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Advanced Engineering Mathematics Lecture Notes: West Virginia Wesleyan College

Jesse Oldroyd

Section 4.3 Green's Theorem

In this section we cover the relationship between line integrals and double integrals. This can be viewed as a generalization of the Fundamental Theorem of Calculus, which relates integrals \(\int_a^b f^\prime(x)\dd{x}\) to values of \(f(x)\) at the boundary points \(x = a\) and \(x = b\text{.}\) The main result of this section, Green's Theorem, relates double integrals of simple, closed and bounded regions \(R\) in the plane to line integrals along their boundaries \(\partial R\text{.}\) This is done for two types of line integrals: the circulation integral and the flux integral.

Subsection Green's Theorem for Circulation Integrals

Given a continuous vector field \(\FF = \mqty[P \amp Q]\) in \(\RR^2\text{,}\) we defined its circulation along a piecewise smooth curve \(C\) as
\begin{equation*} \int_C \FF\cdot\TT\dd{s} = \int_{C}\FF\cdot\dd{\rr}\text{.} \end{equation*}
If we write \(\rr = \mqty[x \amp y]\text{,}\) \(\dd{x} = x^\prime\dd{t}\) and \(\dd{y} = y^\prime\dd{t}\text{,}\) then this formula becomes
\begin{equation*} \int_C\FF\cdot\TT\dd{s} = \int_C P\dd{x} + Q\dd{y}\text{.} \end{equation*}
If \(C\) is also a closed curve, then we often replace \(\int_C\) with \(\oint_C\) to indicate such. Now we can state Green's Theorem.
Green's Theorem has a very nice geometric interpretation. Recall that the integral \(\int_{\partial R} P\dd{x} + Q\dd{y}\) is the circulation of \(\FF\) along \(\partial R\text{.}\) On the right hand side of the identity in Green's Theorem, the quantity \(\pdv{Q}{x} - \pdv{P}{y}\) is the two-dimensional curl of \(\FF\text{,}\) and so \(\iint_R (Q_x - P_y)\dd{A}\) is a measure of the “net (counterclockwise) rotation” of \(\FF\) within the region \(R\text{.}\) Therefore Green's Theorem implies that the net rotation of \(\FF\) inside the region \(R\) is exactly equal to the circulation of \(\FF\) along the boundary \(\partial R\text{.}\)
Beyond its geometric interpretation, Green's Theorem is also useful as a generalization of the Fundamental Theorem of Line Integrals for non-conservative vector fields in the plane, as the double integral that appears in the theorem is often easier to evaluate than the line integral it's equal to.

Example 4.3.2. Using Green's Theorem to Compute a Circulation Integral.

Compute \(\oint_{C}\FF\cdot\dd{\rr}\) where \(\FF = \mqty[-3y \amp 3x]\) and \(C\) is the triangle in \(\RR^2\) with vertices \((0,0), (1,0)\) and \((0,2)\) traversed exactly once counterclockwise.
Solution.
First we'll check to see if \(\FF\) is conservative as this would greatly simplify our calculations. Since its curl is \(6\text{,}\) we see quickly that \(\FF\) is not conservative, and so Theorem 4.2.3 and its related consequences do not apply to \(\FF\text{.}\) We could resort to computing \(\oint_C -3y\dd{x} + 3x\dd{y}\) using Equation (2), but this would require setting up three separate line integrals since \(C\) is piecewise smooth with three smooth components.
To avoid this, we'll make use of Green's Theorem. Let \(R\) denote the triangle contained with the curve \(C\text{.}\) Then
\begin{equation*} \oint_C -3y\dd{x} + 3x\dd{y} = \iint_{R}6\dd{A} = 6\text{,} \end{equation*}
where we have used the fact that
\begin{equation*} \iint_{R}\dd{A} \end{equation*}
is always equal to the area of \(R\text{.}\)
Green's Theorem can also be applied to clockwise-oriented curves, as long as you take care to multiply your integral by a negative.

Example 4.3.3. Green's Theorem and Clockwise Curves.

Find \(\oint_C xy^2\dd{x} + x^3\dd{y}\) where \(C\) is the rectangle with vertices \((0,0), (1,0), (0, 3)\) and \((1,3)\) traversed exactly once clockwise.
Solution.
To use Green's Theorem here, we multiply our integral by a negative. This has the effect of reversing the orientation of the curve \(C\) from clockwise to counterclockwise, allowing us to apply Theorem 4.3.1:
\begin{equation*} \oint_C xy^2\dd{x} + x^3\dd{y} = -\int_0^1\int_0^3 (3x^2 - 2xy)\dd{y}\dd{x}\text{.} \end{equation*}
An interesting use of Green's Theorem is in the computation of areas. Since the area of a region \(R\) is equal to \(\iint_R\dd{A}\text{,}\) then the area of \(R\) can be founded by computing the circulation integral \(\oint_{\partial R}P\dd{x} + Q\dd{y}\) of any vector field \(\FF\) with a curl exactly equal to \(1\text{.}\) An easy choice for this is \(\FF = \frac{1}{2}\mqty[-y \amp x]\text{,}\) which implies that the area of \(R\) can be found by computing
\begin{equation*} \frac{1}{2}\oint_{C} x\dd{y} - y\dd{x}\text{.} \end{equation*}

Example 4.3.4. Area Using Green's Theorem.

Find the area of an ellipse with major axis \(4\) and minor axis \(2\text{.}\)
Solution.
Such an ellipse can be given by the parameterization \(\rr(t) = \mqty[4\cos(t)\amp 2\sin(t)], 0\leq t\leq 2\pi\text{.}\) This gives
\begin{equation*} \frac{1}{2}\oint_C x\dd{y} - y\dd{x} = \frac{1}{2}\int_0^{2\pi} (8\cos^2(t) + 8\sin^2(t))\dd{t} = 16\pi \end{equation*}
for the area contained within the ellipse.

Subsection Green's Theorem for Flux Integrals

Circulation integrals of the form \(\int_C P\dd{x} + Q\dd{y}\) are useful for measuring the flow of a vector field \(\FF = \mqty[P\amp Q]\) along a curve \(C\text{,}\) but to measure the flow across or over \(C\) we instead make use of a flux integral.

Definition 4.3.5. Flux Integral.

Let \(\FF = \mqty[P \amp Q]\) be a continuous vector field defined over a piecewise smooth curve \(C\) with outward facing normal vector \(\mathbf{N}\text{.}\) The flux of \(\FF\) over \(C\) is defined to be
\begin{equation*} \int_C\FF\cdot\mathbf{N}\dd{s} = \int_C P\dd{y} - Q\dd{x}\text{.} \end{equation*}

Example 4.3.6. Computing a Flux Integral.

Let \(\FF = \mqty[x \amp y]\) and let \(C\) denote the line segment from \((0,1)\) to \((1,0)\text{.}\) Find the flux of \(\FF\) across \(C\text{.}\)
Solution.
To build some intuition, we'll begin by graphing both \(\FF\) and \(C\text{.}\)
The vector field F = [x, y] over the line segment C from (0,1) to (1,0).
Figure 4.3.7. The vector field \(\FF = \mqty[x \amp y]\) and the line segment \(C\text{.}\)
Since \(\FF\) appears to be flowing across \(C\) instead of along \(C\text{,}\) we expect that the flux will be nonzero. Furthermore, we'll see that the flux should be negative since \(\FF\) tends to flow to the left across \(C\) relative to the direction of \(C\text{,}\) whereas the outward normal \(\mathbf{N}\) points to the right.
Now we'll compute the flux using Definition 4.3.5. We first need to parameterize \(C\text{,}\) which we can do using \(\rr(t) = \mqty[t \amp 1-t], 0\leq t\leq 1\text{.}\) The flux is then
\begin{align*} \int_C P\dd{y} - Q\dd{x} \amp = \int_0^1[t(-1) - (1-t)]\dd{t} \\ \amp = -1 \end{align*}
Therefore \(\FF\) flows “into” \(C\text{,}\) relative to its orientation.
If \(R\) is a simple, closed and bounded region in the plane, then the normal vector \(\mathbf{N}\) in Definition 4.3.5 is always defined to be the outward normal. In particular, if \(\partial R\) is oreiented counterclockwise then has unit tangent \(\TT\text{,}\) then the region \(R\) will always lie to the left of \(\TT\) and the normal \(\mathbf{N}\) will always point to the right of \(\TT\text{.}\) This convention makes the flux integral a measure of flow outside of the region \(R\text{.}\) As net outflow is also measured by the divergence of a vector field, this suggests the following modification of Green's Theorem.

Example 4.3.9. Measuring Flux Within a Region.

A meteorologist determines that the wind patterns in a certain region are well-represented by the vector field \(\FF = \mqty[x - 2 \amp x + 1]\text{.}\) She wants to determine if the boundary of a rectangular plot of land within this region, represented by
\begin{equation*} R = \{(x, y) : -1\leq x\leq 2, -2 \leq y\leq 2\}\text{,} \end{equation*}
is suitable for the placement of wind turbines. How could she make this determination?
Solution.
One approach is to see if the flux of \(\FF\) over this region is nonzero, which by Theorem 4.3.8 must be equal to the flux of \(\FF\) across the boundary. The flux across the boundary is therefore
\begin{align*} \int_{\partial R} \FF\cdot\mathbf{N}\dd{s} \amp = \int_{-1}^{2}\int_{-2}^{2} \dd{y}\dd{x} \\ \amp = 12 \text{.} \end{align*}
As the flux across the boundary is nonzero, we conclude that it's not unreasonable to place turbines on the boundary of this region.
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