Part 20 - Divergence and Stokes' Theorem

Introduction

In this part we’ll cover surface integrals over vector fields as well as two very important theorems.

Surface integrals over vector fields

Let $S$ be a parametric surface, parameterized by, $\vec{r}: D \rarr \mathbb{R}^2$ and $F: \mathbb{R}^3 \rarr \mathbb{R}^3$ , be a vector field. Then:

\iint_S \vec{F} \cdot dS = \iint_D \vec{F}(\vec{r}(u, v)) \cdot \vec{r_u} \times \vec{r_v}\ du\ dv

We call this the flux of $\vec{F}$ through $S$ .

Example

Given, $\vec{F} = \langle z, y, x \rangle$ , find the flux of this vector field across the unit sphere.

So, we have the surface:

S = \{(x, y, z) | x^2 + y^2 + z^2 \leq 1 \}

As always, good idea to change into polar coordinates, so a good parameterization is:

\vec{r}(\phi, \theta) = \langle sin(\phi)\ cos(\theta), sin(\phi)\ sin(\theta), cos(\phi) \rangle \ | \ 0 \leq \phi \leq \pi, 0 \leq \theta \leq 2\pi

NB: This computation is just tedious (and we’ll see how we can reduce this shortly)

\vec{r_u} \times \vec{r_v} = \langle sin^2(\phi)\ cos(\theta), sin^2(\phi)\ sin(\theta), sin(\phi)\ cos(\phi) \rangle

Now, one can solve this integral by taking the dot product and applying our definition:

\int_0^{2\pi} \int_0^{\pi} \langle cos(\phi), sin(\phi)\ sin(\theta), sin(\phi)\ cos(\theta) \rangle \cdot \langle sin^2(\phi)\ cos(\theta), sin^2(\phi)\ sin(\theta), sin(\phi)\ cos(\phi) \rangle

Let’s see now how we can do the same example, with fewer computations, first some theory.

Orientation

For surfaces, we say that positive orientation is “pointing outwards”. For hollow surfaces this is always the case. If our parameterization is given by a single function, then we always have the correct choice.

However, imagine half the unit sphere with a filled base. The base with parameterization: $D = \{(r, \theta) | 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \}$ . Is going to be a negative integral.

The Divergence Theorem

For a vector field, $\vec{F}: \mathbb{R}^3 \rarr \mathbb{R}^3$ , the divergence is:

div\ \vec{F} = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z}

Let $S$ be a hollow surface and let $E$ be the 3-dimensional region it surrounds ( $\partial E = S$ ).

Then:

\iint_S \vec{F} \cdot dS = \iiint_E div\ \vec{F}\ dV

Equivalently:

\iint_{\partial E} \vec{F} \cdot dS = \iiint_E div\ \vec{F}\ dV

Now let’s take the sample example.

Example

Find the flux of $\vec{F} = \langle z, y, x$ on the unit sphere.

div\ \vec{F} = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} + \dfrac{\partial R}{\partial z} = 0 + 0 + 1 = 1

\iint_S \vec{F} \cdot dS = \iiint_E 1 dV

We could now just transform to polar coordinates, but, this is the volume of the unit sphere, which is just:

\boxed{\dfrac{4\pi}{3}}

Stokes’ Theorem

So far we have seen Green’s Theorem as well as The Divergence theorem. But what is the reason that Green’s Theorem only works in $\mathbb{R}^3$ ?

Well, Stokes Theorem is essentially Green’s Theorem but in $\mathbb{R}^3$ .

For $\vec{F}: \mathbb{R}^3 \rarr \mathbb{R}^3$ , we define its curl as:

NB: it is also called rot and sometimes denoted as $\nabla \times \vec{F}$ .

curl\ \vec{F} = \langle \dfrac{\partial R}{\partial y} - \dfrac{\partial Q}{\partial z}, \dfrac{\partial P}{\partial z} - \dfrac{\partial R}{\partial x}, \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \rangle

If $S \subset \mathbb{R}^3$ is a surface, surrounded by a simple, closed curve in $\mathbb{R}^3$ . Then:

\int_C \vec{F} \cdot dr = \iint_S curl\ F\ dS

Example

Compute $\int_C F \cdot dr$ for $\vec{F} = \langle -y^2, x, z^2 \rangle$ and $C$ being the intersection of $y + z = 2$ with $x^2 + y^1 = 1$ .

Let’s find the curl:

curl\ \vec{F} = \langle 0, 0, 1 + 2y \rangle

We need to find the surface, the intersection of $y + z = 2$ and $x^2 + y^2 = 1$ will just be the unit disk but at a slight tilt.

\iint_S curl\ \vec{F}\ dS

Is just the flux through this region, let’s express this in polar coordinates

D = \{(r, \theta) | 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi \}

Meaning:

\iint_S curl\ \vec{F}\ dS = \iint_D curl\ \vec{F} \cdot \vec{r_u} \times \vec{r_v}\ dA

Let’s find a parameterization of the surface:

\vec{r}(u, v) = \langle u, v, 2 - v \rangle

\vec{r_u}(u, v) = \langle 1, 0, 0 \rangle

\vec{r_v}(u, v) = \langle 0, 1, -1 \rangle

\vec{r_u} \times \vec{r_v} = \langle 0, 1, 1 \rangle

\iint_D curl\ \vec{F} \cdot \vec{r_u} \times \vec{r_v}\ dA = \iint_D \langle 0, 0, 1 + 2y \rangle \cdot \langle 0, 1, 1 \rangle\ dA

\iint_D 1 + 2y\ dA

\int_0^{2\pi} \int_0^1 (1 + 2r sin(\theta))r\ dr\ d\theta

\int_0^{2\pi} \int_0^1 r + 2r^2 sin(\theta)\ dr\ d\theta

\int_0^1 \int_0^{2\pi} r + 2r^2 sin(\theta)\ d\theta\ dr

$sin(\theta)$ has symmetry around $2\pi$ , therefore the integral of that is $0$ .

\int_0^1 2\pi r\ dr = \boxed{\pi}

Compute $\iint_S curl\ \vec{F}\ dS$ where $\vec{F} = \langle xz, yz, xy \rangle$ and $S$ is the part of the sphere of radius 2 around $(0, 0, 0)$ that lies inside the cylinder $x^2 + y^2 \leq 1$ and above the $xy$ -plane.

Let’s actually use Stokes theorem, “backwards”. $C$ will be a circle with:

\begin{cases} x^2 + y^2 = 1 \newline x^2 + y^2 + z^2 = 4 \newline \end{cases}

Which means that $z = \sqrt{3}$

We can parameterize this curve as:

\vec{r}(t) = \langle cos(t), sin(t), \sqrt{3} \rangle \ | \ 0 \leq t \leq 2\pi

\vec{r^\prime}(t) = \langle -sin(t), cos(t), 0 \rangle

\iint_S curl\ \vec{F}\ dS = \int_C \vec{F} \cdot dr

\int_0^{2\pi} \langle \sqrt{3} cos(t), \sqrt{3} sin(t), cos(t)\ sin(t) \rangle \cdot \langle -sin(t), cos(t), 0 \rangle

\int_0^{2\pi} -\sqrt{3} cos(t)\ sin(t) + \sqrt{3} sin(t)\ cos(t)

\int_0^{2\pi} 0 = \boxed{0}