Surface Integrals of Vector Fields Examples 2

# Surface Integrals of Vector Fields Examples 2

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

Evaluate the surface integral $\iint_{\delta} \mathbf{F} \cdot \hat{N} dS$ where $\mathbf{F}(x, y, z) = ze^{xy} \vec{i} - 3ze^{xy} \vec{j} + xy \vec{k}$ and where $\delta$ is the surface obtained by the parametric equations $\left\{\begin{matrix} x(u, v) = u + v\\ y(u, v) = u - v\\ z(u, v) = 1 + 2u + v \end{matrix}\right.$ for $0 ≤ u ≤ 2$ and $0 ≤ v ≤ 1$ oriented upwards.

Let $\vec{r}(u, v) = (x(u, v), y(u, v), z(u, v))$. Let's first compute $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}$:

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

From this we see that $\frac{\partial (y, z)}{\partial (u, v)} = 3$, $\frac{\partial (z, x)}{\partial (u, v)} = 1$ and $\frac{\partial (x, y)}{\partial (u, v)} = -2$. To get a positive orientation, we need to change the sign here so that the last component of the vector $\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = (3, 1, -2)$ is positive.

We also note that the region $D = \{ (u, v) : 0 ≤u ≤ 2, 0 ≤ v ≤ 1 \}$. Thus we have that:

(3)
\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -\iint_D ((1 + 2u + v)e^{(u + v)(u - v)}, -3(1 + 2u + v)e^{(u + v)(u - v)}, (u + v)(u - v) ) \cdot (3, 1, -2) \: du \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -\iint_D \left [ 3(1 + 2u + v)e^{u^2 - v^2} - 3(1 + 2u + v)e^{u^2 - v^2} - 2(u^2 - v^2) \right ] \: du \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = 2 \iint_D (u^2 - v^2) \: du \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = 2 \int_0^1 \int_0^2 (u^2 - v^2) \: du \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = 2 \int_0^1 \left [ \frac{u^3}{3} - uv^2 \right ]_{u=0}^{u=2} \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = 2 \int_0^1 \left ( \frac{8}{3} - 2v^2 \right ) \: dv \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = 2 \left [ \frac{8v}{3} - 2 \frac{v^3}{3} \right ]_{u=0}^{u=1} \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = 2 \left [ \frac{8}{3} - \frac{2}{3} \right ] \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = 4 \end{align}

## Example 2

Evaluate the surface integral $\iint_{\delta} \mathbf{F} \cdot \hat{N} dS$ where $\mathbf{F}(x, y, z) = x \vec{i} - z \vec{j} + y \vec{k}$ and where $\delta$ is the portion of the sphere $x^2 + y^2 = 4$ in the first quadrant oriented towards the origin.

One way to parameterized $\delta$ is with spherical coordinates as:

(4)
\begin{align} \left\{\begin{matrix} x(\phi, \theta) = 2\sin \phi \cos \theta \\ y(\phi, \theta) = 2\sin \phi \sin \theta\\ z(\phi, \theta) = 2\cos \phi \end{matrix}\right. \quad 0 ≤ \phi ≤ \frac{\pi}{2} \: , \: 0 ≤ \theta ≤ \frac{\pi}{2} \end{align}

If we want to compute the surface integral oriented towards the origin, then we will need to consider the negative side of $\delta$. Let's compute $\frac{\partial \vec{r}}{\partial \phi} \times \frac{\partial \vec{r}}{\partial \theta}$:

(5)
\begin{align} \quad \frac{\partial \vec{r}}{\partial \phi} \times \frac{\partial \vec{r}}{\partial \theta} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} & \frac{\partial z}{\partial \phi} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \theta} \end{bmatrix} = \begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 \cos \phi \cos \theta & 2 \cos \phi \sin \theta & -2 \sin \phi \\ -2 \sin \phi \sin \theta & 2 \sin \phi \cos \theta & 0 \end{bmatrix} \end{align}

Therefore we have that $\frac{\partial (y, z)}{\partial (\phi, \theta)} = 4 \sin^2 \phi \cos \theta$, $\frac{\partial (z, x)}{\partial (\phi, \theta)} = 4 \sin^2 \phi \sin \theta$ and $\frac{\partial (x, y)}{\partial (\phi, \theta)} = 4 \sin \phi \cos \phi$. Since we want to compute our surface integral with the negative orientation, then the third component of the vector $(4 \sin^2 \phi \cos \theta, 4 \sin^2 \phi \sin \theta, 4 \sin \phi \cos \phi)$ must be negative, that is, $4 \sin \phi \cos \phi ≤ 0$. Note that for $0 ≤ \phi ≤ \frac{\pi}{2}$ we have that $4 \sin \phi \cos \phi ≥ 0$ so we must include a negative sign, and so for $D = \left \{ (\phi, \theta) : 0 ≤ \phi ≤ \frac{\pi}{2}, 0 ≤ \theta ≤ \frac{\pi}{2} \right \}$ we have that:

(6)
\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -\iint_D (2 \sin \phi \cos \theta, -2 \cos \phi, 2 \sin \phi \sin \theta) \cdot (4 \sin^2 \phi \cos \theta, 4 \sin^2 \phi \sin \theta, 4 \sin \phi \cos \phi) \: dS \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = - \iint_D \left [ 8 \sin^3 \phi \cos^2 \theta - 8 \sin^2 \phi \cos \phi \sin \theta + 8 \sin^2 \phi \cos \phi \sin \theta \right ] \: dS \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = - \iint_D 8 \sin^3 \phi \cos^2 \theta \: dS \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = - 8 \int_0^{\pi/2} \int_0^{\pi/2} \sin^3 \phi \cos^2 \theta \: d \phi \: d \theta \\ \end{align}

Now to evaluate $\int \sin^3 \phi \: d \phi = \int (1 - \cos^2 \phi)\sin \phi \: d \phi$ we need to use substitution. Let $u = \cos \phi$. Then $du = -\sin \phi \: d \phi$ and thus:

(7)
\begin{align} \quad \int (1 - \cos^2 \phi)\sin \phi \: d \phi = - \int (1 - u^2) \: du = - \left [u - \frac{u^3}{3} \right ] = \frac{u^3}{3} - u = \frac{\cos^3 \phi}{3} - \cos \phi \end{align}

Thus we have that:

(8)
\begin{align} \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = - 8 \int_0^{\pi/2} \cos^2 \theta \left [ \frac{\cos^3 \phi}{3} - \cos \phi \right ]_{\phi = 0}^{\phi = \pi/2} \: d \theta \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -8 \int_0^{\pi/2} -\frac{2}{3} \cos^2 \theta \: d \theta \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -\frac{16}{3} \int_0^{\pi/2} \cos^2 \theta \: d \theta \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -\frac{16}{3} \left [ \frac{\cos \theta \sin \theta + \theta}{2} \: d \theta \right ]_{\theta = 0}^{\theta = \pi / 2} \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -\frac{16}{3} \left [ \frac{\pi}{4} \right ] \\ \quad \iint_{\delta} \mathbf{F} \cdot \hat{N} \: dS = -\frac{4 \pi}{3} \end{align}