Mean value theorem for integration in two dimensions The mean value theorem for integration says that, if $G$ is a continuous real-valued function defined over an interval, $G: [a,b] \to \mathbb{R}$, then the mean value of G on the interval is achieved as a certain point of the interval, i.e:
$$\exists x_0\in[a,b]: G(x_0) = \frac{1}{b-a} \int_a^b G(t) \, dt$$
Is this theorem true in two dimensions?
Let $G$ be a continuous, two-dimensional function defined over a connected, convex, closed subset of the plane, e.g the unit disc:  $G: B^2 \to \mathbb{R^2}$. Is this true that the mean value of G on the disc is achieved as a certain point of the disc, i.e:
$$\exists (x_0,y_0)\in B^2: G(x_0,y_0) = \frac{1}{\text{Area}(B^2)} \int_{B^2} G(x,y) \, dxdy$$
? The Wikipedia page on Mean Value Theorem lists some generalizations, but I could not find this exact variant, which seems very intuitive.
 A: I originally read this question incorrectly and provided a proof (see below) for $G : D \to \mathbb{R}$ where $D$ is a connected, convex, closed subset of the plane.  
There is a counter example for a mapping $G : D \to \mathbb{R}^2$.  Let $K_1, K_2$ be two disjoint compact subsets of $D$ and suppose that the components of $G$, namely $G_1$ and $G_2$, are non-negative and compactly supported in $K_1, K_2$ respectively.  Then a point satisfying the equality for $G_1$ must reside in $K_1$.  The same holds for $G_2, K_2$.  Since they are disjoint sets there cannot be a common point satisfying both equalities.
If $G$ was a mapping to the reals then it is true.  Since $G$ is continuous on the closed set $D$, it must attain its minimum at a point $(x_0, y_0)$ and its maximum at a point $(x_1, y_1)$.  Now it is true that
$$G(x_0, y_0) \leq \frac{1}{\text{Area}(B^2)} \int_{B^2} G(x,y) \, dxdy \leq G(x_1, y_1)$$
Let $(x(t), y(t))$ be a line formed from $(x_0, y_0)$ to $(x_1, y_1)$.  This line is in the set $D$ because $D$ is convex.  Now, by the intermediate value theorem $G(x(t), y(t))$ attains all values from $G(x_0, y_0)$ to $G(x_1, y_1)$ so we are done. 
A: In addition to muaddib's answer, I found the following specific counter-example. The domain is the square $[0,2\pi]\times[0,2\pi]$, and the function is:
$$G(x,y)=[sin(x+y), cos(x+y)]$$
Then, by symmetry it is easy to see that the integral of $G$ over the domain is $(0,0)$. 
However, there is no point in which $G(x,y)=(0,0)$, because $|G(x,y)|$ is 1 everywhere.
