The height is constant along each circular ring, whose area element after angular integration is $2\pi rdr$. The volume integral in cylindrical coordinates is thereforeHere $u=r/R$. Using integration by parts, ${\int }_0^{1}u{e}^{-u}du=[-(u+1)e^{-u}{]}_0^1=1-2/e$. HenceThe dimensionless coefficient is positive, and the cube root of a volume has the required physical dimension of length.

Take $r>0$ and select a complex argument $\theta$ for $z$. To specify the complex exponentiation, use $z^{1+i}=\exp ((1+i)(\ln r+i\theta ))$. Expanding its exponent givesIts complex logarithm therefore has valuesThe continuous logarithm obtained by retaining the displayed exponent uses $k=0$. If the outer logarithm is the principal logarithm, choose $k$ to put the imaginary part in $(-\pi ,\pi ]$ instead; the real part is unchanged. In particular one cannot generally assert $\mathrm{Log}(e^w)=w$ without checking its imaginary part.

The condition on the real part gives $\ln r=\theta$, so the required logarithmic spiral has polar coordinatesAs $\theta$ increases, it runs from the negative real axis near radius $e^{-\pi }$ through $-i{e}^{-\pi /2}$, $1$ and $i{e}^{\pi /2}$ to the negative real axis at radius $e^{\pi }$. The angle increases by one complete turn. Using the conventional principal argument $-\pi <\theta \le \pi$, the inner negative-axis endpoint is excluded and the outer endpoint is included. The convention $-\pi \le \theta <\pi$ reverses those endpoint choices. If both limiting arguments are admitted by the paper's wording, both limiting endpoints are drawn; the interior curve is the same. The origin is never included because the complex logarithm is undefined there.

With $z=x+iy$, the expression inside the modulus is $x+3i(y-1)$. Squaring its nonnegative modulus therefore givesThe locus is an ellipse in the complex plane, centred at $(0,1)$, with horizontal semiaxis $2$ and vertical semiaxis $2/3$. Its vertices are $(\pm 2,1)$ and $(0,1\pm 2/3)$.

The complex conjugate is $z^{\ast }=x-iy$. From the complex sine and cosine identities,Divide these expressions and multiply by the complex conjugate of the denominator. Its squared modulus is ${\cos }^{2}x{\cosh }^{2}y+{\sin }^{2}x{\sinh }^{2}y=(\cos 2x+\cosh 2y)/2$. The numerator simplifies using ${\cosh }^{2}y-{\sinh }^{2}y=1$ and the double-angle formulas. ThusThese expressions apply except at the poles $y=0$, $x=\pi /2+k\pi$ with integer $k$, where the original tangent function is undefined.

Put $c=\cos \theta$ and $s=\sin \theta$. Taking the real part of $(c+is{)}^6$ in De Moivre's theorem givesTaking the imaginary parts of the second and fourth powers gives $\sin 2\theta =2cs$ and $\sin 4\theta =4cs(c^2-s^2)$. HenceThe mixed powers cancel exactly.

Apply scalar multiplication and vector addition componentwise:

The Euclidean norm of the vector is the square root of the sum of its squared components:

Use the outward unit normal on the unit sphere and the convention that $\theta$ is measured from the positive vertical axis. Then $\mathbf{F}\cdot \mathbf{n}=\sin \theta$, while the spherical coordinates area element is $dS=\sin \theta d\theta d\varphi$. The flux is therefore

The vector field is radial, while the differential displacement along a centred circle is tangential. These directions are orthogonal, so the dot product vanishes at every point. Hence

Near the initial condition $y=1$, divide by $y^2$ and separate variables: $y^{-2}dy=-x^{2}dx$. Integration gives $-1/y=-x^3/3+C$; the initial value fixes $C=-1$. Thus the solution of the separable differential equation isThe stated interval is the maximal connected interval containing zero: the solution diverges at its left endpoint. The constant solution $y=0$ does not satisfy the initial condition.

The integrating factor for this linear ordinary differential equation is $e^x$. Multiplication gives $(e^{x}y{)}^{\prime }=1$, so $e^{x}y=x+C$. The initial condition sets $C=0$, giving

The vector field has only a vertical component, namely $f(r)$. Its divergence is therefore its vertical partial derivative:

For $r>0$, the partial derivatives of radius are $r_x=x/r$, $r_y=y/r$ and $r_z=z/r$. Applying the chain rule to the outer sine and then the radial profile gives the Cartesian coordinates of the gradient:At the origin these coordinate formulas require a differentiable radial extension; differentiability of $f$ as a one-variable profile alone does not guarantee it.

On a smooth implicit branch, put $Q=x+z^2$ and require $Q\ne 0$. This is exactly the implicit function theorem condition, since the defining equation has vertical partial derivative $3Q$. Differentiating the cubic relation in $x$ and $y$ givesFor a second derivative, the dependence of $Q$ on $z$ must also be included: $Q_x=1+2z{z}_x$ and $Q_y=2z{z}_y$. ThusThe required cancellation follows immediately:The verification is local to smooth branches with $x+z^2\ne 0$; the implicit relation does not guarantee a differentiable function at its singular points.

Write ${ϵ}_k=E_0(k+\frac{1}{2})$ and use Lagrange multipliers $\alpha ,\beta$ for the particle-number and energy constraints. The stationary equations areThus $n_k=e^{-\alpha }{e}^{-\beta {ϵ}_k}$. Finite normalization over infinitely many levels requires $\beta >0$. Put $q=e^{-\beta E_0}$, so $0<q<1$. The geometric series givesThereforeFor the energy constraint, use $n_k=N(1-q)q^k$ and $\sum k{q}^k=q/(1-q{)}^2$, obtained by differentiating the geometric series. This yieldsRecognizing the hyperbolic cotangent givesThe strict positivity of all occupations requires $E>N{E}_0/2$; the paper's positivity assumptions alone do not guarantee this feasibility condition. For admissible energy the multiplier is uniquely $\beta =E_0^{-1}\ln [(2E+N{E}_0)/(2E-N{E}_0)]>0$. Equality $E=N{E}_0/2$ is only the limiting state with all occupation in level zero, outside the stipulated strictly positive family; lower energy is impossible.

For rigor beyond formal multiplier equations, $p_k=n_k/N$ has fixed mean $E/(N{E}_0)-1/2$. The objective differs from $N$ times its information entropy only by the fixed constant $N-N\ln N$. The maximum entropy distribution on the nonnegative integers is uniquely geometric at that mean, confirming the stationary family and its maximum character.

At a fixed radius, Parseval's identity applied to the preceding Fourier series givesIn polar coordinates, $dA=rdrd\theta$. Integrating over radius and interchanging the nonnegative sum and integral givesThe interchange follows from the monotone convergence theorem. This is the special case of the area norm of the Poisson extension in the unit disk for the odd step boundary data.