7. Surface Integrals

7.1. Vector Flux

We can look at surface integrals next, these are the natural extension to line integrals, now we require a two variable parameterisation, here we will use $(u,v)$ as variables which can parameterise the surface $S$ we are interested in.

Lets start by thinking about some surface, as depicted in Fig. 7.1, which we will break up into infinitesimal area elements $\mathrm{d}S$. Just like we can write infinitesimal area elements $\mathrm{d}A=\mathrm{d}x\mathrm{d}y$ in Cartesian coordinates (and in polar coordinates $\mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta$), we can break up this surface using the parameterisation $(u,v)$.

Fig. 7.1 Some general surface $S$ broken up into infinitesimal area elements $\mathrm{d}S$.

We can think about the effect of some vector field crossing this surface, as depicted in Fig. 7.2.

Fig. 7.2 A vector field $\mathbf{F}$ crossing a surface $S$.

To find the total contribution of the vector field actually going through a surface area element, we need to think about the components of the vector field.

We can decompose $\mathbf{F}={\mathbf{F}}_{\mathrm{\perp }}+{\mathbf{F}}_{\parallel }$ components, as depicted in Fig. 7.3

Fig. 7.3 Decomposition of vector field $\mathbf{F}$ into perpendicular components with respect to a surface area element $\mathrm{d}S$.

In order to find the components that cross through the surface, ${\mathbf{F}}_{\mathrm{\perp }}$, we need a surface normal vector. We know that in order to find a surface normal, we can do the cross product of two vectors that live on the surface.

We also that the effect of differentiating vector $\mathbf{r}$ with respect to some parameter produces a vector which points in the direction where $\mathbf{r}$ is maximised with respect to that parameter and crucially lives on the surface $S(\mathbf{r})$, so we use the derivatives with respect to the two parameters $(u,v)$:

$$\mathbf{n}=\left(\frac{\partial \mathbf{r}}{\partial u}\right)\times \left(\frac{\partial \mathbf{r}}{\partial v}\right)$$

This means that combination of the vector field with the area element $|\mathrm{d}\mathbf{S}|=\mathrm{d}u\mathrm{d}v$ will be:

$${\iint }_S\mathbf{F}\cdot \mathrm{d}\mathbf{S}={\iint }_S\mathbf{F}\cdot (\frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v})\mathrm{d}u\mathrm{d}v$$

We call this the flux of the vector field $\mathbf{F}$ crossing the surface $S$.

Definition

We can define:

$${\iint }_S\mathbf{F}\cdot \mathrm{d}\mathbf{S}={\iint }_S\mathbf{F}\cdot (\frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v})\mathrm{d}u\mathrm{d}v$$

as a vector surface integral along some surface $S$, which is parameterised by variables $(u,v)$.

We can also define a scalar surface integral:

$${\iint }_{S}f|\frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v}|\mathrm{d}u\mathrm{d}v$$

along some surface $S$, which is parameterised by variables $(u,v)$.

7.1.1. Calculating Surface Integrals

Similar to line integrals, we must parameterise the surface $S$ in order for this integral to be possible, so we will pick $S=S(\mathbf{r})$ such that $\mathbf{r}=\mathbf{r}(u,v)$ with $a\le u\le b,c\le v\le d$. Hence to actually calculate these we will have:

$$\begin{array}{rl}{I}_f& ={\int }_a^b\mathrm{d}u{\int }_c^d\mathrm{d}vf(\mathbf{r}(u,v)|\frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v}|\\ I_{\mathbf{G}}& ={\int }_a^b\mathrm{d}u{\int }_c^d\mathrm{d}v\mathbf{F}(\mathbf{r}(u,v))\cdot (\frac{\partial \mathbf{r}}{\partial u}\times \frac{\partial \mathbf{r}}{\partial v})\end{array}$$

7.2. Surfaces of the form $z=f(x,y)$

A different sort of example is where we have a 2D scalar field $f(x,y)$ and we plot this as a surface $z=f(x,y)$.
Here the parameterisation of the surface is given for free in terms of $x,y$.

The surface has a coordinate vector of the form:

$$\begin{array}{r}\mathbf{r}(x,y)=\left(\begin{array}{c}x\\ y\\ f(x,y)\end{array}\right)\end{array}$$

and the infinitesimal area element is given by:

$$\begin{array}{rl}\mathrm{d}\mathbf{S}& =(\frac{\partial \mathbf{r}}{\partial x}\times \frac{\partial \mathbf{r}}{\partial y})\mathrm{d}x\mathrm{d}y=\left(\begin{array}{c}1\\ 0\\ \partial z/\partial x\end{array}\right)\times \left(\begin{array}{c}0\\ 1\\ \partial z/\partial y\end{array}\right)\mathrm{d}x\mathrm{d}y\\ & =\left|\begin{array}{ccc}\hat{\mathbf{x}}& \hat{\mathbf{y}}& \hat{\mathbf{z}}\\ 1& 0& f_x\\ 0& 1& f_y\end{array}\right|\mathrm{d}x\mathrm{d}y=\left(\begin{array}{c}-f_x\\ -f_y\\ 1\end{array}\right)\mathrm{d}x\mathrm{d}y\end{array}$$

If we have a scalar surface integral, then area element will involve the magnitude of this vector:

$$\mathrm{d}S=\left(\sqrt{{f_x}^2+{f_y}^2+1}\right)\mathrm{d}x\mathrm{d}y$$

Worked example

1. Lets find the surface area of a sphere with radius $R$. The most natural choice of parameters here would be the angles $(\theta ,\varphi )$:

$$\begin{array}{r}\mathbf{r}(\theta ,\varphi )=\left(\begin{array}{c}R\sin (\theta )\cos (\varphi )\\ R\sin (\theta )\sin (\varphi )\\ R\cos (\theta )\end{array}\right)\end{array}$$

hence we can calculate $\mathrm{d}\mathbf{S}$ here:

$$\begin{array}{rl}\mathrm{d}\mathbf{S}& =(\frac{\partial \mathbf{r}}{\partial \theta }\times \frac{\partial \mathbf{r}}{\partial \varphi })\mathrm{d}\theta \mathrm{d}\varphi \\ & =R^2\left(\begin{array}{c}\cos (\theta )\cos (\varphi )\\ \cos (\theta )\sin (\varphi )\\ -\sin (\theta )\end{array}\right)\times \left(\begin{array}{c}-\sin (\theta )\sin (\varphi )\\ \sin (\theta )\cos (\varphi )\\ 0\end{array}\right)\mathrm{d}\theta \mathrm{d}\varphi \\ & =R^2\left(\begin{array}{c}{\sin }^2(\theta )\cos (\varphi )\\ {\sin }^2(\theta )\sin (\varphi )\\ \sin (\theta )\cos (\theta )\end{array}\right)\mathrm{d}\theta \mathrm{d}\varphi \\ & =R^2\left(\begin{array}{c}\sin (\theta )\cos (\varphi )\\ \sin (\theta )\sin (\varphi )\\ \cos (\theta )\end{array}\right)\sin (\theta )\mathrm{d}\theta \mathrm{d}\varphi \end{array}$$

We can then recall that the radial vector for spherical polar coordinates has the form:

$$\begin{array}{r}\hat{\mathbf{r}}=\left(\begin{array}{c}\sin (\theta )\cos (\varphi )\\ \sin (\theta )\sin (\varphi )\\ \cos (\theta )\end{array}\right)\end{array}$$

and therefore :

$$\mathrm{d}\mathbf{S}=R^2\sin (\theta )\mathrm{d}\theta \mathrm{d}\varphi \hat{\mathbf{r}}$$

Thus the surface area integral takes the form:

$$A={\iint }_A\mathrm{d}S=R^2{\int }_0^{2\pi }\mathrm{d}\varphi {\int }_0^{\pi }\sin (\theta )\mathrm{d}\theta =2\pi R^2[-\cos (\theta ){]}_0^{\pi }=4\pi R^2$$

Further worked examples:

1. Find the surface area for $0\le x\le 3$, $0\le y\le 3$ for a sphere:

$$x^2+y^2+z^2=9$$

We can try to solve this using Cartesian coordinates instead of polar here (spherical polar will also work though!) Lets try to apply $z=f(x,y)=\sqrt{9-x^2-y^2}$,

$$\begin{array}{rl}{\iint }_A\mathrm{d}S& ={\int }_{-R}^R\mathrm{d}x{\int }_{-R}^R\mathrm{d}y\sqrt{{\left(\frac{\partial z}{\partial x}\right)}^2+{\left(\frac{\partial z}{\partial y}\right)}^2+1}\\ & ={\int }_0^3\mathrm{d}x{\int }_0^3\mathrm{d}y\sqrt{\frac{x}{\sqrt{9-x^2-y^2}}+\frac{y}{\sqrt{9-x^2-y^2}}+1}\\ & ={\int }_0^3\mathrm{d}x{\int }_0^3\mathrm{d}y\frac{3}{\sqrt{9-x^2-y^2}}\end{array}$$

If we use a change of variable, $x=\sqrt{9-y^2}\sin (\phi )\Rightarrow \mathrm{d}x=\sqrt{9-y^2}\cos (\phi )\mathrm{d}\phi$ and therefore we have:

$${\iint }_A\mathrm{d}S=3{\int }_0^{\pi /2}\mathrm{d}\phi {\int }_0^3\mathrm{d}y=\frac{9\pi }{2}$$

which corresponds to $\frac{1}{8}$ of the full sphere’s surface area $36\pi$, as expected.

2. Find the surface area of a cone with height $h$ and base radius $R$.

Here we can use the parameterisation $z,\theta$ to describe any point on this surface. This means we describe any point in cylindrical polar coordinates:

$$\begin{array}{r}\mathbf{r}=\left(\begin{array}{c}\frac{R}{h}z\cos (\theta )\\ \frac{R}{h}z\sin (\theta )\\ z\end{array}\right)\end{array}$$

with ranges $0\le z\le h$ and $0\le \theta \le 2\pi$. Then to find the vectors in terms of this parameterisation:

$$\begin{array}{r}\frac{\partial \mathbf{r}}{\partial z}=\left(\begin{array}{c}\frac{R}{h}\cos (\theta )\\ \frac{R}{h}\sin (\theta )\\ 1\end{array}\right)\\ \frac{\partial \mathbf{r}}{\partial \theta }=\left(\begin{array}{c}-\frac{R}{h}z\sin (\theta )\\ \frac{R}{h}z\cos (\theta )\\ 0\end{array}\right)\end{array}$$

which means the integrand takes the form:

$$\begin{array}{rl}\frac{\partial \mathbf{r}}{\partial z}\times \frac{\partial \mathbf{r}}{\partial \theta }& =\left|\begin{array}{ccc}\hat{\mathbf{x}}& \hat{\mathbf{y}}& \hat{\mathbf{z}}\\ \frac{R}{h}\cos (\theta )& \frac{R}{h}\sin (\theta )& 1\\ -\frac{R}{h}z\sin (\theta )& \frac{R}{h}z\cos (\theta )& 0\end{array}\right|\\ & =\left(\begin{array}{c}-\frac{R}{h}z\cos (\theta )\\ -\frac{R}{h}z\sin (\theta )\\ \frac{R^2}{h^2}z\end{array}\right)\\ \Rightarrow |\frac{\partial \mathbf{r}}{\partial z}\times \frac{\partial \mathbf{r}}{\partial \theta }|& =\sqrt{\frac{R^2}{h^2}{z}^2+\frac{R^4}{h^4}{z}^2}=\frac{R}{h}z\sqrt{1+\frac{R^2}{h^2}}\end{array}$$

and so integrating we find:

$$\begin{array}{rl}{\iint }_S\mathrm{d}S& ={\int }_0^{2\pi }\mathrm{d}\theta {\int }_0^h\frac{R}{h}z\sqrt{1+\frac{R^2}{h^2}}\mathrm{d}z\\ & =2\pi \frac{R}{h}\sqrt{1+\frac{R^2}{h^2}}[\frac{1}{2}{z}^2{]}_0^h=2\pi \frac{R}{h}\frac{h^2}{2}\sqrt{1+\frac{R^2}{h^2}}=\pi R\sqrt{h^2+R^2}\end{array}$$