Case III. Complex conjugate roots are of minor practical importance, and we discuss the derivation of real solutions from complex ones just in terms of a typical example
10.4 Green’s Theorem in the Plane
Double integrals over a plane region may be transformed into line integrals over the boundary of the region and conversely. This is of practical interest because it may simplify the evaluation of an integral. It also helps in theoretical work when we want to switch from one kind of integral to the other. The transformation can be done by the following theorem.
T H E O R E M 1 Green’s Theorem in the Plane4
(Transformation between Double Integrals and Line Integrals)
Let R be a closed bounded region(see Sec. 10.3) in the xy-plane whose boundary C consists of finitely many smooth curves(see Sec. 10.1). Let and be functions that are continuous and have continuous partial derivatives
and everywhere in some domain containing R. Then
(1)
Here we integrate along the entire boundary C of R in such a sense that R is on the left as we advance in the direction of integration(see Fig. 234).
冮R冮a00Fx2 ⫺ 0F1
0y b dx dy⫽ 冯C(F1 dx⫹F2 dy).
0F2>0x
0F1>0y F2(x, y) F1(x, y)
4GEORGE GREEN (1793–1841), English mathematician who was self-educated, started out as a baker, and at his death was fellow of Caius College, Cambridge. His work concerned potential theory in connection with electricity and magnetism, vibrations, waves, and elasticity theory. It remained almost unknown, even in England, until after his death.
A “domain containing R” in the theorem guarantees that the assumptions about F1and F2at boundary points of Rare the same as at other points of R.
y
x C1 C2
R
Fig. 234. Region Rwhose boundary Cconsists of two parts:
is traversed counterclockwise, while is traversed clockwise in such a way that Ris on the left for both curves
C2 C1
Setting and using (1)in Sec. 9.9, we obtain (1) in vectorial form,
The proof follows after the first example. For 养see Sec. 10.1.
冮R冮(curl F)•kdx dy⫽ 冯CF•dr.
(1r)
F⫽[F1, F2]⫽F1i⫹F2j
E X A M P L E 1 Verification of Green’s Theorem in the Plane
Green’s theorem in the plane will be quite important in our further work. Before proving it, let us get used to
it by verifying it for and Cthe circle
Solution. In (1) on the left we get
since the circular disk Rhas area
We now show that the line integral in (1) on the right gives the same value, We must orient C
counterclockwise, say, Then and on C,
Hence the line integral in (1) becomes, verifying Green’s theorem,
P R O O F We prove Green’s theorem in the plane, first for a special region Rthat can be represented in both forms
(Fig. 235) and
(Fig. 236) c⬉y⬉d, p(y)⬉x⬉q(y)
a⬉x⬉b, u(x)⬉y⬉v(x)
䊏
⫽0⫹7p⫺0⫹2p⫽9p.
⫽ 冮02p(⫺sin3 t⫹7 sin2 t⫹2 cos2 t sin t⫹2 cos2 t) dt 冯C(F1xr⫹F2yr) dt⫽ 冮02p[(sin2 t⫺7 sin t)(⫺sin t)⫹2(cos t sin t⫹cos t)(cos t)] dt
F1⫽y2⫺7y⫽sin2 t⫺7 sin t, F2⫽2xy⫹2x⫽2 cos t sin t⫹2 cos t.
rr(t)⫽[⫺sin t, cos t], r(t)⫽[cos t, sin t].
9p. p.
冮R冮a00Fx2⫺0F1
0yb dx dy⫽冮R冮[(2y⫹2)⫺(2y⫺7)]dx dy⫽9冮R冮dx dy⫽9p
x2⫹y2⫽1.
F1⫽y2⫺7y, F2⫽2xy⫹2x
y
x v(x)
u(x) b a
R C**
C*
y
x p(y)
c q(y) d
R
Fig. 235. Example of a special region Fig. 236. Example of a special region
Using (3) in the last section, we obtain for the second term on the left side of (1) taken without the minus sign
(2) 冮R冮00Fy1 dx dy⫽ 冮abc冮u(x)v(x)00Fy1 dyd dx (see Fig. 235).
(The first term will be considered later.) We integrate the inner integral:
By inserting this into (2) we find (changing a direction of integration)
Since represents the curve (Fig. 235) and represents the last two integrals may be written as line integrals over and (oriented as in Fig. 235);
therefore, (3)
This proves (1) in Green’s theorem if .
The result remains valid if C has portions parallel to the y-axis (such as and in Fig. 237). Indeed, the integrals over these portions are zero because in (3) on the right we integrate with respect to x. Hence we may add these integrals to the integrals over and
to obtain the integral over the whole boundary Cin (3).
We now treat the first term in (1) on the left in the same way. Instead of (3) in the last section we use (4), and the second representation of the special region (see Fig. 236).
Then (again changing a direction of integration)
Together with (3) this gives (1) and proves Green’s theorem for special regions.
We now prove the theorem for a region Rthat itself is not a special region but can be subdivided into finitely many special regions as shown in Fig. 238. In this case we apply the theorem to each subregion and then add the results; the left-hand members add up to the integral over Rwhile the right-hand members add up to the line integral over Cplus
⫽ 冯CF2(x, y) dy.
⫽ 冮cdF2(q(y), y) dy⫹ 冮dcF2(p(y), y) dy
冮R冮00Fx2 dx dy⫽ 冮cdc冮p(y)q(y)00Fx2 dxd dy
C**
C*
C~ C~ F2⫽0
⫽ ⫺冯CF1(x, y) dx.
冮R冮00Fy1 dx dy⫽ ⫺冮C**F1(x, y) dx⫺ 冮C*F1(x, y) dx
C*
C**
C*, y⫽u(x)
C**
y⫽v(x)
⫽ ⫺冮baF1[x, v(x)] dx⫺ 冮abF1[x, u(x)] dx.
冮R冮00Fy1 dx dy⫽ 冮abF1[x, v(x)] dx⫺ 冮abF1[x, u(x)] dx
冮u(x)v(x)
0F1
0y dy⫽F1(x, y)2
y⫽v(x)
y⫽u(x)
⫽F1[x, v(x)]⫺F1[x, u(x)].
integrals over the curves introduced for subdividing R. The simple key observationnow is that each of the latter integrals occurs twice, taken once in each direction. Hence they cancel each other, leaving us with the line integral over C.
The proof thus far covers all regions that are of interest in practical problems. To prove the theorem for a most general region Rsatisfying the conditions in the theorem, we must approximate R by a region of the type just considered and then use a limiting process.
For details of this see Ref. [GenRef4] in App. 1.
Some Applications of Green’s Theorem
E X A M P L E 2 Area of a Plane Region as a Line Integral Over the Boundary
In (1) we first choose and then This gives
respectively. The double integral is the area Aof R. By addition we have
(4)
where we integrate as indicated in Green’s theorem. This interesting formula expresses the area of Rin terms of a line integral over the boundary. It is used, for instance, in the theory of certain planimeters(mechanical instruments for measuring area). See also Prob. 11.
For an ellipse or we get thus from
(4) we obtain the familiar formula for the area of the region bounded by an ellipse,
E X A M P L E 3 Area of a Plane Region in Polar Coordinates
Let rand be polar coordinates defined by Then
dx⫽cos u dr⫺r sin u du, dy⫽sin u dr⫹r cos u du, x⫽r cos u, y⫽r sin u.
u
䊏
A⫽1
2冮02p(xyr⫺yxr) dt⫽12冮02p[ab cos2 t⫺(⫺ab sin2 t)] dt⫽pab.
xr⫽ ⫺a sin t, yr⫽b cos t;
x⫽a cos t, y⫽b sin t x2>a2⫹y2>b2⫽1
A⫽1 2 冯C
(x dy⫺y dx)
冮R冮dx dy⫽冯Cx dy and 冮R冮dx dy⫽ ⫺冯Cy dx
F1⫽ ⫺y, F2⫽0.
F2⫽x F1⫽0, C**
C*
C C
y
x
y
x
Fig. 237. Proof of Green’s theorem Fig. 238. Proof of Green’s theorem
and (4) becomes a formula that is well known from calculus, namely,
(5)
As an application of (5), we consider the cardioid where (Fig. 239). We find
E X A M P L E 4 Transformation of a Double Integral of the Laplacian of a Function into a Line Integral of Its Normal Derivative
The Laplacian plays an important role in physics and engineering. A first impression of this was obtained in Sec. 9.7, and we shall discuss this further in Chap. 12. At present, let us use Green’s theorem for deriving a basic integral formula involving the Laplacian.
We take a function that is continuous and has continuous first and second partial derivatives in a domain of the xy-plane containing a region Rof the type indicated in Green’s theorem. We set
and Then and are continuous in R, and in (1) on the left we obtain (6)
the Laplacian of w(see Sec. 9.7). Furthermore, using those expressions for and we get in (1) on the right (7)
where sis the arc length of C, and Cis oriented as shown in Fig. 240. The integrand of the last integral may be written as the dot product
(8)
The vector nis a unit normal vector to C, because the vector is the unit tangent vector of C, and , so that nis perpendicular to . Also, nis directed to the exteriorof Cbecause in Fig. 240 the positive x-component of is the negative y-component of n, and similarly at other points. From this and (4) in Sec. 9.7 we see that the left side of (8) is the derivative of win the direction of the outward normal of C. This derivative is called the normal derivativeof wand is denoted by ; that is,
Because of (6), (7), and (8), Green’s theorem gives the desired formula relating the Laplacian to the normal derivative, (9)
For instance, satisfies Laplace’s equation Hence its normal derivative integrated over a closed curve must give 0. Can you verify this directly by integration, say, for the square 0⬉x⬉1, 0⬉y⬉1? 䊏
ⵜ2w⫽0.
w⫽x2⫺y2
冮R冮ⵜ2 w dx dy⫽冯C 00wn ds.
0w>0n⫽(grad w)•n.
0w/0n rr
dx>ds
rr rr•n⫽0
rr(s)⫽dr>ds⫽[dx>ds, dy>ds]
(grad w)•n⫽ c0w 0x, 0w
0yd • cdy ds,⫺dx
dsd ⫽0w 0x
dy ds⫺0w
0y dx ds.
冯C(F1 dx⫹F2 dy)⫽ 冯C aF1
dx ds⫹F2
dy
dsb ds⫽ 冯C a⫺0w 0y
dx ds⫹0w
0x dy dsb ds F2, F1
0F2 0x ⫺0F1
0y ⫽02w 0x2⫹02w
0y2⫽ ⵜ2w, 0F2>0x
0F1>0y F2⫽0w>0x.
F1⫽ ⫺0w>0y w(x, y)
䊏
A⫽a2
2 冮02p(1⫺cos u)2 du⫽32p a2.
0⬉u⬉2p r⫽a(1⫺cos u),
A⫽1 2冯Cr2 du.
y
x
y
x
R r
n C
'
Fig. 239. Cardioid Fig. 240. Example 4
Green’s theorem in the plane can be used in both directions, and thus may aid in the evaluation of a given integral by transforming the given integral into another integral that is easier to solve. This is illustrated further in the problem set. Moreover, and perhaps more fundamentally, Green’s theorem will be the essential tool in the proof of a very important integral theorem, namely, Stokes’s theorem in Sec. 10.9.
1–10 LINE INTEGRALS: EVALUATION BY GREEN’S THEOREM
Evaluate counterclockwise around the boundary Cof the region R by Green’s theorem, where
1. Cthe circle
2. R the square with vertices
3. Rthe rectangle with vertices (0, 0), (2, 0), (2, 3), (0, 3)
4.
5.
6.
7. Ras in Prob. 5
8. Rthe semidisk
9.
10.
Sketch R.
11. CAS EXPERIMENT. Apply (4) to figures of your choice whose area can also be obtained by another method and compare the results.
12. PROJECT. Other Forms of Green’s Theorem in the Plane. Let Rand Cbe as in Green’s theorem, a unit tangent vector, and nthe outer unit normal vector of C(Fig. 240 in Example 4). Show that (1) may be written
(10)
or
(11) 冮R冮(curl F)•kdx dy⫽ 冯CF•rrds 冮R冮div Fdx dy⫽ 冯CF•n ds
rr
y⭌x.
F⫽[x2y2, ⫺x>y2], R: 1⬉x2⫹y2⬉4, x⭌0, F⫽[ey>x, ey ln x⫹2x], R: 1⫹x4⬉y⬉2
x⭌0 x2⫹y2⬉16,
F⫽[⫺eⴚx cos y, ⫺eⴚx sin y], F⫽grad (x3 cos2 (xy)),
F⫽[cosh y, ⫺sinh x], R: 1⬉x⬉3, x⬉y⬉3x F⫽[x2⫹y2, x2⫺y2], R: 1⬉y⬉2⫺x2 F⫽[x cosh 2y, 2x2 sinh 2y], R: x2⬉y⬉x F⫽[x2ey, y2ex],
⫾(2, ⫺2)
⫾(2, 2),
F⫽[6y2, 2x⫺2y4],
x2⫹y2⫽1>4 F⫽[y, ⫺x],
冮CF(r)•dr
where kis a unit vector perpendicular to the xy-plane.
Verify(10) and (11) for and Cthe circle as well as for an example of your own choice.
13–17 INTEGRAL
OF THE NORMAL DERIVATIVE
Using (9), find the value of dstaken counterclockwise over the boundary Cof the region R.
13. Rthe triangle with vertices (0, 0), (4, 2), (0, 2).
14.
15.
16. Confirm the answer
by direct integration.
17.
18. Laplace’s equation.Show that for a solution w(x, y) of Laplace’s equation in a region R with boundary curve Cand outer unit normal vector n,
(12)
19. Show that satisfies Laplace’s equation and, using (12), integrate counter- clockwise around the boundary curve Cof the rectangle 20. Same task as in Prob. 19 when and C the boundary curve of the triangle with vertices (0, 0), (1, 0), (0, 1).
w⫽x2⫹y2 0⬉x⬉2, 0⬉y⬉5.
w(0w>0n) ⵜ2w⫽0
w⫽ex sin y
⫽ 冯Cw 00wn ds.
冮R冮c a00wxb2⫹a00wyb2d dx dy
ⵜ2w⫽0 w⫽x3⫺y3, 0⬉y⬉x2, ƒxƒ ⬉2 W⫽x2⫹y2, C: x2⫹y2⫽4.
w⫽ex cos y⫹xy3, R: 1⬉y⬉10⫺x2, x⭌0 w⫽x2y⫹xy2, R: x2⫹y2⬉1, x⭌0, y⭌0 w⫽ cosh x,
冮C00wn x2⫹y2⫽4
F⫽[7x, ⫺3y]
P R O B L E M S E T 1 0 . 4