If, for a volume $V$ in three dimensions, we have
$$\nabla ^2 u=0 \qquad \text{ in }V \\ u(x,y,z) = ax+by+cz \qquad \text{ on } \partial V$$
then is it possible to find the value of the integral
$$\int_V(u_x^2+u_y^2+u_z^2)\; dxdydz$$
in terms of the volume of $V$, without knowing what $V$ or $u$ is?
I tried the following
\begin{align} & \int_V(u_x^2+u_y^2+u_z^2)\; dxdydz \\ = & \int_V(u_x^2+u_y^2+u_z^2 + u\nabla^2u)\; dxdydz \\ = & \int_V\big[(uu_{xx}+u_x^2)+(uu_{yy}+u_y^2)+(uu_{zz}+u_z^2) \big] \; dxdydz \\ = & \int_V\big[(uu_x)_x+(uu_y)_y+(uu_z)_z \big] \; dxdydz \\ = & \int_V\nabla \cdot (u\nabla u) \; dxdydz \\ = & \int_{\partial V} (u\nabla u) \cdot \mathbf n \; dS && \text{Divergence Theorem} \end{align}
and then I kinda got stuck.
Is it actually possible, or would I have to solve for $u$ first?
${\rm grad}\ u_x=0$ so that $u_x$ is constant. So $u$ is linear function on the whole $V$.
Hence we must find $$ \ast=(a^2+b^2+c^2)\int_{x\in \partial V}\ v\cdot x\ v\cdot n\ dx $$
where $v= \frac{(a,b,c)}{|(a,b,c)|}$
If $h(x)=v\cdot x$, then $h: \partial V\rightarrow \mathbb{R}$ is a height function.
If $\textbf{x} : (a,b)\in \mathbb{R}^2\rightarrow \partial V$ is a parametrization, then $v\cdot n\ dx $ is corresponded to $ \textbf{x}_a\times \textbf{x}_b \cdot v $
(1) Here $\int_{\partial V}\ v\cdot n \ dx=0$ if $\partial V$ is homeomorphic to a sphere.
(2) If $\partial V=\{x \in \mathbb{R}^3||x|=1\}$, then $v\cdot n\ dx$ is area form of unit ball $\{ y\in \mathbb{R}^2||y|=1\}$ and $h$ is height so that $\ast$ is a volume of unit ball in $ \mathbb{R}^3$.