Stokes’ Theorem for Flux Integrals

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Objectives:

1.: Know when Stokes’ theorem can help compute a flux integral.
2.: Understand when a flux integral is surface independent.
3.: Be able to compute flux integrals using Stokes’ theorem or surface independence.

Recap Video

Here is a video highlights the main points of the section.

Here is a second video which gives the steps for using Stokes’ theorem to compute a flux integral.

Example Video

Here is an example of finding the “anti-curl” of a vector field.

If $\textbf {F} = \left \langle x,-2y,y+z \right \rangle$ , is there a vector field $\textbf {G}$ with $\textbf {F} = \text {curl}(\textbf {G})$ ? If so, find one.

The following videos show a worked example of using Stokes’ theorem to compute a flux integral. We compute the answer in two ways, one in each video.

If $\textbf {F} = \left \langle x,-2y,y+z \right \rangle$ and $S$ is the portion of $z = x^2+y^2$ inside the cylinder $x^2+y^2=1$ , oriented outward, evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ .

Approach 1: Anti-curling.
Approach 2: Surface independence.

Problems

Suppose $\textbf {F} = \left \langle e^{z^2}-1+z,e^{z^3}-z+1,x \right \rangle$ and $S$ is the portion of $x^2+y^2+z^2=1$ where $z \geq 0$ , oriented with outward normals. Evaluate $\displaystyle \iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S}$ .

There are two ways to go about solving a problem like this, so we will go through both.

(a): Approach 1: Use Stokes’ theorem directly. Stokes’ theorem says $\iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \int _{C} \textbf {F} \cdot d\textbf {r}.$ In this case, the boundary $C$ is the curve $x^2+y^2 = \answer {1}$ in the plane $z = \answer {0}$ . If $S$ is oriented outward, then $C$ is oriented
when viewed from above. We can parametrize the curve as $\textbf {r}(t) = \left \langle \cos (t),\answer {\sin (t)},\answer {0} \right \rangle ,$ for $0 \leq t \leq \answer {2\pi }$ so that $\textbf {r}'(t) = \left \langle \answer {-\sin (t)},\answer {\cos (t)},\answer {0} \right \rangle .$ Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {0},\answer {2},\answer {\cos (t)} \right \rangle$ and $\textbf {F}(\textbf {r}(t)) \cdot \textbf {r}'(t) = \answer {2\cos (t)}.$ Thus,
$\begin{eqnarray*} \iint_S \text{curl}(\textbf{F}) \cdot d\textbf{S} &=& \int_{C} \textbf{F} \cdot d\textbf{r} \\ &=& \int_{\answer{0}}^{\answer{2\pi}} \answer{2\cos(t)}dt \\ &=& \answer{0}. \end{eqnarray*}$
(b): Approach 2: Use surface independence. Here we want the flux of $\text {curl}(F)$ , which is the curl of some vector field (namely $\textbf {F}$ ), which means the flux integral is surface independent. We can change the surface to be whatever we want as long as $C$ is kept the same. As we saw in the above approach, $C$ is the circle $x^2+y^2=1$ in the plane $z= 0$ , oriented counterclockwise when viewed from above. Also note that $\text {curl}(\textbf {F}) = \left \langle \answer {1-3z^2e^{z^3}},\answer {2ze^{z^2}},\answer {0} \right \rangle .$ This looks ugly, but it works out nicely in the end. A convenient surface to pick is the disc $x^2+y^2 \leq 1$ in the plane $z=0$ , call this $S'$ . Since $C$ is oriented counterclockwise, we need $S'$ to be oriented with
normals. Now we compute the flux. We can parametrize the surface as $\textbf {r}(x,y) = \left \langle x,y,\answer {0} \right \rangle$ with domain $x^2+y^2 \leq \answer {1}$ . Then $\textbf {r}_x = \left \langle \answer {1},\answer {0},\answer {0} \right \rangle ,$ $\textbf {r}_y = \left \langle \answer {0},\answer {1},\answer {0} \right \rangle ,$ and $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {0},\answer {0},\answer {1} \right \rangle ,$ which is the
normal vector. Notice $\text {curl}(\textbf {F})(\textbf {r}(x,y)) = \left \langle \answer {1},\answer {0},\answer {0} \right \rangle$ (which looks much more reasonable), and the flux becomes $\iint _{S'} \text {curl}(\textbf {F}) \cdot d\textbf {S} = \iint _D \answer {0}dA = \answer {0}.$

Let $\textbf {F} = \left \langle 0,-6z^5,1 \right \rangle$ , and let $S$ be the portion of $z = x^2+y^2$ where $z \leq 2$ , oriented with outward normals. Evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ .

This flux integral is slightly unpleasant to do directly from the formula, so we can look for an easier way. Notice that $\text {div}(\textbf {F}) = \answer {0}$ and that the components of $\textbf {F}$ have continuous partials, so we can conclude:

Therefore, Stokes’ theorem and its consequences apply. Finding such a $\textbf {G}$ is a lengthy process, so we will use surface independence instead. A good surface to pick is the disc $x^2+y^2\leq 2$ in the plane $z = \answer {2}$ , call this $S'$ . We must figure out the orientation. To do this, notice that $S$ being oriented with outward normals means that $C$ is when viewed from above. The new surface $S'$ has the same boundary, so if $C$ is clockwise, then $S'$ should be oriented with normals. With this orientation, $\iint _S \textbf {F} \cdot d\textbf {S} = \iint _{S'} \textbf {F} \cdot d\textbf {S}.$ Parametrize the new surface $S'$ as $\textbf {r}(x,y) = \left \langle x,y,\answer {2} \right \rangle$ in the domain $D$ given by $x^2+y^2 \leq \answer {2}$ . Then $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {0},\answer {0},\answer {1} \right \rangle$ which is the normal. So we will take the normal to be $\left \langle 0,0,-1 \right \rangle$ . Therefore, the flux over $S'$ is

$\begin{eqnarray*} \iint_{S'} \textbf{F} \cdot d\textbf{S} &=& \iint_D \textbf{F}(\textbf{r}(x,y)) \cdot -(\textbf{r}_x \times \textbf{r}_y) dA \\ &=& \iint_D \answer{-1}dA \\ &=& -\text{area}(D) \\ &=& \answer{-2\pi}. \end{eqnarray*}$

The first part following problem is not applicable in Math 120 Fall 2019. You can assume $\textbf {G}= \left \langle 0,x,6z^5x \right \rangle$ and evaluate the rest.

With the same $\textbf {F}$ and $S$ as in the previous problem, evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ by finding a $\textbf {G}$ such that $\text {curl}(\textbf {G}) = \textbf {F}$ and computing a line integral instead.

One such $\textbf {G}$ is $\textbf {G} = \left \langle 0,x,6z^5x \right \rangle$ . We know $C$ is the circle $x^2+y^2=2$ in the plane $z=2$ , oriented clockwise, which we can parametrize as $\textbf {r}(t) = \left \langle \sqrt {2}\cos (t),\sqrt {2}\sin (t),2 \right \rangle$ , which goes the wrong way. Stokes’ says

$\begin{eqnarray*} \iint_S \textbf{F} \cdot d\textbf{S} &=& \int_{C} \textbf{G} \cdot d\textbf{S} \\ &=& -\int_0^{2\pi} 2\cos^2(t)dt \\ &=& -2\pi. \end{eqnarray*}$

If $\textbf {F} = \left \langle y,y,-z \right \rangle$ and $S$ is the portion of $x^2+y^2+z^2 = 4$ with $z \geq 1$ , oriented outward, then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = \answer {-3\pi }$ .

Since $\text {div}(\textbf {F}) = \answer {0}$ and the components of $\textbf {F}$ have continuous partials everywhere, we know $\textbf {F}$ is the curl of a vector field. This means we can use surface independence. Change the surface to $x^2+y^2 \leq \answer {3}$ in the plane $z=\answer {1}$ (call this $S'$ ), oriented . You can parametrize this as $\textbf {r}(x,y) = \left \langle x,y,\answer {1} \right \rangle$ in the domain $D$ given by $x^2+y^2 \leq \answer {3}$ . Then $\textbf {r}_x \times \textbf {r}_y = \left \langle 0,0,\answer {1} \right \rangle ,$ which is the correct normal. Then $\textbf {F}(\textbf {r}(x,y)) \cdot (\textbf {r}_x \times \textbf {r}_y) = -1.$ The flux is therefore $\iint _{S'} \textbf {F} \cdot d\textbf {S} = \iint _D -1 dA = -3\pi .$

Let $\textbf {F} = \left \langle z+x,xz^2,yx \right \rangle$ and let $S$ be the portion of $y^2+z^2=9$ with $1 \leq x \leq 9$ , oriented with outward normals, coupled with the disc $y^2+z^2 \leq 9$ in the plane $x = 9$ , oriented with normals pointing in the positive $x$ -direction. Then $\displaystyle \iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \answer {9\pi }$ .

Since $\text {curl}(\textbf {F})$ is the curl of some vector field, we can either parametrize the boundary and use normal Stokes’, or use surface independence. If we use surface independence, we can change the surface to the disc $y^2+z^2 \leq 9$ in the plane $x=1$ , oriented with normals pointing in the positive $x$ -direction. Parametrize this surface as $\textbf {r}(y,z) = \left \langle 1,y,z \right \rangle$ for $D:\,y^2+z^2 \leq 9$ . Then $\textbf {r}_y \times \textbf {r}_z = \left \langle 1,0,0\right \rangle$ , which is the right normal. Since $\text {curl}(\textbf {F}) = \left \langle x-2xz,1-y,z^2 \right \rangle$ , we get $\text {curl}(\textbf {F})(\textbf {r}(y,z)) \cdot \left \langle 1,0,0 \right \rangle = 1+z.$ The flux of the curl becomes $\iint _D (1+z)dA = 9\pi .$

If $\textbf {F} = \left \langle z^8,x^8,y^8 \right \rangle$ and $S$ consists of the cylinder $x^2+y^2=1$ from $0 \leq z \leq 1$ and the discs $x^2+y^2 \leq 1$ at both $z=1$ and $z=0$ , all oriented with inward normals, then $\displaystyle \iint _S \text {curl}(\textbf {F}) \cdot d\textbf {S} = \answer {0}$ .

$\text {curl}(\textbf {F})$ is the curl of a vector field and $S$ is closed, so the flux is $0$ .

$\textbf {F}$ is the curl of a vector field because $\text {div}(\textbf {F}) = 0$ and the components of $\textbf {F}$ have continuous partials everywhere. Since $S$ is closed, the flux is $0$ .

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