Surface Area

Example 1. In all the examples of surface area computations, we take a parametrization \(r(u, v): R \rightarrow S\), then use use that the distortion factor is \(\sqrt{\operatorname{det}(d r^{T} d r)}=|r_{u} \times r_{v}|\).

Example 2. Problem: find the surface area of a sphere \(x^{2}+y^{2}+z^{2}=L^{2}\).
Solution: Parametrize the surface \[r([\theta, \phi])=[L \sin (\phi) \cos (\theta), L \sin (\phi) \sin (\theta), L \cos (\phi)].\] The distortion factor is \(L^{2} \sin (\phi)\). The surface area is \(4 \pi L^{2}\).

Example 3. Problem: find the surface area of surface of revolution given in cylindrical coordinates as \(z=g(\theta)\), \(a \leq z \leq b\).
Solution: Parametrize the surface \[r([\theta, z])=[g(z) \cos (\theta), g(z) \sin (\theta), z].\] The distortion factor is \(g(z) \sqrt{1+g^{\prime}(z)^{2}}\). As an example, we can look at the surface of revolution \(x^{2}+y^{2}=1 / z^{2}\), \(|z|^{2}>1\). The volume of the solid enclosed by the surface is \(\pi\). The surface area is infinite.

Example 4. Problem: find the surface area of the graph of a function \(z=f(x, y)\), \((x, y) \in R\).
Solution: Parametrize the surface as \(r([x, y])=[x, y, f(x, y)]\). The distortion factor is \[|r_{x} \times r_{y}|=\sqrt{1+f_{x}^{2}+f_{y}^{2}}.\]

Example 5. Problem: what is the surface area of the intersection of \(x^{2}+z^{2} \leq 1\), \(6 x+3 y+9 z=12\)?
Solution: The surface is a plane but also a graph over \(R=\{x^{2}+z^{2} \leq 1\}\) in the \(xz\)-plane. The simplest parametrization is \[r([x, z])=[x,(12-6 x-9 z) / 3, z]=[x, 4-2 x-3 z, z].\] It gives \(|r_{x} \times r_{z}|=|[-2,-1,-3]|=\sqrt{14}\). The surface area is \[\iint_{R} \sqrt{14} \,d x \,d y=\sqrt{14} \operatorname{Area}(R)=\sqrt{14} \pi.\]

Example 6. The following hyperspherical coordinates parametrize the \(3\)-dimensional sphere \(x^{2}+y^{2}+z^{2}+w^{2}=1\) in \(\mathbb{R}^{4}\). \[r([\phi, \psi, \theta])=[\cos (\phi), \sin (\phi) \cos (\psi), \sin (\phi) \sin (\psi) \cos (\theta), \sin (\phi) \sin (\psi) \sin (\theta)]\] with \(\theta \in[0,2 \pi]\), \(\phi \in[0, \pi]\), \(\psi \in[0, \pi]\). The distortion factor is \[\sqrt{\operatorname{det}(d r^{T} d r)}=\sqrt{\sin ^{4}(\phi) \sin ^{2}(\psi)}\] so that the surface area of the hypersphere is \[2 \pi \int_{0}^{\pi} \int_{0}^{\pi} \sin ^{2}(\phi) \sin (\psi) \,d \phi \,d \psi=\fbox{$2 \pi^{2}$}.\]

Example 7. In dimension \(n\) what is the volume \(|B_{n}|\) of the \(n\)-dimensional unit ball \(B_{n}\) in \(\mathbb{R}^{n}\) and the volume \(|S_{n}|\) of the \(n\)-dimensional unit sphere \(S_{n}\) in \(\mathbb{R}^{n+1}\)? It starts with \(|B_{0}|=1\), as \(B_{0}\) is a point and \(|S_{0}|=2\), as \(S_{0}\) consists of two points. The \(n\)-ball of radius \(\rho\) has the volume \(|B_{n}| \rho^{n}\) and the \(n\)-sphere of radius \(\rho\) has the volume \(|S_{n}| \rho^{n}\). Because \(|B_{n+1}|=\int_{0}^{1}|S_{n}| \rho^{n} \,d \rho\), we have \(|B_{n+1}|=|S_{n}| /(n+1)\). Because \(S_{n}\) can be written as a union of products \((n-2)\)-spheres with \(S_{1}\) leading to \[|S_{n}|=2 \pi \int_{0}^{\pi / 2}|S_{n-2}| \cos (\phi) \,d \phi=2\pi|B_{n-1}|.\] We know now all: just start with \(|B_{0}|=1\), \(|S_{0}|=2\), \(|B_{1}|=2\), \(|S_{1}|=2 \pi\) and

The \(5\)-ball has maximal volume \(5.26379\ldots\) among all unit balls. The \(6\)-sphere has maximal surface area \(33.0734 \ldots\) among all unit spheres. The volume of the \(30\)-ball is only \(0.00002\ldots\) The surface area of the \(30\)-sphere for example is only \(0.0003\). Compare with an \(\boldsymbol{n}\)-unit cube of volume \(1\) and a boundary surface area \(2n\). High dimensional spheres and balls are tiny!

Example 8. If \(S\) is a cylinder \(x^{2}+y^{2}=1\), \(0, triangulated with each triangle smaller than \(1 / n \rightarrow 0\), does the area converge to the surface area \(A(S)\)? No! A counter example is the Schwarz lantern from 1880. The cylinder is cut into \(m\) slices and \(n\) points are marked on the rim of each slice to get triangles like \(A=(1,0,0)\), \(B=(\cos (4 \pi / n), \sin (4 \pi / n, 0))\), \(C=(\cos (2 \pi / n), \sin (2 \pi / n), 1 / m)\) of area \[\sin (2 \pi / n)(1 / m) \sqrt{2+3 m^{2}-4 m^{2} \cos (2 \pi / n)+m^{2} \cos (4 \pi / n)} / \sqrt{2}.\] The \(n m\) triangles have area \(\sim \sqrt{2+8 m^{2} \pi^{4} / n^{4}} / \sqrt{2}\). For \(m=n^{3}\), the triangulated area diverges.

Example 9. The three dimensional sphere is \(x^{2}+y^{2}+z^{2}+w^{2}=1\) in \(\mathbb{R}^{4}\). The Hopf parametrization is \(r: R \subset \mathbb{R}^{3} \rightarrow S \subset \mathbb{R}^{4}\) is \[r\big([\phi, \theta_{1}, \theta_{2}]\big)=\big[\cos (\phi) \cos (\theta_{1}), \cos (\phi) \sin (\theta_{1}), \sin (\phi) \cos (\theta_{2}), \sin (\phi) \sin(\theta_{2})\big].\] We compute \[|d r|=\sqrt{\operatorname{det}(d r^{T} d r)}=\cos (\phi) \sin (\phi)=\sin (2 \phi) / 2.\] If we fix \(\phi\), we see a two dimensional torus. Their union with \(\phi \in[0, \pi / 2]\) is the Hopf fibration. We can now compute the volume of the three dimensional sphere: \[\int_{0}^{\pi / 2} \int_{0}^{2 \pi} \int_{0}^{2 \pi} \sin (2 \phi) / 2 \,d \phi \,d \theta_{1} \,d \theta_{2}=2 \pi^{2}.\]