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Calculus 3 (Math 221) Notes and Study Guide

Brenton LeMesurier

Section 6.4 Green’s Theorem

Revised on April 11.

References.

Introduction.

The Fundamental Theorem of Calculus in the form
\begin{equation*} \int\limits_{[a,b]} \frac{df}{dx} dx = f(b)-f(a) \end{equation*}
gives the integral of the derivative of a function over an interval in terms of the values of the function itself at the "edges" of that interval.
Green’s Theorem is one of three results that extend this idea to multiple integrals, where the "edge" of a domain becomes a curve for double integrals and a surface for triple integrals, so the values at the edge must be integrated over such curves or surfaces.

Subsection 6.4.1 Simple Closed Curves, Positive Orientation, and Green’s Theorem

The simplest case is for a double integral over a region whose boundary is a simple closed curve, where simple means that the boundary curve does not intersect itself (except that its terminal point is the same as its initial point).
The value of a path integral can depend on the direction of movement along the path, which for a simple closed curve can be described as clockwise or anti-clockwise; to clarify, we specify a default direction of rotation:

Definition 6.4.1. Positive Orientation.

A simple closed curve has positive orientation if it is traversed anti-clockwise (the "trigonometric" direction).
Another way to think of this for a curve \(C\) that is the boundary of a region \(D\) is that when moving forward along the curve, the interior of the region is to the left.

Alternative Notation.

The use of positive orientation for the integral around a simple closed curve is sometimes indicated by \(\oint_C\) and the boundary of \(D\) is sometime denoted \(\partial D\text{,}\) giving the alternative notations
\begin{equation*} \int_C P dx + Q dy = \oint_C P dx + Q dy = \oint_{\partial D} P dx + Q dy \end{equation*}

Subsection 6.4.2 Partial Proof of Green’s Theorem

Green’s Theorem can be verifed in several stages:
  1. Verify Equation (6.4.1) on Type I regions. (This is where most of the work is.)
  2. Verify it on a general region by dividing it into a collection of Type I regions, as done for double integrals in Section 5.2.
  3. Verify (6.4.2) similarly, except starting with Type II regions.
  4. Combine these with \(f = P\) in (6.4.1) and \(f = Q\) in (6.4.2) to get Equation (6.4.3).

Subsection 6.4.2.1 Verifying Equation (6.4.1) on Type I regions

Consider a Type I domain \(D=\{(x,y)| a \leq x \leq b, B(x) \leq y \leq T(x) \}\text{.}\)
Its boundary is the oriented path \(\partial D = C_B+C_R+C_T+C_L\) where
  1. \(C_B = \{(x,B(x)), a \leq x \leq b\}\text{,}\) the bottom of \(D\) going left to right,
  2. \(C_R = \{(b,y), B(x) \leq y \leq T(x)\}\text{,}\) the vertical line at the right, if any,
  3. \(C_T = \{(x,T(x)), b \geq x \geq a\}\text{,}\) the top of \(D\text{,}\) going right to left, and
  4. \(C_L = \{(a,y), T(x) \geq y \geq B(x)\}\text{,}\) the vertical line at the left, if any.
The "side" integrals \(\int_{C_L} f\ dx\) and \(\int_{C_R} f\ dx\) are zero because \(x\) is constant on these paths, so
\begin{align*} \oint_{\partial D} f\ dx \amp= \int_{C_B} f\ dx + \int_{C_T} f\ dx\\ \amp= \int_{x=a}^b f(x,B(x))\ dx + \int_{x=b}^a f(x,T(x))\ dx\\ \amp= -\int_{x=a}^b \left[ f(x,T(x)) - f(x,B(x)) \right]\ dx \end{align*}
The Fundamental Theorem of Calculus gives
\begin{equation*} \int_{y=B(x)}^{T(x)} \frac{\partial f(x,y)}{\partial y}\ dy = f(x,T(x)) - f(x,B(x)) \end{equation*}
leading to
\begin{align*} \oint_{\partial D} f\ dx \amp= -\int_{x=a}^b \int_{y=B(x)}^{T(x)} \frac{\partial f}{\partial y}\ dy\ dx\\ \amp= -\iint\limits_D \frac{\partial f}{\partial y}\ dA \end{align*}
which is Equation (6.4.1).

Subsection 6.4.2.2 Verifying Equation (6.4.1) on more general domains

As discusssed in Section 5.2, integration over a more genereral domain in \(\reals^2\) can be done by cutting the domain into a collection of Type I domains. One method is to cut along each of the vertical lines that pass through points on the boundary where the tangent is vertical (which are the only points where there is a problem descibing the boundary with \(y\) as a function of \(x\text{,}\) as needed for "Type I")
The double integral of the whole domain is just the sum of these double integrals. On the other side of the equation, the boundaries of the new Type I pieces consist of:
  • pieces of the boundary of the original domain, and
  • the new edge pieces which arise in pairs on either side of each cut.
The first collection just add up to the original boundary curve, so give the desired path integral around the original boundary.
The motion along the new "internal" edges is still anticlockwise, so with the respective regions being on the opposite side, keeping the "inside" to the left requires that the motion be in the opposite direction on the two members of a pair, and so the integrals from each pair cancel out as seen in Subsection 6.2.3).

Subsection 6.4.2.3 Verifying Equation (6.4.2)

The second integral is similar; I sketch enough detail to show where the sign change arises.
This time one starts with a Type II domain \(D=\{(x,y)| c \leq y \leq d, L(y) \leq x \leq R(y) \}\text{,}\) whose boundary path is now \(\partial D = C_R+C_T+C_L+C_B\) with
  1. \(C_B = \{(x,c), L(c) \leq x \leq R(c)\}\text{,}\)
  2. \(C_R = \{(R(y),y), c \leq y \leq d\}\text{,}\) the right boundary of \(D\text{,}\) going upward,
  3. \(C_T = \{(x,d), R(d) \geq x \geq L(d)\}\text{,}\) and
  4. \(C_L = \{(L(y),y), d \geq y \geq c\}\text{,}\) the left boundary of \(D\text{,}\) going downward.
The top and bottom integrals vanish as the side integrals did above, leaving
\begin{align*} \oint_{\partial D} Q(x,y)\ dy \amp= \int_{C_R} Q(x,y)\ dy + \int_{C_L} Q(x,y)\ dy\\ \amp= \int_{y=c}^d Q(R(y),y)\ dy + \int_{x=d}^c Q(L(y),y)\ dy\\ \amp= \int_{y=c}^d Q(R(y),y) - Q(L(y),y)\ dy\\ \amp= \int_{y=c}^d \int_{x=L(y)}^{R(y)} \frac{\partial Q(x,y)}{\partial x}\ dx\ dy \quad \text{using the FTC again}\\ \amp= \iint\limits_D \frac{\partial Q}{\partial x}\ dA. \end{align*}

Subsection 6.4.3 An Application of Green’s Theorem: When the Cross-Partials Condition Implies That A Vector Field is Conservative

The cross-partials condition for a vector field to be conservative on a simply connected domain, Theorem 6.3.8, was left unproven in Section 6.3.
The inference in one direction is already given by Theorem 6.3.6 so we only need to verify that the cross-partials condition implies that the field is conservative.
We indicate why this is true by arguing that the cross-partials condition ensures that \(\int_C \vec{F} \cdot d\vec{r}=0\text{,}\) which was seen in Theorem 6.3.5 to be equivalent to being conservative.
If a closed path \(C\) is simple, it surrounds a simply connected domain \(D'\) within the simply connected domain \(D\text{,}\) because if anything inside \(C\) were not part of \(D\text{,}\) there would be a "hole" in \(D\text{.}\) (This is intuitive, but not a rigorous proof).
Green’s theorem then gives
\begin{equation*} \int_C \vec{F} \cdot d\vec{r} = \int_C P dx + Q dy = \int_{D'} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \, dA = \int_{D'} 0 \, dA = 0. \end{equation*}
Next, for a non-simple path, divide it into pieces that are simple closed curves by cutting at each point where it crosses itself (again this is intuitive but not really proved here). The path integral around \(C\) is thus zero because it is the sum of the integrals around each of these simple closed curves, and each of those integrals is zero as argued above.

Subsection 6.4.4 The Flux Form of Green’s Theorem

Proof.

\begin{equation*} \oint_{\partial D} \vec{F} \cdot \N\ ds = \oint_{\partial D} \vector{P, Q} \cdot \vector{dy/ds, -dx/ds}\ ds = \oint_{\partial D} -Q\ dx + P\ dy \end{equation*}
Swapping names \(P \to -Q\text{,}\) \(Q \to P\) in (6.4.3) gives
\begin{equation*} \oint_{\partial D} \vec{F} \cdot \N\ ds = \iint\limits_D \left( \frac{\partial P}{\partial x} - \frac{\partial (-Q)}{\partial y} \right)\ dA = \iint\limits_D \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}\ dA \end{equation*}

Subsubsection 6.4.4.1 Source-free Vector Fields and Their Stream Functions

Added on April 9.
Definition 6.4.4.
A vector field with no flux across any closed curve \(C\) is called source-free.
Just as having no circulation about all closed curves in a conected domain leads to a description of the vector field in terms of a single function (a potential function), any source-free vector field on such a domain can be described by single function:
Definition 6.4.5.
For a vector field \(\vec{F} = \vector{P, Q}\text{,}\) a function \(g\) satisfying
\begin{equation} P = \frac{\partial g}{\partial y} \text{ and } Q = -\frac{\partial g}{\partial x}\tag{6.4.5} \end{equation}
is a stream function for \(\vec{F}\).
Proof.
This can be seen by noting that \(\N\) is got by rotating \(\T\) a quarter turn clockwise, so rotating \(\vec{F} = \vector{P, Q}\) a quarter-turn anticlockwise to define \(\vec{G} = \vector{-Q, P}\) gives
\begin{align*} \vec{G} \cdot \T ds \amp= \vector{-Q, P} \cdot \vector{dx, dy} = P\ dy - Q\ dx = \vector{P, Q} \cdot \vector{dy, -dx}\\ \amp= \vec{F} \cdot \N ds \end{align*}
Thus the source-free condition on \(\vec{F}\) gives \(\oint_{C} \vec{G} \cdot \T\ ds = \oint_{C} \vec{F} \cdot \N\ ds = 0 \text{.}\) That is, by Theorem 6.3.5, \(\vec{G}\) is conservative: there is a function \(g\) with \(\del g = \vector{g_x, g_y} = \vec{G} = \vector{-Q, P}\text{,}\) so
\begin{equation*} P = \frac{\partial g}{\partial y} \text{ and } Q = -\frac{\partial g}{\partial x} \end{equation*}

Subsection 6.4.5 Green’s Theorem for Non-simply Connected Domains

If a domain \(D\) has "holes", it can intuitively be considered as coming from a larger domain \(E\) by removing one or more simply connected pieces \(H_1\text{,}\) \(H_2\text{,}\) etc., so that \(E = D \cup H_1 \dots\text{.}\)
Then intuitively
\begin{equation*} \iint\limits_E \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\ dA = \iint\limits_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\ dA + \iint\limits_{H_1} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\ dA + \cdots \end{equation*}
or
\begin{equation*} \iint\limits_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\ dA = \iint\limits_E \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\ dA - \iint\limits_{H_1} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\ dA \ \cdots \end{equation*}
All the integrals at right are over simply connected regions, so Green’s Theorem in the form of Equation (6.4.3) allows each to be written as a circulation.
This motivates

Proof.

The more careful proof of this is done by cutting with curves \(C_1\) and so on from the outer boundary of \(D\) to each of the holes, giving a simply connection region whose boundary consists of
  • The outer boundary \(\partial E\text{,}\) cut into several pieces.
  • The boundary of each hole, but traversed clockwise, so that the pieces are \(-\partial H_1\) and so on.
  • The two sides of each of the cut curves: \(C_1\text{,}\) \(-C_1\) and so on.
The cuts do not change what is in the region, so the double integral is unchanged. On the other hand, the integrals along the two sides of each cut cancel out, leaving the boundary circulation integral as
\begin{gather*} \oint_{\partial D} P\ dx + Q\ dy + \oint_{-\partial H_1} P\ dx + Q\ dy\ ...\\ = \oint_{\partial D} P\ dx + Q\ dy - \oint_{\partial H_1} P\ dx + Q\ dy\ ... \end{gather*}
Because of this, the boundary of such a non-simply connected region is sometimes denoted
\begin{equation*} \partial D = \partial E - \partial H_1 \dots \end{equation*}
and then Equation Theorem 6.4.2 still holds.

Study Guide.

Study Section 6.4 of OSC3
 2 
openstax.org/books/calculus-volume-3/pages/6-4-greens-theorem
; in particular
  • The concept of a positively oriented curve in the plane (anti-clockwise rotation)
  • Both versions of Green’s Theorem.
  • All Examples and Checkpoints.
  • One or several exercises from each of the following ranges: 146–151, 152–160, 161–169, 174–179.
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