TITLE: Does the domain of integration have to be convex?
QUESTION [6 upvotes]: The following theorem is straightforward to prove if we ignore the convexity of the domain : 

Suppose that $\text{R}$ is a convex region in the plane and that the function $g : \text{R} \to \mathbb{R}$ has continuous bounded partial derivatives. Show that the surface $S = {(x, y, g(x, y)) \ | \ (x, y) \ \text{in} \ \text{R}}\}$ has area equal to that of $\text{R}$ if and only if the function $g : \text{R} \to \mathbb{R}$ is constant. 

For the proof I used the fact that $\iint_D \Big(\sqrt{1+f_x^2+f_y^2} -1\Big)\,dA = 0 \iff f_x=0 =f_y \iff f = \text{const.}$
But why it is supposed that the region to be convex? I can't see why it is necessary to use convexity of the region in the proof. Is there a counterexample i.e. a non-convex region that the mentioned theorem fails to hold?

REPLY [2 votes]: The key property you need is that $R$ is connected. Indeed, you use this to conclude your proof when you write:
$$f_x = f_y = 0 \Leftrightarrow f = \mathit{const}$$
The general fact that we are using here is:

Theorem A. Let $f \colon \Omega \to \mathbb{R}$ be a differentiable function, where $\Omega$ is an open connected set
  in $\mathbb{R}^n$. Then $\nabla f = 0$ in $\Omega$ if and only if $f$
  is constant.

It is easy to give a counter-example to this when $\Omega$ is not connected: take a function which is piecewise constant, with a different value for the constant on each connected component of $\Omega$. For instance take $\Omega = D_1 \cup D_2$ where $D_1$ and $D_2$ are two disjoint open disks, and define $f \colon \Omega \to \mathbb{R}$ by $f = 1$ on $D_1$ and $f = 2$ on $D_2$.
Your problem probably assumes that $R$ is convex instead of merely connected because it is easier to prove Theorem A. In fact, the outline of the proof of the theorem is as follows:

Prove the theorem in the case where $R$ is convex. Hint: use the mean value theorem.
Prove the theorem in the general case where $R$ is connected (and nonempty): let $x_0 \in \Omega$, show that $\{x\in \Omega ~\colon~ f(x) = f(x_0)\}$ is nonempty, open and closed in $\Omega$. Hint: to show openness, use the fact that $\Omega$ is locally convex, and use 1. 

Remark: Note that you can also relax the hypothesis that $f$ has continuous and bounded partial derivatives: it is enough to assume that $f$ is differentiable. The reason that your problem makes this extra assumption is probably to easily guarantee that the area of the surface is well-defined and finite. In fact they probably forgot to also say that $R$ is bounded.