Stokes' Theorem Examples 1

# Stokes' Theorem Examples 1

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}

## Example

Let $\mathbf{F} (x, y, z) = x^2 z^2 \vec{i} + y^2 z^2 \vec{j} + xyz \vec{k}$, and let $\delta$ be the portion of the paraboloid $z = x^2 + y^2$ inside the cylinder $x^2 + y^2 = 4$ that is oriented upwards. Using Stokes' theorem evaluate $\iint_{\delta} \mathrm{curl} ( \mathbf{F}) \cdot d \vec{S}$.

We note that all of the conditions of Stokes' theorem hold.

Note that the paraboloid $z = x^2 + y^2$ intersects the cylinder $x^2 + y^2 = 4$ as a circle $x^2 + y^2 = 4$ on the plane $z = 4$. As viewed from above, $C$ should be oriented counterclockwise. This curve $C$ can be parameterized for $0 ≤ u ≤ 2\pi$ as:

(3)
\begin{align} \quad \vec{r}(u) = (2 \cos u, 2 \sin u, 4) \end{align}

Noting that $P(x, y, z) = x^2 z^2$, $Q(x, y, z) = y^2 z^2$ and $R(x, y, z) = xyz$, then in applying Stokes' theorem we have that:

(4)
\begin{align} \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \oint_C x^2 z^2 \: dx + y^2 z^2 \: dy + xyz \: dz \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \int_0^{2\pi} \left [ (2 \cos u)^2 (4)^2 (-2 \sin u) + (2 \sin u)^2 (4)^2 (2 \cos u) + (2 \cos u)(2 \sin u)(4)(0) \right ] \:du \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \int_0^{2\pi} \left [ -128 \cos^2 u \sin u + 128 \sin^2 u \cos u \right ] \: du \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = -128 \left [ \int_0^{2\pi} \cos^2 u \sin u \: du - \int_0^{2\pi} \sin^2 u \cos u \: du \right ] \end{align}

Let $m = \cos u$. Then $dm = -\sin u \: du$. Therefore we have that:

(5)
\begin{align} \quad \int \cos^2 u \sin u \: du = - \int m^2 \: dm = -\frac{m^3}{3} = -\frac{\cos^3 u}{3} \end{align}

Let $n = \sin u$. Then $dn = \cos u \: du$. Therefore we have that:

(6)
\begin{align} \quad \int \sin^2 u \cos u \: du = \int n^2 \: dn = \frac{n^3}{3} = \frac{\sin^3 u}{3} \end{align}

Thus:

(7)
\begin{align} \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = -128 \left [- \left ( \frac{\cos^3}{3} \right )_{u=0}^{u=2\pi} - \left ( \frac{\sin^3 u}{3} \right )_{u=0}^{u=2\pi} \right ] \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = -128 \left ( -\frac{1}{3} + \frac{1}{3} \right ) \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = 0 \end{align}

## Example 2

Let $\mathbf{F}(x, y, z) = xyz \vec{i} + xy \vec{j} + x^2yz \vec{k}$ and let $\delta$ be the surface of the top and sides of the cube with vertices $(\pm 1, \pm 1, \pm 1) \in \mathbb{R}^3$ oriented outward. Use Stokes' theorem to evaluate $\iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S}$.

We note that the boundary curve $C$ of this surface is the square formed with the vertices $(1, 1, -1)$, $(1, -1, -1)$, $(-1, 1, -1)$ and $(-1, -1, -1)$ lying on the plane $z = -1$.

Thus $C$ is comprised of $4$ lines, $C_1$, $C_2$, $C_3$, and $C_4$ defined parametrically below:

(8)
\begin{align} \quad C_1 : \vec{r}(t) = \{ (1, t , -1) \} \quad -1 ≤ t ≤ 1 \\ \quad C_2 : \vec{r}(t) = \{ (-t, 1 , -1) \} \quad -1 ≤ t ≤ 1 \\ \quad C_3 : \vec{r}(t) = \{ (-1, -t , -1) \} \quad -1 ≤ t ≤ 1 \\ \quad C_4 : \vec{r}(t) = \{ (t, -1 , -1) \} \quad -1 ≤ t ≤ 1 \\ \end{align}

We apply Stokes' theorem to get that:

(9)
\begin{align} \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \: d\vec{S} = \oint_C \mathbf{F} \cdot d \vec{r} \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \: d\vec{S} = \left ( \int_{C_1} + \int_{C_2} + \int_{C_3} + \int_{C_4} \right ) (\mathbf{F} \cdot d \vec{r}) \end{align}

Let's compute each of these integrals separately:

(10)
\begin{align} \quad \int_{C_1} \mathbf{F} \cdot d \vec{r} = \int_{-1}^{1} (-t, t, -t) \cdot (0, 1, 0) \: dt = \int_{-1}^{1} t \: dt = 0 \end{align}
(11)
\begin{align} \quad \int_{C_2} \mathbf{F} \cdot d \vec{r} = \int_{-1}^{1} (t, -t, -t^2) \cdot (-1, 0, 0) \: dt \end{align}
(12)
\begin{align} \quad \int_{C_3} \mathbf{F} \cdot d \vec{r} = \int_{-1}^{1} (-t, t, -t) \cdot (0, -1, 0) \: dt = \int_{-1}^{1} -t \: dt = 0 \end{align}
(13)
\begin{align} \quad \int_{C_4} \mathbf{F} \cdot d \vec{r} = \int_{-1}^{1} (t, -t, t^2) \cdot (1, 0, 0) \: dt = \int_{-1}^{1} t \: dt = 0 \end{align}

Therefore $\iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = 0$.

Alternatively, we can simplify this problem without computing four line integrals. Let $\gamma$ be the bottom face of the cube described above. Then this cube has the same boundary curve $C$. By Stokes' theorem we have that:

(14)
\begin{align} \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \oint_C \mathbf{F} \cdot d \vec{r} = \iint_{\gamma} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} \end{align}

Let $\hat{N} = \vec{k}$. Then note that this vector is normal the square face $\gamma$ and induces the same orientation on the curve $C$ as did $\delta$. Furthermore, we have that the region $\gamma$ can be described as

(15)
\begin{align} \quad \gamma = \{ (x, y, z) : -1 ≤ x ≤ 1 , -1 ≤ y ≤ 1, z = -1 \} \end{align}

And we have that:

(16)
\begin{align} \quad \mathrm{curl} (\mathbf{F}) \cdot \hat{N} \: dS = \mathrm{curl} (\mathbf{F}) \cdot \vec{k} dS = \left ( \frac{\partial}{\partial x} (xy) - \frac{\partial}{\partial y} (xyz) \right ) \: dS = (y - xz) \: dS \end{align}

Therefore

(17)
\begin{align} \quad \iint_{\gamma} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \iint_{\gamma} (y - xz) \: dS \\ \quad \iint_{\gamma} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \int_{-1}^{1} \int_{-1}^{1} (y + x) \: dx \: dy \\ \quad \iint_{\gamma} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = 0 \end{align}