Math 344: Calculus III

16.8 Stokes' Theorem

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16.8.1 Stokes' Theorem

In Section 16.4 , we discussed Green's Theorem which relates an area integral to a path integral. Stokes' Theorem, the focus of this section, is similar in that it relates the surface integral over a vector field to a path integral. To do this, suppose we have an orientable surface $S$ that is bounded by a simple closed curve $C$. We need to have the proper (positive) orientations for both the surface and the boundary curve. The normal vector $\mathbf{n}$ for the boundary curve will match the orientation of the surface if we imagine the positive direction being determined the same way that we did for Green's Theorem, that is if we move in the positive direction for $C$ then the surface will be to the left.

This can be tracked with the right hand rule. If you extend the thumb, index, and middle fingers of your right hand so that they are orthogonal to one another, you point your index finger in the positive direction of $C$, your middle figure toward the interior of the surface, and then your thumb will be the normal vector $\mathbf{n}$.

Stokes' Theorem

Let $S$ be a positively oriented piecewise-smooth surface bounded by a positively oriented simple closed piecewise-smooth curve $C$. If $\mathbf{F}$ is a vector field with continuous partial derivatives on an open region in $\mathbb{R}^3$ containing the surface $S$, then

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pzi}[2][]{\dfrac{\partial #1}{\partial #2}} \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \left(\nabla\times\mathbf{F}\right)\cdot d\mathbf{S} $$

In words, this states that the surface integral over $S$ of the normal component of a vector field $\mathbf{F}$'s curl is equal to the path integral over boundary curve $C$ of the tangential component of $\mathbf{F}$. As was the case with Green's Theorem, this relates the surface integral for the rotational element of a vector field to an integral over a boundary curve. In fact, if $S$ is a flat surface in the $xy$-plane bounded by a positively oriented plane curve $C$, then the normal to $S$ is $\khat$ and we recover the curl form of Green's Theorem

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \left(\text{curl}\,\mathbf{F}\right)\cdot\khat\, dA $$
so this is a special case for 2D flows, such as what you would see on the surface of a river.

16.8.2 Computations Involving Stokes' Theorem

Example 1

Use Stokes' Theorem to compute $\iint_S \text{curl}\,\mathbf{F}\cdot d\mathbf{S}$ when

$$ \mathbf{F}(x,y,z) = x^2\sin^2 z\,\ihat + y^2\jhat + xy\,\khat $$
and $S$ is the part of the paraboloid $z = 1 - x^2 - y^2$ above the $xy$-plane with upward orientation.

Solution

Since this is a closed surface bounded by a simple close curve, we can use Stokes' Theorem and compute the path integral around the boundary instead. On the $xy$-plane, $z = 0$ so the boundary curve $C$ obeys the equation $x^2 + y^2 = 1$. We want the parametrization $\mathbf{r}(t) = \cos t\,\ihat + \sin t\,\jhat$ for $t\in[0,2\pi]$ since it will guarantee the required positive orientation. The tangent vector function is $\mathbf{r}'(t) = -\sin t\,\ihat + \cos t\,\jhat$. The last step prior to applying Stokes' Theorem is to find

$$ \mathbf{F}(\mathbf{r}(t)) = (\cos t)^2 \sin^2 (0)\,\ihat + (\sin t)^2\jhat + (\cos t)(\sin t)\,\khat = \sin^2 t\,\jhat + \cos t \sin t\,\khat $$

Now we apply the theorem,

$$ \begin{align*} \iint_S \text{curl}\,\mathbf{F}\cdot d\mathbf{S} & = \int_C \mathbf{F}\cdot d\mathbf{r} \\ \\ &= \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)\,dt \\ \\ &= \int_0^{2\pi} \left(\sin^2 t\,\jhat + \cos t\sin t\,\khat\right)\cdot \left(-\sin t\,\ihat + \cos t\,\jhat\right) dt \\ \\ &= \int_0^{2\pi} \left(0 + \cos t\sin^2 t + 0\right) dt \\ \\ &= \left. \frac{1}{3}\sin^3 t\, \right|_{\,0}^{\,2\pi} = 0 \end{align*} $$

Example 2

Compute $\iint_S \text{curl}\,\mathbf{F}\cdot d\mathbf{S}$ again, except this time the surface $S$ is the upward oriented hemisphere above the $xy$-axis given by $z = \sqrt{1 - x^2 - y^2}$.

Solution

Note that the boundary curve $C$ is given by $z = 0$, so its equation is $0 = \sqrt{1 - x^2 - y^2}$ which implies that $x^2 + y^2 = 1$. This is same boundary curve as the example we just computed, so by Stokes' Theorem the result must be the same and so

$$ \iint_S \text{curl}\,\mathbf{F}\cdot d\mathbf{S} = \int_C \mathbf{F}\cdot d\mathbf{r} = 0 $$

This is an extremely important result. Using Stoke's Theorem, we can identify the integral

$$ \iint_S \text{curl}\,\mathbf{F}\cdot d\mathbf{S} $$
for any positively oriented surfaces $S$ that have the same boundary curve $C$.

Example 3

Use Stokes' Theorem to compute $\int_C \mathbf{F}\cdot d\mathbf{r}$ for

$$ \mathbf{F}(x,y,z) = 2y\,\ihat + xz\,\jhat + (x+y)\khat $$
where $C$ is the curve of intersection between the plane $z = y + 2$ and the cylinder $x^2 + y^2 = 1$.

Solution

Here our curve is an ellipse and we are not given its parameterization, so it will be easier to compute the surface integral in this case. We begin by computing

$$ \begin{align*} \nabla\times\mathbf{F} &= \left(\pzi{y}\left(x+y\right) - \pzi{z}\left(xz\right) \right)\ihat - \left(\pzi{x}\left(x+y\right) - \pzi{z}\left(2y\right)\right)\jhat + \left(\pzi{x}\left(xz\right) - \pzi{y}\left(2y\right)\right)\khat \\ \\ &= \left(1-x\right)\ihat -\jhat + \left(z-y\right)\khat \end{align*} $$
The surface $S$ is given by the graph $z(x,y) = y + 2$, so we may use this formula from Section 16.7 when we apply Stokes' Theorem with $z_x = 0$ and $z_y = 1$. Once we simplify the integrand, we will see that switch to polar coordinates is natural.

$$ \begin{align*} \oint_C \mathbf{F}\cdot d\mathbf{r} &= \iint_S \left(\nabla\times\mathbf{F}\right)\cdot d\mathbf{S} \\ \\ &= \iint_\limits{x^2 + y^2 \leq 1} \left[ -(1-x)(0) - (-1)(1) + \left((y+2) - 2 \right)\right]\, dA \\ \\ &= \iint_\limits{x^2 + y^2 \leq 1} \left(y + 1\right)\, dA \\ \\ &= \int_0^{2\pi} \int_0^1 \left(r\sin\theta + 1\right) rdrd\theta \\ \\ &= \int_0^{2\pi} \left.\left[\frac{1}{3}r^3\sin\theta + \frac{1}{2}r^2\right]\right|_{\,0}^{\,1} d\theta \\ \\ &= \int_0^{2\pi} \frac{1}{3}\sin\theta + \frac{1}{2}\, d\theta \\ \\ &= \left.\left[-\frac{1}{3}\cos\theta + \frac{1}{2}\theta\right]\right|_{\,0}^{\,2\pi} \\ \\ &= \pi \end{align*} $$

16.8.3 Verifying Stokes' Theorem

Stokes' Theorem states that the integrals over the surface and the boundary should be equivalent, let us verify that for an example.

Example 4

Suppose that

$$ \mathbf{F}(x,y,z) = -y\,\ihat + x\,\jhat -2\,\khat $$
and $S$ is the cone $z^2 = x^2 + y^2$ for $z\in[0,4]$ and oriented downward.

Solution

Since our surface is oriented downward, we need our boundary curve $C$ to have the same orientation. $C$ is given by the circle $x^2 + y^2 = 16$ for $z = 4$ with clockwise orientation to have the normal point downward. The parameterization is $\mathbf{r}(t) = 4\cos (-t)\,\ihat + 4\sin (-t)\,\jhat + 4\,\khat$ for $t\in[0,2\pi]$ (the negative sign on $t$ is to enforce clockwise movement in $C$). The tangent vector function is $\mathbf{r}'(t) = -4\sin t\,\ihat - 4\cos t\,\jhat$. This allows us to write

$$ \mathbf{F}(\mathbf{r}(t)) = 4\sin t\,\ihat + 4\cos t\,\jhat -2\,\khat $$
and
$$ \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = -16\sin^2 t - 16\cos^2 t = -16 $$
Therefore, the integral over the close boundary curve is given by

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^{2\pi} -16\, dt = -32\pi $$
Next, we compute the surface integral of $\mathbf{F}$ over $S$. The first step is finding

$$ \text{curl}\,\mathbf{F} = \left(\pzi{y}\left(-2\right) - \pzi{z}\left(x\right) \right)\ihat - \left(\pzi{x}\left(-2\right) - \pzi{z}\left(-y\right)\right)\jhat + \left(\pzi{x}\left(x\right) - \pzi{y}\left(-y\right)\right)\khat = 2\,\khat $$
Our cone surface may be described by $z(x,y) = \sqrt{x^2 + y^2}$ and the integration domain $D$ is given by the projection of $S$ onto the $xy$-plane, which in this case is the disk $x^2 + y^2 \leq 16$. We combine this with the shortcut equation from the previous section and a minus sign for downward orientation to write the integral

$$ \iint_S \text{curl}\,\mathbf{F}\cdot d\mathbf{S} = -\iint_D (-0 - 0 + 2)\,dA = -2\left[\pi\left(4^2\right)\right] = -32\pi $$
where the integral over $D$ is given by the area of the integration domain since the integrand is a constant.

Our integrals are equal, even with the correct sign for the orientation.

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