Stokes' Theorem
 Table of Contents

# Stokes' Theorem

Recall from the Green's Theorem page that if $D$ is a regular and closed region in the $xy$-plane whose boundary is a positively oriented, piecewise smooth, simple, and closed curve $C$ and if $\mathbf{F}(x, y) = P(x, y) \vec{i} + Q(x, y) \vec{j}$ is a smooth vector field in $\mathbb{R}^2$ then:

(1)
\begin{align} \quad \oint_C P(x, y) \: dx + Q(x, y) \: dy = \iint_D \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \: dA \end{align}

In essence, Green's theorem relates the line integral around a simple and closed curve $C$ and the double integral over the region $D$ bounded by $C$.

We will now look at higher-dimensional version of Green's Theorem known as Stokes' Theorem which relates a surface integral over a surface $S$ to a line integral of the boundary of $S$, which is a space curve.

 Theorem 1 (Stokes' Theorem): Let $\delta$ be an oriented surface that is piecewise-smooth and bounded by the simple, closed, and positively oriented piecewise-smooth boundary curve $C$. Let $\mathbf{F}(x, y, z) = P(x, y, z) \vec{i} + Q(x, y, z) \vec{j} + R(x, y, z) \vec{k}$ be a vector field on $\mathbb{R}^3$ and let $P$, $Q$, and $R$ all have continuous partial derivatives in a region that contains $\delta$. Then $\oint_C \mathbf{F} \cdot d \vec{r} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S}$.

Note that $\oint_C \mathbf{F} \cdot d \vec{r} = \oint_C \mathbf{F} \cdot \hat{T} ds$ and $\iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot \hat{N} dS$. Therefore the line integral of the tangential component of $\mathbf{F}$ around the boundary curve $C$ of the surface $\delta$ is equal to the surface integral of the normal component of the curl of $\mathbf{F}$ over $\delta$.

• Proof: It will be sufficient to show that if $\delta$ is the union of a finite number of non-overlapping subsurfaces $\delta_1$, $\delta_2$, …, $\delta_n$, then the formula given in Theorem 1 holds for each of these subsurfaces.
• Assume that $\delta$ has a one-to-one normal projection onto the $xy$-plane. Let $\hat{N}$ be a corresponding normal field that points upward for the surface $\delta$. On the surface $\delta$ we will thus define a function $z = g(x, y)$ over a region $R$ of the $xy$-plane. The region $R$ will also have a boundary curve, call it $C^*$, and $C^*$ will be positively oriented as viewed from above.
• Now since the surface $\delta$ is generated by the function $z = g(x, y)$ then as seen on the Orientable Surfaces page, we will have that the natural normal field on $\delta$ is:
(2)
\begin{align} \hat{N} = \frac{-\frac{\partial g}{\partial x} \vec{i} - \frac{\partial g}{\partial y} \vec{j} + \vec{k}}{\sqrt{\left ( \frac{\partial g}{\partial x} \right )^2 + \left ( \frac{\partial g}{\partial y} \right )^2 + 1}} \end{align}
• Furthermore, the surface area element of $\delta$ will be:
(3)
\begin{align} \quad dS = \sqrt{\left ( \frac{\partial g}{\partial x} \right)^2 + \left ( \frac{\partial g}{\partial y} \right)^2 + 1} \: dA \end{align}
• Therefore we have that:
(4)
\begin{align} \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot \hat{N} \: dS \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \iint_R \left [ \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \vec{i} + \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) \vec{j} + \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k} \right ] \cdot \left [ \frac{-\frac{\partial g}{\partial x} \vec{i} - \frac{\partial g}{\partial y} \vec{j} + \vec{k}}{\sqrt{\left ( \frac{\partial g}{\partial x} \right )^2 + \left ( \frac{\partial g}{\partial y} \right )^2 + 1}} \right ] \sqrt{\left ( \frac{\partial g}{\partial x} \right)^2 + \left ( \frac{\partial g}{\partial y} \right)^2 + 1} \: dA \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \iint_R \left [ \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \vec{i} + \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) \vec{j} + \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \vec{k} \right ] \cdot \left [ \left ( - \frac{\partial g}{\partial x} \right ) \vec{i} + \left ( - \frac{\partial g}{\partial y} \right ) \vec{j} + \vec{k} \right ] \: dA \\ \quad \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} = \int_R \left [ \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \left ( - \frac{\partial g}{\partial x} \right ) + \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) \left ( \frac{\partial g}{\partial y} \right )+ \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \right ] \: dA \end{align}
• We will now evaluate the other side of the equation. We note that since $z = g(x, y)$ on the curve $C$ that then in applying the chain rule and differentiating both sides of $z = g(x, y)$ with respect to $t$ and we have that:
(5)
\begin{align} \quad \frac{dz}{dt} = \frac{\partial g}{\partial x} \frac{dx}{dt} + \frac{\partial g}{\partial y} \frac{dy}{dt} \\ \quad dz = \frac{\partial g}{\partial x} dx + \frac{\partial g}{\partial y} dy \end{align}
• We will use this substitution in evaluating the line integral below:
(6)
\begin{align} \quad \oint_C \mathbf{F} \cdot d \vec{r} = \int_{C^*} P(x, y, z) \: dx + Q(x, y, z) \: dy + R(x, y, z) \: dz \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = \int_{C^*} P(x, y, z) \: dx + Q(x, y, z) \: dy + R(x, y, z) \left ( \frac{\partial g}{\partial x} dx + \frac{\partial g}{\partial y} dy \right ) \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = \int_{C^*} \left [ P(x, y, z) + R(x, y, z) \frac{\partial g}{\partial x} \right ] \: dx + \left [ Q(x, y, z) + R(x, y, z) \frac{\partial g}{\partial y} \right ] \: dy \end{align}
• We will now apply Green's Theorem to get that:
(7)
\begin{align} \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_R \left [ \frac{\partial}{\partial x} \left (Q(x, y, z) + R(x, y, z) \frac{\partial g}{\partial y} \right ) - \frac{\partial}{\partial y} \left (P(x, y, z) + R(x, y, z) \frac{\partial g}{\partial x} \right ) \right ] \: dA \\ \quad \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_R \left [ \left ( \frac{\partial Q}{\partial x} + \frac{\partial Q}{\partial z} \frac{\partial g}{\partial x} + \frac{\partial R}{\partial x} \frac{\partial g}{\partial y} + \frac{\partial R}{\partial z} \frac{\partial g}{\partial x} \frac{\partial g}{\partial y} + R \frac{\partial^2 g}{\partial x \partial y} \right ) - \left ( \frac{\partial P}{\partial y} + \frac{\partial P}{\partial z} \frac{\partial g}{\partial y} + \frac{\partial R}{\partial y} \frac{\partial g}{\partial x} + \frac{\partial R}{\partial z} \frac{\partial g}{\partial y} \frac{\partial g}{\partial x} - R \frac{\partial^2 g}{\partial y}{\partial x} \right ) \right ] \: dA \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_R \left ( \frac{\partial Q}{\partial x} + \frac{\partial Q}{\partial z} \frac{\partial g}{\partial x} + \frac{\partial R}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial P}{\partial y} - \frac{\partial P}{\partial z} \frac{\partial g}{\partial y} - \frac{\partial R}{\partial y} \frac{\partial g}{\partial x} \right ) \: dA \\ \quad \oint_C \mathbf{F} \cdot d \vec{r} = \int_R \left [ \left ( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right ) \left ( - \frac{\partial g}{\partial x} \right ) + \left ( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right ) \left ( \frac{\partial g}{\partial y} \right )+ \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right ) \right ] \: dA \end{align}
• Therefore we have that:
(8)
\begin{align} \quad \oint_C \mathbf{F} \cdot d \vec{r} = \iint_{\delta} \mathrm{curl} (\mathbf{F}) \cdot d \vec{S} \quad \blacksquare \end{align}
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