Use of the divergence theorem I have the following problem:

Let $\Omega \subset \mathbb{R}^3$ be an open bounded set with a smooth boundary $\partial \Omega$ and the unit normal $v$. Calculate for the vector field $a(x,y,z)=(0,0,-pz)$ with $p>0$ the value of -$\int_{\partial\Omega}\langle a,v\rangle d\mu_{\partial\Omega}$.

I don't really know how to start with this problem. So I used the divergence theorem: -$\int_{\partial\Omega}\langle a,v\rangle d\mu_{\partial\Omega}$=$\int_\Omega \text{div}(a)d\mu_M$=$\int_\Omega pd\mu_M$, but I don't know how to proceed from here.
Can someone help me please? Thanks in advance.
 A: Not giving the explicit form for $\Omega$, we at most can say:
$$\int_\Omega pd\mu_M=p\int_\Omega d\mu_M=pV$$
Being $V$ the volume of $\Omega$ (Because bonded $\vert x-y\vert\lt M$ for some $M$ and $x,y\in\Omega$)
A: You're almost there: Since $p$ is constant, you can just pull it out of the integral, and have $\int_{\Omega} d\mu_M$. But this is just the volume of $\Omega$ (that's the point of the measure!).
You can think of this as a generalisation of the idea that the area under the graph is 
$$\iint_A dx \times dy = \int_{\partial A} (0,y) \cdot \nu \, dl = \int_{\alpha}^{\beta} y(t) \frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}} \sqrt{x'(t)^2+y'(t)^2} \, dt = \int_a^b y \, dx. $$
A: We have 
$$
\begin{align*}
-\int_{\partial\Omega}\langle a,v\rangle d\mu_{\partial\Omega}
&= -\int_{\Omega} \text{div}(a) \: d\mu_{\Omega} \\
&= -\int_{\Omega}\dfrac{\partial}{\partial z}(-pz) d\mu_{\Omega} \\
&= \int_{\Omega}p \: d\mu_{\Omega} \\
&= p \int_{\Omega} d\mu_{\Omega} \\
&= p \cdot \text{Vol}(\Omega). 
\end{align*}
$$
