A ten-cell isoparametric mesh exactly covers the unit disk

$g^{\mathrm{\prime }}(t)=a(\pi \mathrm{/}4){\sec }^2(\pi t\mathrm{/}4)>0$, so $H$ maps $[-1,1{]}^2$ bijectively onto $[-a,a{]}^2$, since $g(-1)=-a$ and $g(1)=a$. Its Gram determinant gives the same volume factor by [F2]. On the top sector, ${\theta }^{\mathrm{\prime }}=-\pi \mathrm{/}4$ and $r_s=1-a\mathrm{/}\sin \theta \ge 1-a\sqrt{2}>0$ because $\sin \theta \ge \sqrt{2}\mathrm{/}2$. Polar-coordinate differentiation gives $\det DF=r{r}_s(-{\theta }^{\mathrm{\prime }})>0$; the Gram determinant satisfies $\sqrt{\det (D{F}^{T}DF)}=\det DF$. Rotations preserve these determinants.

At $s=0$, $F(t,0)=(a\cot \theta (t),a)=(g(t),a)$, which is exactly the trace of $H$ on its top edge; at $s=1$, $r=1$, so the outer trace is the unit-circle arc from angle $3\pi \mathrm{/}4$ to $\pi \mathrm{/}4$. The two radial side traces match those of adjacent rotated sectors. Hence all ten pieces meet only along identical mapped edges.

For each angle $\theta \in [\pi \mathrm{/}4,3\pi \mathrm{/}4]$, the square boundary on that ray has radius $a\mathrm{/}\sin \theta$ and the unit-circle boundary has radius $1$. The sector map covers exactly the radii between them. Four rotated angular ranges cover the annulus outside the central square, with shared corner rays only; $H$ covers the central square. Therefore the ten mapped triangles cover the unit disk without interior overlap.

The central area is $4a^2=1$. By [F3], the sector area is the polar integral $\frac{1}{2}{\int }_{\pi \mathrm{/}4}^{3\pi \mathrm{/}4}(1-a^2{\csc }^2\theta )d\theta$. Since $[-\cot \theta {]}_{\pi \mathrm{/}4}^{3\pi \mathrm{/}4}=2$, this equals $\pi \mathrm{/}4-a^2=(\pi -1)\mathrm{/}4$. Four rotated sectors have total area $\pi -1$, so the union area is $\pi$, consistent with the unit disk.