \nabla^2 \phi\ \mathrm dA= \oint_{\mathcal{C}}\nabla \phi \cdot \widehat{n}\ \mathrm ds.$$ However, I don't understand what the three-dimensional version of this would look like.
For part (b), I think you are supposed to use the Mean Value Theorem.
For part (c), I think it would be best to use spherical coordinates.