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Vector Calculus: An Introduction

Daniel Daners

Section 9.5 Surface Integrals

In this section we define integrals of functions on surfaces.

Subsection 9.5.1 The Definition of Surface Integrals

In Theorem 5.19 we derived a formula telling us how to integrate a function over a deformed plane domain. We called it the transformation formula. The original domain, and its image, the deformed domain, were both lying in the plane. A surface, at least locally, is the image of a plane domain lying not in the plane but in space. Hence to construct surface integrals we proceed in exactly the same way as we derived the transformation formula. Assume now that \(S\) is a smooth surface, and for simplicity we assume that it is parametrised by one function \(\vect g\colon D\to S\text{,}\) where \(D\subset\mathbb R^2\text{.}\) We again look at the image, \(\vect g(R)\text{,}\) of a small rectangle \(R\subset D\) as shown in Figure 9.18
Figure 9.18. Deformation of a small rectangle by \(\vect g\)
The idea is similar to Figure 5.16, with the only difference that \(\vect g(R)\subset\mathbb R^3\text{.}\) The vectors \(\vect v_1\) and \(\vect v_2\) are given exactly the same way:
\begin{equation} \vect v_1=\frac{\partial\vect g}{\partial y_1}(\vect y)\Delta y_1 \qquad\text{and}\qquad \vect v_2=\frac{\partial\vect g}{\partial y_2}(\vect y)\Delta y_2\text{.}\tag{9.6} \end{equation}
The difference is that they have three and not two components, which means that we need a different formula to compute the area of the parallelogram they span. To compute that area we use Theorem 1.16. Note that the matrix with columns
\begin{equation*} \frac{\partial\vect g}{\partial y_1}(\vect y) \qquad\text{and}\qquad \frac{\partial\vect g}{\partial y_2}(\vect y) \end{equation*}
is the Jacobian matrix, \(J_{\vect g}(\vect y)\text{,}\) of \(\vect g\) at \(\vect y\text{.}\) Hence the approximate area of \(\vect g(R)\) in our present situation is
\begin{equation*} \area(\vect g(R)) \approxeq\sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g}(\vect y)\bigr)}\,\Delta y_1\Delta y_2\text{.} \end{equation*}
We encountered the above determinant already when defining smooth surfaces, and called it the Jacobian of the parametrisation \(\vect g\text{.}\) The Jacobian is the factor by which the area of a small rectangle is distorted when mapped by \(\vect g\) onto the surface. To define surface integrals we replace \(|\det(J_{\vect g}(\vect y))\bigr|\) by the Jacobian of the parametrisation \(\vect g\text{.}\) This motivates the following definition.

Definition 9.19. Surface integral of a scalar function.

Suppose that \(S\) is a smooth surface parametrised by \(\vect g\colon D\to S\text{.}\) If \(f\) is a continuous function on \(S\) then we define
\begin{equation} \int_{S}f\,dS :=\int_Df(\vect g(\vect y)) \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g}(\vect y)\bigr)}\,d\vect y\text{.}\tag{9.7} \end{equation}
If \(S\) is piecewise smooth, or has to be parametrised by more than one function, we just add up the corresponding integrals.

Remark 9.20.

Note that we can compute the Jacobian of \(\vect g\) by using Proposition 9.17.

Observation 9.21.

If we set \(f=1\) on \(S\) then the surface integral yields the area of the surface. Hence, if \(S\) is parametrised by \(\vect g\) then
\begin{equation*} \area(S) =\int_{S}1\,dS =\int_D \sqrt{\det\bigl(\bigl(J_{\vect g}(\vect y)\bigr)^T J_{\vect g}(\vect y)\bigr)}\,d\vect y\text{.} \end{equation*}

Remark 9.22.

Let us note that the above definition of surface integrals can be generalised to to \(k\)-dimensional “surfaces” in \(\mathbb R^N\text{.}\) In particular it is consistent with the definition of integrals over a one dimensional “surface,” that is, a curve, where \(k=1\text{.}\) If we parametrise a curve \(C\) by \(\vect\gamma(t)\text{,}\) \(t\in I\text{,}\) then it was shown in Remark 9.4 that
\begin{equation*} \bigl(J_{\gamma}(t)\bigr)^T J_{\gamma}(t) =\vect\gamma'(t)^T\vect\gamma'(t) =\|\vect\gamma'(t)\|^2\text{.} \end{equation*}
If we apply (9.7) to the present situation we get
\begin{align*} \int_C f\,ds \amp:=\int_I f(\vect\gamma(t)) \sqrt{\det\bigl(\bigl(J_{\gamma}(t)\bigr)^T J_{\gamma}(t)\bigr)}\,dt\\ \amp=\int_I f(\vect\gamma(t))\|\vect\gamma'(t)\|\,dt, \end{align*}
which coincides with the formula given in Definition 7.10.

Remark 9.23.

Let us emphasise that Definition 9.19 reduces to the transformation formula in Theorem 5.19 if \(S\) is flat, that is, lies in \(\mathbb R^2\text{!}\) One could ask why the transformation formula was a theorem, and the above is a definition. The reason is that, if \(\vect g(D)\) is in \(\mathbb R^2\text{,}\) then the integral over \(\vect g(D)\) is already defined. The theorem tells us that we can compute it using the transformation formula. For surfaces, however, the integral is not already defined, so we use (9.7) to do so.
It was not a trivial task to define the surface area for general surfaces such that it has the properties we expect, and some attempts failed. To define the length of a curve we approximated the given curve by polygons. Mathematicians tried to generalise this idea to surfaces, approximating them by polyhedra. We choose a grid of points and form triangles as shown in Figure 9.24, add up their areas and try to pass to the limit! However, as an example due to Hermann Amandus Schwarz (1843-1921) shows, the limit could be infinite for a surface as simple as a finite cylinder! (see [4] pages~75-77 for details). Hence it is not possible to define surface area by approximation by polyhedra. In our approach we approximated the surface by small scales like on the skin of a fish. Doing so we get a “surface area” having the expected properties. In particular it behaves well under transformation of coordinates.
Figure 9.24. Approximating a surface by a polyhedron
We next want to compute the Jacobians for two important cases, namely for spherical coordinates, and for surfaces given as the graph of a function.

Subsection 9.5.2 The Jacobian for spherical coordiates

We can parametrise a sphere by spherical coordinates by
\begin{equation*} \vect g(\theta,\varphi) :=\bigl(R\cos\varphi\sin\theta,R\sin\varphi\sin\theta,R\cos\theta\bigr), \end{equation*}
where \((\theta,\varphi)\in D:=[0,\pi]\times[0,2\pi)\text{.}\) Applying Example 9.3 we have the following proposition.
Note that if we measure the angle from the \(xy\) plane then we need to replace \(\sin\theta\) by \(\cos\theta\) in the above formula. We can use this to compute the surface area of the sphere.

Example 9.26. Surface area of sphere.

We can use the above to compute the surface area of a sphere of radius \(R\text{.}\) It is most convenient to use spherical coordinates as introduced in Example 9.3. We deduce from Example 9.3 and Proposition 9.17 that the surface area is
\begin{align*} \int_{S}1\,dS &=\int_0^{2\pi}\int_0^\pi R^2\sin\theta\,d\varphi\,d\theta\\ &=2\pi R^2[-\cos\theta]_0^\pi\\ &=2\pi R^2(-(-1)+1)=4\pi R^2 \end{align*}
which is the well known result.

Subsection 9.5.3 The Jacobian for a graph

We next consider the Jacobian for a surface given by a graph of a function. Suppose that \(S\) is the graph of a smooth function \(h\colon D\to\mathbb R\text{.}\) We want to compute Jacobian of the corresponding parametrisation
\begin{equation*} \vect g(x_1,x_2):=(x_1,x_2,h(x_1,x_2)),\qquad x\in D \end{equation*}
of \(S\text{.}\) We first compute the Jacobian matrix of \(\vect g\text{:}\)
\begin{equation*} J_{\vect g}(\vect x)= \begin{bmatrix} 1 & 0 \\ 0&1\\ \dfrac{\partial h}{\partial x_1}(\vect x) & \dfrac{\partial h}{\partial x_2}(\vect x) \end{bmatrix}\text{.} \end{equation*}
Then
\begin{align*} \bigl(J_{\vect g}(\vect x)\bigr)^TJ_{\vect g}(\vect x) &= \begin{bmatrix} 1 & 0 & \dfrac{\partial h}{\partial x_1}(\vect x) \\ 0 & 1 & \dfrac{\partial h}{\partial x_2}(\vect x) \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0&1\\ \dfrac{\partial h}{\partial x_1}(\vect x) & \dfrac{\partial h}{\partial x_2}(\vect x) \end{bmatrix}\\ &= \begin{bmatrix} 1+\Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr)^2 & \Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr) \Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr) \\ \Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr) \Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr) & 1+\Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr)^2 \end{bmatrix} \end{align*}
Now we compute the determinant of the above matrix:
\begin{align*} \det& \begin{bmatrix} 1+\Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr)^2 & \Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr) \Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr) \\ \Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr) \Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr) & 1+\Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr)^2 \end{bmatrix}\\ &=\Bigl(1+\Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr)^2\Bigr) \Bigl(1+\Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr)^2\Bigr) -\Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr)^2 \Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr)^2\\ &=1+\Bigl(\dfrac{\partial h}{\partial x_1}(\vect x)\Bigr)^2 +\Bigl(\dfrac{\partial h}{\partial x_2}(\vect x)\Bigr)^2\\ &=1+\|\grad h(\vect x)\|^2 \end{align*}
Hence we have the following proposition.
Let us make use of the formula in an example.

Example 9.28. Surface area of a graph.

Calculate the surface area of the graph of \(f(x,y)=xy\) over the disc \(D\) given by \(x^2+y^2\leq 1\text{.}\)
Answer.
The surface area is \(\dfrac{2\pi}{3}\left(2\sqrt{2}-1\right)\)
Solution.
We have \(\grad f(x,y)=(y,x)\text{,}\) and so by the above proposition the Jacobian is \(1+x^2+y^2\text{.}\) Hence the surface area is
\begin{equation*} \iint_D\sqrt{1+x^2+y^2}\,dx\,dy\text{.} \end{equation*}
Changing to polar coordinates, using (5.8), we get
\begin{align*} \iint_D\sqrt{1+x^2+y^2}\,dx\,dy &=\int_0^{2\pi}\int_0^1\sqrt{1+r^2}\;r\,dr\,d\varphi\\ &=2\pi\Bigl[\frac{1}{2}\frac{2}{3}(1+r^2)^{3/2}\Bigr]_0^1\\ &=\frac{2\pi}{3}\bigl(2\sqrt{2}-1\bigr)\text{.} \end{align*}
Hence the surface area is \(\frac{2\pi}{3}(2\sqrt{2}-1)\) square units.
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