Surface integrals
Preliminary: Multiple integrals
You may not have seen integrals before which look like this: \[ \int_a^b dx \int_c^d dy\, f(x,y). \]
Many people write these integrals as \(\int_a^b \int_c^d f(x,y)\, dx\, dy,\) or \(\int_a^b \int_c^d f(x,y)\, dy\, dx\). You may think that the elements \(dx\) and \(dy\) have to go on the right, but they don’t. These notations don’t make it clear which limits go with which integration, so I always write the elements on the left with the integral sign.
To calculate these multiple integrals, you just have to do two normal integrals. Actually the order you do them in shouldn’t matter, it’s like a multiplication, but sometimes the limits of one depend on the other.
Example
Calculate \[ \int_0^1 dx \int_0^x dy (x^2y + 2y) \]
The limits of the \(y\) integral depend on \(x\), so we need to do the \(y\) integration first. We rewrite it as \[ \int_0^1 dx \left(\int_0^x (x^2y + 2y) dy \right) \] The inner integral here is just a normal integration with respect to \(y\). When we are integrating with respect to \(y\), we just treat \(x\) as a constant. So we have \[ \int_0^1 dx \left(\int_0^x (x^2y + 2y) dy \right) = \int_0^1 dx \left[ \frac{1}{2} x^2 y^2 + y^2\right]_{y=0}^{y=x} = \int_0^1 dx \left[ \left(\frac{1}{2} x^4 + x^2\right) - \left(\frac{1}{2} x^2\cdot 0 + 0\right)\right]. \] Now we just have one integration in \(x\), which we know how to do \[ \int_0^1 \left(\frac{1}{2} x^4 + x^2\right) dx = \left[\frac{1}{10}x^5 + \frac{1}{3}x^3\right]_0^1 = \frac{13}{30}. \]
Surface integrals of scalar fields
\[ \int_S f({\mathbf{r}}) dS \] There are two different \(S\) symbols here, and they mean different things. \(S\) is a surface, a particular object in space, but \(dS\) is the general surface area element. What this means is, add up lots of infinitely small bits of the surface, weighted by the function \(f\).
The steps to do these integrals are exactly the same as the four steps we saw for line integrals, with two key differences: firstly, we will have to do a multiple integral, and secondly, we need to know that the surface area element is \[dS = \left|{\frac{\partial {\mathbf{r}}}{\partial u}}\times{\frac{\partial {\mathbf{r}}}{\partial v}}\right|du\,dv\] when the surface is parameterised by \(u\) and \(v\).
Why is it this? It’s complicated, but we’ll go through a derivation in class if we have time.
Example
Calculate surface integral \[ \int_S (x + 4xz)\, dS \] where \(S\) is the surface \(\{(x,y,z)\in\mathbb{R}^3\,|\,z=x^2,\,0\leq x\leq 1, 0\leq y\leq 1\}\).
Let’s solve the surface integral step by step.
Step 1: Parameterise the surface
The surface \(S\) is given by \(z = x^2\) for \(0 \leq x \leq 1\), \(0 \leq y \leq 1\). The easiest way to parameterise this is to let \(u=x\) and \(v=y\). Then \[ {\mathbf{r}}(u, u) = [u, v, u^2], \qquad 0 \leq u \leq 1,\ 0 \leq v \leq 1. \]
Step 2: Calculate \(dS\)
Compute the partial derivatives: \[ {\frac{\partial {\mathbf{r}}}{\partial u}} = [1, 0, 2u], \qquad {\frac{\partial {\mathbf{r}}}{\partial v}} = [0, 1, 0] \] Take the cross product: \[ {\frac{\partial {\mathbf{r}}}{\partial u}} \times {\frac{\partial {\mathbf{r}}}{\partial v}} = [-2u, 0, 1] \] The magnitude is: \[ \left|[-2u, 0, 1]\right| = \sqrt{(-2u)^2 + 0^2 + 1^2} = \sqrt{4u^2 + 1} \] So, \[ dS = \sqrt{4u^2 + 1}\ du\ dv \]
Step 3: Rewrite the integration
The function is \(f(x, y, z) = x + 4xz = u + 4u^3\).
The surface integral becomes: \[ \int_S (x^2 + z^2)\, dS = \int_0^1 du \int_0^1 dv\ (u + 4u^3)\ \sqrt{4u^2 + 1} \]
Step 4: Compute the answer
No \(v\) appears in the integrand so this integral just simplifies to 1. We are left with \[ \int_S (x^2 + z^2)\, dS = \int_0^1 (u + 4u^3)\ \sqrt{4u^2 + 1} \, du = \int_0^1 u(1+4u^2)^{3/2} \, du. \]
By inspection, the integral is \[ \int_0^1 u(1+4u^2)^{3/2} \, du = \left[\frac{1}{20}(1+4u^2)^{5/2}\right]_0^1 = \frac{5^{5/2}-1}{20}. \]
Special case: surface area
As promised, we can now calculate the area of surfaces. This is simply the empty integral \[ \int_S dS = \int du \int dv \left|{\frac{\partial {\mathbf{r}}}{\partial u}}\times{\frac{\partial {\mathbf{r}}}{\partial v}}\right| \]
Example: surface area of a sphere
Calculate the surface area of a sphere of radius \(R\).
We’re not told the exact surface, but let’s assume the sphere is centred at the origin. Then we can parameterise by taking some inspiration from spherical polar co-ordinates: \[[x,y,z] = [R\cos u\sin v,\,R\sin u\sin v,\,R\cos v], \qquad 0\leq u < 2\pi,\, 0\leq v\leq \pi.\] This isn’t exactly polar co-ordinates because \(R\) is a constant not a variable, and \(u\) and \(v\) just parameterise the surface not the whole of 3D space.
Now we can just plug this into the formula above: \[ \int_S dS = \int_0^{2\pi} du \int_0^{\pi} dv \left|[-R\sin u \sin v,\,R\cos u\sin v,\,0]\times[R\cos u\cos v,\,R\sin u\cos v,-R\sin v]\right| \] Pro-tip: take out as many factors from the cross product as you can before calculating it: \[ \begin{aligned} \int_S dS &= \int_0^{2\pi} du \int_0^{\pi} dv \left|R^2 \sin v\right|\cdot\left|[-\sin u,\,\cos u,\,0]\times[\cos u\cos v,\,\sin u\cos v,-\sin v]\right| \\ &= \int_0^{2\pi} du \int_0^{\pi} dv \left|R^2 \sin v\right|\cdot\left|[-\cos u \sin v,\,-\sin u\sin v,\,-\sin^2 u \cos v -\cos^2 u \cos v]\right| \\ &= \int_0^{2\pi} du \int_0^{\pi} dv \left|R^2 \sin v\right|\cdot\left|[\cos u \sin v,\,\sin u\sin v,\,\cos v]\right| \\ &= \int_0^{2\pi} du \int_0^{\pi} dv \left|R^2 \sin v\right| \end{aligned} \] where we used the fact that \(\left|[\cos u \sin v,\,\sin u\sin v,\,\cos v]\right| = 1\) (this is good to remember).
We don’t need to worry about the absolute value of \(\sin v\) because \(v\in[0,\pi]\) where sine is non-negative. There is no \(u\) in this expression so we can just mupltiply by \(2\pi\) to give \[ \begin{aligned} \int_S dS = 2\pi R^2 \int_0^{\pi} dv \sin v = 2\pi R^2 [-\cos u]_0^{\pi} = 4\pi R^2. \end{aligned} \] But you knew that already.
Flux integrals: surface integrals of vector fields
This integral has a special name and comes up a lot in physics. It is the rate at which a vector field ‘goes through’ a surface.
Like line integrals of scalar fields, the direction matters. In this case, it is the direction of the normal (in or out, up or down, …) which affects the sign of the final answer.
There are two equivalent notations \[ \int_S {\mathbf{F}}({\mathbf{r}}) \cdot d{\mathbf{S}} = \int_S {\mathbf{F}}({\mathbf{r}}) \cdot \hat{\mathbf{n}} dS \] and the second one gives a hint how to calculate these things: we define the vector surface element \(d{\mathbf{S}}\) in terms of the scalar element \(dS\) as \[ d{\mathbf{S}} = \hat{\mathbf{n}} dS \] where \(\hat{\mathbf{n}}\) is the unit normal to the surface.
For a parameterisation \((u,v)\), recall that the general normal is \[ {\mathbf{n}}(u,v) = \frac{\partial{\mathbf{r}}}{\partial u} \times \frac{\partial{\mathbf{r}}}{\partial v} \] so the unit normal is \[ \hat{\mathbf{n}}(u,v) = \frac{\frac{\partial{\mathbf{r}}}{\partial u} \times \frac{\partial{\mathbf{r}}}{\partial v}}{\left|\frac{\partial{\mathbf{r}}}{\partial u} \times \frac{\partial{\mathbf{r}}}{\partial v}\right|} \] and substituting \(dS\), we can simplify to \[ d{\mathbf{S}} = \hat{\mathbf{n}} dS = \frac{\partial{\mathbf{r}}}{\partial u} \times \frac{\partial{\mathbf{r}}}{\partial v} du\,dv. \]
Learn this last formula. The rest is really just derivation.
Example: Flux through a surface
Calculate the total flux of the vector field \({\mathbf{F}}(x,y,z) = [z^2,z,0]\) through the hemisphere \(x^2+y^2+z^1=1\), \(x\geq 0\), with an outward-facing normal.
Step 1: Parameterise the surface
The hemisphere means spherical co-ordinates seem natural here, as we can take \(r=1\) and use \[ [x,y,z] = [\cos u \sin v,\,\sin u \sin v,\,\cos v]. \] We need to be careful here with the constraint that \(x\geq 0\). With the usual limit for \(v=\theta\), \(0\leq v \leq \pi\), we know that \(\sin v\geq 0\). So for \(x\geq 0\) we just need \(\cos u \geq 0\), so let’s take \(-\pi/2 \leq u \leq \pi/2\).
Step 2: Calculate \(d{\mathbf{S}}\)
Hopefully this is starting to become familiar: \[ \begin{aligned} d{\mathbf{S}} &= {\frac{\partial {\mathbf{r}}}{\partial u}} \times {\frac{\partial {\mathbf{r}}}{\partial v}} \\ &= [-\sin u \sin v,\,\cos u\sin v,\,0]\times[\cos u\cos v,\,\sin u\cos v,\,-\sin v]\\ &= \sin v\,[-\sin u,\,\cos u,\,0]\times[\cos u\cos v,\,\sin u\cos v,\,-\sin v]\\ &= \sin v\,[-\cos u \sin v,\,-\sin u\sin v,\,-\sin^2 u \cos v -\cos^2 u \cos v]\\ &= \sin v\,[-\cos u \sin v,\,-\sin u\sin v,\,-\cos v] \end{aligned} \] We need to check this is the outward normal, as specified by the question. Actually this is \(\sin v\) times \(-{\mathbf{r}}\), so it’s inward pointing!
Not to worry - we can just flip the signs: \[d{\mathbf{S}} = \sin v\,[\cos u \sin v,\,\sin u\sin v,\,\cos v]\].
Step 3: Rewrite the integration
In our local co-ordinates, \({\mathbf{F}}(u,v) = [z^2,z,0] = \cos v\,[\cos v,\,1,\,0]\) (remembering to take out the common factor), so the integral becomes \[ \int_S {\mathbf{F}}\cdot d{\mathbf{S}} = \int_0^\pi dv \int_{-\pi/2}^{\pi/2} du \cos v[\cos v,\,1,\,0] \cdot \sin v\,[\cos u \sin v,\,\sin u\sin v,\,\cos v] = \int_0^\pi dv \int_{-\pi/2}^{\pi/2} du \sin v \cos v \left(\cos u \cos v \sin v + \sin u \sin v\right) \]
Step 4: Compute the answer
Here it’s the sum of two different things, so it’s easiest to split it up: \[\begin{aligned}\int_S {\mathbf{F}}\cdot d{\mathbf{S}} &= \int_0^\pi dv \int_{-\pi/2}^{\pi/2} du \left(\cos u \cos^2 v \sin^2 v + \sin u \cos v \sin^2 v\right)\\ &=\int_0^\pi dv \int_{-\pi/2}^{\pi/2} du\cos u \cos^2 v \sin^2 v + \int_0^\pi dv \int_{-\pi/2}^{\pi/2} du\sin u \cos v \sin^2 v.\end{aligned}\]
For both parts, the \(u\) and \(v\) integrals just separate into multiplication, giving \[ \begin{aligned} \int_0^\pi dv \int_{-\pi/2}^{\pi/2} du\cos u \cos^2 v \sin^2 v &= \left(\int_{-\pi/2}^{\pi/2} \cos u\,du\right)\left(\int_0^\pi \cos^2 v \sin^2 v\,dv\right) \\&= 2\left(\int_0^\pi \frac{1}{2}\sin^2 2v\,dv\right) \\&= 2\times \frac{1}{4} \\&= \frac{1}{2}. \end{aligned} \] For the second part, note that \(\int_{-\pi/2}^{\pi/2} \sin u\, du = 0\) (since sine is an odd function over a symmetric interval), so \[ \int_0^\pi dv \int_{-\pi/2}^{\pi/2} du\sin u \cos v \sin^2 v = 0. \]
Therefore, the total flux is \(\frac{1}{2}\).
Other surface integrals
Occasionally you might see things like \[ \int_S {\mathbf{F}}({\mathbf{r}}) \times d{\mathbf{S}}. \] This means exactly what you think: take the cross product with the surface normal instead of the dot product, so that the final answer is a vector instead of a scalar.