Surface Integrals of Vector Fields Examples 1

# Surface Integrals of Vector Fields Examples 1

Recall from the Surface Integrals of Vector Fields page that if $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ is a continuous vector field and that if $\delta$ is a smooth orientable surface then the surface integral of $\mathbf{F}$ over $\delta$ (also called the flux of $\mathbf{F}$ across $\delta$) is given by the following formula:

(1)
\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = \pm \iint_D \left (P(x, y, z) \frac{\partial (y, z)}{\partial (u, v)} + Q(x, y, z) \frac{\partial (z, x)}{\partial (u, v)} + R(x, y, z) \frac{\partial (x, y)}{\partial (u, v)} \right ) \: du \: dv \end{align}

The sign is dependent on the orientation of $\delta$.

We will now look at some examples of computing surface integral integrals over vector fields.

## Example 1

If an orientable surface $\delta$ is given as the function $z = f(x, y)$ then derive a formula for the surface integral of the continuous vector field $\mathbf{F}$ over $\delta$.

We can easily parameterize our surface $z = f(x, y)$ by setting $x = x$, $y = y$, and $z = f(x, y)$ to get $\vec{r} (x, y) = (x, y, f(x, y))$. We then compute $\frac{\partial \vec{r}}{\partial x} \times \frac{\partial \vec{r}}{\partial y}$ to get:

(2)
\begin{align} \quad \frac{\partial \vec{r}}{\partial x} \times \frac{\partial \vec{r}}{\partial y} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial x}{\partial x} & \frac{\partial y}{\partial x} & \frac{\partial z}{\partial x} \\ \frac{\partial x}{\partial y} & \frac{\partial y}{\partial y} & \frac{\partial z}{\partial y} \end{bmatrix} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & \frac{\partial z}{\partial x} \\ 0 & 1 & \frac{\partial z}{\partial y} \end{bmatrix} \end{align}

From this we immediately see that $\frac{\partial (y, z)}{\partial (x, y)} = - \frac{\partial z}{\partial x}$, $\frac{\partial (z, x)}{\partial (x, y)} = - \frac{\partial z}{\partial y}$ and $\frac{\partial (x, y)}{\partial (x,y)} = 1$, and so:

(3)
\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = \pm \iint_D \left (P(x, y, f(x,y)) \frac{\partial z}{\partial x} + Q(x, y, f(x,y)) \frac{\partial z}{\partial y} - R(x, y, f(x, y)) \right ) \: dx \: dy \end{align}

## Example 2

Compute the flux of the vector field $\mathbf{F}(x, y, z) = \frac{2x}{x^2 + y^2} \vec{i} + \frac{2y}{x^2 + y^2} \vec{j} + \vec{k}$ downward through the surface $\delta$ given parametrically as $\vec{r}(u, v) = (u \cos v, u \sin v, u^2)$ for $0 ≤ u ≤ 1$ and $0 ≤ v ≤ 2 \pi$.

We must first compute the vector area element $d \vec{S}$. We have that:

(4)
\begin{align} \quad d \vec{S} = \pm \left ( \frac{\partial (y, z)}{\partial (u, v)} \vec{i}+ \frac{\partial (z, x)}{\partial (u, v)} \vec{j} + \frac{\partial (x, y)}{\partial (u, v)} \vec{k} \right ) \: du \: dv \\ \quad d \vec{S} = \pm \left ( \begin{vmatrix} \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \end{vmatrix} \vec{i} + \begin{vmatrix} \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v}\\ \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \end{vmatrix} \vec{j} + \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \vec{k} \right ) \: du \:dv \\ \quad d \vec{S} = \pm \left ( \begin{vmatrix} \sin v & u \cos v \\ 2u & 0 \end{vmatrix} \vec{i} + \begin{vmatrix} 2u & 0 \\ \cos v & -u \sin v \end{vmatrix} \vec{j} + \begin{vmatrix} \cos v & - u \sin v\\ \sin v & u \cos v \end{vmatrix} \vec{k} \right ) \: du \:dv \\ d\vec{S} = \pm \left ( - 2u^2 \cos v \vec{i} + -2u^2 \sin v \vec{j} + u \vec{k} \right ) \: du \: dv \end{align}

We want to compute the flux of $\mathbf{F}$ downward, so we will want the last component of $d \vec{S}$ to be negative. We note that $u ≥ 0$ on $\delta$ (since $0 ≤ u ≤ 1$) and so if we want the last component to be negative, then we will choose the negative sign in the plus/minus to get that:

(5)
\begin{align} \quad d \vec{S} = \left (2u^2 \cos v \vec{i} + 2u^2 \sin v \vec{j} - u \vec{k} \right ) \: du \: dv \end{align}

Thus the flux of $\mathbf{F}$ over $\delta$ where $D = \{ (u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2\pi \}$ is given by:

(6)
\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \iint_D \left ( \frac{2u \cos v}{u^2}, \frac{2u \sin v}{u^2}, 1 \right ) \cdot \left (2u^2 \cos v, 2u^2 \sin v, - u \right ) \: du \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \int_0^{2\pi} \int_0^1 \left ( 4u \cos^2 v + 4u \sin^2 v - u \right ) \: du \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \int_0^{2\pi} \int_0^1 3u \: du \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \int_0^{2\pi} \left [ \frac{3u^2}{2} \right ]_{u=0}^{u=1} \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \int_0^{2\pi} \frac{3}{2} \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = \frac{3}{2} \left [ v \right ]_{v = 0}^{v = 2\pi} \\ \quad \iint_{\delta} \mathbf{F} \cdot d \vec{S} = 3 \pi \end{align}

Therefore the flux of $\mathbf{F}$ downward through $\delta$ is $3 \pi$.