Hello @TedShifrin
Let $u(x) \geq 0$ be harmonic in $\Omega$ and $u|_{\partial{\Omega'}}=0$ where the space $\Omega'$ is a proper subset of $\Omega$. I want to prove that $u(x) \equiv 0$ in $\Omega$.

If we suppose that $u \in C^0(\overline{\Omega'})$ we can deduce that $u(x)=0$ in $\Omega'$.

But how can we deduce that $u(x) \equiv 0$ in $\Omega$ ?

Could be, but I don't know. Did you see this? https://gist.github.com/Croxed/f7f451608143f253b50803785f7ce3f0

Might help ya

the problem consisting calculation of the normalization constant K from a function described as:
psi=K(x+y+2z)exp(-alpha*r)

K is the normalization cst alpha is real

i done the following transformation (catesian to spherical coords.)

psi=Krexp(-2alpha*r)(sin(theta)*cos(fi)+sin(theta)*sin(fi)+2cos(theta))

then to find K

integral of |psi|^2=1 the last expression gives

|K|^2* integral r^3*exp(-alpha*r)dr * integral d(fi) * integral (sin^2(theta)*cos(fi)+sin^2(theta)*sin(fi)+2cos(theta))