$\int_\Omega \leq C \Rightarrow \int_{\partial \Omega}?$ Let $\Omega$ be a compact smooth subset of $\mathbb{R}^n$ and $f$ a smooth real function. Imagine I know that $$\int_\Omega f(u(x))\ dx\leq C.$$
I wonder if I could say something about the integral $$\int_{\partial \Omega} f(u(x))\ dx$$ in terms of $C$?
 A: I'm supposing that $u(x)$ is actually $x$.
Answer: nothing. Consider $\Omega = $ the unit ball of $\Bbb R^n$, $f_{M,p}(x) = M\|x\|^{2p}$. Now, fixing $M$, make $p$ big.
A: Yes, literally nothing. What I had in mind when I guessed no allows you to basically pick any $C$ and any value of the surface integral $-\!-$ Let $\Omega=B_1$ be the unit closed ball. Fix any interval containing $C>0$ and $D\in\mathbb R$. Note, $C,D$ are completely arbitrary, the adjustments for $C\le 0$ are trivial. Consider a $C^\infty$ function $F:[0,1]\to I$  with two piecewise constant parts  ,
$$ F(x) = \begin{cases} C/2 & 0\le x\le 1-\epsilon \\ \text{smooth} & 1-\epsilon \le x < 1-\epsilon/2 \\ D & 1-\epsilon/2 \le x\le 1    \end{cases} $$
Extend $F$ to be smooth $\mathbb R\to \mathbb R$ by extending $F$ to be piecewise constant on $\mathbb R\setminus I$. Now set $$f:\mathbb R^n \to \mathbb R, \quad f(x) = \frac{F(|x|)}{|B_1|}.$$ This is a smooth map. By choosing $0< \epsilon \ll 1$, $|\int_{B_1} f(x)dx - C/2|$ can be made arbitrarily small so in particular $\int_{B_1} f(x)dx < C$. But but $F|_{\partial B_1} = D/|B_1|$. So the integral over the boundary is $\frac{c_n D }{|B_1|}$ where $c_n=\int_{\partial B_1} d\sigma $ is the surface measure of the unit $n$-sphere.  
