Unit-sphere area form and outward radial flux

The rectangle $D$ is a compact Jordan parameter region. Since ${\sin }^{\mathrm{\prime }}=\cos$ and ${\cos }^{\mathrm{\prime }}=-\sin$, repeated differentiation shows that every derivative of sine and cosine exists and is one of $\pm \sin ,\pm \cos$; these derivatives are continuous, so $\mathrm{\Phi }$ is smooth on all of ${\mathbb{R}}^2$. Every point of $S^2$ is in its image: choose $\theta \in [0,\pi ]$ with $\cos \theta =p_3$; when $\sin \theta >0$, the pair $(p_1\mathrm{/}\sin \theta ,p_2\mathrm{/}\sin \theta )$ is a unit-circle point and [F7] supplies $\varphi \in [0,2\pi ]$, while at a pole any $\varphi$ works. More precisely, the interior image is $W=S^2\setminus \mathrm{\Phi }(\mathrm{\partial }D)$: for every point not on the poles or the longitude seam, the latitude is unique by strict decrease of cosine on $(0,\pi )$, and the longitude is unique by the bijection in [F7]. Direct differentiation gives ${\mathrm{\Phi }}_{\theta }=(\cos \theta \cos \varphi ,\cos \theta \sin \varphi ,-\sin \theta )$ and ${\mathrm{\Phi }}_{\varphi }=(-\sin \theta \sin \varphi ,\sin \theta \cos \varphi ,0)$; substituting into the coordinate cross-product formula [F3] yields ${\mathrm{\Phi }}_{\theta }\times {\mathrm{\Phi }}_{\varphi }=({\sin }^2\theta \cos \varphi ,{\sin }^2\theta \sin \varphi ,\sin \theta \cos \theta )=\sin \theta \mathrm{\Phi }$. [F7] makes this nonzero for $0<\theta <\pi$, and its direction is outward by the stated orientation. In any smooth chart on $S^2$, the coordinate derivative of $\mathrm{\Phi }$ is therefore invertible; [F6] gives a smooth local inverse. The bijection onto $W$ and these local inverses make $\mathrm{\Phi }{\mathrm{\mid }}_{D^{\circ }}$ an orientation-preserving diffeomorphism onto the open set $W$. Thus [F1] applies, including the permitted seam and pole degeneracy on $\mathrm{\partial }D$.

By [F2] and step 1.1, the flux integrand is $f(\theta ,\varphi )=\mathrm{\Phi }\cdot (\sin \theta \mathrm{\Phi })=\sin \theta$, and the same coefficient is $\omega ({\mathrm{\Phi }}_{\theta },{\mathrm{\Phi }}_{\varphi })$. This function is continuous on the compact Jordan rectangle $D$, so [F11] makes it Riemann integrable there; [F7] gives $0\le f\le 1$. The bounded open rectangle $D^{\circ }$ has boundary $\mathrm{\partial }D$, which has two-dimensional content zero by [F1], so [F10] makes $D^{\circ }$ Jordan measurable. Define $g:D\to \mathbb{R}$ by $g=f$ on $D^{\circ }$ and $g=0$ on $\mathrm{\partial }D$. The functions $f,g$ are bounded and differ only on the content-zero set $\mathrm{\partial }D$; [F9] therefore gives that $g$ is Riemann integrable on $D$ and ${\int }_{D}g={\int }_{D}f$. By the zero-extension definition in [F10], ${\int }_{D}g={\int }_{D^{\circ }}f$. This supplies the Riemann-integral consequence of the null boundary explicitly. [F4] defines the oriented-sphere integral and ensures compact support, and [F5] identifies it with ${\int }_{D^{\circ }}f$. Therefore [F8] gives ${\int }_D(X\circ \mathrm{\Phi })\cdot ({\mathrm{\Phi }}_{\theta }\times {\mathrm{\Phi }}_{\varphi })d\theta d\varphi ={\int }_0^{2\pi }\text{\hspace{0.17em}⁣}{\int }_0^{\pi }\sin \theta d\theta d\varphi =2\pi (-\cos \theta ){\mid }_0^{\pi }=4\pi$. The vanishing cross product at the poles and the repeated longitude seam occur only on $\mathrm{\partial }D$.