= Solution
In a steady state, <Maxwell's equations> imply $\nabla\times\mathbf E=0$, since the magnetic field is time independent. The divergence of the steady Ampère equation gives $\nabla\cdot\mathbf j=0$. With uniform <electrical conductivity>, <Ohm's law> is $\mathbf j=\sigma\mathbf E$, so
$$
\nabla\times\mathbf j=0,\qquad \nabla\cdot\mathbf j=0.
$$
For the usual straight-wire model, the cylindrical side is insulating and the two flat end faces are equipotentials. Put $\mathbf E=-\nabla V$. The <electric potential> solves <Laplace's equation>, with constant end values and zero normal derivative on the side. The linear axial potential $V(z)=V(0)-\Delta V\,z/l$ satisfies all these conditions. It is unique: the difference $h$ of two solutions has zero end values and zero side normal derivative, and <integration by parts> gives $\int|\nabla h|^2\,dV=0$. Thus
$$
\boxed{\mathbf j=\frac{\sigma\Delta V}{l}\mathbf e_z
=\frac{I}{\pi a^2}\mathbf e_z.}
$$
This proves <uniform current in a straight homogeneous wire> with the usual contact conditions. Integrating the <electric current density> over a cross-section and using <Ohm's law> then gives
$$
\boxed{R=\frac{l}{\sigma\pi a^2},\qquad
\dot Q=\int\mathbf j\cdot\mathbf E\,dV
=\frac{I^2l}{\sigma\pi a^2}=I^2R.}
$$
The second expression is the rate of <Ohmic heating>.
Strictly, steady current and uniform conductivity alone do not force uniformity; the contact conditions are implicit in the wire model. For example, let $\chi$ be a nonconstant transverse <Neumann eigenfunction> on the disk, satisfying $\Delta_\perp\chi=-\lambda^2\chi$ and zero normal derivative at its edge. Then $V=-E_0z+C\chi\cosh(\lambda(z-l/2))$ is harmonic and preserves an insulating cylindrical side, but its nonconstant end potentials produce nonuniform current. Its added axial current has zero cross-sectional mean. This supplies a counterexample if arbitrary contacts are permitted.
A bend generally redistributes the <electric current density>: the inner side of the bend carries more current than the outer side, as illustrated by $j_\theta\propto1/r$ in an annular conductor. Exact resistance and local <Ohmic heating> can therefore change. For a slender, gently curved wire the straight-wire formulas remain the leading approximation; the identity $\dot Q=I^2R$ remains valid for the total steady dissipation.
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