# Divergence as a surface integral

I had a shot at the final problem of section 16.4 in 'Calculus a Complete Course' by Adams, I knew full well that I wasn't even going to come close to a correct answer so my question relates to the answer provided in the solution manual.

Note: I can't get the surface integral to show up as a closed surface integral, so where ever you see a double integral please assume that i'm referring to the integral over a closed surface.

Problem statement:

Let $P_{0}$ be a fixed point and for each $\epsilon>0$ let $D_\epsilon$ and $S_\epsilon$ define the the domain and boundary respectively. (such that the conditions of the Divergence Theorem are satisfied) Suppose that the maximum distance from $P_0$ to some other point $P\in D_\epsilon$ approach zero as $\epsilon\rightarrow 0^+$. If $D_\epsilon$ has volume $vol(D_\epsilon)$, show that $$\lim_{\epsilon\rightarrow 0^+} \frac{1}{vol(D_\epsilon)} \iint_{S_\epsilon} \boldsymbol{F\cdot\hat{N}} dS = \nabla\cdot\boldsymbol{F}(P_0)$$

My Problem:

So the proof given by Adams starts out by stating the divergence theorem and dividing both sides of the equation by the volume of the domain. Some algebraic manipulation brings me to this point. $$\begin{array}{lcr} \left|\frac{1}{vol(D_\epsilon)} \iint_{S_\epsilon}\boldsymbol{F\cdot\hat{N}}\, dS - \nabla\cdot\boldsymbol{F}(P_0)\right| & \leq &\frac{1}{vol(D_\epsilon)} \iiint_{D_\epsilon}\left|\nabla\cdot\boldsymbol{F}-\nabla\cdot\boldsymbol{F}(P_0)\right|dV \\& \leq & \max\limits_{P\in D_\epsilon} \left|\nabla\cdot\boldsymbol{F}-\nabla\cdot\boldsymbol{F}(P_0)\right|\end{array}$$

Now the first line seems to result from some algebraic manipulation and application of the triangle inequality, but the jump to the second line baffles me. (Hopefully i'm not overlooking something that's painfully obvious)

Ps. cannot figure out how to get the expression on the second line to align properly.

The integral of a constant equals that constant times the volume integrated over, and integrals respect inequality (so if $f<g$, the integral of $f$ is also less than the integral of $g$). By definition of max, the integrand is less than or equal to the max everywhere in your domain $D_\epsilon$, so you can replace the integral by max times volume and maintain the inequality.