Vector Surface Integrals

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Objectives:

1.: Know what a flux integral $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ represents geometrically.
2.: Understand what an oriented surface is and how it relates to flux integrals.
3.: Understand the steps to compute a flux integral.

Recap Video

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

To summarize:

Given an oriented surface $S$ and a vector field $\textbf {F}$ , we compute the flux integral $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ as follows:

Parametrize the surface $S$ as $\textbf {r}(u,v)$ , where the domain of $(u,v)$ is $D$ .
Compute $\textbf {r}_u\times \textbf {r}_v$ and check whether this normal matches the orientation given for $S$ (if not, replace it by $-(\textbf {r}_u \times \textbf {r}_v)$ ).
Compute $\iint _S \textbf {F} \cdot d\textbf {S} = \iint _D \textbf {F}(\textbf {r}(u,v)) \cdot \pm (\textbf {r}_u \times \textbf {r}_v) dA,$ where the $\pm$ is determined by the second step.

Example Video

The following video shows a worked example of a flux integral.

If $S$ is the portion of $x = \sqrt {y^2+z^2}$ oriented with outward normals, and $\textbf {F} = \left \langle x,0,1 \right \rangle$ , evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ .

Problems

If $S$ is the portion of $x+y+z=1$ in the first octant, oriented downward, and $\textbf {F} = \left \langle x,0,y-z \right \rangle$ , evaluate $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S}$ .

We can parametrize this surface as $\textbf {r}(x,y) = \left \langle x,y,\answer {1-x-y} \right \rangle ,$ where $x$ and $y$ lie in the triangle bounded by $x=0$ , $y=0$ , and $x+y=1$ .

We compute: $\textbf {r}_x = \left \langle \answer {1},\answer {0},\answer {-1} \right \rangle$ $\textbf {r}_y = \left \langle \answer {0},\answer {1},\answer {-1} \right \rangle ,$ and $\textbf {r}_x \times \textbf {r}_y = \left \langle \answer {1},\answer {1},\answer {1} \right \rangle .$ Does this match the given orientation?

Since the orientation is incorrect, we can correct this by taking $-(\textbf {r}_x \times \textbf {r}_y)$ , so our normal will be $-\left \langle 1,1,1 \right \rangle = \left \langle -1,-1,-1 \right \rangle .$ We now need to do

$\begin{eqnarray*} \iint_S \textbf{F} \cdot d\textbf{S}&=& \iint_D \textbf{F}(\textbf{r}(x,y)) \cdot -(\textbf{r}_x \times \textbf{r}_y)dA \\ &=& \int_0^1 \int_0^{1-x} \left\langle \answer{x},\answer{0},\answer{x+2y-1} \right\rangle \cdot \left\langle -1,-1,-1 \right\rangle dy\,dx \\ &=& \int_0^1 \int_0^{1-x} \answer{-2x-2y+1} dy\,dx \\ &=&-\frac{1}{6}. \end{eqnarray*}$

Find the flux of $\textbf {F} = \left \langle z,z,x \right \rangle$ through the portion of $z = 9-x^2-y^2$ in the first octant, oriented with outward pointing normals.

We can parametrize this surface as $\textbf {r}(r,\theta ) = \left \langle r\cos (\theta ),r\sin (\theta ),\answer {9-r^2} \right \rangle ,$ where the bounds for $r$ and $\theta$ are: $\answer {0} \leq r \leq \answer {3},\quad 0 \leq \theta \leq \answer {\pi /2}.$ We compute the partials: $\textbf {r}_r = \left \langle \answer {\cos (\theta )},\answer {\sin (\theta )},\answer {-2r} \right \rangle ,$ $\textbf {r}_{\theta } = \left \langle \answer {-r\sin (\theta )},\answer {r\cos (\theta )},\answer {0} \right \rangle ,$ and $\textbf {r}_r \times \textbf {r}_{\theta } = \left \langle \answer {2r^2\cos (\theta )},\answer {2r^2\sin (\theta )},\answer {r} \right \rangle .$ We need to check the orientation. In this case, an outward normal for the paraboloid means the $z$ -component for the normal needs to be . Does our normal vector match the given orientation?

Since this is the correct orientation, we can just compute:

$\begin{eqnarray*} \iint_S \textbf{F} \cdot d\textbf{S} &=& \int_0^{\pi/2} \int_0^3 \textbf{F}(\textbf{r}(r,\theta)) \cdot (\textbf{r}_r \times \textbf{r}_{\theta})drd\theta \\ &=& \int_0^{\pi/2} \int_0^3 \left\langle \answer{9-r^2},\answer{9-r^2},\answer{r\cos(\theta)} \right\rangle \cdot \left\langle 2r^2\cos(\theta),2r^2\sin(\theta),r \right\rangle drd\theta \\ &=& \int_0^{\pi/2} \int_0^3 [(9-r^2)2r^2\cos(\theta) + (9-r^2)2r^2\sin(\theta)+r^2\cos(\theta)]drd\theta \\ &=& \frac{693}{5}. \end{eqnarray*}$

If $\textbf {F} = \left \langle x,y,e^z \right \rangle$ and $S$ is the portion of $x^2+y^2=4$ where $1\leq z \leq 5$ , oriented with outward normals, then $\iint _S \textbf {F} \cdot d\textbf {S} = \answer {32\pi }.$ (The hint shows the explanation.)

Parametrize $S$ as, for example, $\textbf {r}(\theta ,z) = \left \langle 2\cos (\theta ),2\sin (\theta ),z \right \rangle$ for $0 \leq \theta \leq 2\pi$ and $1 \leq z \leq 5$ . The cross product $\textbf {r}_{\theta } \times \textbf {r}_z = \left \langle \answer {2\cos (\theta )},\answer {2\sin (\theta )},\answer {0} \right \rangle ,$ which is the normal. We now compute

$\begin{eqnarray*} \iint_S \textbf{F} \cdot d\textbf{S} &=& \int_0^{2\pi} \int_1^5 \textbf{F}(\textbf{r}(\theta,z)) \cdot (\textbf{r}_{\theta} \times \textbf{r}_z)dzd\theta \\ &=& \int_0^{2\pi} \int_1^5 \left\langle 2\cos(\theta),2\sin(\theta),e^z \right\rangle \cdot \left\langle 2\cos(\theta),2\sin(\theta),0 \right\rangle dz d\theta \\ &=& \int_0^{2\pi} \int_1^5 4\cos^2(\theta) + 4\sin^2(\theta)dz d\theta \\ &=& 4 \cdot 4 \cdot 2\pi. \end{eqnarray*}$

If $\textbf {F} = \left \langle xz,yz,z^{-1} \right \rangle$ and $S$ is the portion of $z=4$ where $x^2+y^2 \leq 9$ , oriented with downward normals, then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = \answer {-9\pi /4}$ .

Parametrize the surface as $G(x,y) = \left \langle x,y,\answer {4} \right \rangle$ for $x^2+y^2 \leq 9$ . Then $G_x \times G_y = \left \langle 0,0,\answer {1} \right \rangle ,$ which is the normal, so we will take $-(G_x \times G_y) = \left \langle 0,0,-1\right \rangle$ as the normal. Notice $\textbf {F}(\textbf {r}(x,y)) = \left \langle 4x,4y,\frac {1}{4} \right \rangle ,$ so that $\textbf {F}(\textbf {r}(x,y)) \cdot -(G_x \times G_y) =- \frac {1}{4}.$ We therefore get $\iint _D -\frac {1}{4}dA.$ From here, either see this is $-1/4$ times the area of $D$ , which is $9\pi$ , giving $-9\pi /4$ . Alternatively, go to polar: $\int _0^{2\pi } \int _0^3 -\frac {1}{4}rdrd\theta .$

The flux of $\textbf {F} = \left \langle x,y,z \right \rangle$ through the sphere $x^2+y^2+z^2=1$ , oriented with outward normals, is: $\answer {4\pi }$ .

Parametrize the sphere as $\textbf {r}(\theta ,\phi ) = \left \langle \cos (\theta )\sin (\phi ),\sin (\theta )\sin (\phi ),\cos (\phi ) \right \rangle .$ You should get $\textbf {r}_{\theta } \times \textbf {r}_{\phi } = \left \langle \cos (\theta )\sin ^2(\phi ),\sin (\theta )\sin ^2(\phi ),\cos (\phi )\sin (\phi ) \right \rangle .$ Plugging in, for example, $\phi = \pi /2$ and $\theta = 0$ shows this is the correct normal. Then check $\textbf {F}(\textbf {r}(\theta ,\phi )) \times (\textbf {r}_{\theta } \times \textbf {r}_{\phi }) = \sin (\phi ).$ We then integrate $\int _0^{2\pi } \int _0^{\pi } \sin (\phi )d\phi d\theta = 4\pi .$

True/False

Let $\textbf {F}$ be a vector field. If $S$ is an oriented surface and $S'$ is the same surface but with the opposite orientation, then $\iint _S \textbf {F} \cdot d\textbf {S} = \iint _{S'} \textbf {F} \cdot d\textbf {S}.$

If $\textbf {F}$ is a vector field and $S$ is an oriented surface with $\textbf {F}$ tangent to $S$ at all points on $S$ , then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = 0$ .

If $\textbf {F}$ is a conservative vector field and $S$ is a closed, oriented surface (e.g. a sphere with outward normals), then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} = 0$ .

If $S$ is an oriented surface and $\textbf {F}$ is a vector field that, at all points on $S$ , makes an acute angle with the normal given by the orientation on $S$ , then $\displaystyle \iint _S \textbf {F} \cdot d\textbf {S} > 0$ .

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