= Solution
\b[Equilibrium radiation and cavity <frequencies>.] A <blackbody> absorbs all incident radiation at every <frequency> and direction. In <thermal equilibrium> its emission spectrum depends only on its <temperature>. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes <black-body radiation> to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.
In a <rectangular electromagnetic cavity>, the <wave equation> and <perfect conductor> wall <boundary conditions> select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the <vacuum> <dispersion relation> $\omega=c|\mathbf k|$ and $\nu=\omega/(2\pi)$ gives
$$
\boxed{\nu_{lmn}=\frac{c}{2\pi}\sqrt{\left(\frac{\pi l}{L_x}\right)^2+
\left(\frac{\pi m}{L_y}\right)^2+\left(\frac{\pi n}{L_z}\right)^2}.}
$$
For example, electric-field components can have the spatial dependence
$$
\begin{aligned}
E_x&=A_x\cos(k_xx)\sin(k_yy)\sin(k_zz),\\
E_y&=A_y\sin(k_xx)\cos(k_yy)\sin(k_zz),\\
E_z&=A_z\sin(k_xx)\sin(k_yy)\cos(k_zz).
\end{aligned}
$$
For three positive indices, <Gauss's law> requires $k_xA_x+k_yA_y+k_zA_z=0$, leaving two independent <polarizations of an electromagnetic wave>.
\b[Index qualification.] The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one <polarization of an electromagnetic wave>. For instance, with $n=0$, $l,m\geq1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional <power law>.
\b[Three-dimensional mode count.] A lattice cell in positive $\mathbf k$ space has <volume (thermodynamics)> $\pi^3/V$, where $V=L_xL_yL_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:
$$
dN=2\frac{V}{\pi^3}\frac18\,4\pi k^2\,dk
=\frac{V}{\pi^2}k^2\,dk.
$$
Using $k=2\pi\nu/c$ gives
$$
\boxed{dN=V\frac{8\pi\nu^2}{c^3}\,d\nu.}
$$
For a finite cavity the exact <density of states> is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.
\b[<Mean> <photon number>.] A mode is a <quantum harmonic oscillator> with thermal excitation <energies> $s h\nu$, $s=0,1,2,\ldots$. The <zero-point energy> does not enter its thermal occupation. Put $q=e^{-h\nu/(k_BT)}$. Its <partition function> and <mean> occupation are
$$
Z_\nu=\sum_{s=0}^\infty q^s=\frac1{1-q},\qquad
\langle s\rangle=q\frac{\partial\ln Z_\nu}{\partial q}
=\frac{q}{1-q}
=\boxed{\frac{1}{e^{h\nu/(k_BT)}-1}.}
$$
This is the <Bose-Einstein distribution> with zero <photon> <chemical potential>: equilibrium absorption and emission do not conserve <photon number>.
\b[Thermal <energy>.] Multiply the <density of states> by the <mean> excitation <energy> $h\nu\langle s\rangle$ and integrate:
$$
U=V\int_0^\infty\frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/(k_BT)}-1}\,d\nu.
$$
The substitution $x=h\nu/(k_BT)$ isolates all the dependence on cavity dimensions and <temperature>:
$$
U=\frac{8\pi V(k_BT)^4}{h^3c^3}
\int_0^\infty\frac{x^3}{e^x-1}\,dx.
$$
Expanding $1/(e^x-1)=\sum_{r=1}^\infty e^{-rx}$ gives an <integral> $\sum_r6/r^4=6\zeta(4)=\pi^4/15$. Hence
$$
\boxed{U=\frac{8\pi^5k_B^4}{15h^3c^3}VT^4
=\frac{\pi^2k_B^4}{15\hbar^3c^3}L_xL_yL_zT^4.}
$$
This is the thermal <energy> only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_BT$ large compared with the geometric level spacings in all three directions.
\b[Spectral sketch.] The spectral <energy density> per unit ordinary <frequency> is <Planck's law>,
$$
u_\nu(T)=\frac{8\pi h\nu^3}{c^3}\frac{1}{e^{h\nu/(k_BT)}-1}.
$$
At small <frequency> it rises as $8\pi k_BT\nu^2/c^3$, while at large <frequency> it decays exponentially. Raising <temperature> increases the spectrum at every fixed positive <frequency>, moves the peak to larger <frequency> in proportion to $T$, and increases the integrated <area> in proportion to $T^4$. A plot per unit <wavelength> has a different peak position because the integration variable changes.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-2-2018-blackbody-spectra.png]
{title=Planck spectral energy density at three temperatures on common frequency and energy-density scales}
{description=Original plots of the frequency spectrum, with temperatures in the ratios 1:1.5:2. The dimensionless horizontal axis is hν/(kBTref), and the vertical scale is common to all curves. The higher-temperature peak moves right and the area grows as T to the fourth power.}
\b[Thin-cavity radiation.] Let $A=L_xL_y$. The relevant low-temperature window is
$$
\frac{\pi\hbar c}{L_x},\ \frac{\pi\hbar c}{L_y}\ll k_BT\ll\frac{\pi\hbar c}{L_z}.
$$
Nonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one <polarization of an electromagnetic wave>. Its allowed in-plane wavenumbers form a lattice of cell <area> $\pi^2/A$; a quarter of a circular shell therefore contains
$$
dN=\frac{A}{\pi^2}\frac14(2\pi k\,dk)
=\frac{A}{2\pi}k\,dk
=\frac{2\pi A\nu}{c^2}\,d\nu.
$$
Thus the <dimensional crossover of thermal cavity radiation> gives
$$
U_{2D}=\frac{2\pi A(k_BT)^3}{h^2c^2}\int_0^\infty\frac{x^2}{e^x-1}\,dx
=\boxed{\frac{4\pi\zeta(3)k_B^3}{h^2c^2}L_xL_yT^3\ \propto\ L_xL_yT^3.}
$$
The <integral> is $2\zeta(3)$ by the same geometric-series argument. A model with a different number of independent <polarizations of an electromagnetic wave> changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to0$ thermal <energy> is exponentially suppressed. The requested <area> law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.
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