Stokes' theorem relates a surface integral of a the curl of the vector field to a line integral of the vector field around the boundary of the surface. After reviewing the basic idea of Stokes' theorem and how to make sure you have the orientations of the surface and its boundary matched, try your hand at these examples to see Stokes' theorem in action.
Example 1
Let $\dlc$ be the closed curve illustrated below.
For $\dlvf(x,y,z) = (y,z,x)$, compute\begin{align*} \dlint\end{align*}using Stokes' Theorem.
Solution:Since we are given a line integral and told to use Stokes' theorem, we need to compute a surface integral\begin{align*} \sint{\dls}{\curl \dlvf},\end{align*}where $\dls$ is a surface with boundary $\dlc$. We have freedom to chooseany surface $\dls$, as long as we orient it so that $\dlc$ is a positivelyoriented boundary.
In this case, the simplest choice for $\dls$ is clear. Let $\dls$ be the quarter disk in the $yz$-plane.
Given the orientation of the curve $\dlc$, we need to choose the surfacenormal vector $\vc{n}$ to point in which direction? By theright hand rule criterion, the normal vectorshould point toward the negative side of the $x$-axis.
We need to calculate the curl of $\dlvf$. We can calculate the curl asusing the notation\begin{align*} \curl(\dlvf) &= \nabla \times \dlvf = \nabla \times (y,z,x)\\ & = \left| \begin{array}{ccc} \vc{i} & \vc{j} & \vc{k}\\ \displaystyle \pdiff{}{x} & \displaystyle \pdiff{}{y} & \displaystyle \pdiff{}{z}\\ y & z & x \end{array} \right|\\ &= \vc{i} \left(\pdiff{}{y} x - \pdiff{}{z} z\right) -\vc{j} \left(\pdiff{}{x} x - \pdiff{}{z}y\right)\\ &\quad+\vc{k} \left(\pdiff{}{x} z - \pdiff{}{y}y\right)\\ &= \vc{i} (-1) - \vc{j} (1) + \vc{k} (-1)\\ &= (-1, -1, -1)\end{align*}
Next, parameterize the surface (the quarter disk) by\begin{align*} \dlsp(r,\theta) = (0, r\cos\theta, r\sin\theta)\end{align*}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 unitnormal vector $\vc{n}$):\begin{align*} \pdiff{\dlsp}{r} &= (0, \cos\theta, \sin\theta)\\ \pdiff{\dlsp}{\theta} &= (0, -r\sin\theta, r\cos\theta)\\ \pdiff{\dlsp}{r} \times \pdiff{\dlsp}{\theta} &= \vc{i} (r \cos^2\theta + r \sin^2\theta) = r \vc{i}\end{align*}
Is the surface oriented properly? The normal vector points in thepositive x-direction. But we need it to point it negativex-direction. Therefore, the surface is not oriented properly if wewere to choose this normal vector.
To orient the surface properly, we must instead use the normal vector $\displaystyle\pdiff{\dlsp}{\theta} \times \pdiff{\dlsp}{r} = -r \vc{i}$.
At this point, we can already see that the integral$\sint{\dls}{\curl \dlvf}$ should be positive. The vectorfield $\curl \dlvf = (-1,-1,-1)$ and the normal vector $(-r,0,0)$ arepointing in a similar direction.
Now, we have all pieces together to compute the integral.\begin{align*} \dlint &=\sint{\dls}{\curl \dlvf}\\ &= \psintrnro{0}{1}{0}{\pi/2}{\curl \dlvf}{\dlsp}{r}{\theta}\\ &=\int_0^1 \int_0^{\pi/2} (-1,-1,-1) \cdot (-r, 0, 0) d\theta\, dr\\ &=\int_0^1 \int_0^{\pi/2} r d\theta\,dr = \frac{\pi}{4}\end{align*}
Double-check example 1
Just for verification, we can compute the line $\dlint$ directly.
We need to parametrize $\dlc$. We'll do it by dividing $\dlc$ intothree parts.
We'll use the fact that\begin{align*} \dlint =\lint{\dlc_1}{\dlvf} + \lint{\dlc_2}{\dlvf} + \lint{\dlc_3}{\dlvf}\end{align*}
Recall $\dlvf(x,y,z) = (y,z,x)$
First we'll compute the integral over $\dlc_1$. Parameterize it by\begin{align*} \dllp(t) &= (0,0,t), \qquad 0 \le t \le 1.\end{align*}Since $\dllp'(t)=(0,0,1)$, we compute that \begin{align*} \dlvf(\dllp(t)) \cdot \dllp'(t) &= \dlvf(0,0,t) \cdot (0,0,1)\\ &= (0,t,0) \cdot (0,0,1)\\ &=0\end{align*}Therefore,\begin{align*} \lint{\dlc_1}{\dlvf} &= \plint{0}{1}{\dlvf}{\dllp} =0.\end{align*}
The integral for $\dlc_3$ is similar.\begin{align*} \lint{\dlc_3}{\dlvf} = 0\end{align*}
Last, we'll compute the integral over $\dlc_2$. Parameterize $\dlc_2$ as\begin{align*} \dllp(t) &= (0,\sin t,\cos t), \qquad 0 \le t \le \pi/2,\end{align*}so that $\dllp'(t) = (0, \cos t, -\sin t)$. We then compute\begin{align*} &\lint{\dlc_2}{\dlvf} = \plint{0}{\pi/2}{\dlvf}{\dllp} \\ &=\int_0^{\pi/2}\dlvf(0,\sin t,\cos t) \cdot (0, \cos t,-\sin t)dt\\ &=\int_0^{\pi/2} (\sin t, \cos t,0) \cdot (0, \cos t,-\sin t)dt\\ &=\int_0^{\pi/2} \cos^2t \, dt\\ &=\int_0^{\pi/2} \frac{1+\cos 2t}{2} dt\\ &= \frac{t}{2} + \left.\left.\frac{\sin 2t}{4} \right|_0^{\pi/2}\right. = \frac{\pi}{4}.\end{align*}
Therefore,\begin{align*} \dlint = \frac{\pi}{4}\end{align*}in agreement with our Stokes' theorem answer.
Example 2
We often present Stoke's theorem problems as we did above. We give acurve $\dlc$ and expect you to compute the surface integral over somesurface $\dls$ with boundary $\dlc$. In general, one can pick manysurfaces. (See this applet.) But, sometimes, there is a surface that is “obviously” thebest one.
One special case where this is relatively easy is when $\dlc$ lies in aplane. This is especially easy when that plane is parallel to acoordinate plane, as in the following example. (We won't work outthis example, but just discuss how to choose $\dls$).
Let's say you want to use Stokes' theorem to compute $\dlint$ where $\dlc$ is polygon path connecting the followingpoints: (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 figurebelow is just the plane $y=1$.
If one coordinate is constant, then curve is parallel to a coordinateplane. (The $xz$-plane for above example). For Stokes' theorem, usethe surface in that plane. For our example, the natural choice for$\dls$ is the surface whose $x$ and $z$ components are inside the aboverectangle and whose $y$ component is 1.
Example 3
In other cases, a surface is given explicitly in the problem.
Compute $\dlint$, where $\dlc$ is the curve inwhich the cone $z^2=x^2+y^2$ intersects the plane $z=1$. (Orientedcounter clockwise viewed from positive $z$-axis).
\begin{align*} \dlint = \sint{\dls}{\curl \dlvf}\end{align*}for what surface $\dls$?
In this case, there are two natural choices for the surface. Youcould use the portion of the plane or the portion of the coneillustrated below.
Let $P$ be the portion of the plane $z=1$ with $x^2+y^2 < 1$ withupward 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 $\sint{P}{\curl \dlvf}$ and $\sint{Q}{\curl \dlvf}$ compare?They are the same. (For both surfaces, $\dlc$ is a positive oriented boundary.)
Continue example: Let\begin{align*} \dlvf(x,y,z) = \left(\sin x - \frac{y^3}{3}, \cos y + \frac{x^3}{3}, xyz \right)\end{align*}Compute $\dlint$.
One can show that $\curl(\dlvf) = (xz,-yz,x^2+y^2)$.
Use surface $P$, parameterized by\begin{align*} \dlsp(r,\theta) = (r \cos\theta, r\sin\theta, 1) \end{align*}for $0 \le r \le 1, 0 \le \theta \le 2\pi$. Then normal vector is\begin{align*} \pdiff{\dlsp}{r} \times \pdiff{\dlsp}{\theta} = (0,0,r),\end{align*}which points in the correct direction, as mentioned above.\begin{align*} \sint{P}{\curl \dlvf} &=\int_0^1 \int_0^{2\pi} \curl(\dlvf(r\cos\theta,r\sin\theta,1)) \cdot (0,0,r) d\theta dr \\ &=\int_0^1 \int_0^{2\pi} (r\cos\theta, -r\sin\theta, r^2) \cdot (0,0,r) d\theta\, dr \\ &= \int_0^1 \int_0^{2\pi} r^3 d\theta\, dr\\ &= \int_0^1 2\pi r^3 dr = \frac{\pi}{2}\end{align*}
