Surface Area on a Sphere

11 Oct 2023

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§ Differential Mapping

Let $X(u, v) \triangleq \langle u, v, f(x, y)\rangle$ denote the differentiable mapping of the surface $S,$ and its partial derivatives w.r.t. $u$ and $v$ be

$$X_u = \langle 1, 0, f_u(u, v)\rangle\\ X_v = \langle 1, 0, f_v(u, v)\rangle.$$

The first fundamental form of $S$ is given by

$$\begin{align*} \mathrm{d}s^2 &= E\mathrm{d}u^2 + 2F\mathrm{d}u\mathrm{d}v + G\mathrm{d}v^2\\ E &= X_u \cdot X_u = 1 + f^2_u\\ F &= X_u \cdot X_v = f_u \cdot f_v\\ G &= X_v \cdot X_v = 1 + f^2_v \end{align*}$$

Finally, the surface area of $S$ is given by

$$\begin{align*} \mathrm{area}(S) &= \int_D\sqrt{EG-F^2}\ \mathrm{d}u\,\mathrm{d}v\\ &= \int_D\sqrt{(1 + f^2_u)(1+f^2_v)(f_uf_v)^2}\ \mathrm{d}u\,\mathrm{d}v \end{align*}$$

§ Problem

§ Area of Cylindrical Slices

Assume, wlog. $z_2 > z_1$. The area of $\hat{C}$ is the same as an open-ended cylinder with height $z_2 - z_1$, which is the same as the integral

$$\mathrm{area}(\hat{C}) = \int_{z_1}^{z_2} 2\pi \mathrm{d}t = 2\pi(z_2 - z_1).$$

§ Area of Spherical Slices

To find the area of $S$, we can express it in terms of a parametric equation $S(\theta, \phi) = (\cos\theta \sin\phi, \sin \theta\sin\phi, \cos\phi)$ where $\theta \in [0, 2\pi)$ is the azimuthal angle and $\phi \in [0, \pi]$ is the polar angle.

The differential arclength is given by

$$\begin{align*} \mathrm{d}s &= \|S_\theta\,\mathrm{d}\theta + S_\phi\,\mathrm{d}\phi\|,\\ \mathrm{d}s^2 &= (S_\theta\,\mathrm{d}\theta + S_\phi\,\mathrm{d}\phi) \cdot (S_\theta\,\mathrm{d}\theta + S_\phi\,\mathrm{d}\phi)\\ &= (S_\theta \cdot S_\theta)\,\mathrm{d}\theta^2 + 2(S_\theta \cdot S_\phi)\,\mathrm{d}\theta\,\mathrm{d}\phi + (S_\phi \cdot S_\phi)\,\mathrm{d}\phi^2\\ &= \sin^2\phi\,\mathrm{d}\theta^2 + 0\,\mathrm{d}\theta\,\mathrm{d}\phi + 1\,\mathrm{d}\phi^2 \end{align*}$$

yielding the first fundamental form of $S$,

$$E = (S_\theta \cdot S_\theta) = \sin^2\phi,\quad F = 0,\quad G = 1.$$

The area of $\hat{S}$ is given by

$$\begin{align*} \mathrm{area}(\hat{S}) &= \int_{\hat{D}} \|S_\theta \times S_\phi\|\mathrm{d}\theta \,\mathrm{d}\phi\\ &= \int_{\hat{D}} \sqrt{EG - F^2}\mathrm{d}\theta \,\mathrm{d}\phi\\ &= \int_{\hat{D}} \sin\phi\ \mathrm{d}\theta \,\mathrm{d}\phi. \end{align*}$$

§ Finding the Domain

We found a general expression for $\hat{S}$, however, to get the area we must find $\hat{D}$. As $\theta$ is the azimuthal angle within the $xy$-plane, it is independent of $z$ and unaffected by the cutting planes $z_1$ and $z_2$.

Then, we want to limit the polar angle $\phi$ such that $z = \cos\phi \in [z_1, z_2]$; this is given by $\phi \in [\cos^{-1}(z_2), \cos^{-1}(z_1)]$.

Note that if we flip our assumption such that $-1 \leq z_1 < z_2 \leq 1$, then $\cos^{-1}(z_2) < \cos^{-1}(z_1)$.

Finally, we evaluate the $\mathrm{area}(\hat{S})$ integral over $\hat{D}$,

$$\begin{align*} \int_{(\theta, \phi) \in \hat{D}} \sin\phi\ \mathrm{d}\theta\,\mathrm{d}\phi &= \int_{\cos^{-1}z_2}^{\cos^{-1}z_1}\int_{0}^{2\pi}\sin\phi\ \mathrm{d}\theta\,\mathrm{d}\phi\\ &= \left(\int_{\cos^{-1}z_2}^{\cos^{-1}z_1}\sin\phi\ \mathrm{d}\phi\right)\left(\int_{0}^{2\pi}\ \mathrm{d}\phi\right)\\ &= 2\pi \left[\cos\phi\right]_{\phi=\cos^{-1}z_1}^{\cos^{-1}z_2}\\ &= 2\pi (z_2 - z_1). \end{align*}$$

Therefore, the strips $\hat{S}$ and $\hat{C}$ have equal area.