= Solution
For a real-valued function on a bounded regular region, apply the <divergence theorem> to $\phi\nabla\phi$. The <product rule> yields $\nabla\cdot(\phi\nabla\phi)=|\nabla\phi|^2+\phi\nabla^2\phi$. The prescribed zero boundary values and <Laplace equation> therefore imply
$$
\int_V|\nabla\phi|^2dV=\int_{\partial V}\phi\,\partial_n\phi\,dS-\int_V\phi\nabla^2\phi\,dV=0.
$$
The continuous nonnegative integrand must vanish everywhere. Hence $\phi$ is constant on each connected component, and its boundary value makes each constant zero. This proves <zero-boundary harmonic uniqueness>. If complex-valued functions are allowed, apply the same argument separately to their real and imaginary parts. If $\psi_1,\psi_2$ solve the same <Poisson equation> and <Dirichlet boundary data>, their difference is harmonic with zero boundary value, so is zero. Thus \b[there is at most one solution]. An exterior unbounded region would require an additional condition at infinity; the preceding bounded-domain proof does not assert uniqueness without one.
For the radial calculation use Cartesian coordinates with $r=(x_ix_i)^{1/2}>0$. The <chain rule> gives $\partial_ir=x_i/r$ and
$$
\boxed{\nabla\psi=\frac{\psi'(r)}r\mathbf x.}
$$
Differentiate again and sum the three coordinates:
$$
\begin{aligned}
\nabla^2\psi&=\partial_i\left(\psi'(r)\frac{x_i}r\right)\\
&=\psi''(r)\frac{x_ix_i}{r^2}+\psi'(r)\left(\frac3r-\frac{x_ix_i}{r^3}\right)\\
&=\psi''(r)+\frac2r\psi'(r)=\boxed{\frac1r\frac{d^2(r\psi)}{dr^2}}.
\end{aligned}
$$
This derives the <radial Laplacian> directly in Cartesian coordinates. For $\nabla^2\psi=c$, set $u=r\psi$, so $u''=cr$. Integrating twice gives $u=cr^3/6+ar+b$, and consequently
$$
\boxed{\psi(r)=\frac{cr^2}{6}+a+\frac br,\qquad r>0.}
$$
If the solution must extend boundedly through the origin, $b=0$. The remaining polynomial is smooth there.
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