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Given a bounded, compact, closed surface $\Gamma\subset\mathbb{R}^n$, I'm searching $$ \max_{y\in\mathbb{R}^n, \|y\|=1} \int_\Gamma \langle y, x\rangle^2. $$ Without the square, and with the enclosed volume $\Omega$, $$ \max_{y\in\mathbb{R}^n, \|y\|=1} \int_\Omega \langle y, x\rangle, $$ I'm guessing $y_\text{max}$ points towards the centroid of $\Omega$. Not sure how to prove that though.
Any hints?
Let's do the second one first. Let \begin{align*} f(y_1, \dots, y_n) &= \int_\Omega (y_1x_1 + \dots y_n x_n)\,dV \\ g(y_1, \dots, y_n) &= y_1^2 + y_2^2 + \dots + y_n^2 \end{align*} We want the critical values of $f$ subject to the constraint that $g=1$. By the method of Lagrange multipliers, there is a constant $\lambda$ such that, for each $i$ from $1$ to $n$, $\frac{\partial f}{\partial y_i} = \lambda \frac{\partial g}{\partial y_i}$. That is, $$ \int_\Omega x_i \,dV = 2\lambda y_i $$ for each $i$. By squaring and summing all $n$ of these equations, we get $$ 4\lambda^2 = \sum_{i=1}^n \left(\int_\Omega x_i \,dV\right)^2 $$
The only way the right-hand side is zero is if $\int_\Omega x_i \,dV = 0$ for each $i$. In that case $f$ is identically zero and any $y$ will do. Otherwise, $\lambda \neq 0$, so $$ y_i = \frac{1}{2\lambda} \int_\Omega x_i \,dV $$ Since the $i$th coordinate of the centroid of $\Omega$ is $\bar x_i =\frac{1}{\operatorname{Vol}(\Omega)}\int_\Omega x_i \,dV$, I agree that $y$ is a multiple of $\bar x$.
The first one is trickier, at least, to me. Let $$ h(y_1,\dots,y_n) = \int_\Gamma (x_1y_1+\dots+x_ny_n)^2\,ds $$ Again, we want to maximize $h$ subject to $g=1$. We have $$ 2\int_\Gamma (x_1y_1+\dots+x_ny_n)x_i\,ds = 2\lambda y_i \tag{$*$} $$ for each $i$. If we set $$ M_{ij} = \int_\Gamma x_i x_j \,ds $$ then $(*)$ is equivalent to $$ \sum_{j=1}^n M_{ij} y_j = \lambda y_i $$ In other words, $y$ is an eigenvector of the matrix $M$, corresponding to the eigenvalue $\lambda$.
The matrix $M$ is symmetric, and I think, positive definite. I am guessing the latter can be shown by squaring each equation and adding them up like before. So there is an orthonormal basis of eigenvectors. Therefore the maximum value of $f$ is the maximum of the eigenvalues of $M$.
I'm not sure if we can say more than that. This is a very interesting problem but I've already spent way too much time that I should be doing something else! \smiley
