Multivariable calculus and vector analysis

Stokes’ theorem examples

Example 1

Let $C$ be the closed curve illustrated below.

For $\mathbf{F}\left(x,y,z\right)=\left(y,z,x\right)$, compute

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& & \hfill \end{array}$$

using Stokes’ Theorem.

Solution: By Stokes’ theorem, we need to compute

$$\begin{array}{lll}\hfill {\iint }_S\mathrm{curl}\mathbf{F}\cdot d\mathbf{S}& & \hfill \end{array}$$

where $S$ is a surface with boundary $C$. We have freedom to choose any surface $S$, as long as we orient it so that $C$ is a positively oriented boundary.

In this case, the simplest choice for $S$ is clear. Let $S$ be the quarter disk in the $yz$-plane.

Given the orientation of the curve $C$, we need to choice the surface normal vector $\mathbf{n}$ to point in which direction? By our right hand rule criterion, the normal vector should point toward the negate negative side of the $x$-axis.

We need to calculate the curl of $\mathbf{F}$. We could use the formula

$$\begin{array}{lll}\hfill \mathrm{curl}\mathbf{F}=\left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z},\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x},\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right),& & \hfill \end{array}$$

but I’m not very good at memorizing that formula. I tend to use the mnemonic that $\mathrm{curl}\mathbf{F}=\nabla \times \mathbf{F}$, where I pretend that $\nabla$ is the vector

$$\begin{array}{lll}\hfill \nabla =\left(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\right).& & \hfill \end{array}$$

Since I remember the cross product using a $3\times 3$ determinant, by mnemonic device for remembering $\mathrm{curl}\mathbf{F}$ is

$$\begin{array}{llll}\hfill \mathrm{curl}\mathbf{F}& =\nabla \times \mathbf{F}& \hfill & \\ \hfill & =\left|\begin{array}{ccc}\hfill \mathbf{i}\hfill & \hfill \mathbf{j}\hfill & \hfill \mathbf{k}\hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill F_1\hfill & \hfill F_2\hfill & \hfill F_3\hfill \end{array}\right|.& \hfill & \end{array}$$

But use whatever method works for you.

For our vector field $\mathbf{F}=\left(y,z,x\right)$, we can calculate the curl as

$$\begin{array}{llll}\hfill \mathrm{curl}\left(\mathbf{F}\right)& =\nabla \times \mathbf{F}=\nabla \times \left(y,z,x\right)& \hfill & \\ \hfill & =\left|\begin{array}{ccc}\hfill \mathbf{i}\hfill & \hfill \mathbf{j}\hfill & \hfill \mathbf{k}\hfill \\ \hfill \frac{\partial }{\partial x}\hfill & \hfill \frac{\partial }{\partial y}\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill y\hfill & \hfill z\hfill & \hfill x\hfill \end{array}\right|& \hfill & \\ \hfill & =\mathbf{i}\left(\frac{\partial }{\partial y}x-\frac{\partial }{\partial z}z\right)-\mathbf{j}\left(\frac{\partial }{\partial x}x-\frac{\partial }{\partial z}y\right)& \hfill & \\ \hfill & +\mathbf{k}\left(\frac{\partial }{\partial x}z-\frac{\partial }{\partial y}y\right)& \hfill & \\ \hfill & =\mathbf{i}\left(-1\right)-\mathbf{j}\left(1\right)+\mathbf{k}\left(-1\right)& \hfill & \\ \hfill & =\left(-1,-1,-1\right)& \hfill & \end{array}$$

Next, parameterize the surface (the quarter disk) by

$$\begin{array}{lll}\hfill \mathbf{\Phi }\left(r,\theta \right)=\left(0,rcos\theta ,rsin\theta \right)& & \hfill \end{array}$$

for $0\le r\le 1$ and $0\le \theta \le \pi ∕2$.

Calculate the normal vector (we don’t need to normalize it to the unit normal vector $\mathbf{n}$):

$$\begin{array}{llll}\hfill \frac{\partial \mathbf{\Phi }}{\partial r}& =\left(0,cos\theta ,sin\theta \right)& \hfill & \\ \hfill \frac{\partial \mathbf{\Phi }}{\partial \theta }& =\left(0,-rsin\theta ,rcos\theta \right)& \hfill & \\ \hfill \frac{\partial \mathbf{\Phi }}{\partial r}\times \frac{\partial \mathbf{\Phi }}{\partial \theta }& =\mathbf{i}\left(r{cos}^2\theta +r{sin}^2\theta \right)=r\mathbf{i}& \hfill & \end{array}$$

Is the surface oriented properly? The normal vector points in the positive x-direction. But we need it to point it negative x-direction. Therefore, the surface is not oriented properly if we were to choose this normal vector.

To orient the surface properly, we must instead use the normal vector $\frac{\partial \mathbf{\Phi }}{\partial \theta }\times \frac{\partial \mathbf{\Phi }}{\partial r}=-r\mathbf{i}$.

(Note, at this point, we can already see that the integral ${\iint }_S\mathrm{curl}\mathbf{F}\cdot d\mathbf{S}$ should be positive. The vector field $\mathrm{curl}\mathbf{F}=\left(-1,-1,-1\right)$ and the normal vector $\left(-r,0,0\right)$ are pointing more-or-less in the same direction.)

Now, we have all pieces together to compute the integral.

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}& ={\iint }_S\mathrm{curl}\mathbf{F}\cdot d\mathbf{S}& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{\pi ∕2}\mathrm{curl}\mathbf{F}\left(\mathbf{\Phi }\left(r,\theta \right)\right)\cdot \left(\frac{\partial \mathbf{\Phi }}{\partial \theta }\left(r,\theta \right)\times \frac{\partial \mathbf{\Phi }}{\partial r}\left(r,\theta \right)\right)d\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{\pi ∕2}\left(-1,-1,-1\right)\cdot \left(-r,0,0\right)d\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{\pi ∕2}rd\theta dr=\frac{\pi }{4}& \hfill & \end{array}$$

Double-check example

Just for verification, we can compute the line ${\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}$ directly.

We need to parametrize $C$. We’ll do it by dividing $C$ into three parts.

We’ll use the fact that

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot d\mathbf{s}+{\int \; <!--nolimits\; -->}_{C_2}\mathbf{F}\cdot d\mathbf{s}+{\int \; <!--nolimits\; -->}_{C_3}\mathbf{F}\cdot d\mathbf{s}& & \hfill \end{array}$$

Recall $\mathbf{F}\left(x,y,z\right)=\left(y,z,x\right)$

First we’ll compute the integral over $C_1$. Parameterize it by

$$\begin{array}{llll}\hfill \mathbf{c}\left(t\right)& =\left(0,0,t\right),0\le t\le 1.& \hfill & \end{array}$$

Since ${\mathbf{c}}^{\text{'}}\left(t\right)=\left(0,0,1\right)$, we compute that

$$\begin{array}{llll}\hfill \mathbf{F}\left(\mathbf{c}\left(t\right)\right)\cdot {\mathbf{c}}^{\text{'}}\left(t\right)& =\mathbf{F}\left(0,0,t\right)\cdot \left(0,0,1\right)& \hfill & \\ \hfill & =\left(0,t,0\right)\cdot \left(0,0,1\right)& \hfill & \\ \hfill & =0& \hfill & \end{array}$$

Therefore,

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_{C_1}\mathbf{F}\cdot d\mathbf{s}& ={\int \; <!--nolimits\; -->}_0^1\mathbf{F}\left(\mathbf{c}\left(t\right)\right)\cdot {\mathbf{c}}^{\text{'}}\left(t\right)dt=0.& \hfill & \end{array}$$

The integral for $C_3$ is similar.

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_{C_3}\mathbf{F}\cdot d\mathbf{s}=0& & \hfill \end{array}$$

Last, we’ll compute the integral over $C_2$. Parameterize $C_2$ as

$$\begin{array}{llll}\hfill \mathbf{c}\left(t\right)& =\left(0,sint,cost\right),0\le t\le \pi ∕2,& \hfill & \end{array}$$

so that ${\mathbf{c}}^{\text{'}}\left(t\right)=\left(0,cost,-sint\right)$. We then compute

$$\begin{array}{llll}\hfill & {\int \; <!--nolimits\; -->}_{C_2}\mathbf{F}\cdot d\mathbf{s}={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\mathbf{F}\left(\mathbf{c}\left(t\right)\right)\cdot {\mathbf{c}}^{\text{'}}\left(t\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\mathbf{F}\left(0,sint,cost\right)\cdot \left(0,cost,-sint\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\left(sint,cost,0\right)\cdot \left(0,cost,-sint\right)dt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}{cos}^{2}tdt& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}\frac{1+cos2t}{2}dt& \hfill & \\ \hfill & =\frac{t}{2}+{\left.\frac{sin2t}{4}\right|}_0^{\pi ∕2}=\frac{\pi }{4}.& \hfill & \end{array}$$

Therfore,

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}=\frac{\pi }{4}& & \hfill \end{array}$$

in agreement with our Stokes’ theorem answer.

Example 2

We often present Stoke’s theorem problems as we did above. We give a curve $C$ and expect you to compute the surface integral over some surface $S$ with boundary $C$. In general, one can pick many surfaces. (See demo by Jon Rogness). But, sometimes, there is a surface that is “obviously” the best one.

One special case where this is relatively easy is when $C$ lies in a plane. This is especially easy when that plane is parallel to a coordinate plane, as in the following example. (We won’t work out this example, but just discuss how to choose $S$).

Let’s say you want to use Stokes’ theorem to compute ${\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}$ where $C$ is polygon path connecting the following points: (1,1,0), (3,1,4), (1,1,5), (-1,1,1)

Does this curve lie in a plane? Yes, in the plane $y=1$. The figure below is just the plane $y=1$.

If one coordinate is constant, then curve is parallel to a coordinate plane. (The $xz$-plane for above example). For Stokes’ theorem, use the surface in that plane. For our example, the natural choice for $S$ is the surface whose $x$ and $z$ components are inside the above rectangle and whose $y$ component is 1.

Example 3

In other cases, a surface is given explicitly in the problem.

Compute ${\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}$, where $C$ is the curve in which the cone $z^2=x^2+y^2$ intersects the plane $z=1$. (Oriented counter clockwise viewed from positive $z$-axis).

$$\begin{array}{lll}\hfill {\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}={\iint }_S\mathrm{curl}\mathbf{F}\cdot d\mathbf{S}& & \hfill \end{array}$$ for what surface $S$?

In this case, there are two natural choices for the surface. You could use the portion of the plane or the portion of the cone illustrated below.

Let $P$ be the portion of the plane $z=1$ with $x^2+y^2<1$ with upward pointing normal. Let $Q$ be the portion of the cone $z^2=x^2+y^2$ with $0<z<1$ with upward angling normal.

How do ${\iint }_P\mathrm{curl}\mathbf{F}\cdot d\mathbf{S}$ and ${\iint }_Q\mathrm{curl}\mathbf{F}\cdot d\mathbf{S}$ compare? They are the same. (For both surfaces, $C$ is a positive oriented boundary.)

Continue example: Let

$$\begin{array}{lll}\hfill \mathbf{F}\left(x,y,z\right)=\left(sinx-\frac{y^3}{3},cosy+\frac{x^3}{3},xyz\right)& & \hfill \end{array}$$

Compute ${\int \; <!--nolimits\; -->}_C\mathbf{F}\cdot d\mathbf{s}$.

One can show that $\mathrm{curl}\left(\mathbf{F}\right)=\left(xz,-yz,x^2+y^2\right)$.

Use surface $P$, parameterized by

$$\begin{array}{lll}\hfill \mathbf{\Phi }\left(r,\theta \right)=\left(rcos\theta ,rsin\theta ,1\right)& & \hfill \end{array}$$

for $0\le r\le 1,0\le \theta \le 2\pi$. Then normal vector is

$$\begin{array}{lll}\hfill \frac{\partial \mathbf{\Phi }}{\partial r}\times \frac{\partial \mathbf{\Phi }}{\partial \theta }=\left(0,0,r\right),& & \hfill \end{array}$$

which points in the correct direction, as mentioned above.

$$\begin{array}{llll}\hfill {\iint }_P\mathrm{curl}\mathbf{F}\cdot d\mathbf{S}& ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }\mathrm{curl}\left(\mathbf{F}\left(rcos\theta ,rsin\theta ,1\right)\right)\cdot \left(0,0,r\right)d\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }\left(rcos\theta ,-rsin\theta ,r^2\right)\cdot \left(0,0,r\right)d\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }{r}^{3}d\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{1}2\pi r^{3}dr=\frac{\pi }{2}& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

5a. Green's theorem

5b. The theorem for conservative vector fields

5c. Stokes' theorem

• The idea of Stokes' theorem
• Stokes' theorem examples

5d. The divergence theorem

6. Review material