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Calculus 3 (Math 221) Notes and Study Guide

Brenton LeMesurier

Section 6.7 Stokes’ Theorem

References.

Introduction.

Green’s Theorem (Circulation Form) says that for \(D\) a region in the plane with boundary \(\C = \partial D\) that is a positively-oriented simply connected curve,
\begin{equation} \oint_{C} \vec{F} \cdot d\vec{r} = \iint\limits_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\ dA\tag{6.7.1} \end{equation}
Considering \(D\) as a surface in \(\reals^3\) that lies in the plane \(z=0\text{,}\) it has unit normal vector \(\N = \vec{k}\text{,}\) and in the notation of the previous section, \(\vec{k}\ dS = \vec{N}\ dS = d\vec{S}\text{.}\) Also, the integrand at right is the third component of the curl of \(\vec{F} = P\veci + Q\vecj\text{.}\) Thus, Equation (6.7.1) can be expressed as
\begin{equation} \oint_{C} \vec{F} \cdot d\vec{r} = \iint\limits_D ( \Curl~\vec{F} \cdot \veck ) dS = \iint\limits_D \Curl~\vec{F} \cdot d\vec{S}\text{.}\tag{6.7.2} \end{equation}
Stokes’ Theorem extends this to the case where \(D\) is replaced by an oriented surface \(S\) in \(\reals^3\) with continuous unit normal \(\N\) and bounded by the piecewise smooth curve \(\C = \partial S\text{,}\) with the integral around \(\C\) done with an orientation that is compatible with the choice of the unit normal for \(S\text{.}\)

Topics.

Subsection 6.7.1 Orientation of a Surface and its Boundary Curve

First, we need to extend the idea of "anti-clockwise" or the "positive" orientation of a planar closed curve.

Definition 6.7.1.

For an oriented surface with chosen continuous unit normal \(\N\) and whose boundary \(\C = \partial S\) is a smooth curve with unit tangent \(\T\text{,}\) the boundary is positively oriented with respect to the surface if \(\N \times \T\) points into the surface.
For example, for a planar region with upward normal \(\N = \veck\) and boundary curve parameterized with \(\T = \vector{dx/ds, dy/ds}\text{,}\) the cross-product \(\N \times \T = \vector{-dy/ds, dx/ds}\text{,}\) which is the vector got by rotating \(\T\) a quarter turn anti-clockwise, or to the left relative to the direction \(\T\text{:}\) this is indeed "into the surface" if the curve is traversed anti-clockwise.
Intuitively, if one imagines walking around the edge \(C\) of region \(S\) in the direction of the parameterization and with one’s “up” direction as indicated by \(\N\text{,}\) then the surface \(S\) is to one’s left.
Alternatively, if one views the surface from the side indicated by \(\N\text{,}\) the boundary is traversed anti-clockwise, matching our description of positive orientation for plane curves.
In the case where normal \(\N\) has positive \(z\) component, as when the surface is the graph of a function, positive orientation is moving anti-clockwise around \(C\) as seen from "above", meaning from higher \(z\) values.

Subsection 6.7.2 The Theorem, With a Partial Proof

Proof for the surface as the graph of a function.

We start by proving this in the case that the surface is a graph \(z=g(x,y)\text{,}\) \(x=g(y,z)\) or \(y=g(x,z)\)), and then deal with the general case by dividing the surface into a union of "easy pieces" of these three types. This mimics the strategy used to prove Theorem 6.4.2, where the easy pieces were regions of Type I or Type II.
First, using Equation (6.6.15) for \(d\vec{S}\) on the right-hand side,
\begin{align} \amp\iint\limits_S \Curl~\vec{F} \cdot \N\ dS\notag\\ \amp= \iint\limits_S \Curl~\vec{F} \cdot \vector{-{\partial g}/{\partial x}, -{\partial g}/{\partial y},1}\ dA\notag\\ \amp= \iint\limits_S \vector{\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}} \cdot \vector{-{\partial g}/{\partial x}, -{\partial g}/{\partial y},1}\ dA\notag\\ \amp= \iint\limits_S -\left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \frac{\partial g}{\partial x} -\left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \frac{\partial g}{\partial y} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \ dA\notag\\ \amp= \iint\limits_S \frac{\partial Q}{\partial z} \frac{\partial g}{\partial x} + \frac{\partial R}{\partial x} \frac{\partial g}{\partial y} +\frac{\partial Q}{\partial x} - \frac{\partial R}{\partial y}\frac{\partial g}{\partial x} - \frac{\partial P}{\partial z}\frac{\partial g}{\partial y} - \frac{\partial P}{\partial y} \ dA\tag{6.7.4} \end{align}
Next the contour integral along a parameterization \(x = x(t), y=y(t), z = g(x(t), y(t))\text{,}\) \(a \leq t \leq b\) of the boundary curve, with \((x(t), y(t))\) going anti-clockwise in the \(x\)–\(y\) plane, is
\begin{align*} \amp \int_{\partial D} \vec{F} \cdot d\vec{r}\\ \amp= \int_a^b \left( P \frac{dx}{dt} + Q \frac{dy}{dt} + R \frac{dz}{dt} \right) dt\\ \amp= \int_a^b \left( P \frac{dx}{dt} + Q \frac{dy}{dt} + R \left(\frac{\partial g}{\partial x}\frac{dx}{dt} + \frac{\partial g}{\partial y}\frac{dy}{dt} \right) \right) dt\\ \amp= \int_a^b \left( \left( P + R\frac{\partial g}{\partial x} \right) \frac{dx}{dt} + \left( Q + R\frac{\partial g}{\partial y} \right)\frac{dy}{dt} \right) dt\\ \amp= \int_a^b \left( P + R\frac{\partial g}{\partial x} \right) dx + \left( Q + R\frac{\partial g}{\partial y} \right) dy \end{align*}
and by Green’s theorem this gives
\begin{align*} \amp\int_{\partial D} \vec{F} \cdot d\vec{r}\\ \amp= \iint\limits_D \left[ \frac{\partial}{\partial x} \left( Q + R\frac{\partial g}{\partial y} \right) - \frac{\partial}{\partial y} \left( P + R\frac{\partial g}{\partial x} \right) \right] dA\\ \amp=\iint\limits_D \bigg[ \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)\\ \amp- \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) \bigg] dA \end{align*}
The two pairs of terms involving \(\displaystyle \frac{\partial R}{\partial z}\frac{\partial g}{\partial x}\frac{\partial g}{\partial y}\) and \(\displaystyle R \frac{\partial^2 g}{\partial x \partial y}\) cancel, leaving
\begin{align} \amp\int_{\partial D} \vec{F} \cdot d\vec{r}\notag\\ \amp=\iint\limits_D \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} dA\tag{6.7.5} \end{align}
The right-hand side here is is the same as in (6.7.4), completing the verification for surfaces \(z = g(x,y)\text{:}\) the same verification strategy clearly works for the two other "graph directions" \(x = g(y,z)\) and \(y = g(x,z)\text{.}\)

Proof sketch for the general case.

For other surfaces, the basic idea is to cut the surface up into pieces, each of which is a graph in one of the three directions. This will be described here for a parametric surface, or a collection of parametric surface pieces.
As noted in Subsection 6.6.4, Oriented Surfaces, a smooth level surface \(G(x,y,z) = 0\) with unit normal \(\N = \del G/\|\del G\|\) defined everywhere can be described as a collection of parametric surface pieces, (indeed, graphs of functions), so the result works there too.
Writing the normal as \(\N = \vector{n_x, n_y, n_z}\text{,}\) The Implicit Function Theorem in 3D can be used to show that if \(n_z \neq 0\) at a point of the surface \(\vec{r}(u,v) = \vector{x(u,v), y(u,v), z(u,v)}\text{,}\) one can solve nearby for the parameters \(u\) and \(v\) in terms of \(x\) and \(y\text{,}\) and thus get \(z\) in terms of \(x\) and \(y\text{:}\) this part of the surface is a graph \(z = g(x,y)\text{,}\) as handled in the first part of this proof.
Similarly, having either \(n_x \neq 0\) or \(n_y \neq 0\) at a point allows the surface to be descrbed as a graph in one of the other two directions.
Since \(\N \neq 0\) anywhere, at least one of these three conditions holds at every point on the surface, so every part of the surface can be covered by at least one of these types of graph.
This huge collection of "graph pieces" covers the surface many times over; then one can cut the collection and the domains down to cover each point once, except for overlap along the edges of their domains.
Then the sum of the surface integrals over each piece is the total surface integral, while the sum of the boundary integrals over the pieces is
  • the integral around the boundary of the whole surface, plus
  • various path integrals along the internal edges produced by cutting.
These internal edge path integrals come in pairs going in opposite directions (along the "opposite sides" of each cut), and so cancel out, leaving just that outside boundary integral, as needed.

Subsection 6.7.3 Velocity Fields, Circulation, and Curl

Stokes’ Theorem helps with the interpretion of the curl in the context of the velocity field of a fluid, \(\vec{v}\text{.}\)
The circulation of a velocity field \(\vec{v}\) around a curve \(\C\) is the net flow around this curve in the direction of the curve’s orientation, given by integrating the tangential part \(\vec{v} \cdot \vec{T}\) of the velocity with respect to arc length: \(\oint_C \vec{v} \cdot \vec{T}\ ds\text{.}\)
Stokes’ Theorem says that for any oriented surface \(S\) with this curve as boundary, positively oriented, this is
\begin{equation*} \oint_\C \vec{v} \cdot \vec{T}\ ds = \oint_\C \vec{v} \cdot d\vec{r} = \iint\limits_S \Curl~\vec{v} \cdot \vec{N} dS, = \iint\limits_S (\del \times \vec{v}) \cdot d\vec{S} \end{equation*}
so the circulation is given by integrating the part of the curl normal to the surface. (This shows a sense in which the curl measures anti-clockwise rotation in a velocity vector field.)

Subsection 6.7.4 The Connection Between “Curl-free” and Conservative

We can now at least indicate the validity of the second, converse, half of Theorem 6.5.4 in Section 6.5; the part that requires a simply connected domain.
The hardest part is showing that for any closed curve \(\C\) in the simply connected domain \(D\text{,}\) one can find a surface \(S\) with \(\C\) as its boundary. This takes some work, with the main idea being to shrink the curve down to a point, sweeping out a surface \(S\) as you go, so that \(\C = \partial S\text{.}\) Being simply connected is the condition which ensures that this shrinking can be done without the shrinking curve getting stuck around a hole in the domain.
Then Equation (6.7.3) gives
\begin{equation*} \oint_\C \vec{F} \cdot d\vec{r} = \oint_{\partial S} \vec{F} \cdot d\vec{r} = \iint\limits_S \Curl \vec{F} \cdot d\vec{S} = 0\text{,} \end{equation*}
making \(\vec{F}\) conservative as seen in Section 6.3.

Study Guide.

Study Section 6.7 of OSC3
 5 
openstax.org/books/calculus-volume-3/pages/6-7-stokes-theorem
; in particular
  • Stokes’ Theorem (of course).
  • Examples 73–75, and the Checkpoints following each.
  • One or several exercises from each of the ranges 326–331 and 332–335.
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