The subcritical Kepler conic is an ellipse with semiaxes $a=p\mathrm{/}(1-e^2)$, $b=a\sqrt{1-e^2}$, focus at the origin and area $\pi ab$; its area is also $\frac{1}{2}{\int }_0^{2\pi }{\varrho }^2$

Clause 1: $\varrho$ is a quotient of $C^1$ functions with denominator $1+e\cos \phi \ge 1-e>0$, hence $C^1$ by [L4]; it is $2\pi$-periodic because $\cos$ is; and $p\mathrm{/}(1+e)\le \varrho \le p\mathrm{/}(1-e)$ because $\cos \phi \in [-1,1]$.

Clause 2, the identity $E_{\mathrm{e}\mathrm{l}\mathrm{l}}=T[\bar{B}]$: by [L1], for $(u,v)\in {\mathbb{R}}^2$, $Q(T(u,v))=a^2(1-e^2)(u^2+v^2-1)$, so $Q(T(u,v))\le 0$ if and only if $u^2+v^2\le 1$, that is, $(u,v)\in \bar{B}$; hence $(x,y)\in E_{\mathrm{e}\mathrm{l}\mathrm{l}}$ exactly when $(x,y)=T(u,v)$ for some $(u,v)\in \bar{B}$, since $T$ is a bijection of ${\mathbb{R}}^2$ onto itself (its linear part is invertible, $\det \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(a,b)=ab>0$).

Clause 3, the radial description: for $s>0$ put $(x,y):=s(\cos \phi ,\sin \phi )$. Using [L4], $Q(x,y)=s^2(1-e^2{\cos }^2\phi )+2pes\cos \phi -p^2$, a quadratic in $s$ with positive leading coefficient (since $1-e^2{\cos }^2\phi \ge 1-e^2>0$) and negative constant term $-p^2<0$; hence the set of $s>0$ with $Q\le 0$ is the interval $(0,\varrho (\phi )]$: indeed $s=\varrho (\phi )$ is the unique positive root of $Q=0$, because substituting $\varrho =p\mathrm{/}(1+e\cos \phi )$ and multiplying by $(1+e\cos \phi {)}^2$ gives $p^2(1-e^2{\cos }^2\phi )+2p^{2}e\cos \phi (1+e\cos \phi )-p^2(1+e\cos \phi {)}^2=0,$ a direct expansion — and $Q(0)=-p^2<0$ while $Q(s)\to +\mathrm{\infty }$ as $s\to \mathrm{\infty }$, so the two roots have opposite signs and the positive one is $\varrho (\phi )$. Therefore $E_{\mathrm{e}\mathrm{l}\mathrm{l}}=\{s(\cos \phi ,\sin \phi ):0\le \phi <2\pi ,0\le s\le \varrho (\phi )\}$, the boundary is the displayed profile, and each ray meets it exactly once.

Clause 2, the area: $T$ is an affine map with invertible linear part, hence a $C^1$ diffeomorphism of ${\mathbb{R}}^2$ with $\mathrm{\mid }\det DT(u,v)\mathrm{\mid }=ab$ everywhere; $E_{\mathrm{e}\mathrm{l}\mathrm{l}}=\{Q\le 0\}$ is closed (so Lebesgue measurable with indicator in $L^1$), and $T[\bar{B}]=E_{\mathrm{e}\mathrm{l}\mathrm{l}}$ by step 1.2. The change-of-variables formula of [L2] applied to the indicator of $E_{\mathrm{e}\mathrm{l}\mathrm{l}}$ gives ${\lambda }_2(E_{\mathrm{e}\mathrm{l}\mathrm{l}})={\int }_{\bar{B}}ab\mathrm{d}{\lambda }_2=ab{\lambda }_2(\bar{B})$; the closed unit disc has ${\lambda }_2(\bar{B})=\pi$ by [L3], so ${\lambda }_2(E_{\mathrm{e}\mathrm{l}\mathrm{l}})=\pi ab=\pi a^2\sqrt{1-e^2}$.

Clause 3, the polar area formula: let $W:=\{(\phi ,s):0<\phi <2\pi ,0<s\le \varrho (\phi )\}\subseteq D:=(0,2\pi )\times (0,\mathrm{\infty })$ and $S:=\mathrm{\Phi }(W)$, where $\mathrm{\Phi }(\phi ,s)=(s\cos \phi ,s\sin \phi )$ is the restriction of the polar map of [L2]. The set $W$ is Borel because $\varrho$ is continuous, and $S$ is Borel because $\mathrm{\Phi }$ is a homeomorphism from $D$ onto its image. Since $S\subseteq E_{\mathrm{e}\mathrm{l}\mathrm{l}}$ by step 1.3 and $E_{\mathrm{e}\mathrm{l}\mathrm{l}}$ has finite measure by step 2.1, ${\mathbf{1}}_S$ is integrable. The change-of-variables formula [L2] applied on $D$ gives ${\lambda }_2(S)={\int }_{W}s\mathrm{d}{\lambda }_2(\phi ,s)={\int }_0^{2\pi }\text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}{\int }_0^{\varrho (\phi )}s\mathrm{d}s\mathrm{d}\phi =\frac{1}{2}{\int }_0^{2\pi }\varrho (\phi {)}^2\mathrm{d}\phi ,$ the middle equality by Tonelli [L2] applied to the nonnegative function $(\phi ,s)\mapsto s{\mathbf{1}}_W$. By step 1.3, $E_{\mathrm{e}\mathrm{l}\mathrm{l}}\setminus S$ is contained in the closed positive $x$-axis, including the origin, a subset of a line and hence null by [L3]. Therefore ${\lambda }_2(E_{\mathrm{e}\mathrm{l}\mathrm{l}})={\lambda }_2(S)=\frac{1}{2}{\int }_0^{2\pi }\varrho (\phi {)}^2\mathrm{d}\phi$, which is clause 3.