Surface areas and surface integrals

This chapter tries to explain the background of the ComputeMetricTensorForSurface function that can be found at various places in 4C.

Motivation and definition.

1. Let \(a^{1},\ldots,a^{k}\in\mathbb{R}^{n}\). Let

\[A:=\bigl(a^{1}\; a^{2}\;\cdots\; a^{k}\bigr)\in\mathbb{R}^{n\times k}\]

be the matrix with columns \(a^{1},\ldots,a^{k}\). Then

\[A\bigl([0,1]^{k}\bigr)=\bigl\{Ax;\,x\in\mathbb{R}^{k},\; x_{j}\in[0,1]\text{ for all }j=1,\ldots,k\bigr\}\subseteq\mathbb{R}^{n}\]

is the parallelepiped spanned by the vectors \(a^{1},\ldots,a^{k}\). If \(k=2\), this is the parallelogram spanned by the vectors \(a^1\) and \(a^2\).

(b) In the case \(k=n\), it holds that \(\operatorname{vol}_{n}\bigl(A\bigl([0,1]^{n}\bigr)\bigr)=\lvert\det A\rvert\). If \(a^{1},\ldots,a^{n}\) are linearly dependent, then both sides are \(=0\), for \(A\bigl([0,1]^{k}\bigr)\) has width \(0\) in (at least) one direction.

(c) The Gramian matrix of the vectors \(a^{1},\ldots,a^{k}\) is defined as

\[A^{\top}A = \bigl(\langle A^{\top}Ae_{i},e_{j}\rangle\bigr)_{i,j=1,\ldots,n} = \bigl(\langle Ae_{i},Ae_{j}\rangle\bigr)_{i,j=1,\ldots,n} = \bigl(\langle a_{i},a_{j}\rangle\bigr)_{i,j=1,\ldots,n} \in\mathbb{R}^{n\times n}\]

(with \(\langle\,\cdot\,,\,\cdot\,\rangle\) being the standard scalar product in \(\mathbb{R}^{k}\) and \(e_{i}\) the \(i\)th standard basis vector of \(\mathbb{R}^{k}\)). It is positive-semidefinite (because \(\langle A^{\top}Ax,x\rangle=\lvert Ax\rvert^{2}\ge0\) for all \(x\in\mathbb{R}^{k}\)) and therefore \(\det A^{\top}A\ge0\).

Now we define

\[\gamma(A):=\sqrt{\det(A^{\top}A)}=\sqrt{\det\bigl(\langle Ae_{i},Ae_{j}\rangle\bigr)}.\]

Notice that \(\gamma(A)=\lvert\det A\rvert\) if \(k = n\).

(d) We want to motivate why \(\gamma(A)\) is the \(k\)-dimensional volume of \(A\bigl([0,1]^{k}\bigr)\). Here is an example: Let \(n=k+1\) and \(A\bigl([0,1]^{k}\bigr)\subseteq\mathbb{R}^{k}\times\{0\}\), e.g. the parallelepiped has width \(0\) in the direction of \(e_{n}=e_{k+1}\). Let \(Q:\mathbb{R}^{k+1}\to\mathbb{R}^{k}\) be the projection onto the first \(k\) coordinates (thus \(Q(x)=Q\bigl((x_{1},\ldots,x_{n})\bigr)=(x_{1},\ldots,x_{k})\)), then

\[(QA)^{\top}QA=A^{\top}Q^{\top}QA=A^{\top}A,\]

and therefore \(\operatorname{vol}_{k}(QA\bigl([0,1]^{k}\bigr))=\lvert\det(QA)\rvert=\gamma(A)\), using part (b). So in this case, \(\gamma(A)\) is indeed the \(k\)-dimensional volume of \(A\bigl([0,1]^{k}\bigr)\).

Integration on submanifolds.

Now let \(M\subseteq\mathbb{R}^{n}\) be a \(k\)-dimensional submanifold of \(\mathbb{R}^{n}\), with global parameterization. While we won’t give the exact definition of submanifolds here, this basically means that there is some open set \(\Omega\subseteq\mathbb{R}^{k}\) and a parameterization \(\Phi:\Omega\to M\) that is, among other things, smooth and one-to-one. Also, let \(f:M\to\mathbb{R}\) be a suitable function (again, we don’t give the exact requirements here). Then we define the surface integral

\[\int_{M}f(x)\,\mathrm{d}S(x) := \int_{\Omega}f(\Phi(\xi))\,\gamma(\Phi'(\xi))\,\mathrm{d}\xi.\]

In particular, we define

\[\operatorname{vol}_{k}(M):=\int_{M}1\,\mathrm{d}S(x) =\int_{\Omega}\gamma(\Phi'(\xi))\,\mathrm{d}\xi.\]

Example.

A parameterization [1] of the two-dimensional unit sphere \(S_{2}:=\{x\in\mathbb{R}^{3};\,\lvert x\rvert=1\}\) in \(\mathbb{R}^{3}\) is

\[\begin{split}\begin{gathered} \Phi:(-\pi,\pi)\times(-\tfrac{\pi}{2},\tfrac{\pi}{2})\to\mathbb{R}^{3},\\ \Phi(\varphi_{1},\varphi_{2}):=\begin{pmatrix}\cos\varphi_{1}\cos\varphi_{2}\\ \sin\varphi_{1}\cos\varphi_{2}\\ \sin\varphi_{2}\end{pmatrix}. \end{gathered}\end{split}\]

Its derivative (the Jacobian matrix) is

\[\begin{split}\Phi'(\varphi_{1},\varphi_{2})=\begin{pmatrix}-\sin\varphi_{1}\cos\varphi_{2} & -\cos\varphi_{1}\sin\varphi_{2}\\ \cos\varphi_{1}\cos\varphi_{2} & -\sin\varphi_{1}\sin\varphi_{2}\\ 0 & \cos\varphi_{2}\end{pmatrix},\end{split}\]

and thus

\[\begin{split}\Phi'(\varphi_{1},\varphi_{2})^{\top}\Phi'(\varphi_{1},\varphi_{2})=\begin{pmatrix}\cos^{2}\varphi_{2} & 0\\ 0 & 1\end{pmatrix},\end{split}\]

so we see that \(\gamma(\Phi'(\varphi_{1},\varphi_{2}))=\cos\varphi_{2}\). Now we compute the 2-dimensional volume, i.e. the surface area of the unit sphere, by

\[\operatorname{vol}_{2}(S_{2})=\int_{S_{2}}1\,\mathrm{d}S(x)=\int_{\varphi_{1}=-\pi}^{\pi}\int_{\varphi_{2}=-\pi/2}^{\pi/2}\cos\varphi_{2}\,\mathrm{d}\varphi_{2}\,\mathrm{d}\varphi_{1}=\left.2\pi\sin\varphi_{2}\right|_{-\pi/2}^{\pi/2}=4\pi.\]

Remark.

For vectors \(a,b\in\mathbb{R}^{3}\) in the three-dimensional space \(\mathbb{R}^{3}\) it holds that

\[\begin{split}\lvert a\times b\rvert^{2}=\langle a\times b,a\times b\rangle=\langle a,a\rangle\langle b,b\rangle-\langle a,b\rangle^{2}=\det\begin{pmatrix}\langle a,a\rangle & \langle a,b\rangle\\ \langle b,a\rangle & \langle b,b\rangle\end{pmatrix},\end{split}\]

and thus if \(\Phi(t)=\Phi(t_{1},t_{2})\), then

\[\lvert\partial_{t_{1}}\Phi\times\partial_{t_{2}}\Phi\rvert^{2}=\det\bigl(\langle\partial_{t_{i}}\Phi,\partial_{t_{j}}\Phi\rangle_{i,j}\bigr)=\det(\Phi'^{\top}\Phi').\]

The matrix \(\Phi'^{\top}\Phi'\), i.e. the Gramian matrix of the vectors \(\partial_{t_{1}}\Phi\) and \(\partial_{t_{2}}\Phi\), is also called the metric tensor, and the expression

\[\lvert\partial_{t_{1}}\Phi\times\partial_{t_{2}}\Phi\rvert\,\mathrm{d}t_{1}\,\mathrm{d}t_{2}=\gamma(\Phi')\,\mathrm{d}S\]

is known in engineering as the area element.

[1]

Actually, this is not a parameterization of all of the unit sphere: The half-plane \(\{x\in\mathbb{R}^{3};\,x_{2}=0,x_{1}\le 0\}\) is missing. A global parameterization of the unit sphere doesn’t exist, and the missing half-plane is a set of \(2\)-dimensional measure zero, so we can disregard it.