6.4 Green's theorem and change of variables

Green's theorem says the circulation of a field around a closed curve equals the total spin inside, because the circulations around tiny cells add up and the inner edges cancel; it gives areas by line integrals, handles regions with holes through the winding number, and its proof for the area of an image gives the change of variables formula with the Jacobian determinant.

A circle of wind blows round a hill. How much does it push a cyclist who rides once round the foot of the hill? You can add up the push along the road, the line integral of 6.1. Or you can cut the area inside the road into tiny squares, measure how much the wind swirls in each, and add. Green’s theorem says the two answers are the same. It turns a one-dimensional integral into a two-dimensional one, and back, and the same idea turns the problem ‘what factor multiplies the area when you change coordinates’ into the Jacobian determinant.

Before you start

6.1: the circulation around a closed path. 6.3: type I and type II regions and iterated integrals. 5.4: the Jacobian determinant. 2.7: the determinant of a matrix.

Quick checkThe circulation of

F = (-y, x)

counterclockwise around the unit circle is

2\pi

. What is it clockwise?

Quick checkIn

\displaystyle\int_0^1\!\int_0^xf\,dy\,dx

, what is the region of integration?

Quick checkThe matrix

A = \begin{bmatrix}2 & 0\\ 0 & 3\end{bmatrix}

sends the unit square to a rectangle. By what factor does it multiply areas?

The picture: spin in every cell, cancellation on the edges

Each little cell has its own circulation: the push of the field around its four edges. Pull the cells apart and look at two neighbours: they run along their shared edge in opposite directions, so that edge contributes $+$ in one cell and $-$ in the other, and cancels in the sum. Only the outer edges are left. So the sum of all cell circulations equals the circulation around the whole boundary. And for a small cell the circulation is about the spin $Q_x - P_y$ times its area, so the sum is a Riemann sum for $\iint (Q_x - P_y)\,dA$ . The three numbers in the explanation line agree.

The idea, in words

For a field $F = (P, Q)$ and a tiny square of side $h$ , go once counterclockwise around it. The bottom edge pushes by about $P(x, y)\,h$ . The top edge, run backwards, contributes $-P(x, y + h)\,h$ . The two together are $-\big(P(x, y + h) - P(x, y)\big)h \approx -P_y\,h^2$ . The right and left edges give $+Q_x\,h^2$ in the same way. The circulation of the square is about $(Q_x - P_y)\,h^2$ : the field’s local spin, times area. Adding up the squares gives the theorem.

Why· Why the theorem is true, in one sentence, and why it needs a hole-free field

Inside the region everything cancels in pairs, so the total is carried by the boundary. That argument breaks exactly when the field is not defined, or not differentiable, at some point inside: then the small cell around that point has no honest spin. The angle field of 6.2 has $Q_x - P_y = 0$ everywhere it is defined, yet its circulation around the unit circle is $2\pi$ , all of it carried by the one cell around the origin that we are not allowed to cut.

Where this came from· Green, Riemann and the planimeter

George Green’s 1828 essay on electricity and magnetism contains an identity between an integral over a region of space and one over its boundary surface, in the form his theory of the potential needed. It was published privately in Nottingham and almost nobody noticed until William Thomson (Kelvin) found it in the 1840s and had it reprinted. The statement for a plane region, the one in this chapter, was first printed without proof by Cauchy in 1846, and Riemann proved it in his 1851 doctoral dissertation on functions of a complex variable. Meanwhile engineers had a machine that computes an area from the boundary: Johann Martin Hermann is credited with inventing the first planimeter in 1814, and in 1854 Jakob Amsler built the polar planimeter, in which you trace the boundary of a drawn region with a pointer and read the area off a wheel. The area formula $\frac12\oint(x\,dy - y\,dx)$ below is the mathematics behind that idea.

The statement

Definition T

Jordan curve

A curve $C \subset \mathbb R^2$ is a Jordan curve if it is traced by a path $\alpha: [a, b] \to \mathbb R^2$ with $\alpha(a) = \alpha(b)$ and $\alpha$ one-to-one on $[a, b)$ : a closed simple path (6.1). The Jordan curve theorem (stated in the lecture without proof) says that such a curve splits the plane into exactly two connected open pieces, a bounded interior and an unbounded exterior. The region $R$ in Green’s theorem is the interior together with $C$ .

In one lineA closed curve that does not cross or touch itself: $\alpha: [a, b] \to \mathbb R^2$ with $\alpha(a) = \alpha(b)$ and $\alpha$ one-to-one on $[a, b)$ .

Why it existsIt is the shape of a boundary: a loop with a clear inside and outside, which is what Green's theorem integrates around and over.

Definition T

Positive orientation

A Jordan curve $C$ is positively oriented (traversed counterclockwise) if, walking along $C$ in the direction of the parametrization, the interior lies on your left. The circle $\alpha(t) = (\cos t, \sin t)$ , $0 \le t \le 2\pi$ , is positively oriented. A rigorous definition through the winding number is given at the end of the section on holes.

In one lineA Jordan curve is positively oriented (counterclockwise) if the region it bounds lies on your left as you walk along it.

Why it existsGreen's theorem is stated for this direction; the opposite direction gives the opposite sign.

Theorem T MATH 376 lecture 4.20; Apostol Thm 11.10 (11.19)

Green's theorem

Let $C$ be a piecewise smooth Jordan curve, positively oriented, and $R$ the interior of $C$ together with $C$ . Let $S$ be an open set with $R \subset S$ and $F = (P, Q): S \to \mathbb R^2$ of class $C^1$ (continuous first partial derivatives). Then

\oint_C P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA.

In one lineIf $C$ is a positively oriented piecewise smooth Jordan curve bounding $R$ , and $P$ and $Q$ are $C^1$ on an open set containing $R$ , then $\oint_C P\,dx + Q\,dy = \iint_R (Q_x - P_y)\,dA$ .

Why it is trueIt trades a line integral for an area integral (or the reverse), whichever is easier, and it is the plane case of the big theorems of 6.5 and 6.6.

What goes wrong without each hypothesis. Check all of them every time.

Orientation. Traverse $C$ clockwise and the line integral changes sign. A problem that says ‘counterclockwise’ is giving you the positive orientation.
Closed. An open curve encloses no region.
Simple. If $C$ crosses itself, for example a figure eight, the two loops run in opposite senses, and the line integral equals the double integral over one loop minus the double integral over the other, not over the whole enclosed set.
Piecewise smooth. The line integral needs a velocity that is continuous on pieces (6.1).
$C^1$ on an open set containing all of $R$ , the inside included. The angle field $F = \frac{(-y, x)}{x^2 + y^2}$ is $C^1$ on the plane minus the origin and has $Q_x - P_y = 0$ there, so the right side of Green over the unit disc would be $0$ if the theorem applied; but the circulation around the unit circle is $2\pi$ . The theorem fails because $F$ is not defined at the origin, which is inside the circle: no open set containing $R$ is available.

Watch out· Four checks before you use Green

The curve is closed and simple. It is counterclockwise. The field is $C^1$ everywhere inside and on it. Then replace the line integral by $\iint (Q_x - P_y)\,dA$ .

Proof for a region that is of type I and type II

The lecture skipped the proof for time; worksheet 8 problem 6 proves it for a region of both types (6.3).

ProofMATH 376 worksheet 8 problem 6

The idea: Split the line integral into its P part and its Q part. For the P part use the type I description: only the bottom and top curves contribute, and the fundamental theorem of calculus turns their difference into the integral of P_y. For the Q part use the type II description, and add.

Let

R

be of both types.

R = \{a \le x \le b,\ g_1(x) \le y \le g_2(x)\} = \{c \le y \le d,\ h_1(y) \le x \le h_2(y)\}

with all four functions continuous. We prove

\oint P\,dx = -\iint P_y\,dA

and

\oint Q\,dy = \iint Q_x\,dA

and add.

The curve has four pieces: the bottom graph

y = g_1(x)

(traversed with

x

increasing), the right side, the top graph

y = g_2(x)

(traversed with

x

decreasing) and the left side. The sides may be single points.

Counterclockwise keeps the region on the left: along the bottom curve you move right, along the top curve you move left.

On the vertical sides

x

is constant, so they add nothing to

\oint P\,dx

dx = x'(t)\,dt

and

x' = 0

on a vertical segment.

\oint_CP\,dx = \displaystyle\int_a^bP(x, g_1(x))\,dx - \int_a^bP(x, g_2(x))\,dx = -\int_a^b\big[P(x, g_2(x)) - P(x, g_1(x))\big]\,dx

Parametrize each by

x

. Reversing the direction of the top piece flips the sign of its integral (6.1).

P(x, g_2(x)) - P(x, g_1(x)) = \displaystyle\int_{g_1(x)}^{g_2(x)}P_y(x, y)\,dy

Fundamental theorem of calculus in the variable

y

P_y

is continuous, so

P(x, \cdot)

is an antiderivative of

P_y(x, \cdot)

\oint_CP\,dx = -\displaystyle\int_a^b\!\int_{g_1(x)}^{g_2(x)}P_y\,dy\,dx = -\iint_RP_y\,dA

The type I integral formula of 6.3 for the continuous function

P_y

\oint_CQ\,dy = \displaystyle\int_c^dQ(h_2(y), y)\,dy - \int_c^dQ(h_1(y), y)\,dy

Now describe

R

as type II. Counterclockwise: up the right curve

x = h_2(y)

, down the left curve

x = h_1(y)

. The horizontal pieces have

dy = 0

= \displaystyle\int_c^d\!\int_{h_1(y)}^{h_2(y)}Q_x\,dx\,dy = \iint_RQ_x\,dA

Fundamental theorem of calculus in

x

for the continuous

Q_x

, then the type II integral formula.

\oint_CP\,dx + Q\,dy = \iint_R(Q_x - P_y)\,dA

\blacksquare

Add the two results.

Proof workshopRebuild the proof of Green's theorem for a region of both typesMATH 376 worksheet 8 problem 6

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

A more general region, such as an annulus, is cut by interior segments into finitely many pieces of this kind. Along each cut the two neighbouring pieces traverse the cut in opposite directions, so the line integrals along it cancel, and only the original boundary is left: this is the cancellation of the picture at the top of the chapter.

What is proved here, and what is not. The proof above covers regions of both types, and, by cutting, every region that can be cut into such pieces (discs, polygons, regions between curves, the annulus), which is every region met in practice. Green’s theorem for an arbitrary piecewise smooth Jordan curve is true, and Apostol 11.19 states it in that generality, but its complete proof is a topic for analysis (it rests on the Jordan curve theorem, which Veblen proved completely in 1905).

Worked example MATH 376 worksheet 8 problem 3

Compute

\oint_C xy^2\,dx + x^2y\,dy

where

C

is the boundary of the rectangle with vertices

(0, 0)

(2, 0)

(2, 3)

(0, 3)

, counterclockwise.

Solution.

Check the hypotheses. $C$ is closed, simple, piecewise smooth (four corners) and counterclockwise as stated. $P = xy^2$ and $Q = x^2y$ are polynomials, hence $C^1$ on the whole plane. Green applies.
Compute the spin. $Q_x = 2xy$ and $P_y = 2xy$ , so $Q_x - P_y = 0$ .
Integrate. $\oint_C = \iint_R0\,dA = 0$ .

Answer. $0$ .

Worked example MATH 376 worksheet 8 problem 5 and worksheet 9 problem 5

Find the work of

F = (y + \sqrt x,\ 2x + \cos(y^2))

around the triangle with vertices

(0, 0)

(1, 0)

(0, 1)

, counterclockwise. Then compute

\oint_C(xy + e^{x^2})\,dx + (x^2 - \ln(y^2 + 1))\,dy

around the same triangle.

Solution.

First field, spin. $P = y + \sqrt x$ , $Q = 2x + \cos(y^2)$ , so $Q_x = 2$ and $P_y = 1$ , and $Q_x - P_y = 1$ .
First field, integrate. The work is $\iint_R1\,dA$ , the area of the triangle: $\frac12$ . One subtlety: $\partial P/\partial x = \frac{1}{2\sqrt x}$ is not continuous at $x = 0$ and $\sqrt x$ is not defined for $x < 0$ , so no open set containing the triangle works and the theorem as stated here does not literally apply. Only $P_y$ and $Q_x$ appear in the formula, and they are fine, so the worksheet’s answer comes from Green, but the direct computation along the three edges is what proves it. It confirms $\frac23 + (-\frac16 + K) + (-K) = \frac12$ , where $K = \int_0^1\cos(t^2)\,dt$ cancels.
Second field, spin. $Q_x = 2x$ and $P_y = x$ , so $Q_x - P_y = x$ . Both components are $C^1$ on the plane ( $y^2 + 1 > 0$ ).
Second field, integrate. The triangle is $0 \le x \le 1$ , $0 \le y \le 1 - x$ : $\int_0^1\!\int_0^{1-x}x\,dy\,dx = \int_0^1x(1 - x)\,dx = \frac12 - \frac13 = \frac16$ .

Answer. $\dfrac12$ and $\dfrac16$ .

Worked example MATH 376 worksheet 9 problem 4 and worksheet 27 problem 3

Compute

\oint_Cy^3\,dx + (x^3 + 3xy^2)\,dy

where

C

is the circle

x^2 + y^2 = 4

, counterclockwise. Then compute the circulation of

F = (-y, x)

around the unit circle both directly and by Green.

Solution.

Spin. $P = y^3$ , $Q = x^3 + 3xy^2$ are polynomials. $Q_x = 3x^2 + 3y^2$ and $P_y = 3y^2$ , so $Q_x - P_y = 3x^2$ .
Polar coordinates over the disc of radius $2$ . $\iint3x^2\,dA = \int_0^{2\pi}\!\int_0^23r^2\cos^2\theta\cdot r\,dr\,d\theta = 3\int_0^{2\pi}\cos^2\theta\,d\theta\int_0^2r^3\,dr = 3\cdot\pi\cdot4 = 12\pi$ . The integrand factors into a function of $\theta$ times a function of $r$ over constant limits. Parametrizing the circle would give a harder integral.
Second problem, directly. $\alpha = (\cos t, \sin t)$ , $F(\alpha) = (-\sin t, \cos t)$ , $F \cdot \alpha' = 1$ , so the circulation is $\int_0^{2\pi}1\,dt = 2\pi$ .
Second problem, by Green. $P = -y$ , $Q = x$ , so $Q_x - P_y = 1 - (-1) = 2$ ; the field is $C^1$ on the plane; $\iint_{\text{disc}}2\,dA = 2\pi$ .

Answer. $12\pi$ and $2\pi$ (both ways).

Your answerCompute

\oint_C(-y^3)\,dx + x^3\,dy

around the unit circle, counterclockwise.

Quick checkGreen's theorem is applied to the unit circle traversed clockwise with

F = (-y, x)

. What does

\oint F \cdot dr

equal?

Quick check

F = \dfrac{(-y, x)}{x^2 + y^2}

has

Q_x - P_y = 0

, and

C

is the unit circle counterclockwise. Why is

\oint_C F \cdot dr = 2\pi

and not

0

Area by a line integral

To get an area, choose a field with $Q_x - P_y = 1$ ; then the right side of Green is $\iint_R1\,dA = \text{area}(R)$ .

Theorem T MATH 376 lecture 4.22; Apostol 11.20

Area as a line integral

Let $C$ be a positively oriented piecewise smooth Jordan curve $\alpha(t) = (X(t), Y(t))$ , $a \le t \le b$ , bounding $S$ . Then for any real $c$ , $\text{area}(S) = \int_a^b\big[(c - 1)Y(t)X'(t) + cX(t)Y'(t)\big]\,dt$ .

In one lineFor a positively oriented piecewise smooth Jordan curve $\alpha(t) = (X(t), Y(t))$ bounding $S$ , and any real $c$ , $\text{area}(S) = \int_a^b\big[(c - 1)YX' + cXY'\big]\,dt$ ; with $c = 1, 0, \frac12$ : $\text{area}(S) = \oint x\,dy = -\oint y\,dx = \frac12\oint(x\,dy - y\,dx)$ .

Why it is trueIt computes an area by walking once around the boundary, which is what the planimeter does and what a surveyor does with the coordinates of the corners.

ProofMATH 376 lecture 4.22

The idea: Apply Green's theorem to the field ((c-1) y, c x), whose spin is c minus (c-1), which is 1.

Take the field

F = ((c - 1)y,\ cx)

P = (c - 1)y

and

Q = cx

are polynomials, so

C^1

on the plane; the curve satisfies the other hypotheses.

Q_x - P_y = c - (c - 1) = 1

Q_x = c

and

P_y = c - 1

\oint_C F \cdot dr = \iint_R1\,dA = \text{area}(S)

, and the left side is

\int_a^b\big[(c - 1)YX' + cXY'\big]\,dt

\blacksquare

Green's theorem. The left side in terms of the parameter:

P\,x' + Q\,y' = (c - 1)YX' + cXY'

Worked example MATH 376 lecture example 4.23; worksheet 8 problem 2; worksheet 27 problem 4; homework 4 problem 4

Use the area formula with

c = 1

and with

c = \frac12

to find the area of the unit disc, then the area of the ellipse

x = a\cos t

y = b\sin t

using

\oint x\,dy

, and again using

\frac12\oint(x\,dy - y\,dx)

Solution.

Disc, $c = 1$ . $\alpha = (\cos t, \sin t)$ , so $XY' = \cos^2t$ and $\int_0^{2\pi}\cos^2t\,dt = \big[\frac t2 + \frac{\sin2t}{4}\big]_0^{2\pi} = \pi$ . The term with $(c - 1)$ vanishes.
Disc, $c = \frac12$ . The integrand is $-\frac12\sin t\cdot(-\sin t) + \frac12\cos t\cos t = \frac12$ , and $\frac12\cdot2\pi = \pi$ . Both agree with $\pi\cdot1^2$ .
Ellipse with $\oint x\,dy$ . The curve is a Jordan curve and increasing $t$ goes counterclockwise. $x\,dy = (a\cos t)(b\cos t)\,dt = ab\cos^2t\,dt$ , and $\int_0^{2\pi}ab\cos^2t\,dt = \pi ab$ .
Ellipse with $\frac12\oint(x\,dy - y\,dx)$ . $dx = -a\sin t\,dt$ and $dy = b\cos t\,dt$ , so $x\,dy - y\,dx = ab(\cos^2t + \sin^2t)\,dt = ab\,dt$ and the area is $\frac12\cdot ab\cdot2\pi = \pi ab$ .

Answer. The disc has area $\pi$ both ways, and the ellipse has area $\pi ab$ both ways.

Worked example MATH 376 homework 4 problem 5

A thin plate of uniform density occupies the triangle with vertices

(-1, 0)

(0, 1)

(1, 0)

. Use Green's theorem to find its centre of mass.

Solution.

Area. Base $2$ , height $1$ , so $A = 1$ . The centre of mass has $\bar x = \frac1A\iint x\,dA$ and $\bar y = \frac1A\iint y\,dA$ .
Choose fields with the right spin. For $\iint y\,dA$ take $P = -\frac{y^2}{2}$ , $Q = 0$ (spin $y$ ). For $\iint x\,dA$ take $P = 0$ , $Q = \frac{x^2}{2}$ (spin $x$ ). Both are $C^1$ on the plane.
Boundary, counterclockwise. $(-1, 0) \to (1, 0) \to (0, 1) \to (-1, 0)$ .
$\iint y\,dA = \oint-\frac{y^2}{2}\,dx$ . Base: $y = 0$ , so $0$ . Right edge from $(1, 0)$ to $(0, 1)$ : $x = 1 - t$ , $y = t$ , $dx = -dt$ , giving $\int_0^1\frac{t^2}{2}\,dt = \frac16$ . Left edge from $(0, 1)$ to $(-1, 0)$ : $x = t - 1$ , $y = t$ , $t$ from $1$ to $0$ , giving $\int_1^0-\frac{t^2}{2}\,dt = \frac16$ . Sum $\frac13$ , so $\bar y = \frac13$ .
$\iint x\,dA = \oint\frac{x^2}{2}\,dy$ . The base has $dy = 0$ . Right edge: $\int_0^1\frac{(1 - t)^2}{2}\,dt = \frac16$ . Left edge: $\int_1^0\frac{(t - 1)^2}{2}\,dt = -\frac16$ . Sum $0$ , so $\bar x = 0$ , as symmetry in the $y$ -axis says.

Answer. $(0, \frac13)$ : one third of the way up. Check: the width of the triangle at height $y$ is $2(1 - y)$ , so $\iint y\,dA = \int_0^1y\cdot2(1 - y)\,dy = \frac13$ .

Worked example Apostol 11.22 (exercise style)

Find the area of the astroid

x = a\cos^3t

y = a\sin^3t

0 \le t \le 2\pi

Solution.

Setup. The curve is simple and counterclockwise. $x' = -3a\cos^2t\sin t$ and $y' = 3a\sin^2t\cos t$ .
Choose the symmetric formula. $x\,y' - y\,x' = a\cos^3t\cdot3a\sin^2t\cos t + a\sin^3t\cdot3a\cos^2t\sin t = 3a^2\sin^2t\cos^2t\,(\cos^2t + \sin^2t) = 3a^2\sin^2t\cos^2t$ .
Integrate. Area $= \frac12\int_0^{2\pi}3a^2\sin^2t\cos^2t\,dt = \frac{3a^2}{2}\int_0^{2\pi}\frac{\sin^22t}{4}\,dt = \frac{3a^2}{2}\cdot\frac{\pi}{4}$ . $\sin t\cos t = \frac12\sin2t$ and $\int_0^{2\pi}\sin^22t\,dt = \pi$ .

Answer. $\dfrac{3\pi a^2}{8}$ .

Your answerA land parcel has corners, in order counterclockwise, at

(0, 0)

(120, 0)

(150, 80)

(40, 110)

(metres). Use the formula

\frac12\sum(x_iy_{i+1} - x_{i+1}y_i)

for the area in square metres.

Regions with holes, and the winding number

Beyond the course: for understanding· Apostol 11.21 to 11.25 (label A)

The course stated only the simplest version of Green’s theorem. This section is the rest of Apostol’s treatment: what happens when the region has holes.

Theorem A Apostol Thm 11.12 (11.23)

Green's theorem for regions with holes

Let $C_1, \dots, C_n$ be piecewise smooth Jordan curves such that no two intersect, $C_2, \dots, C_n$ lie in the interior of $C_1$ , and each $C_i$ lies in the exterior of $C_j$ for $i \ne j$ , $i, j > 1$ . Let $R$ be $C_1$ together with the part of its interior outside $C_2, \dots, C_n$ . If $P$ and $Q$ are $C^1$ on an open set containing $R$ , then

\iint_R\Big(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Big)dA = \oint_{C_1}P\,dx + Q\,dy - \sum_{k=2}^n\oint_{C_k}P\,dx + Q\,dy,

with every curve traversed counterclockwise. The minus signs are the clockwise orientation of the inner curves: the region stays on your left.

In one lineIf $C_1$ is a Jordan curve and $C_2, \dots, C_n$ are disjoint Jordan curves inside it, all counterclockwise, and $R$ is the region inside $C_1$ and outside the others, then $\iint_R(Q_x - P_y)\,dA = \oint_{C_1} - \oint_{C_2} - \dots - \oint_{C_n}$ of $P\,dx + Q\,dy$ , for $P, Q$ of class $C^1$ on an open set containing $R$ .

Why it is trueIt handles an annulus or a plate with holes: the outer curve counts positively, the inner curves negatively, and the field may be undefined inside the holes.

ProofApostol 11.23

The idea: For n = 2, cut the annulus by two crosscuts into two simply connected pieces, apply Green's theorem to each piece, and add. The crosscuts are traversed once in each direction and cancel.

Cut the region into an upper piece bounded by the Jordan curve

K_1

(upper halves of

C_2

and

C_1

and the two crosscuts) and a lower piece bounded by

K_2

Introduce two crosscuts

AB

and

CD

joining the inner and outer curves.

\iint_{R_i}(Q_x - P_y)\,dA = \oint_{K_i}P\,dx + Q\,dy

for

i = 1, 2

Green's theorem for each piece, which has no holes.

Adding, the crosscut integrals disappear, and the remaining parts of

K_1

and

K_2

are

C_1

counterclockwise and

C_2

clockwise.

Each crosscut is traversed once in each direction, so its two integrals cancel.

\iint_R(Q_x - P_y)\,dA = \oint_{C_1} - \oint_{C_2}

. The general

n

follows by induction on

n

\blacksquare

Additivity of the double integral over the two pieces.

Corollary A Apostol Thm 11.13 (11.25)

Invariance of a line integral under deformation

Let $P, Q$ be $C^1$ on an open connected set $S$ with $P_y = Q_x$ everywhere on $S$ . Let $C_1, C_2$ be piecewise smooth Jordan curves in $S$ with $C_2$ inside $C_1$ , and the points inside $C_1$ and outside $C_2$ in $S$ . Then $\oint_{C_1}P\,dx + Q\,dy = \oint_{C_2}P\,dx + Q\,dy$ (both traversed in the same direction).

In one lineIf $P_y = Q_x$ on an open connected set $S$ and $C_2$ lies inside $C_1$ with the region between them in $S$ , then $\oint_{C_1}P\,dx + Q\,dy = \oint_{C_2}P\,dx + Q\,dy$ , both counterclockwise.

Why it is trueIt lets you replace a complicated curve around a hole by a small circle, and it shows that for such fields the integral around a hole counts how often the curve winds.

The proof is the previous theorem with $Q_x - P_y = 0$ : the double integral is $0$ , so $\oint_{C_1} - \oint_{C_2} = 0$ .

Worked example MATH 376 (the argument for the angle field); Apostol 11.25, exercise style

Let

F = \dfrac{(-y, x)}{x^2 + y^2}

and let

C

be any counterclockwise Jordan curve with the origin inside it. Show

\oint_CF \cdot dr = 2\pi

. Then let

G = F_{(1, 0)} + 3F_{(-1, 0)}

, where

F_p(x) = F(x - p)

is the angle field about the point

p

, and compute

\oint G \cdot dr

around the circle of radius

3

about the origin and around small circles about each point.

Solution.

Cut a hole. Choose a small circle $C_2$ of radius $\rho$ about the origin lying inside $C$ . Let $R$ be the region between them. $F$ is $C^1$ on the plane minus the origin, and the origin is not in $R$ , so the hole version applies with $C_1 = C$ .
Apply the hole version. $\oint_CF \cdot dr - \oint_{C_2}F \cdot dr = \iint_R(Q_x - P_y)\,dA = \iint_R0\,dA = 0$ . $Q_x - P_y = 0$ was computed in 6.2.
Compute on the small circle. $\alpha(t) = (\rho\cos t, \rho\sin t)$ , $F(\alpha) = \frac{(-\rho\sin t, \rho\cos t)}{\rho^2}$ , and $F \cdot \alpha' = 1$ , so $\oint_{C_2}F \cdot dr = 2\pi$ . Hence $\oint_CF \cdot dr = 2\pi$ for every such $C$ .
The two-hole field. $G$ has spin $0$ away from the two points $(\pm1, 0)$ . Around the circle of radius $3$ , which contains both points, the hole version with two small circles gives $\oint G = 2\pi\cdot1 + 2\pi\cdot3 = 8\pi$ . Around a small circle about $(1, 0)$ that excludes $(-1, 0)$ : $2\pi$ . Around a small circle about $(-1, 0)$ : $6\pi$ . Each hole contributes its coefficient times $2\pi$ .

Answer. $2\pi$ ; and $8\pi$ , $2\pi$ , $6\pi$ for the three curves. This is also why a circulation of $0$ around circles about the missing point implies a potential for a field with $P_y = Q_x$ (the instructor’s remark on Final problem 1(a)): every closed curve is a whole number of laps around the hole.

Your answerWith

G = F_{(1, 0)} + 3F_{(-1, 0)}

as above, what is the circulation of

G

counterclockwise around the circle of radius

3

centred at the origin?

Definition A

Winding number

Let $r(t) = (X(t), Y(t))$ , $a \le t \le b$ , be a closed piecewise smooth path and $p = (x_0, y_0)$ a point not on it. The winding number of $r$ about $p$ is

W(r; p) = \frac1{2\pi}\int_a^b\frac{[X(t) - x_0]Y'(t) - [Y(t) - y_0]X'(t)}{[X(t) - x_0]^2 + [Y(t) - y_0]^2}\,dt = \frac1{2\pi}\oint\frac{-(y - y_0)\,dx + (x - x_0)\,dy}{(x - x_0)^2 + (y - y_0)^2}.

It is always an integer (the total change of the angle of $r(t) - p$ , divided by $2\pi$ ). If $C$ is a Jordan curve it is $0$ for $p$ outside $C$ , and for $p$ inside it is $+1$ at every inside point or $-1$ at every inside point; the first case is the positive orientation.

In one lineThe winding number $W(r; p)$ of a closed piecewise smooth path $r$ about a point $p$ not on it is $\frac1{2\pi}\oint\frac{-(y - y_0)\,dx + (x - x_0)\,dy}{(x - x_0)^2 + (y - y_0)^2}$ : the number of times $r$ goes counterclockwise around $p$ .

Why it existsIt makes 'goes around' exact: it is the circulation of the angle field about $p$ divided by $2\pi$ , and it is always a whole number.

Worked example Apostol 11.24 (definition and Figure 11.20); the limacon is new

Find the winding number of (a) the unit circle run twice counterclockwise about the origin; (b) the circle of radius

1

about

(3, 0)

, about the origin; (c) the limacon

r = 1 + 2\cos\theta

0 \le \theta \le 2\pi

, about the points

(1.5, 0)

(0.5, 0)

and

(-0.2, 0)

Solution.

(a) The circulation of the angle field is $2\pi$ per lap, so after two laps it is $4\pi$ and $W = 2$ . The lab in 6.2 gave $12.566 = 4\pi$ .
(b) The origin is outside the circle about $(3, 0)$ , which lies in the half plane $x > 2$ , where the angle $\theta$ is a single valued function, so the angle returns to its start without winding: $W = 0$ .
(c) Shape. The limacon has a big outer loop through $(3, 0)$ and a small inner loop through the origin and $(1, 0)$ (where $r = -1$ ).
(c) Count. A point inside the outer loop but outside the inner loop, such as $(1.5, 0)$ , has $W = 1$ ; a point inside the inner loop, such as $(0.5, 0)$ , is circled twice, $W = 2$ ; a point outside everything, such as $(-0.2, 0)$ , has $W = 0$ . Checked numerically by following the angle of $r(t) - p$ and dividing the total change by $2\pi$ : the values $1, 2, 0$ .

Answer. (a) $2$ ; (b) $0$ ; (c) $1$ , $2$ and $0$ .

Python labGreen's theorem, numerically, and where it fails

import numpy as np
c, s = np.cos, np.sin

def circulation(F, r, dr, n=20000):                  # line integral of F around the closed curve r(t), 0 <= t <= 2 pi
    t = np.linspace(0, 2 * np.pi, n + 1); m = (t[:-1] + t[1:]) / 2
    (x, y), (u, v) = r(m), dr(m); P, Q = F(x, y)
    return np.sum(P * u + Q * v) * 2 * np.pi / n

def double(curl, inside, box, n=1500):               # double integral of the curl over the region, midpoint grid
    x0, x1, y0, y1 = box
    X, Y = np.meshgrid(x0 + (np.arange(n) + .5) * (x1 - x0) / n, y0 + (np.arange(n) + .5) * (y1 - y0) / n)
    return np.sum(np.where(inside(X, Y), curl(X, Y), 0)) * (x1 - x0) * (y1 - y0) / n**2

ell = (lambda t: (2 * c(t), s(t)), lambda t: (-2 * s(t), c(t)))          # the ellipse x^2/4 + y^2 = 1, counterclockwise
inside = lambda X, Y: X**2 / 4 + Y**2 <= 1
box = (-2, 2, -1, 1)
F1 = lambda x, y: (-y**3, x**3)                                          # spin 3 x^2 + 3 y^2
F2 = lambda x, y: (y**2, x)                                              # spin 1 - 2 y
F3 = lambda x, y: (-y / (x**2 + y**2), x / (x**2 + y**2))                # the angle field, spin 0 away from the origin
for name, F, curl in [("(-y^3, x^3)", F1, lambda X, Y: 3 * X**2 + 3 * Y**2), ("(y^2, x)", F2, lambda X, Y: 1 - 2 * Y),
                      ("angle field", F3, lambda X, Y: 0 * X)]:
    print(f"F = {name:12s} line integral {circulation(F, *ell):9.4f}    integral of the spin {double(curl, inside, box):9.4f}")

Try this. Run it. For the ellipse x^2/4 + y^2 = 1 and F = (-y^3, x^3), the line integral around the boundary and the double integral of Q_x - P_y over the inside agree to about three decimals (both are 15*pi/2 = 23.562). For F = (y^2, x) the spin is 1 - 2y, whose integral over the symmetric ellipse equals the area 2*pi. For the angle field the spin is 0 but the line integral is 2*pi, because the field is not defined at the origin inside the ellipse. Change the curve to the circle of radius 0.5 centred at (2, 0), that is (2 + 0.5 cos t, 0.5 sin t), and the box and the inside test to match: the angle field now gives 0, because the origin is outside the curve and Green's theorem applies with spin 0.

Change of variables

In one variable, if $g: [a, b] \to \mathbb R$ is $C^1$ and $f$ is continuous on $g([a, b])$ , then $\int_{g(a)}^{g(b)}f(x)\,dx = \int_a^bf(g(u))\,g'(u)\,du$ : substitution. The factor $g'(u)$ says by how much $g$ stretches a tiny interval. In the plane, a transformation is a map $g(u, v) = (X(u, v), Y(u, v))$ from a region $T$ in the $(u, v)$ -plane to a region $S = g(T)$ in the $(x, y)$ -plane. Its derivative is the Jacobian matrix (5.3), and its determinant, the Jacobian determinant (5.4),

J(u, v) = \frac{\partial(x, y)}{\partial(u, v)} = \det\begin{bmatrix}X_u & X_v\\ Y_u & Y_v\end{bmatrix} = X_uY_v - X_vY_u

should be the area-stretching factor, the analogue of $g'$ . The columns are the partial derivatives with respect to $u$ and $v$ .

Slide the cell around and shrink its size $h$ . The purple patch is the true image of the little square; the green parallelogram is what the Jacobian matrix at the corner does to it. The ratio of areas is the Jacobian, up to a small error that goes to zero with $h$ .

Why· Why the weight is the absolute value of the Jacobian determinant

Put $f = 1$ ; the formula says $\text{area}(g(T)) = \iint_Tw\,du\,dv$ for some weight $w$ , which must be the local area stretching factor. A translation keeps areas, so $w = 1$ , and its Jacobian matrix is the identity. A linear map with matrix $A$ sends the unit square to the parallelogram spanned by the columns of $A$ , of area $|\det A|$ (2.7). A general $C^1$ map is close to its linear approximation $g(a + h) \approx g(a) + Dg(a)\,h$ on a tiny square, so there the factor is $|\det Dg(a)|$ . Area is positive, so we need the absolute value: $J$ is negative for a map that flips orientation.

Where this came from· Euler, Lagrange and Jacobi

Leonhard Euler studied double integrals in 1769 and was apparently the first to ask systematically how they change under a change of variables. A few years later Joseph-Louis Lagrange, working on the attraction of a solid body, needed a triple integral, found Euler’s method messy in three variables, and derived a general change of variables for triple integrals. The determinant that appears in the formula was studied for its own sake by Carl Jacobi in his 1841 paper on functional determinants (see 5.3), and carries his name. The best single account is V. Katz, Change of Variables in Multiple Integrals: Euler to Cartan (1982).

Theorem T MATH 376 lecture Thm 4.27 and the lecture of 16 February; Apostol 11.26 and 11.29 to 11.30

Change of variables in a double integral

Let $T = [a, b] \times [c, d]$ (or a more general region) and $g = (X, Y): T \to \mathbb R^2$ one-to-one and $C^1$ with $J(u, v) \ne 0$ for every $(u, v) \in T$ . If $f$ is continuous on $g(T)$ , then

\iint_{g(T)}f(x, y)\,dx\,dy = \iint_Tf\big(X(u, v), Y(u, v)\big)\,|J(u, v)|\,du\,dv.

In $\mathbb R^n$ the statement is the same with $T$ a box and $|\det Dg|$ .

In one lineIf $g: T \to S$ is one-to-one and $C^1$ with $J \ne 0$ (allowed to fail on a set of content zero) and $f$ is continuous on $S$ , then $\iint_Sf\,dx\,dy = \iint_Tf(X(u, v), Y(u, v))\,|J(u, v)|\,du\,dv$ .

Why it is trueIt lets you trade a region or integrand that is awkward in $(x, y)$ for a simple one in $(u, v)$ , at the price of one extra factor: the absolute Jacobian determinant.

What fails without each hypothesis.

One-to-one. If $g$ covers part of the image twice, the right side counts it twice. Example: $g(u, v) = (u^2, v)$ on $[-1, 1] \times [0, 1]$ has image $[0, 1]^2$ of area $1$ , but $\iint_T|2u|\,du\,dv = 2$ . Polar coordinates with $\theta \in [0, 4\pi]$ cover the disc twice.
$J \ne 0$ . Where $J = 0$ the map squashes a direction. For polar coordinates $J = r$ vanishes at $r = 0$ , and $g$ also fails to be one-to-one on the line $r = 0$ and on $\theta = 0$ versus $\theta = 2\pi$ ; these bad sets have area zero, so the formula still holds.
$C^1$ . The linear approximation needs continuous partial derivatives.
The absolute value. Dropping it is a common error (worksheet 9 problems 3 and 6, worksheet 27 problem 6).
The direction of the Jacobian. The formula needs $\partial(x, y)/\partial(u, v)$ : the old variables expressed through the new. If you solved for $u, v$ in terms of $x, y$ , what you computed is $\partial(u, v)/\partial(x, y)$ , the reciprocal.

The recipe

Choose a substitution $(u, v)$ that makes the region or the integrand simple.
Map the boundary. For a polygon, send each vertex to the $uv$ -plane and check that the sides map to the sides.
Write $x, y$ in terms of $u, v$ , compute $J = X_uY_v - X_vY_u$ and take $|J|$ .
Rewrite the integrand in $u, v$ .
Integrate over the new region with the right limits.

Map	$x, y$ in terms of the new variables	$\lvert J\rvert$
polar	$x = r\cos\theta,\ y = r\sin\theta$	$r$
stretched polar	$x = ar\cos\theta,\ y = br\sin\theta$	$abr$
scaling	$x = 2u,\ y = 3v$	$6$
sum and difference	$u = x - y,\ v = x + y$ , so $x = \frac{u + v}{2},\ y = \frac{v - u}{2}$	$\frac12$
cylindrical	$x = r\cos\theta,\ y = r\sin\theta,\ z = z$	$r$
spherical	$x = \rho\sin\varphi\cos\theta,\ y = \rho\sin\varphi\sin\theta,\ z = \rho\cos\varphi$	$\rho^2\sin\varphi$

Quick checkYou substitute

u = x + y

v = x - y

and compute

\partial(u, v)/\partial(x, y) = -2

. What factor multiplies the integrand in the

du\,dv

integral?

Worked example MATH 376 lecture 4.28; practice Final problem 3, see [A.5](/vol2/a5-practice-exams/)

Find the Jacobian determinant of polar coordinates and of

g(r, \theta) = r\big(\cos2\pi\theta,\ \tfrac12\sin2\pi\theta\big)

. Then compute the area of the elliptic ring

E = \{(x, y): 1 \le x^2 + 4y^2 \le 11\}

Solution.

Polar. $\det\begin{bmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{bmatrix} = r\cos^2\theta + r\sin^2\theta = r$ . Columns are $\partial/\partial r$ and $\partial/\partial\theta$ .
The practice map. $x = r\cos2\pi\theta$ , $y = \frac r2\sin2\pi\theta$ . Then $x_r = \cos2\pi\theta$ , $x_\theta = -2\pi r\sin2\pi\theta$ , $y_r = \frac12\sin2\pi\theta$ , $y_\theta = \pi r\cos2\pi\theta$ . Chain rule: the inner function $2\pi\theta$ has derivative $2\pi$ .
Determinant. $J = x_ry_\theta - x_\theta y_r = \pi r\cos^22\pi\theta + \pi r\sin^22\pi\theta = \pi r$ . The second product is $2\pi r\sin(\cdot)\cdot\frac12\sin(\cdot) = \pi r\sin^2(\cdot)$ , with a plus sign after subtracting the negative.
Turn the ring into a rectangle. $x^2 + 4y^2 = r^2\cos^22\pi\theta + r^2\sin^22\pi\theta = r^2$ , so the ring is $1 \le r \le \sqrt{11}$ , $\theta \in [0, 1)$ . $g$ is the polar map followed by a squeeze of $y$ by $\frac12$ , one-to-one for $r > 0$ and $C^1$ with $J = \pi r \ne 0$ for $r \ge 1$ : all hypotheses hold.
Integrate. $\text{Area}(E) = \iint_E1\,dx\,dy = \int_0^1\!\int_1^{\sqrt{11}}\pi r\,dr\,d\theta = \pi\cdot\frac{11 - 1}{2} = 5\pi$ .
Check. The ellipse $x^2 + 4y^2 \le \rho^2$ has half-axes $\rho$ and $\rho/2$ , area $\frac{\pi\rho^2}{2}$ , and $\frac{11\pi}{2} - \frac{\pi}{2} = 5\pi$ .

Answer. $J = r$ for polar, $J = \pi r$ for the practice map, and $\text{Area}(E) = 5\pi$ . (With the polar $J = r$ the disc of radius $R$ has area $\int_0^{2\pi}\!\int_0^Rr\,dr\,d\theta = \pi R^2$ , the lecture’s first use of the theorem.)

Worked example MATH 376 homework 4 problem 3; worksheet 9 problem 6

Find the area of the ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1

with

x = ar\cos\theta

y = br\sin\theta

. Then compute

\iint_R(x^2 + y^2)\,dA

over

\frac{x^2}{4} + \frac{y^2}{9} \le 1

with

x = 2u

y = 3v

Solution.

Jacobian. $\begin{bmatrix}a\cos\theta & -ar\sin\theta\\ b\sin\theta & br\cos\theta\end{bmatrix}$ has determinant $abr\cos^2\theta + abr\sin^2\theta = abr$ . $\frac{x^2}{a^2} + \frac{y^2}{b^2} = r^2$ , so the ellipse is $0 \le r \le 1$ .
Area. $\int_0^{2\pi}\!\int_0^1abr\,dr\,d\theta = \pi ab$ . One-to-one except on a set of area zero ( $r = 0$ and the cut at $\theta = 0$ ).
Second problem: scaling. The map $x = 2u$ , $y = 3v$ is linear with $|J| = 6$ , and the condition becomes $u^2 + v^2 \le 1$ .
Rewrite. $\iint_R(x^2 + y^2)\,dA = \iint_{u^2 + v^2 \le 1}(4u^2 + 9v^2)\cdot6\,du\,dv$ .
Polar in the $uv$ -plane. $6\int_0^{2\pi}\!\int_0^1(4r^2\cos^2\theta + 9r^2\sin^2\theta)\,r\,dr\,d\theta = 6\cdot\frac14(4\pi + 9\pi) = \frac{39\pi}{2}$ . $\int_0^1r^3\,dr = \frac14$ and $\int_0^{2\pi}\cos^2\theta\,d\theta = \int_0^{2\pi}\sin^2\theta\,d\theta = \pi$ .

Answer. $\pi ab$ and $\dfrac{39\pi}{2}$ .

Your answerUse

x = 2u

y = 3v

to find the area of the ellipse

\frac{x^2}{4} + \frac{y^2}{9} \le 1

Worked example MATH 376 worksheet 9 problem 3

Compute

\iint_R\dfrac{x - y}{x + y}\,dA

where

R

is the square with vertices

(0, 1)

(1, 2)

(2, 1)

(1, 0)

, using

u = x - y

v = x + y

Solution.

Map the vertices. $(x, y) \mapsto (u, v)$ : $(0, 1) \to (-1, 1)$ , $(1, 2) \to (-1, 3)$ , $(2, 1) \to (1, 3)$ , $(1, 0) \to (1, 1)$ . These are the corners of the rectangle $[-1, 1] \times [1, 3]$ . The sides of $R$ lie on the lines $x - y = \pm1$ and $x + y = 1, 3$ .
Solve for the old variables. Adding the definitions, $u + v = 2x$ ; subtracting, $v - u = 2y$ . So $x = \frac{u + v}{2}$ , $y = \frac{v - u}{2}$ . The formula needs $\partial(x, y)/\partial(u, v)$ , so $x, y$ must be written in terms of $u, v$ .
Jacobian. $\det\begin{bmatrix}\frac12 & \frac12\\ -\frac12 & \frac12\end{bmatrix} = \frac14 + \frac14 = \frac12$ , so $|J| = \frac12$ .
Rewrite the integrand. $\frac{x - y}{x + y} = \frac uv$ , continuous since $v \ge 1$ .
Integrate. $\int_1^3\!\int_{-1}^1\frac uv\cdot\frac12\,du\,dv = \frac12\int_1^3\frac1v\Big[\frac{u^2}{2}\Big]_{-1}^1dv = 0$ . The integrand is odd in $u$ over the symmetric interval $[-1, 1]$ .

Answer. $0$ .

Worked example MATH 376 worksheet 27 problem 6

Compute

\iint_D(x + y)\,dA

where

D

is the parallelogram that corresponds to

0 \le u \le 2

0 \le v \le 4

under

u = x + y

v = x - y

Solution.

Solve. $x = \frac{u + v}{2}$ , $y = \frac{u - v}{2}$ .
Jacobian. $\det\begin{bmatrix}\frac12 & \frac12\\ \frac12 & -\frac12\end{bmatrix} = -\frac14 - \frac14 = -\frac12$ . The Jacobian is negative, so $|J| = \frac12$ . The absolute value is required: area is positive, and this map reverses orientation.
Integrand. $x + y = u$ .
Integrate. $\int_0^2\!\int_0^4u\cdot\frac12\,dv\,du = \int_0^22u\,du = 4$ .

Answer. $4$ .

Worked example MATH 376 homework 4 problem 6; Topic 5 table

Compute the Jacobian determinants of cylindrical and spherical coordinates, and use the second to verify

dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta

Solution.

Cylindrical. For $(r, \theta, z) \mapsto (r\cos\theta, r\sin\theta, z)$ the matrix is $\begin{bmatrix}\cos\theta & -r\sin\theta & 0\\ \sin\theta & r\cos\theta & 0\\ 0 & 0 & 1\end{bmatrix}$ . Expand along the third column, which has a single non-zero entry $1$ with sign $+$ : $\det = 1\cdot(r\cos^2\theta + r\sin^2\theta) = r$ .
Spherical matrix. Rows $x, y, z$ , columns $\rho, \varphi, \theta$ : $\begin{bmatrix}\sin\varphi\cos\theta & \rho\cos\varphi\cos\theta & -\rho\sin\varphi\sin\theta\\ \sin\varphi\sin\theta & \rho\cos\varphi\sin\theta & \rho\sin\varphi\cos\theta\\ \cos\varphi & -\rho\sin\varphi & 0\end{bmatrix}$ .
Expand along the third row, with signs $+, -, +$ ; the third entry is $0$ . The first entry $\cos\varphi$ has minor $\rho^2\cos\varphi\sin\varphi(\cos^2\theta + \sin^2\theta) = \rho^2\sin\varphi\cos\varphi$ . The second entry $-\rho\sin\varphi$ (sign $-$ ) has minor $\rho\sin^2\varphi$ .
Combine. $\det = \cos\varphi\cdot\rho^2\sin\varphi\cos\varphi - (-\rho\sin\varphi)\cdot\rho\sin^2\varphi = \rho^2\sin\varphi(\cos^2\varphi + \sin^2\varphi) = \rho^2\sin\varphi$ .
Absolute value. On $0 \le \varphi \le \pi$ , $\sin\varphi \ge 0$ , so $|J| = \rho^2\sin\varphi$ .

Answer. $r$ and $\rho^2\sin\varphi$ .

Worked example Apostol 11.27, example 2

Compute

\iint_Se^{(y-x)/(y+x)}\,dx\,dy

where

S

is the triangle bounded by

x + y = 2

and the coordinate axes.

Solution.

Choose the substitution from the integrand. $u = y - x$ , $v = y + x$ , so $x = \frac{v - u}{2}$ , $y = \frac{v + u}{2}$ and $J = -\frac12$ , $|J| = \frac12$ .
Map the boundary. $x = 0$ becomes $u = v$ ; $y = 0$ becomes $u = -v$ ; $x + y = 2$ becomes $v = 2$ . The new region is the triangle $0 \le v \le 2$ , $-v \le u \le v$ .
Inner integral. $\int_{-v}^ve^{u/v}\,du = \big[ve^{u/v}\big]_{u=-v}^{u=v} = v(e - e^{-1})$ . Treat $v$ as a constant; the antiderivative of $e^{u/v}$ in $u$ is $ve^{u/v}$ .
Outer integral. $\iint e^{u/v}\cdot\frac12\,du\,dv = \frac12\int_0^2v(e - e^{-1})\,dv = \frac12(e - e^{-1})\cdot2$ .

Answer. $e - e^{-1}$ .

Worked example the Gaussian integral (label B); used in 8.1

Show that

\displaystyle\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt\pi

Solution.

Square it. $\big(\int e^{-x^2}\,dx\big)^2 = \iint_{\mathbb R^2}e^{-(x^2 + y^2)}\,dx\,dy$ . Separable integrand: the product of two single integrals (Fubini, extended to the plane as a limit of integrals over larger and larger discs).
Polar coordinates. $= \int_0^{2\pi}\!\int_0^\infty e^{-r^2}\,r\,dr\,d\theta = 2\pi\Big[-\tfrac12e^{-r^2}\Big]_0^\infty = \pi$ . The factor $r$ of the Jacobian is what makes the antiderivative elementary.
Square root. The integral is positive, so it equals $\sqrt\pi$ .

Answer. $\sqrt\pi$ . This is the normalising constant of the Gaussian density in 10.1.

Why the formula is true

Three layers, in the order Apostol gives them. The first layer is the one a good proof sketch should name; the second is a complete proof of the case that matters; the third is a sketch.

Theorem A Apostol 11.29

Area under a change of variables

Let $g = (X, Y)$ be one-to-one with continuous second partial derivatives and $J = X_uY_v - X_vY_u > 0$ on a region $T$ bounded by a piecewise smooth Jordan curve $C^*$ (positively oriented), and $S = g(T)$ with boundary $C = g(C^*)$ . Then $\text{area}(S) = \iint_TJ(u, v)\,du\,dv$ .

In one lineIf $g = (X, Y)$ has continuous second partials and $J > 0$ on a region $T$ bounded by a piecewise smooth Jordan curve, then the area of $g(T)$ equals $\iint_TJ\,du\,dv$ .

Why it is trueThis is the formula with $f = 1$ ; the rest of the change of variables theorem is deduced from it by a step-function argument, and its proof is Green's theorem.

ProofApostol 11.29

The idea: Write the area of S as a line integral of x dy around its boundary with Green's theorem. Parametrize the boundary through the (u, v) plane. Then apply Green's theorem again in the (u, v) plane; the mixed second partials cancel and what is left is the Jacobian.

\text{area}(S) = \oint_Cx\,dy

Green's theorem with

P = 0

Q = x

(spin

1

), for the curve

C

, positively oriented because

J > 0

preserves orientation (Apostol states this).

\oint_Cx\,dy = \oint_{C^*}X\,(Y_u\,du + Y_v\,dv)

Parametrize

C^*

(u(t), v(t))

; then

C

(X(u(t), v(t)), Y(u(t), v(t)))

, and the chain rule gives

dy = Y_uu'\,dt + Y_vv'\,dt

= \iint_T\big[(XY_v)_u - (XY_u)_v\big]\,du\,dv

Green's theorem in the

(u, v)

-plane with

P = XY_u

and

Q = XY_v

(these are

C^1

because

X, Y

are

C^2

(XY_v)_u - (XY_u)_v = X_uY_v - X_vY_u + X(Y_{vu} - Y_{uv})

Product rule:

(XY_v)_u = X_uY_v + XY_{vu}

and

(XY_u)_v = X_vY_u + XY_{uv}

So the integrand is

X_uY_v - X_vY_u = J

, and

\text{area}(S) = \iint_TJ\,du\,dv

\blacksquare

Equality of mixed partials for the

C^2

function

Y

(5.5):

Y_{vu} = Y_{uv}

Proof workshopGive the reason for each line: the area of an image is the integral of JApostol 11.29

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\text{area}(S) = \oint_C x\,dy$ .
$\oint_C x\,dy = \oint_{C^*} X\,(Y_u\,du + Y_v\,dv)$ .
$= \iint_T [(XY_v)_u - (XY_u)_v]\,du\,dv$ .
$(XY_v)_u - (XY_u)_v = X_uY_v - X_vY_u + X(Y_{vu} - Y_{uv})$ .
$Y_{vu} = Y_{uv}$ , so the integrand is $J$ .

From $f = 1$ to every $f$ . Apply the formula to a rectangle $R$ in the $xy$ -plane and its image $R^*$ in the $uv$ -plane under the inverse map: $\iint_Rdx\,dy = \iint_{R^*}|J|\,du\,dv$ . Multiply by a constant $c_{ij}$ and sum over the cells of a partition: this proves the formula for step functions $s$ on $R$ . Given an integrable $f$ and step functions $s \le f \le t$ , composing with $g$ keeps the inequalities, and multiplying by $|J| \ge 0$ keeps them too; so the number $\iint_{R^*}f(g(u, v))|J|\,du\,dv$ lies between $\iint s$ and $\iint t$ for all such $s, t$ , hence equals $\iint_Rf$ . Finally extend $f$ by zero from a general region to a rectangle containing it (Apostol 11.30). This is the sandwich argument of 6.3 again.

The lecture’s proof (label T, a sketch). The lecture of 16 February proved the theorem on a rectangle by a different route. By uniform continuity of the partial derivatives, for every $\varepsilon$ there is $\delta$ such that $\lVert g(p + h) - g(p) - Dg(p)\,h \rVert_\infty \le \varepsilon\lVert h \rVert_\infty$ for all $p$ and all $\lVert h \rVert_\infty < \delta$ . Cut the rectangle into cells of side below $\delta$ . On each cell the true image $g(R_k)$ lies within a band of width about $\varepsilon\delta$ of the parallelogram $Dg(p_k)(R_k)$ , whose area is $|J(p_k)|\,\text{area}(R_k)$ ; since $|J|$ has a positive minimum, the area error is a small multiple of $\varepsilon$ times the area of the cell. Summing, the integral of $f$ over $g(Q)$ is within $O(\varepsilon)$ of $\sum f(g(p_k))|J(p_k)|\,\text{area}(R_k)$ , which lies between the step integrals of the continuous function $(f \circ g)|J|$ . Since $\varepsilon$ is arbitrary, the two integrals are equal. This is a sketch: several constants are not tracked.

Where it is used

Used inside maths· Inside maths: Cauchy's theorem, Stokes, and densities under a change of variables

Green’s theorem is the plane case of Stokes’ theorem. The 376 midterm asked to deduce Green’s theorem from Stokes’ theorem for the flat disc $D \times \{0\}$ with normal $(0, 0, 1)$ and the field $(P, Q, 0)$ , whose curl is $(0, 0, Q_x - P_y)$ ; the full-credit solution is in A.1, where Stokes’ theorem is the given and this chapter’s statement is the target. Another consequence is a ‘normal form’ of Green’s theorem: applying it to $(-Q, P)$ shows that the flux of a field $F = (P, Q)$ out of a region equals $\iint(P_x + Q_y)\,dA$ , the divergence, which becomes the divergence theorem in 3-space. In complex analysis, $\oint f(z)\,dz = 0$ for analytic $f$ follows from Green’s theorem and the Cauchy–Riemann equations, and the winding number appears in the residue theorem. In probability, if $(X, Y)$ has density $f_{X,Y}$ and $(U, V) = h(X, Y)$ is one-to-one and smooth, then $f_{U,V}(u, v) = f_{X,Y}(h^{-1}(u, v))\,|\det Dh^{-1}(u, v)|$ : this chapter’s theorem, read as a statement about densities.

Used in the real world· Used in the real world: the planimeter, the surveyor's formula and the bell curve

A polar planimeter measures the area of a region on a map or an indicator diagram by tracing its boundary once with a pointer; the reading of its wheel is a line integral of the motion of the pointer along the boundary, and by Green’s theorem that line integral is proportional to the enclosed area. Land surveyors use the polygon version: for a parcel with corners $(x_i, y_i)$ listed counterclockwise, the area is $\frac12\sum(x_iy_{i+1} - x_{i+1}y_i)$ , because on the straight edge from $(x_i, y_i)$ to $(x_{i+1}, y_{i+1})$ the integral of $x\,dy - y\,dx$ is exactly $x_iy_{i+1} - x_{i+1}y_i$ . For corners $(0, 0)$ , $(120, 0)$ , $(150, 80)$ , $(40, 110)$ metres the area is $\frac12(0 + 9600 + 13300 + 0) = 11{,}450$ m $^2$ . The same formula is what graphics libraries use to find the signed area of a polygon, and a negative sign tells them the corners were listed clockwise.

The bell curve: the density of a standard normal variable is $\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ , and the constant $\frac1{\sqrt{2\pi}}$ comes from the Gaussian integral $\int e^{-x^2/2}\,dx = \sqrt{2\pi}$ , which is the polar-coordinate computation above (with $x^2/2$ in place of $x^2$ ). Without the factor $r$ from the Jacobian, no one would have found the integral in closed form. And when the heat in a circular plate is computed in polar coordinates, or the gravitational pull of a ball in spherical coordinates, the Jacobians $r$ and $\rho^2\sin\varphi$ are the factors that make the integrals tractable.

Traps

Watch out· Deduce means: say what each ingredient becomes

On the 376 midterm, a student were asked in problem 4(b) to ‘Deduce Green’s theorem from Stokes’ theorem’. They wrote Stokes’ theorem and the determinant formula for the curl, and the rubric item was ‘Stated Green’s theorem or Stokes’ theorem, otherwise no helpful steps’: 0.5 of 3. Quoting the big theorem is the first line of the argument. The rest: name the surface ( $D \times \{0\}$ ), its normal ( $(0, 0, 1)$ ), the field $(P, Q, 0)$ and its curl $(0, 0, Q_x - P_y)$ , check the orientation of the boundary, and conclude. The full solution is in A.1. A ‘deduce’ question is a checklist.

Watch out· Check all the hypotheses of Green, especially the inside

Orientation (counterclockwise, region on your left), closed and simple, and a $C^1$ field on an open set containing the inside as well as the boundary. The designed counterexample is the angle field around the origin: spin $0$ , circulation $2\pi$ . If the curve encloses a point where $F$ is undefined, cut it out with the hole version.

Watch out· Keep the absolute value, and use the right Jacobian

The factor is $|J|$ , never $J$ (worksheet 9 problems 3 and 6, worksheet 27 problem 6 all test this). And it is $\partial(x, y)/\partial(u, v)$ , the old variables in terms of the new. If you solved $u = x + y$ , $v = x - y$ and computed the matrix of $u, v$ in terms of $x, y$ (determinant $-2$ ), invert: the factor is $\frac12$ .

Watch out· Map the region, not only the integrand

Change the boundary as well as the integrand. For a polygon, map its vertices and check that the images of the sides are the sides of the new region (worksheet 9 problem 3 turns a tilted square into a rectangle). A substitution that simplifies the integrand but leaves a messy region is usually the wrong substitution.

Watch out· Do not forget r in polar coordinates

$dA = r\,dr\,d\theta$ . The same goes for $r$ in cylindrical and $\rho^2\sin\varphi$ in spherical coordinates. Polar limits are constants only for a disc, a sector or a ring centred at the origin.

Watch out· A curve that crosses itself

For a figure eight, Green’s theorem does not give the integral over the whole enclosed set: the loops run in opposite senses and the line integral is the integral over one loop minus the integral over the other. Split the curve into its simple loops, or use the winding number.

Problems

checkMATH 376 worksheet 8 problem 3 style

Compute $\oint_C x^2y\,dx + xy^2\,dy$ around the boundary of the square $[0, 1]^2$ , counterclockwise.

Solution

$P = x^2y$ , $Q = xy^2$ are polynomials. $Q_x - P_y = y^2 - x^2$ . $\iint_{[0,1]^2}(y^2 - x^2)\,dA = \frac13 - \frac13 = 0$ .

check

The Jacobian determinant of $(r, \theta) \mapsto (r\cos\theta, r\sin\theta)$ is $r$ . By what factor does polar coordinates multiply the area of a tiny cell at $r = 5$ ?

Solution

By $5$ . A cell of sides $dr$ and $d\theta$ is sent to a nearly rectangular patch with sides $dr$ and $5\,d\theta$ , of area $5\,dr\,d\theta$ .

check

True or false: if $Q_x - P_y = 0$ at every point of the plane minus the origin, then $\oint_CP\,dx + Q\,dy = 0$ for every closed curve in that set.

Solution

False. The angle field has $Q_x - P_y = 0$ there and circulation $2\pi$ around the unit circle. The statement is true for curves that do not wind around the origin (their inside stays in the domain).

checkMATH 376 worksheet 9 problem 3, step 1 only

The linear map $u = x - y$ , $v = x + y$ sends the triangle with vertices $(0, 0)$ , $(1, 0)$ , $(0, 1)$ to a triangle in the $uv$ -plane. Find its vertices and the Jacobian $\partial(x, y)/\partial(u, v)$ .

Solution

$(0, 0) \mapsto (0, 0)$ , $(1, 0) \mapsto (1, 1)$ , $(0, 1) \mapsto (-1, 1)$ . With $x = \frac{u + v}{2}$ , $y = \frac{v - u}{2}$ , the Jacobian is $\det\begin{bmatrix}\frac12 & \frac12\\ -\frac12 & \frac12\end{bmatrix} = \frac12$ . (Check: the new triangle has base $2$ and height $1$ , area $1$ , which is twice the area $\frac12$ of the old; so areas in the $(x, y)$ -plane are half those in the $(u, v)$ -plane, matching $|J| = \frac12$ .)

usenew (MATH 376 worksheet 9 problem 4 style)

Use Green’s theorem to compute the work of $F = (e^x - y^3,\ \sin y + x^3)$ around the circle $x^2 + y^2 = 1$ , counterclockwise.

Solution

$F$ is $C^1$ on the plane. $Q_x - P_y = 3x^2 + 3y^2$ . In polar coordinates $\iint3r^2\,dA = \int_0^{2\pi}\!\int_0^13r^3\,dr\,d\theta = \frac{3\pi}{2}$ .

useMATH 376 worksheet 27 problem 4 style

Find the area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{16} = 1$ by $\frac12\oint(x\,dy - y\,dx)$ .

Solution

$x = 3\cos t$ , $y = 4\sin t$ , $x\,dy - y\,dx = 12(\cos^2t + \sin^2t)\,dt = 12\,dt$ , so the area is $\frac12\cdot12\cdot2\pi = 12\pi$ ( $= \pi ab$ ).

useApostol 11.28 style

Compute $\iint_D\sqrt{x^2 + y^2}\,dA$ where $D$ is the half disc $x^2 + y^2 \le 4$ , $y \ge 0$ .

Solution

Polar: $\int_0^\pi\!\int_0^2r\cdot r\,dr\,d\theta = \pi\cdot\frac83 = \frac{8\pi}{3}$ .

use

Use $u = x + y$ , $v = y - 2x$ to compute $\iint_T(x + y)\,dA$ over the triangle $T$ with vertices $(0, 0)$ , $(1, 0)$ , $(0, 1)$ .

Solution

The vertices map to $(0, 0)$ , $(1, -2)$ , $(1, 1)$ . Solving: $u - v = 3x$ , so $x = \frac{u - v}{3}$ , and $y = u - x = \frac{2u + v}{3}$ . $J = \det\begin{bmatrix}\frac13 & -\frac13\\ \frac23 & \frac13\end{bmatrix} = \frac19 + \frac29 = \frac13$ . The new triangle is $0 \le u \le 1$ , $-2u \le v \le u$ , and the integrand is $u$ : $\int_0^1\!\int_{-2u}^uu\cdot\frac13\,dv\,du = \frac13\int_0^13u^2\,du = \frac13$ . Check directly: $\iint_T(x + y)\,dA = \int_0^1\!\int_0^{1-x}(x + y)\,dy\,dx = \frac13$ .

proveMATH 376 lecture 4.22 / Apostol 11.20

Use Green’s theorem to prove that $\frac12\oint(x\,dy - y\,dx)$ is the area of the region bounded by a positively oriented piecewise smooth Jordan curve.

Solution

Take $P = -\frac y2$ , $Q = \frac x2$ . They are $C^1$ on the plane, and $Q_x - P_y = \frac12 + \frac12 = 1$ . By Green’s theorem, $\oint P\,dx + Q\,dy = \iint_R1\,dA = \text{area}(R)$ , and the left side is $\frac12\oint(x\,dy - y\,dx)$ .

proveMATH 376 Midterm 2 problem 4(b) style (the Green side)

Using Green’s theorem, show that the area of a polygon with counterclockwise vertices $(x_1, y_1), \dots, (x_n, y_n)$ is $\frac12\sum_{i=1}^n(x_iy_{i+1} - x_{i+1}y_i)$ , with $(x_{n+1}, y_{n+1}) = (x_1, y_1)$ .

Solution

By the previous problem the area is $\frac12\oint(x\,dy - y\,dx)$ around the polygon. On the edge from $(x_i, y_i)$ to $(x_{i+1}, y_{i+1})$ parametrize $x = x_i + t(x_{i+1} - x_i)$ , $y = y_i + t(y_{i+1} - y_i)$ , $0 \le t \le 1$ . Then $dx = (x_{i+1} - x_i)\,dt$ and $dy = (y_{i+1} - y_i)\,dt$ , so $x\,dy - y\,dx = \big[(x_i + t\Delta x)\Delta y - (y_i + t\Delta y)\Delta x\big]\,dt$ with $\Delta x = x_{i+1} - x_i$ , $\Delta y = y_{i+1} - y_i$ . The terms in $t$ cancel ( $t\Delta x\Delta y - t\Delta y\Delta x = 0$ ), leaving $(x_i\Delta y - y_i\Delta x)\,dt = (x_iy_{i+1} - x_{i+1}y_i)\,dt$ , because $-x_iy_i + y_ix_i = 0$ . Integrating over $[0, 1]$ and summing over the edges gives the formula.

proveMATH 376 lecture 4.25 / Remark 4.26

Prove that a linear map $g(u, v) = A(u, v)^{\mathsf T}$ with invertible $A$ multiplies areas by $|\det A|$ , by finding the area of the image of the unit square.

Solution

$A(u, v)^{\mathsf T} = u\,c_1 + v\,c_2$ where $c_1, c_2$ are the columns of $A$ , so the image of $[0, 1]^2$ is the parallelogram spanned by $c_1$ and $c_2$ . Its area is $|\det(c_1, c_2)| = |\det A|$ (2.7); for example, with sides $(2, 0)$ and $(0, 3)$ it is a $2 \times 3$ rectangle of area $6 = \det\begin{bmatrix}2 & 0\\ 0 & 3\end{bmatrix}$ . The absolute value is needed because area is not negative. By translation invariance and additivity the same factor applies to every square, hence (by exhaustion with squares) to every region.

prove

Let $F = \dfrac{(-y, x)}{x^2 + y^2}$ . Prove that $\oint_CF \cdot dr = 0$ for any counterclockwise Jordan curve $C$ that does not enclose the origin.

Solution

The origin is outside $C$ , so it is not in the region $R$ bounded by $C$ , and $F$ is $C^1$ on the open set (plane minus origin) containing $R$ . By Green’s theorem, $\oint_CF \cdot dr = \iint_R(Q_x - P_y)\,dA = \iint_R0\,dA = 0$ .

proveMATH 376 homework 4 problem 6 and Topic 5 table

Show that a map $g: \mathbb R^2 \to \mathbb R^2$ of the form $g(r, \theta) = (r\cos\theta, r\sin\theta)$ restricted to $(0, \infty) \times [0, 2\pi)$ is one-to-one, and explain why the change of variables formula still applies on the whole disc, where $J = 0$ at the origin.

Solution

If $r_1(\cos\theta_1, \sin\theta_1) = r_2(\cos\theta_2, \sin\theta_2)$ with $r_i > 0$ , take lengths: $r_1 = r_2$ ; then $(\cos\theta_1, \sin\theta_1) = (\cos\theta_2, \sin\theta_2)$ forces $\theta_1 = \theta_2$ in $[0, 2\pi)$ . So $g$ is one-to-one there. On the closed rectangle $[0, R] \times [0, 2\pi]$ the map fails to be one-to-one on the sides $r = 0$ (all go to the origin) and $\theta = 0$ versus $\theta = 2\pi$ , and $J = r = 0$ on $r = 0$ . These sets are segments of the boundary of the rectangle, which have content zero, and the theorem allows $g$ to fail to be one-to-one, and $J$ to vanish, on a set of content zero (Apostol 11.26). So $\iint_{\text{disc}}f\,dA = \int_0^{2\pi}\!\int_0^Rf(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$ .

stretchApostol 11.25 (exercise style)

A curve goes twice counterclockwise around the unit circle and then once clockwise around the circle of radius $3$ about the origin (a closed path in two pieces, connected by a radial segment traversed both ways). Find its winding number about (a) the origin; (b) the point $(0, 2)$ ; (c) the point $(5, 0)$ .

Solution

The winding number adds over pieces, and the radial segment traversed both ways contributes nothing. (a) Origin: inside both circles, $2\cdot(+1) + 1\cdot(-1) = 1$ . (b) $(0, 2)$ : inside the circle of radius $3$ but outside the unit circle, so $2\cdot0 + (-1) = -1$ . (c) $(5, 0)$ : outside both, $0$ .

stretchnew (change of variables to a density)

The random point $(X, Y)$ is uniform on the unit disc, with density $\frac1\pi$ there. Use polar coordinates to find the density of the distance $R = \sqrt{X^2 + Y^2}$ and the expected distance $E[R]$ .

Solution

$P(R \le \rho) = \iint_{r \le \rho}\frac1\pi\,dA = \int_0^{2\pi}\!\int_0^\rho\frac1\pi r\,dr\,d\theta = \rho^2$ for $0 \le \rho \le 1$ . So the density of $R$ is $2\rho$ on $[0, 1]$ (the Jacobian factor $r$ is where the $2\rho$ comes from), and $E[R] = \int_0^1\rho\cdot2\rho\,d\rho = \frac23$ .

Summary

Green’s theorem: for a positively oriented (counterclockwise) piecewise smooth Jordan curve $C$ bounding $R$ , and $P, Q$ of class $C^1$ on an open set containing $R$ , $\oint_CP\,dx + Q\,dy = \iint_R(Q_x - P_y)\,dA$ . Check closed, simple, counterclockwise and $C^1$ inside. The proof for a region of both types integrates $P_y$ in $y$ and $Q_x$ in $x$ and uses the fundamental theorem of calculus; in general, the inner edges of small cells cancel.
Area: $\text{area}(R) = \oint x\,dy = -\oint y\,dx = \frac12\oint(x\,dy - y\,dx)$ (take a field with spin $1$ ). The same method gives centroids and the surveyor’s polygon formula.
Holes: for a region with holes, the outer curve counts positively and the inner curves negatively. For a field with $P_y = Q_x$ , a curve around a hole can be replaced by a small circle, so the angle field gives $2\pi$ times the winding number.
Change of variables: $\iint_{g(T)}f\,dx\,dy = \iint_Tf(g(u, v))\,|J|\,du\,dv$ with $J = \partial(x, y)/\partial(u, v)$ , for $g$ one-to-one and $C^1$ with $J \ne 0$ (except on a set of content zero). Use the absolute value, the old variables in terms of the new, and map the region.
Standard Jacobians: $r$ (polar and cylindrical), $abr$ (stretched polar), $\rho^2\sin\varphi$ (spherical), $|\det A|$ (linear map). They are the area or volume stretching factors.
Why it is true: a linear map multiplies areas by $|\det A|$ ; for $f = 1$ the area of an image equals $\iint J$ , proved by applying Green’s theorem twice (using equality of mixed partials); general $f$ follows by a step-function sandwich.
Uses: the Gaussian integral $\sqrt\pi$ , planimeters, densities of transformed random variables, and Cauchy’s theorem in complex analysis.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

6.4 Green's theorem and change of variables

The picture: spin in every cell, cancellation on the edges

The idea, in words

The statement

Jordan curve

Positive orientation

Green's theorem

Proof for a region that is of type I and type II

Area by a line integral

Area as a line integral

Regions with holes, and the winding number

Green's theorem for regions with holes

Invariance of a line integral under deformation

Winding number

Change of variables

Change of variables in a double integral

The recipe

Why the formula is true

Area under a change of variables

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere