Consider a convex surface $\mathcal S$ defined by $z = f(x,y)$. The volume between the surface and its tangent plane at point $P$, enclosed within exterior $\mathcal S$ is minimal when $P$ is the centroid of region $\mathcal R$.
Note:There is a region $\mathcal R$ whose outline is any general curve lying along the $xy$-plane, and it extends along the $z$-axis. For example, if $\mathcal R$ is a circle in the $xy$-plane, extending it along the $z$-axis results in a cylinder, denoted as exterior $\mathcal S$.
I have proven this for $P(x,y) = (0,0)$ when region $\mathcal R$ is a circle of radius $r$, as $P$ is its centroid for any convex surface.
However, I am struggling to prove why this result holds for any convex surface $z = f(x,y)$ and any region $\mathcal R$.
Some of the tools I used to tackle the problem include Lagrange multipliers, the Euler-Lagrange equation, and since the volume under the surface is constant, the problem is equivalent to showing that the volume under the tangent plane at point $P$ is minimum for $P$, where $P$ is the centroid of any region $\mathcal R$ and for any surface $z= f(x,y)$.
First of all, to clarify what I think the question is:
We have a curve $\mathcal{C}$ in the $xy$ plane, which encloses a region $\mathcal{R}$. Now consider a function $f:\mathcal{R}\rightarrow\mathbb{R}_{\geq0}$, defining a surface $S=\{(x,y,f(x,y)\,:\,(x,y)\in\mathcal{R}\}$ lying "above" $\mathcal{R}$, and suppose that the region between the $xy$-plane and $S$ is convex. We will also be interested in the "cylinder" $\mathcal{R}_1=\{(x,y,z)\,:\,(x,y)\in\mathcal{R},z\geq0\}$.
For any point $P=(x,y)\in\mathcal{R}$ we can then find the plane $T_P$ tangent to $S$ at $P$, and we want to choose $P$ to minimize the volume of the region inside $\mathcal{R}_1$, above $S$, and below $T_P$.
Well, we can simplify this a bit: if we add the volume of $S$ to this -- which doesn't change where the minimum is, because the volume of $S$ doesn't depend on $P$ -- then we find we're just looking for the volume of the region inside $\mathcal{R}_1$ and below $T_P$.
If $f(X,Y)=z+a(X-x)+b(Y-y)$ to first order near $(x,y)$ -- so $z=f(x,y)$ and $a,b$ are the first partial derivatives of $f$ at $(x,y)$ -- then $T_P$ is just the plane $Z=z+a(X-x)+b(Y-y)$. Then the volume inside $\mathcal{R}_1$ and below $T_P$ is $\int_\mathcal{R}z+a(X-x)+b(Y-y)\,dx\,dy$.
(I am assuming that the intention is that when the plane $T_P$ goes below the $xy$ plane we include the resulting negative volume.)
We are interested in where this is minimal, so let's think about its derivatives. In order to reduce visual clutter I've written e.g. $f_x$ for $\partial f/\partial x$ and $f_{xy}$ for $\partial^2 f/\partial x\partial y$. I haven't distinguished between $f_{xy}$ and $f_{yx}$ because these are equal for well-behaved functions. So, we have
$$V(x,y)=\int_{\mathcal{R}}f(x,y)+f_x(x,y)(X-x)+f_y(x,y)(Y-y)\,dX\,dY$$
and so
$$\begin{eqnarray} V_x(x,y)&=&\int_{\mathcal{R}}f_x(x,y)+\frac{\partial}{\partial x}\bigl[f_x(x,y)(X-x)\bigr]+\frac{\partial}{\partial x}f_y(x,y)(Y-y)\,dX\,dY \\ &=&\int_{\mathcal{R}}f_x(x,y)+\bigl[f_{xx}(x,y)(X-x)-f_x(x,y)\bigr]+f_{xy}(x,y)(Y-y)\,dX\,dY \\ &=&\int_{\mathcal{R}}f_{xx}(x,y)(X-x)+f_{xy}(x,y)(Y-y)\,dX\,dY \\ &=&f_{xx}(x,y)\int_{\mathcal{R}}(X-x)\,dX\,dY+f_{xy}(x,y)\int_{\mathcal{R}}(Y-y)\,dX\,dY. \end{eqnarray}$$
If we write $I=\int_{\mathcal{R}}dX\,dY$ then our centroid's coordinates are $\bar x=\frac1I\int_{\mathcal{R}}X\,dX\,dY$ and $\bar y=\frac1I\int_{\mathcal{R}}Y\,dX\,dY$, so we can rewrite that last line as
$$f_{xx}(x,y)I(\bar x-x)+f_{xy}(x,y)I(\bar y-y).$$
We have a similar expression for $V_y$. So both these partial derivatives will be zero iff
$$\begin{pmatrix}f_{xx}(x,y) & f_{xy}(x,y) \\ f_{xy}(x,y) & f_{yy}(x,y)\end{pmatrix}\begin{pmatrix}\bar x-x\\ \bar y-y\end{pmatrix}=0$$
which, provided our function is strictly convex, is true only when $\bar x-x=\bar y-y=0$ because that matrix is negative definite.