Stokes' Theorem Examples 2

# Stokes' Theorem Examples 2

Recall from the Stokes' Theorem page that if $\delta$ is an oriented surface that is piecewise-smooth, and that $\delta$ is bounded by a simple, closed, positively oriented, and piecewise-smooth boundary curve $C$, and 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 vector field on $\mathbb{R}^3$ such that $P$, $Q$, and $R$ have continuous partial derivatives in a region containing $\delta$ then:

(1)
\begin{align} \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} \end{align}

Or equivalently:

(2)
\begin{align} \quad \oint_C P(x, y, z) \: dx + Q(x, y, z) \: dy + R(x, y, z) \: dz = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot \hat{N} \: dS \end{align}

We will now look at some more examples involving Stokes' theorem.

## Example 1

Let $\mathbf{F} (x, y, z) = yz \vec{i} + 2xz \vec{j} + e^{xy} \vec{k}$ and let $C$ be the circle $x^2 + y^2 = 16$ lying on the plane $z = 5$ that is oriented counterclockwise when viewed from above. Evaluate $\oint_C \mathbf{F} \cdot d \vec{r}$ using Stokes' theorem.

This circle bounds a disk $\delta$ given parametrically for $0 ≤ r ≤ 4$ and $0 ≤ u ≤ 2 \pi$:

(3)
\begin{align} \quad \vec{r}(u, v) = (r \cos (u), r \sin (u), 5) \end{align}

Let $\hat{N} = \vec{k}$. Note that $\hat{N}$ is perpendicular to this disk and induces the correct orientation onto the boundary curve $C$. We have that:

(4)
\begin{align} \quad \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \mathrm{curl} (\mathbf{F}) \cdot \hat{N} dS = \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) dS = \left ( 2z - z \right ) dS = z \: dS \end{align}

Applying Stokes' theorem and we have that:

(5)
\begin{align} \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_{\delta} 5 \: dS \quad \oint_C \mathbf{F} \cdot d \vec{r} = 5 \iint_{\delta} \: dS \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = 5 \mathrm{Area \: of \: \delta} \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = 5 (16 \pi) \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = 80 \pi \end{align}

## Example 2

Let $\mathbf{F} (x, y, z) = x^2 z \vec{i} +xy^2 \vec{j} + z^2 \vec{k}$ and let $C$ be the curve obtained from the intersection of the plane $x + y + z = 1$ and the cylinder $x^2 + y^2 = 9$ that is oriented counterclockwise when viewed from above. Evaluate $\oint_C \mathbf{F} \cdot d \vec{r}$ using Stokes' theorem.

We note that the curve $C$ bounds an ellipse $\delta$ lying on the plane $x + y + z = 1$. The normal to this plane, and therefore the ellipse $\delta$ is $(1, 1, 1)$, and the unit normal is $\hat{N} = \left ( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right )$.

We therefore have that:

(6)
\begin{align} \quad \mathrm{curl} (\mathbf{F}) \cdot \hat{N} \: dS = \left [ \frac{1}{\sqrt{3}} \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) + \frac{1}{\sqrt{3}} \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) + \frac{1}{\sqrt{3}} \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \right ] \: dS \\ \quad \mathrm{curl} (\mathbf{F}) \cdot \hat{N} \: dS = \left [ \frac{1}{\sqrt{3}} (0 - 0) + \frac{1}{\sqrt{3}}x^2 + \frac{1}{\sqrt{3}} y^2 \right ] \: dS \\ \quad \mathrm{curl} (\mathbf{F}) \cdot \hat{N} \: dS = \frac{1}{\sqrt{3}} (x^2 + y^2) \: dS \end{align}

Now the ellipse $\delta$ projects a circle $x^2 + y^2 = 9$ onto the $xy$-plane (since this ellipse is contained in the circular cylinder $x^2 + y^2 = 9$). Let $D$ be this region. Then in polar-coordinates, we have that:

(7)
\begin{align} \quad D = \{ (r, \theta) : 0 ≤ r ≤ 3 , 0 ≤ \theta ≤ 2\pi \} \end{align}

Therefore in applying Stokes' theorem and using polar coordinates to evaluate the resulting surface integral, we have that

(8)
\begin{align} \quad \oint \mathbf{F} \cdot d \vec{r} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} \\ \quad \oint \mathbf{F} \cdot d \vec{r} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot \hat{N} \: dS \\ \quad \oint \mathbf{F} \cdot d \vec{r} = \frac{1}{\sqrt{3}} \iint_{\delta} (x^2 + y^2) \: dS \\ \quad \oint \mathbf{F} \cdot d \vec{r} = \frac{1}{\sqrt{3}} \iint_D r^2 r \: dr \: d \theta \\ \quad \oint \mathbf{F} \cdot d \vec{r} = \frac{1}{\sqrt{3}} \int_0^{2\pi} \int_0^3 r^3 \: dr \: d \theta \\ \quad \oint \mathbf{F} \cdot d \vec{r} = \frac{1}{\sqrt{3}} \int_0^{2\pi} \left [ \frac{r^4}{4} \right ]_{r=0}^{r=3} \: d \theta \\ \quad \oint \mathbf{F} \cdot d \vec{r} = \frac{81}{4\sqrt{3}} \int_0^{2\pi} \: d \theta \\ \quad \oint \mathbf{F} \cdot d \vec{r} = \frac{81 \pi}{2 \sqrt{3}} \end{align}