Green’s Theorem will allow us to convert between integrals over regions in \(\R^2\), and line integrals over their boundaries. The region and boundary need to satisfy certain hypotheses. They allow a wide range of possible sets, so their purpose here is to avoid pathologies.
A set \(S\subset \R^n\) is a regular region if it is compact and \(S=\overline{S^{int}}\), that is, \(S\) is the closure of its interior.
A curve \(C\subset \R^n\) is a simple, closed curve if it has a parametrization \(\mathbf x = \mathbf g(t)\), \(a\le t \le b\) (with \(a<b\)) such that \(\mathbf g\) is continuous, \(\mathbf g(a) = \mathbf g(b)\) and if \(s<t\) and \(\mathbf g(s) = \mathbf g(t)\), then \(s=a\) and \(t=b\).
That is, the parametrization starts and ends at the same point but otherwise does not intersect itself.
A simple, closed curve \(C\) is piecewise smooth if it has a parametrization \(\mathbf g\) as above, and there exists a finite (possibly empty) set of points \(\left\{ t_1,\ldots, t_K\right\}\subset [a,b]\) such that
\(\mathbf g\) is continuously differentiable with \(\mathbf g'(t) \ne {\bf 0}\), except at points in \(\left\{t_1,\ldots, t_K\right\}\), and
\(\lim_{t\to t_k^+} \mathbf g'(t)\) and \(\lim_{t\to t_k^-} \mathbf g'(t)\) exist for every \(k\in \{1,\ldots, K\}\).
Note that the limits are vectors, so a curve in \(\mathbb R^2\) with a vertical tangent has a limit \([0,\lambda]\), even though \(\frac{dy}{dx}\) does not exist.
A piecewise smooth curve can have a finite number of corners or cusps.
A set \(S\subset \R^2\) has piecewise smooth boundary if \(\partial S\) can be written as a finite union of one or more disjoint, piecewise smooth, simple closed curves.
Let \(S\) be a set with piecewise smooth boundary. For any curve in \(\partial S\), the positive orientation is the direction along the curve which keeps \(S^{int}\) on the left.
If \(S\) does not have any holes, that is if \(\partial S\) consists of only one curve, this means that the positive orientation circles \(S\) is a direction that is (on the whole) counterclockwise.
If \(S\) is a regular region with piecewise smooth boundary and \(\partial S = \cup_{j=1}^J C_j\) (where each \(C_j\) is a simple, closed curve) then \[
\int_{\partial S} P\, dx + Q\, dy = \sum_{j=1}^J \int_{C_j} P\, dx + Q\, dy
\qquad\text{ where each }C_j \text{ is positively oriented.}
\]
Let \(S\subset \R^2\) be a regular region with a piecewise smooth boundary, and let \(\mathbf F\) be a \(C^1\) vector field on an open set that contains \(S\). \[
\int_{\partial S} \mathbf F\cdot d\mathbf x = \iint_S (\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2})\, dA.
\] Using the notation \(\mathbf F = (P, Q)\), \[
\int_{\partial S} P\, dx+ Q\, dy \ = \ \iint_S ( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\, dA.
\]
Sketch of the proof First, we will prove the theorem for regular regions of a special form. Then we will indicate how to prove the general case.
We will say that a regular region \(S\subset \R^2\) is \(x\)-simple if it can be defined by inequalities \[
\begin{aligned}a\le x\le b, \qquad \psi(x) \le y \le \phi(x)
\end{aligned}
\] for some \(C^1\) functions \(\psi, \phi: [a,b]\to \R\) with \(\psi(x)\le \phi(x)\) for \(x\in (a,b)\) (and maybe at the endpoints as well). Such a domain lies between the graphs of \(\psi\) and \(\phi\), as pictured below (with the positive orientation indicated by the red arrows):
We will show that if \(S\) is \(x\)-simple, then \[\begin{equation}
\int_{\partial S} P(x,y)\, dx = \iint_S -\frac {\partial P}{\partial y}(x,y)\, dA.
\label{greenx}\end{equation}\] The same argument will show that if \(S\) is \(y\)-simple (defined in a similar way, with the roles of \(x\) and \(y\) reversed) then \[\begin{equation}
\int_{\partial S} Q(x,y)\, dy = \iint_S \frac {\partial Q}{\partial x}(x,y)\, dA.
\label{greeny}\end{equation}\] The full theorem, for a general region \(S\), follows from these special cases via an argument that involves decomposing the region into pieces that are both \(x\)- and \(y\)-simple, applying the previous results to each piece. We will describe this argument below.
For an \(x\)-simple set \(S\), we can compute \(\int_{\partial S} P\,dx\) by breaking \(\partial S\) into 4 pieces, parametrized as follows (consistent with the positive orientation):
\((t, \psi(t))\), for \(t\) increasing from \(a\) to \(b\).
\((b, t)\) for \(t\) increasing from \(\psi(b)\) to \(\phi(b)\).
for the top of \(S\), the orientation goes from right to left, but it easier to parametrize with the opposite order, that is, by \((t,\phi(t))\) for \(a\le t\le b\), then to multiply by \(-1\), to correct for the orientation. (See Example 5 in Section 5.1.).
with the wrong orientation, \((a, t)\) for \(\psi(a)\le t\le \phi(a)\).
It is easy to see that the vertical pieces of the boundary contribute nothing (since on these segments, \(x\) is constant, so \(dx = x'(t)dt = 0\).) Thus \[
\begin{aligned}
\int_{\partial S} P(x,y)\, dx
&=
\int_a^b P(t, \psi(t))\,dt -
\int_a^b P(t, \phi(t))\,dt
\end{aligned}
\] On the other hand, by the fundamental Theorem of Calculus \[
\begin{aligned}
\iint_S \frac{\partial P}{\partial y}(x,y) dA
&=
\int_a^b \int_{\psi(x)}^{\phi(x)}\frac{\partial P}{\partial y}(x,y)dy\, dx\\
&=
\int_a^b P(x,\phi(x)) - P(x, \psi(x)) dx.
\end{aligned}
\] By comparing the last two equations, we conclude that holds if \(S\) is \(x\)-simple.
Exactly the same argument shows that if \(S\) is \(y\)-simple, then holds. To see why the sign is different, you should do it yourself; see the problems below.
In particular, if \(S\) is both\(x\)- and \(y\)-simple, then Green’s Theorem is valid on \(S\).
To illustrate the general case, consider the domain (call it \(S\)) pictured below, which is neither \(x\)- nor \(y\)-simple. The orientation is indicated by the red arrows.
We can split this up into two regions that are both \(x\)-simple and \(y\)-simple, as shown below (with orientations).
Let’s call the two halves \(S_1\) (above the diagonal line) and \(S_2\) (below). We already know that \[
\iint_S \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\, dA =
\iint_{S_1} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\, dA +
\iint_{S_2} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\, dA .
\] The key point is that \[\begin{equation}
\int_{\partial S}P\,dx+Q\,dy =
\int_{\partial S_1} P\,dx+Q\,dy +
\int_{\partial S_2}P\,dx+Q\,dy.
\label{keypoint}\end{equation}\] Using these two facts, and knowing that the theorem holds for both \(S_1\) and \(S_2\), we can deduce that it also holds for \(S\). The argument is the same for more complicated regions; they may simply have to be cut up into more pieces.
To see why holds, note that when we compute \(\int_{\partial S_1}P\, dx+Q\, dy\), we integrate along
the line segment from A to C
the line segment from C to D,
the line segment from D to A,
and add up the results. Similarly, when we compute \(\int_{\partial S_2}P\, dx+Q\, dy\), we integrate along
the line segment from A to B
the line segment from B to C,
the line segment from C to A.
Now we add all of these together. The key point is that the diagonal line is counted twice, with opposite orientations (A to C and C to A), and when these are added, the contributions cancel. What is left exactly adds up to \(\int_{\partial S} P\, dx+Q\, dy\).
This discussion omits numerous details but contains the main ideas of the proof.
Uses of Green’s Theorem
Green’s Theorem can be used to prove important theorems such as \(2\)-dimensional case of the Brouwer Fixed Point Theorem (in Problem Set 8). It can also be used to complete the proof of the 2-dimensional change of variables theorem, something we did not do (although this would be circular, because we required Change of Variables to show that line integrals did not depend on parameterization).
Green’s Theorem is frequently used to simplify computations, by transforming complicated integrals to simpler integrals. For example, it can happen that \(P,Q\) are quite complicated functions, and hard to integrate, but that \(\frac {\partial Q}{\partial x} -\frac{\partial P}{\partial y}\) is much simpler.
Example 1
Green’s Theorem implies that \[
\int_{\partial S} x\, dy = - \int_{\partial S} y\,dx = \int_{\partial S} \frac 12(x\, dy - y\, dx)
= \iint_S 1\, dA = \text{area}(S).
\]
Example 2
Let \(S\) be the region in the first quadrant of \(\R^2\) bounded by the curve \(y = 3-x^2+2x\). Compute \[
\int_{\partial S}(xy + \sin(e^y)) dx + x e^y \cos(e^y) dy
\]
Let’s write \(P(x,y) =xy + \sin(e^y)\) and \(Q(x,y) = x e^y \cos(e^y)\). Then \[
\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = e^y \cos(e^y) - \left[
x + e^y \cos(e^y)
\right] = -x.
\] Thus \[
\int_{\partial S}(xy + \sin(e^y)) dx + x e^y \cos(e^y) dy = -\iint_S x\, dA,
\] and the integral on the right-hand side is straightforward to compute.
Example 3
Suppose that \(f\) is a \(C^2\) function, and let \(S\subset \R^2\) be a regular region with piecewise smooth boundary. Consider \[
\int_{\partial S} \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy.
\] According to Green’s Theorem, \[
\int_{\partial S} \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy
=\iint_S \left( \frac{\partial}{\partial x} \frac{\partial f}{\partial y} - \frac{\partial }{\partial y} \frac{\partial f}{\partial x}\right)dA \ = \ \iint_S 0 \, dA = 0
\] using \(\frac{\partial^2 f}{\partial x\ \partial y} = \frac{\partial^2 f}{\partial y\ \partial x}\) by Clairault’s theorem, since \(f\) is \(C^2\).
Thus, if someone wanted to make up a integration problem that looks difficult but is completely feasible, they can
Start with some moderately complicated function, such as \(f(x,y) = y e^{x^3}\) or \(g(x,y) = \ln( 2+\sin(xy))\)
Make up a complicated-looking set, which is a regular region with piecewise smooth boundary, for example \[
S = \left\{ (x,y)\in \R^2 : x\ge 0, \ x^2 \le y \le 8 \sqrt x, \ \ x+y \le 3\right\} \ ;
\]
Then compute \[
\int_{\partial S} 3x^2 y e^{x^3}\, dx + e^{x^3}\, dy,
\] or \[
\int_{\partial S} \frac{y\cos(xy) }{2+\sin(xy) }\, dx + \frac{x\cos(xy) }{2+\sin(xy) }\, dy.
\]
Green’s Theorem will convert this difficult line integral into \(\iint_S 0 \, dA = 0\).
Example 4: “Moving the curve”
A variant of the above example: Let \(C_1\) be the set \[
\left\{ (x,y) : x^2+y^2 = 1, \ \ x\ge 0\right\}
\] oriented counterclockwise, and compute \[
\int_{C_1} 3x^2 y e^{x^3} dx + e^{x^3}\, dy.
\] To do this, let \(S= \left\{ (x,y) : x^2+y^2\le 1, \ \ x\ge 0\right\}\), Then \[
\partial S = C_1 \cup C_2\qquad\text{ for }C_2= \left\{(0,y) : -1\leq y\leq 1\right\} .
\] So \(C_2\) is parametrized by \(\gamma:[-1,1]\to \R^2\), \(\gamma(t)=(0,-t)\), which is positively oriented. Then Green’s Theorem says that \[\begin{align*}
\int_{C_1} 3x^2 y e^{x^3} dx + e^{x^3}\,dy +\int_{C_2} 3x^2 y e^{x^3} dx + e^{x^3}\,dy
&=
\int_{\partial S} 3x^2 y e^{x^3} dx + e^{x^3} \,dy \\
&=
\iint_S \left(\frac{\partial}{\partial x} e^{x^3} - \frac{\partial}{\partial y} 3x^2 y e^{x^3}\right)dA\nonumber \\
&=0.
\end{align*}\] Thus \[
\int_{C_1} 3x^2 y e^{x^3} dx + e^{x^3}\,dy
= -
\int_{C_2} 3x^2 y e^{x^3} dx + e^{x^3}\,dy.
\] But the integral on the right is easy to evaluate. (Do it!) So again we have succeeded in using Green’s Theorem to convert an impossible problem to a straightforward problem.
Example 5.
Next, another example in which we “move the curve”.
Let \(C\) be the curve consisting of 2 line segments: the segment from \((0,0)\) to \((3,4)\), followed by the segment from \((3,4)\) to \((2,0)\). For this \(C\), compute \[
\int_C (y^2 e^{xy^2}+y)\, dx + 2xy e^{xy^2}\, dy.
\]
Since this looks hard to evaluate, let’s again see whether Green’s Theorem can help. Set \(Q = 2xy e^{xy^2}\) and \(P= y^2 e^{xy^2}+y\). Then \[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -1.
\] Now let \(C'\) be the line segment from \((2,0)\) to \((0,0)\), and let \(S\) be the triangle with vertices \((0,0), (2,0)\), and \((3,4)\). Then paying attention to orientation, we see that \[
\int_{\partial S} P\, dx + Q\, dy = - \left(\int_{C} P\, dx + Q\, dy + \int_{C'} P\, dx + Q\, dy\right)
\] (since if we follow first \(C\) and then \(C'\) around the triangle \(S\), we circle it in a clockwise direction, whereas the positive orientation used for \(\partial S\) is counterclockwise.) Rearranging and using Green’s Theorem, we find that \[
\int_{C} P\, dx + Q\, dy =
-\int_{C'} P\, dx + Q\, dy - \iint_S(- 1)\, dA
\] Again, the integrals on the right-hand side are easily evaluated. For the second one, we can just compute the area of the triangle \(S\).
A reformulation of Green’s Theorem
Let \(S\) be a regular region in \(\R^2\), and at a point \(\mathbf x\in \partial S\), let \(\mathbf t(\mathbf x) = (t_1, t_2)\) denote the unit tangent vector to \(\partial S\), positively oriented. That is, if we rotate \(\mathbf t(\mathbf x)\) by 90 degrees counterclockwise, the resulting vector points into \(S\).
The vector \(\mathbf n(\mathbf x)\) defined by \(\mathbf n(\mathbf x) = (t_2, -t_1)\) is orthogonal to \(\partial S\) (that is, it’s orthogonal to every tangent vector to \(\partial S\) at \(\mathbf x\), since \(\mathbf n(\mathbf x)\cdot \mathbf t(\mathbf x) = 0\), and every tangent vector at \(\mathbf x\) is a multiple of \(\mathbf t(\mathbf x)\)), and our rule for “positive orientation” implies that \(\mathbf n(\mathbf x)\) points outward, since we have obtained \(\mathbf n(\mathbf x)\) from \(\mathbf t(\mathbf x)\) by rotating by 90 degrees in the clockwise direction. We therefore call \(\mathbf n(\mathbf x)\) the “outer unit normal” to \(\partial S\) at \(\mathbf x\). In this context, “normal” means the same thing as “orthogonal”, except that “normal” can be either a noun or an adjective, whereas “orthogonal” is always an adjective.
Green’s Theorem can be reformulated in terms of the outer unit normal, as follows:
Let \(S\subset \R^2\) be a regular domain with piecewise smooth boundary. If \(\mathbf F\) is a \(C^1\) vector field defined on an open set that contains \(S\), then \[
\iint_S \left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y}\right)\, dA
= \int_{\partial S} \mathbf F \cdot \mathbf n \, ds.
\]
Sketch of the proof We will prove the theorem for a set \(S\) whose boundary consists of a single piecewise-smooth curve.
Let \(\mathbf g:[a,b]\to \R^2\) be a positively oriented parametrization of \(\partial S\) such that \(|\mathbf g'(t)|\) never equals zero, and satisfying the conditions in the definitions of “piecewise smooth simple closed curve”.
Then the positively oriented unit tangent is given by \(\mathbf t = \frac {\mathbf g'(t)}{|\mathbf g'(t)|}\) at \(\mathbf g(t)\in \partial S\). (There may be finitely many points where \(\mathbf g'\) does not exist, but this is a set of zero content hence negligible for integration.) It follows that at the same point, \(\mathbf n = \frac 1{|\mathbf g'(t)|}\left[g_2'(t), -g_1'(t)\right]\).
Thus \[
\mathbf n (\mathbf g(t))\cdot \mathbf F(\mathbf g(t)) =
\frac{(g_2'(t), -g_1'(t))}{|\mathbf g'(t)|} \cdot (F_1, F_2)(\mathbf g(t))= \frac{\mathbf g'(t)}{|\mathbf g'(t)|}\cdot\tilde{\mathbf F}(\mathbf g(t)), \text{ for }\
\tilde{\mathbf F} = (-F_2, F_1).
\] It follows that \[
\int_{\partial S}\mathbf F\cdot \mathbf n \, ds =
\int_a^b \tilde{\mathbf F}(\mathbf g(t))\cdot \mathbf g'(t) \, dt =
\int_{\partial S} -F_2\, dx + F_1\, dy.
\] Now the conclusion follows from Green’s Theorem.
Problems
Basic
Compute \(\int_{\partial S} (x^2 +3xy)\, dx + (e^y - 2x^2) dy\) where \(S\) is the unit square \(S = [0,1]\times [0,1]\) in the \(xy\)-plane.
Compute \(\int_{\partial S}(x e^{x^2} + xe^y)dx + \cos x \, e^y\, dy\) where \(S\) is the triangle with vertices \((-2,0), (2,0)\) and \((0,3)\).
HintYou may be able to use parity considerations (even/odd functions) to simplify some integrals.
Let \(C\) be the unit circle, oriented counterclockwise, and evaluate \(\int_C (1+x+x^2+y^3)dx + (1- y+y^2-x^3)dy\).
Let \(C\) be the set in the \(xy\) plane where \(x^2+y^2 = 1\) and \(y\ge 0\), oriented counterclockwise. Evaluate \(\int_C -(3y^5 + 5x^2y^3)dx + (3x^5+5y^2x^3)dy\).
Compute \(\iint_S 1\, dA\), where \(S\) is the region between the \(x\)-axis and the curve \(\mathbf g(t) = \binom{x(t)}{y(t)} = \binom {(t-\sin t)}{1-\cos t}\) for \(0 \le t \le 2\pi\). (This curve is called a cycloid.)
Hint Convert to a line integral of a vector field. Example 1 may be helpful.
Compute \(\int_C (4x+ e^{\sin x}) dy + y\cos x e^{\sin x}\, dx\), where \(C\) is the curve consisting of the line segment from \((0,0)\) to \((1,3)\), followed by the line segment from \((1,3)\) to \((0,7)\)
Advanced
Find the simple closed curve \(C\) (oriented counterclockwise) that maximizes the line integral \[
\int_C y^3 \,dx+ (12x-x^3)dy
\]
Suppose that \(S\) is a set of the form\(\left\{(x,y) : c\le y \le d, \ \psi(y)\le x \le \phi(y)\right\}\) for some \(C^1\) functions \(\phi, \psi\) such that \(\psi(y) < \phi(y)\) for \(y\in (c,d)\). (That is \(S\) is what we called “\(y\)-simple”.) Modify arguments from the sketch of the proof of Green’s Theorem to show that \[
\int_{\partial S} Q(x,y) \, dy = \iint_S \frac {\partial Q}{\partial x}\, dA.
\]
Suppose that \(f:\R^2 \to \R\) is a \(C^2\) function and that \[
\frac{\partial^2 f}{\partial x^2} +
\frac{\partial^2 f}{\partial y^2} =0
\] everywhere in \(\R^2\). Prove that if \(S\) is any regular region with piecewise smooth boundary, then \[
\int_{\partial S} \nabla f \cdot \mathbf n\, ds = 0
\]
