Finiteness of surface integral vs volume integral

Question

Suppose we're given that the integral over $\mathbb {R}^3$ of, say, $|f|$ is finite. Then can we say that the surface integral of $|f|$ over any (2D) surface in $\mathbb {R}^3$ is finite and bounded by the value of the volume integral of $|f|$ over $\mathbb {R}^3$? I'm assuming $f $ is smooth or whatever.

How about if we restrict ourselves to compact surfaces? Then I guess finiteness isn't an issue, but what about the bound? Can we bound it (strictly, not modulo constant) by the volume integral?

At first I thought the answers were all obviously "yes". Now I'm starting to think that they're niether true nor obvious.

Answer 1

You will need to have a compact surface and more restrictions on $f$. Let $f: \mathbb{R}^2 \to \mathbb{R}$ be the map $f(x,y) = y$. Then we have,

$$\int_{\mathbb{R}^2} |f| = \int_{\mathbb{R}^2} |y| $$

which is not finite. If one now allows $\Sigma$ to be compact then we have,

$$\int_{\Sigma} |f| \leq \textbf{max} (|f|) \cdot \textbf{Vol}(\Sigma)$$

In the above, $|f|$ may not be smooth as in teh example, but it is still continuous which is all we need to get the above bound.

Answer 2

False even if $f = 1$ in a ball $B$, $f = 0$ in the complementary on a greater ball. You can take $S\subset B$ of area arbitrarily great and the surface integral can't be bound for the triple integral.