$\newcommand{\R}{\mathbb R }$ $\newcommand{\N}{\mathbb N }$ $\newcommand{\Z}{\mathbb Z }$ $\newcommand{\bfa}{\mathbf a}$ $\newcommand{\bfb}{\mathbf b}$ $\newcommand{\bfc}{\mathbf c}$ $\newcommand{\bfe}{\mathbf e}$ $\newcommand{\bft}{\mathbf t}$ $\newcommand{\bfn}{\mathbf n}$ $\newcommand{\bff}{\mathbf f}$ $\newcommand{\bfF}{\mathbf F}$ $\newcommand{\bfk}{\mathbf k}$ $\newcommand{\bfg}{\mathbf g}$ $\newcommand{\bfG}{\mathbf G}$ $\newcommand{\bfh}{\mathbf h}$ $\newcommand{\bfu}{\mathbf u}$ $\newcommand{\bfv}{\mathbf v}$ $\newcommand{\bfx}{\mathbf x}$ $\newcommand{\bfp}{\mathbf p}$ $\newcommand{\bfy}{\mathbf y}$ $\newcommand{\ep}{\varepsilon}$
The statement of Green's Theorem require a lot of definitions, in order to state the hypotheses. In practice, these hypotheses will always be satisfied in this class.
a regular region is a compact set in $\R^n$ that is the closure of its interior.
a simple closed curve is a curve $C$ that has a parametrization $\bfx = \bfg(t)$, $a\le t \le b$ (with $a<b$) such that $\bfg$ is continuous and
A simple closed curve is piecewise smooth if it has a parametrization $\bfg$ as above, and there exists a finite (possibly empty) set of points $\{ t_1,\ldots, t_K\}\subset [a,b]$ such that
For $S\subset \R^2$, we say that $S$ has piecewise smooth boundary if $\partial S$ consists of one or more disjoint, peicewise smooth, simple closed curves
Now suppose $S$ is a regular region with piecewise smooth boundary consisting of one or more piecewise smooth curves. For one of these curves, the positive orientation is the orientation such that, is we stand at any point on the curve facing in the positively oriented direction, then $S^{int}$ is on our 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 $$ \int_{\partial S} P\, dx + Q\, dy := \sum_{j=1}^J \int_{C_j} P\, dx + Q\, dy \qquad\mbox{ where each }C_j \mbox{ is positively oriented.} $$
Theorem 1. (Green's Theorem) Let $S\subset \R^2$ be a regular region with a piecewise smooth boundary, and let $\bfF$ be a $C^1$ vector field on an open set that contains $S$. $$ \int_{\partial S} \bfF\cdot d\bfx = \iint_S (\frac{\partial F_2}{\partial x_1} - \frac{\partial F_1}{\partial x_2})\, dA. $$ In different notation, $$ \int_{\partial S} P\, dx+ Q\, dy \ = \ \iint_S ( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\, dA. $$
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 pictued 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):
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 \eqref{greenx} holds if $S$ is $x$-simple.
Exactly the same argument shows that if $S$ is $y$-simple, then \eqref{greeny} 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 \eqref{keypoint} holds, note that when we compute $\int_{\partial S_1}P\, dx+Q\, dy$, we integrate along
and add up the results. Similarly, when we compute $\int_{\partial S_2}P\, dx+Q\, dy$, we integrate along
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.
Green's Theorem can be used to prove important theorems such as $2$-dimensional case of the Brouwer Fixed Point Theorem. It can also be used to complete the proof of the 2-dimensional change of variables theorem, something we did not do. (You proved half of the theorem in a homework assignment.) These sorts of applications will explored in tutorials and the remaining homework assignment.
Green's Theorem is also often 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 $$ \int_{\partial S} x\, dy = - \int_{\partial S} y\,dx = \int_{\partial S} \frac 12(x\, dy - y\, dx) = \iint_S 1\, dA = \mbox{area}(S). $$
Example 2. Let $S$ be the region in the first quadrant of $\R^2$ bounded by the curve $y = 3-x^2+2x $, and 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: One way to make up a question for a math test.
Now assume that $f$ is a function of class $C^2$, and let $S\subset \R^2$ be a regular regioun 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, $$ \int_{\partial S} \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy =\iint_S \left( \frac{\partial}{\partial x} \frac{\partial f}{\partial y} - \frac{\partial }{\partial y} \frac{\partial f}{\partial x}\right)dA \ = \ \iint_S 0 \, dA = 0 $$ since $\frac{\partial^2 f}{\partial x\ \partial y} = \frac{\partial^2 f}{\partial y\ \partial x}$ when $f$ is of class $C^2$.
Thus, if math professors want to make up a question that looks difficult but is completely feasible, they can
If the professors have not made any mistakes in computing derivatives or typing the question, then 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
$$
\{ (x,y) : x^2+y^2 = 1, \ \ x\ge 0\}
$$
oriented counterclockwise, and compute
$$
\int_{C_1} 3x^2 y e^{x^3} dx + e^{x^3}\, dy.
$$
To do this, let $S:= \{ (x,y) : x^2+y^2\le 1, \ \ x\ge 0\}$,
Then
$$
\partial S = C_1 \cup C_2\qquad\mbox{ for }C_2:= \{(0,y) : |y|\le 1\} .
$$
So $C_2$ is the line segment connecting $(0,1)$ to $(0,-1)$ and oriented
from up to down, so to speak.
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\nonumber\\
&=
\iint_S \left(\frac{\partial}{\partial x} e^{x^3} - \frac{\partial}{\partial y} 3x^2 y e^{x^3}\right)dA\nonumber \\
&=0. \nonumber
\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 $$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \cdots = -1. $$ 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 \begin{align} \int_{C} P\, dx + Q\, dy&= -\int_{C'} P\, dx + Q\, dy - \iint_S(- 1)\, dA \end{align} Again, the integrals on the right-hand side are easily evaluated. (For the second one, we can just compute the area of the triangle $S$.)
Let $S$ be a regular region in $\R^2$, and at a point $\bfx\in \partial S$, let $\bft(\bfx) = (t_1, t_2)$ denote the unit tangent vector to $\partial S$, positively oriented. (That is, if we rotate $\bft(\bfx)$ by 90 degrees counterclockwise, the resulting vector points into $S$.)
The vector $\bfn(\bfx)$ defined by $\bfn(\bfx) := (t_2, -t_1)$ is orthogonal to $\partial S$ (that
is, it's orthogonal to every tangent vector to $\partial S$ at $\bfx$, since $\bfn(\bfx)\cdot \bft(\bfx) = 0$,
and every tangent vector at $\bfx$ is a multiple of $\bft(\bfx)$), and
our rule for positive orientation
implies that $\bfn(\bfx)$ points
outward, since we have obtained $\bfn(\bfx)$ from $\bft(\bfx)$ by rotating by 90 degrees in the clockwise direction. We therefore call $\bfn(\bfx)$ the outer unit normal
to $\partial S$ at $\bfx$.
(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:
Theorem 2. Let $S\subset \R^2$ be a regular domain with piecewise smooth boundary. If $\bfF$ is a $C^1$ vector field defined on an open set that contained $S$, then $$ \iint_S \left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y}\right)\, dA = \int_{\partial S} \bfF \cdot \bfn \, ds. $$
Sketch of the proof.
We will prove the theorem for a set $S$ whose boundary consists of a single piecewise smooth simple closed curve.
Let $\bfg:[a,b]\to \R^2$ be a positively oriented
parametrization of $\partial S$ such that $|\bfg'(t)|$ 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
$\bft = \frac {\bfg'(t)}{|\bfg'(t)|}$ at $\bfg(t)\in \partial S$.
(There may be finitely many points where $\bfg'$ does not
exist, but this is a set of zero content hence negligible for
integration.)
It follows that at the same point,
$\bfn = \frac 1{|\bfg'(t)|}(g_2'(t), -g_1'(t))$.
Thus
$$
\bfn (\bfg(t))\cdot \bfF(\bfg(t)) =
\frac{(g_2'(t), -g_1'(t))} {|\bfg'(t)|} \cdot (F_1, F_2)(\bfg(t))= \frac{\bfg'(t)}{|\bfg'(t)|}\cdot\tilde \bfF(\bfg(t)), \ \ \mbox{ for }\
\tilde \bfF := (-F_2, F_1).
$$
It follows that
$$
\int_{\partial S}\bfF\cdot \bfn \, ds =
\int_a^b \tilde \bfF(\bfg(t))\cdot \bfg'(t) \, dt =
\int_{\partial S} -F_2\, dx + F_1\, dy.
$$
Now the conclusion follos from Green's Theorem.
You should be able to use Green's Theorem to evaluate integrals,
either by changing a complicated line integral to a simple area integral $\iint_S \cdots dA$, or by moving the curve
as in Examples 4 and 5.
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 $(0,3)$.
Hint. You 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, and 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 $\bfg(t) = \binom{x(t)}{y(t)} = \binom {(t-\sin t)}{1-\cos t}$ 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 $(0,7)$
Find the simple closed curve $C$ (oriented counterclockwise) that maximizes the line integral $$ \int_C y^3 \,dx+ (12x-x^3)dy $$
Assume that $S$ is a set of the form$\{(x,y) : c\le y \le d, \ \psi(y)\le x \le \phi(y)\}$ 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
$$
\int_{\partial S} Q(x,y) \, dy = \iint_S \frac {\partial Q}{\partial x}\, dA.
$$
Assume 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 $$ \int_{\partial S} \nabla f \cdot \bfn\, ds = 0 $$