Stokes' Theorem (Fully Explained w/ Step-by-Step Examples!)

Were you ever relegated to the kid’s table at family functions?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching stoke's theorem

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

You were in that in-between stage in your life where you were no longer a child but also not quite an adult…

…so you got stuck at the kid’s table and were left fancifully wondering about all the things you were missing out on at the far more exciting adult table.

Well, thanks to Stoke’s theorem, we are officially unstuck and free to pursue those more interesting things as the adult table… ahem…more interesting surfaces.

Say what?!

Green’s Theorem

Recall that Green’s theorem relates a double integral over a plane region $D$ to a line integral around its plane boundary curve.

Given a line integral of a vector field $\vec{F} = ⟨ P, Q ⟩$ over a planar closed curve $C$ with positive orientation, the line integral is equal to the double integral of $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ over the planar region bounded by $C$ . Then,

$\oint_{C} P d x + Q d y = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d A$

Greens Theorem Geometric Representation

So, Green’s theorem measures circulation along the boundary curve $C$ , but Stoke’s theorem helps us measure circulation for more interesting surfaces!

Stoke’s Theorem

In fact, Stoke’s theorem relates a surface integral over a surface $S$ to a line integral around the boundary curve of $S$ .

Let $S$ be a smooth oriented surface in $R^{3}$ with a smooth boundary curve $C$ whose orientation is consistent with that of $S$ . Assume that $\vec{F} = ⟨ P, Q, R ⟩$ is a vector field whose components have continuous first-order partial derivatives on $S$ . Then,
$\oint_{C} \vec{F} \cdot d \vec{r} = \iint_{S} (\nabla \times \vec{F}) \cdot \vec{n} d S = \iint_{S} curl \vec{F} \cdot d \vec{S}$

Stokes Theorem Geometric Interpretation

And this is why Stoke’s theorem is more generalized in that our surface is no longer restricted to the xy-plane and can be regarded as a higher-dimensional version of Green’s theorem with the curl.

Curl (An Aside)

As a matter of fact, Stokes’ theorem provides insight into a physical interpretation of the curl.

In a vector field, the rotation of the vector field is at a maximum when the curl of the vector field and the normal vector have the same direction.

In other words, while the tendency to rotate will vary from point to point on the surface, Stoke’s theorem says that the collective measure of this rotational tendency taken over the entire surface is equal to the tendency of a fluid to circulate around the boundary curve.

Example

Let’s put all of this new information, along with our previously learned skills, to work with an example.

Suppose $\vec{F} = ⟨ x^{2}, 2 x y + x, z ⟩$ .

Let $C$ be the circle $x^{2} + y^{2} = 1$ in the plane $z = 0$ oriented counterclockwise, and let $S$ be the disk $x^{2} + y^{2} \leq 1$ oriented with the normal vector $\vec{k}$ .

Verify Stoke’s theorem by evaluating the integral of $\nabla \times \vec{F}$ over $S$ .

Okay, so we are being asked to find $\iint_{S} (\nabla \times \vec{F}) \cdot \vec{n} d S$ given the oriented surface $S$ .

So, the first thing we need to do is compute $\nabla \times \vec{F}$ .

$\nabla \times \vec{F} = | \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{array} |$

$= | \begin{array}{ccc} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^{2} & 2 x y + x & z \end{array} |$

$= ⟨ 0, 0, 2 y + 1 ⟩$

Next, we need to find our unit normal vector $\vec{n}$ , which we were told is our $\vec{k}$ vector, $\vec{k} = ⟨ 0, 01 ⟩$ .

Thus…

$(\nabla \times \vec{F}) \cdot \vec{n} = (\nabla \times \vec{F}) \cdot \vec{k} = ⟨ 0, 0, 2 y + 1 ⟩ \cdot ⟨ 0, 0, 1 ⟩ = 2 y + 1$

But because $S$ is the disk $x^{2} + y^{2} \leq 1$ we want to utilize polar coordinates.

$\iint_{S} (2 y + 1) d S = \iint_{S} (2 r \sin θ + 1) r d r d θ$

$where 0 \leq r \leq 1 and 0 \leq θ \leq 2 π$

Therefore,

$\int_{0}^{2 π} \int_{0}^{1} (2 r \sin θ + 1) r d r d θ$

$= \int_{0}^{2 π} (\int_{0}^{1} (2 r^{2} \sin θ + r) d r) d θ$

$= \int_{0}^{2 π} (\frac{2}{3} \sin θ + \frac{1}{2}) d θ$

$= π$

And that’s it.

Don’t worry. We’re going to spend our entire lesson making sure you understand the importance of Stoke’s theorem and learning how to evaluate this incredible double integral.

So, together we will begin by comparing Green’s theorem with Stoke’s theorem and learn how to verify that Stoke’s theorem equals a line integral. Using Stoke’s theorem, we will learn how to find the total net flow in or out of a closed surface for liquids, electric charge, or temperature.

It’s going to be great, so let’s get to it!

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

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