Line Integral - Tích phân đường

Gii tch hm nhiu binChng: TCH PHN NG

u Th Phit

Ngy 24 thng 4 nm 2014

u Th Phit () Gii tch hm nhiu bin Chng: TCH PHN NGNgy 24 thng 4 nm 2014 1 / 53

Ni dung

1 Tch phn ng loi mt

2 Tch phn ng loi hai

3 Mt s tnh cht ca tch phn ng

4 nh l Green


Tch phn ng loi mt

Tch phn ng loi mt


Tch phn ng loi mt

nh ngha

Tng t tch phn hm mt bin trn on [a, b], ta xy dng tch phntrn ng cong C , tch phn c gi l tch phn ng.Xt ng cong C cho bi phng trnh tham s

x = x(t), y = y(t) a t b

hay trong khng gian R2, ta c vector r(t) = x(t)i + y(t)j = (x(t), y(t)).Ta gi s C l mt ng cong trn (o hm r lin tc v r (t) 6= 0).Ta chia on [a, b] thnh n on nh u nhau [ti1, ti ], tng ng ccim Pi (x(ti ), y(ti )) chia ng cong C thnh n ng cong con.Ta gi chiu di ca cc ng cong con ny tng ng l s1, . . . ,sn.


Tch phn ng loi mt

LINE INTEGRALS

In this section we define an integral that is similar to a single integral except that insteadof integrating over an interval , we integrate over a curve . Such integrals are calledline integrals, although curve integrals would be better terminology. They were inventedin the early 19th century to solve problems involving fluid flow, forces, electricity, andmagnetism.

We start with a plane curve given by the parametric equations

or, equivalently, by the vector equation , and we assume that is asmooth curve. [This means that is continuous and . See Section 13.3.] If wedivide the parameter interval into n subintervals of equal width and we let

and , then the corresponding points divide into subarcswith lengths (See Figure 1.) We choose any point in the thsubarc. (This corresponds to a point in .) Now if is any function of two vari-ables whose domain includes the curve , we evaluate at the point , multiply bythe length of the subarc, and form the sum

which is similar to a Riemann sum. Then we take the limit of these sums and make the fol-lowing definition by analogy with a single integral.

DEFINITION If is defined on a smooth curve given by Equations 1, thenthe line integral of f along C is

if this limit exists.

In Section 10.2 we found that the length of is

A similar type of argument can be used to show that if is a continuous function, then thelimit in Definition 2 always exists and the following formula can be used to evaluate theline integral:

The value of the line integral does not depend on the parametrization of the curve, pro-vided that the curve is traversed exactly once as t increases from a to b.

yC

f !x, y" ds ! yba f (x!t", y!t")#$dxdt %2 ! $dydt %2 dt 3

f

L ! yba

#$dxdt %2 ! $dydt %2 dtC

yC f !x, y" ds ! lim

n l " &

n

i!1 f !xi*, yi*" #si

Cf2

&n

i!1 f !xi*, yi*"#si

#si!xi*, yi*"fC

f'ti$1, ti(ti*iPi*!xi*, yi*"#s1, #s2, . . . , #sn.

nCPi !xi, yi "yi ! y!ti"xi ! x!ti "'ti$1, ti('a, b(

r%!t" " 0r%Cr!t" ! x!t" i ! y!t" j

a & t & by ! y!t"x ! x!t"1

C

C'a, b(

16.2

1034 | | | | CHAPTER 16 VECTOR CALCULUS

FIGURE 1

t i-1

PPP

C

a b

x0

y

tt i

t*i

Pi-1 Pi

Pn

P*i (x*i ,y*i )

Trn mi ng cong con ta chn im Pi (xi , yi ) bt k (tng ng vi

im ti trn on [a, b]).


Tch phn ng loi mt

Cho f l hm theo hai bin (x , y) xc nh trn min cha ng cong C ,ta tnh tng

ni=1

f (xi , yi )si

Ta thy tng trn c dng tng t tng Riemann, ly gii hn khi n tinti v cng, ta c tch phn ng tng t tch phn mt bin

nh ngha

Nu f c nh ngha trn ng cong trn C cho bi phng trnh thams (x(t), y(t)) vi a t b, th tch phn ng ca f theo C cho bi

Cf (x , y)ds = lim

n

ni=1

f (xi , yi )si

nu gii hn trn tn ti.


Tch phn ng loi mt

Ta bit, di ng cong C cho bi

L =

ba

(dx

dt

)2+

(dy

dt

)2dt

Do , nu f l hm s lin tc th gii hn trong nh ngha trn luntn ti, ng thi ta c th tnh tch phn trn bi cng thc

Cf (x , y)ds =

ba

f (x(t), y(t))

(dx

dt

)2+

(dy

dt

)2dt

If is the length of C between and , then

So the way to remember Formula 3 is to express everything in terms of the parameter Use the parametric equations to express and in terms of t and write ds as

In the special case where is the line segment that joins to , using as theparameter, we can write the parametric equations of as follows: , ,

. Formula 3 then becomes

and so the line integral reduces to an ordinary single integral in this case.Just as for an ordinary single integral, we can interpret the line integral of a positive

function as an area. In fact, if , represents the area of one side ofthe fence or curtain in Figure 2, whose base is and whose height above the point

is .

EXAMPLE 1 Evaluate , where is the upper half of the unit circle.

SOLUTION In order to use Formula 3, we first need parametric equations to represent C.Recall that the unit circle can be parametrized by means of the equations

and the upper half of the circle is described by the parameter interval (See Figure 3.) Therefore Formula 3 gives

M

Suppose now that is a piecewise-smooth curve; that is, is a union of a finite num-ber of smooth curves where, as illustrated in Figure 4, the initial point of

is the terminal point of Then we define the integral of along as the sum of theintegrals of along each of the smooth pieces of :

yC f !x, y" ds ! y

C1 f !x, y" ds ! y

C2 f !x, y" ds ! " " " ! y

Cn f !x, y" ds

CfCfCi .Ci!1

C1, C2, . . . , Cn,CC

! 2# ! 23

! y#0

!2 ! cos2t sin t" dt ! #2t $ cos3t3 $0#

! y#0

!2 ! cos2t sin t"ssin2 t ! cos2 t dt

yC !2 ! x 2y" ds ! y#

0 !2 ! cos2t sin t"%&dxdt '2 ! &dydt '2 dt

0 % t % #.

y ! sin tx ! cos t

x 2 ! y 2 ! 1CxC !2 ! x 2y" ds

f !x, y"!x, y"C

xC f !x, y" dsf !x, y" & 0

yC f !x, y" ds ! yb

a f !x, 0" dx

a % x % by ! 0x ! xC

x!b, 0"!a, 0"C

ds !%&dxdt '2 ! &dydt '2 dtyx

t:

dsdt!%&dxdt '2 ! &dydt '2

r!t"r!a"s!t"

SECTION 16.2 LINE INTEGRALS | | | | 1035

N The arc length function is discussed in Section 13.3.

s

FIGURE 2

f(x,y)

(x,y)

C y

z

x

0

FIGURE 3

0

+=1(y0)

x

y

1_1

FIGURE 4A piecewise-smooth curve

0

CC

C

CC

x

y


Tch phn ng loi mt

V d

Tnh tch phnC (2 + x

2y)ds vi C l na trn ca vng trn n vx2 + y2 = 1.

Ta vit phng trnh tham s cho ng cong C

x = cos t y = sin t

Do C l na trn ca vng trn n v, 0 t pi.







is .




M



yC f !x, y" ds ! y

C1 f !x, y" ds ! y

C2 f !x, y" ds ! " " " ! y

Cn f !x, y" ds

CfCfCi .Ci!1

C1, C2, . . . , Cn,CC

! 2# ! 23

! y#0


! y#0


yC !2 ! x 2y" ds ! y#


0 % t % #.

y ! sin tx ! cos t

x 2 ! y 2 ! 1CxC !2 ! x 2y" ds

f !x, y"!x, y"C

xC f !x, y" dsf !x, y" & 0

yC f !x, y" ds ! yb

a f !x, 0" dx

a % x % by ! 0x ! xC

x!b, 0"!a, 0"C


t:


r!t"r!a"s!t"



s

FIGURE 2

f(x,y)

(x,y)

C y

z

x

0

FIGURE 3

0

+=1(y0)

x

y

1_1


0

CC

C

CC

x

y


Tch phn ng loi mt

C

(2 + x2y)ds =

pi0

(2 + cos2 t sin t)

(dx

dt

)2+

(dy

dt

)2dt

=

pi0

(2 cos2 t sin t)

sin2 t + cos2 tdt

=

pi0

(2 + cos2 t sin t)dt

=

[2t cos

3 t

3

]pi0

= 2pi +2

3


Tch phn ng loi mt

Ta ch rng ng cong C c gi thit l trn. Xt trng hp C lng cong trn tng khc (C l hp hu hn cc on ng cong trnC1, . . . ,Cn), ta c th nh ngha tch phn ng ca f trn ng congC l tng cc tch phn ng ca f trn cc ng cong Ci

Cf (x , y)ds =

Ci

f (x , y)ds + . . .+

Ci

f (x , y)ds







is .




M



yC f !x, y" ds ! y

C1 f !x, y" ds ! y

C2 f !x, y" ds ! " " " ! y

Cn f !x, y" ds

CfCfCi .Ci!1

C1, C2, . . . , Cn,CC

! 2# ! 23

! y#0


! y#0


yC !2 ! x 2y" ds ! y#


0 % t % #.

y ! sin tx ! cos t

x 2 ! y 2 ! 1CxC !2 ! x 2y" ds

f !x, y"!x, y"C

xC f !x, y" dsf !x, y" & 0

yC f !x, y" ds ! yb

a f !x, 0" dx

a % x % by ! 0x ! xC

x!b, 0"!a, 0"C


t:


r!t"r!a"s!t"



s

FIGURE 2

f(x,y)

(x,y)

C y

z

x

0

FIGURE 3

0

+=1(y0)

x

y

1_1


0

CC

C

CC

x

y


Tch phn ng loi mt

V d

Tnh tch phnC 2xds vi C l ng cong cha parabol y = x

2 t (0, 0)

ti (1, 1), v on thng C2 t (1, 1) ti (1, 2).

ng cong C1 l th ca hm s theo bin x , ta c th chn x ltham s, C1 biu din bi

x = x y = x2 0 x 1Trn on thng C2 ta chn y l tham s, C2 cho bi

x = 1 y = y 1 y 2 EXAMPLE 2 Evaluate , where consists of the arc of the parabola from to followed by the vertical line segment from to .SOLUTION The curve is shown in Figure 5. is the graph of a function of , so we canchoose as the parameter and the equations for become

Therefore

On we choose as the parameter, so the equations of are

and

Thus M

Any physical interpretation of a line integral depends on the physical inter-pretation of the function . Suppose that represents the linear density at a point

of a thin wire shaped like a curve . Then the mass of the part of the wire from to in Figure 1 is approximately and so the total mass of the wire is approx-imately . By taking more and more points on the curve, we obtain the mass

of the wire as the limiting value of these approximations:

[For example, if represents the density of a semicircular wire, then theintegral in Example 1 would represent the mass of the wire.] The center of mass of thewire with density function is located at the point , where

Other physical interpretations of line integrals will be discussed later in this chapter.

EXAMPLE 3 A wire takes the shape of the semicircle , , and isthicker near its base than near the top. Find the center of mass of the wire if the lineardensity at any point is proportional to its distance from the line .

SOLUTION As in Example 1 we use the parametrization , , ,and find that . The linear density is

!!x, y" ! k!1 " y"

ds ! dt0 # t # $y ! sin tx ! cos t

y ! 1

y % 0x 2 & y 2 ! 1V

y !1m

yC y !!x, y" dsx !

1m

yC x !!x, y" ds4

!x, y"!

f !x, y" ! 2 & x 2y

m ! limn l '

#n

i!1 !!xi*, yi*" (si ! y

C !!x, y" ds

m

$ !!xi*, yi*"(si

!!xi*, yi*"(siPiPi"1C!x, y"

!!x, y"fxC f !x, y" ds

yC 2x ds ! y

C1 2x ds & y

C2 2x ds !

5s5 " 16

& 2

yC2

2x ds ! y21 2!1"%&dxdy'2 & &dydy'2 dy ! y21 2 dy ! 2

1 # y # 2y ! yx ! 1

C2yC2

! 14 !23 !1 & 4x 2 "3(2]01 ! 5s5 " 16

! y10 2xs1 & 4x 2 dxy

C1 2x ds ! y1

0 2x%&dxdx'2 & &dydx'2 dx

0 # x # 1y ! x 2x ! x

C1xxC1C

!1, 2"!1, 1"C2!1, 1"!0, 0"y ! x 2C1CxC 2x ds


FIGURE 5C=C " C

(0,0)

(1,1)

(1,2)

C

C

x

y


Tch phn ng loi mt

C1

2xds =

10

2x

(dx

dx

)2+

(dy

dx

)2dx

=

10

2x

1 + 4x2dx

=1

4

2

3(1 + 4x2)

32

]10

=5

5 16

C2

2xds =

21

2

(dx

dy

)2+

(dy

dy

)2dy

=

21

2dy = 2

Vy ta c C

2xds =

C1

2xds +

C2

2xds =5

5 16

+ 2


Tch phn ng loi mt

Tch phn ng loi mt trong khng gian

Tng t, ta nh ngha tch phn ng trong khng gian.Xt f (x , y , z) xc nh trn ng cong trn C trong khng gian Oxyz .Vi C cho bi phng trnh tham s

x = x(t)

y = y(t)

z = z(t)

a t b

Khi tch phn ng ca f trn C cho bi cng thc

Cf (x , y , z)ds =

ba

f (x(t), y(t), z(t))

(dx

dt

)2+

(dy

dt

)2+

(dz

dt

)2dt


Tch phn ng loi mt

V d

Tnh tch phnC (x + y)ds vi C l ng trn x

2 + y2 + z2 = 4, x = y

Phng trnh tham s ca ng cong C .Ta thy 2x2 + z2 = 4,(hnh ellipse). t{

x = y =

2r cos

z = 2r sin

V x2 + y2 + z2 = 4 nn r = 1. Phngtrnh tham s ca C{

x = y =

2 cos

z = 2 sin 0 2pi

2pi0

(

2 cos +

2 cos )

(

2 sin )2 + (

2 sin )2 + (2 cos )2d


Tch phn ng loi hai

Tch phn ng loi hai


Tch phn ng loi hai

Nu ta thay vi phn si bng xi = xi xi1 v yi = yi yi1 trongtch phn ng loi mt, ta thu c hai tch phn ng. Cc tch phntrn l tch phn ng ca f trn ng cong C tng ng vi x v y

C

f (x , y)dx = limn

ni=1

f (xi , yi )xi

C

f (x , y)dy = limn

ni=1

f (xi , yi )yi

Tch phn ng tng ng vi x v y c th c biu din di dngtham s t nh sau

C

f (x , y)dx =

ba

f (x(t), y(t))x (t)dt

C

f (x , y)dy =

ba

f (x(t), y(t))y (t)dt


Tch phn ng loi hai

nh ngha tch phn ng loi hai

nh ngha

Gi s trn ng C xc nh hai hm s P(x , y) v Q(x , y). Tch phnng loi hai ca P(x , y) v Q(x , y) trn cung C xc nh bi cng thc

I =

C

P(x , y)dx + Q(x , y)dy

Nu ng cong C xc nh theo phng trnh tham s t trn on [a, b]th tch phn ng loi hai c tnh theo cng thc

I =

C

P(x , y)dx + Q(x , y)dy

=

ba

(P(x(t), y(t))x (t) + Q(x(t), y(t))y (t))

)dt


Tch phn ng loi hai

Tnh cht

1 Tch phn ng loi hai ph thuc chiu ly tch phn trn ngcong C

_AB

Pdx + Qdy = _BA

Pdx + Qdy

2 Nu ng cong_AB c chia thnh

_AC v

_CB v P,Q kh tch

trn_AB th ta c

_AB

Pdx + Qdy =

_AC

Pdx + Qdy +

_CB

Pdx + Qdy


Tch phn ng loi hai

V d 1.

Tnh I =C

(x2 + 3y)dx + 2ydy trong C l cnh tam gic OAB vi

O(0, 0),A(1, 1),B(0, 2) theo chiu ngc chiu kim ng h.

Ta c I =C

=OA

+AB

+BO

.

Phng trnh on OA:x = t, y = t 0 t 1

OA

=

10

(t2 + 3t).1.dt + 2t.1.dt

=

10

(t2 + 5t)dt =17

6-1.5 -1 -0.5 0 0.5 1 1.5 2

0.5

1

1.5

2

Text

A

B

O


Tch phn ng loi hai

V d 1. (cont)

Phng trnh on AB: x = 1 t, y = 1 + t 0 t 1AB

=

10

((1 t)2 + 3(1 + t)).(1).dt + 2(t + 1).1.dt

=

10

(t2 + t 2)dt = 116

Phng trnh on BO: x = 0, y = 2 t 0 t 2BO

=

20

((0)2 + 3(2 t)).0.dt + 2(2 t).(1).dt =2

0

2(t 2)dt = 4

Vy I =17

6 11

6 4 = 3


Tch phn ng loi hai

V d 1.(cch 2)

Tnh I =C

(x2 + 3y)dx + 2ydy trong C l cnh tam gic OAB vi

O(0, 0),A(1, 1),B(0, 2) theo chiu ngc chiu kim ng h.

Ta c I =C

=OA

+AB

+B

O.

Phng trnh on OA: y = x x t 0 n 1

OA

=

10

(x2 + 3x).1.dx + 2x .1.dx

=

10

(x2 + 5x)dx =17

6

-1.5 -1 -0.5 0 0.5 1 1.5 2

0.5

1

1.5

2

Text

A

B

O


Tch phn ng loi hai

V d 1. (cch 2)

Phng trnh on AB: y = 2 x x t 1 n 0AB

=

01

(x2 + 3(2 x))dx + 2(2 x).(1).dx = 116

Phng trnh on BO: x = 0.y y t 2 n 0

BO

=

20

((0)2 + 3y).0.dy + 2ydy = 4

Vy I =17

6 11

6 4 = 3


Tch phn ng loi hai

V d 2.

Tnh I =C

ydx + xdy trong C l cung x2 + y2 = 2x t O(0, 0) n

A(1, 1) theo chiu kim ng h.

Cung C c phng trnh tham s{x = 1 + cos t

y = sin tt t pi ti

pi

2

I =

pi/2pi

(sin t)( sin t)dt + (1 + cos t) cos tdt

= pi2

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

-1.5

-1

-0.5

0.5

1

1.5

A

O


Tch phn ng loi hai

Bi tp

Tnh tch phn ng loi 1.

1C

y3ds, C : x = t3, y = t, 0 t 2.2C

xyds, C : x = t2, y = 2t, 0 t 1.3C

xy4ds, C l na bn phi ca ng trn x2 + y2 = 16.

4C

x sin yds, C l on thng ni (0, 3) ti (4, 6).

5C

x2 + y2ds vi C l na ng trn x2 + y2 = 2x v x 1.

6C

(x

43 + y

43

)ds vi C c phng trnh tham s

x = 2 cos3 t, y = 2 sin3 t, 0 t 2pi.7C

(x2 + y2 + z2)ds vi C l ng xon c x = a cos t, y = a sin t,

z = bt, 0 t 2pi, a, b dngu Th Phit () Gii tch hm nhiu bin Chng: TCH PHN NGNgy 24 thng 4 nm 2014 24 / 53

Tch phn ng loi hai

Bi tp

Tnh tch phn ng loi hai

1C

(x2y3 x)dy vi C l ng cong y = x t (1, 1) ti (4, 2).

2C

xydx + (x y)dy vi C l cha cc ng thng t (0, 0) n(2, 0) v t (2, 0) n (3, 2).

3C

sin xdx + cos ydy vi C l na di ng trn x2 + y2 = 1 t

(1, 0) n (1, 0) v on thng t (1, 0) n (2, 3).4C

ydx (x + y)2dy vi C l cung parabol y = 2x x2 nm phay 0 v theo chiu ngc kim ng h.

5C

xydx + ydy yzdz trong C l ng cong cho bi phng trnhx = t, y = t2, z = t vi t t 0 n 1.


Mt s tnh cht ca tch phn ng

Mt s tnh cht ca tch phnng



Lin h gia tch phn ng loi mt v loi hai

Xt ng cong_AB c phng trnh tham s

r(t) = (x(t), y(t), z(t)) a t bKhi vector ~r (t) = x (t)~i + y (t)~j + z (t)~k l vector tip tuyn vi

ng cong_AB v ~T (t) = r

(t)|r (t)| l vector tip tuyn n v.

Gi F = (P,Q,R) = P(x , y , z)~i + Q(x , y , z)~j + R(x , y , z)~k l trngvector xc nh trn ng cong CCPdx + Qdy + Rdz =

ba

(P(r(t))x (t) + Q(r(t))y (t) + R(r(t))z (t)

)dt

=

ba

(P(r(t))

x (t)|r (t)| + Q(r(t))

y (t)|r (t)| + R(r(t))

z (t)|r (t)|

)|r (t)|dt

=

CF T (t)ds



nh l cn bn

Trong tch phn hm mt bin, ta c tnh cht

ba

F (x)dx = F (b) F (a)

vi F lin tc trn [a, b]. Trong khng gian hu hn chiu, ta dng vectorgradient f thay cho F v c nh l saunh l

Cho C l ng cong trn cho bi hm vector r(t), a t b. Cho f lhm kh vi vi vector gradient f lin tc trn C . Khi

Cf dr = f (r(b)) f (r(a))



T nh l trn, ta c th tnh tch phn ng loi 2 ca f , ta ch quantm n gi tr ca f ti im u v im cui ca ng cong C .

Chng minh.

T nh ngha ta c

Cf dr =

ba

f (r(t)) r (t)dt

=

ba

(f

x

dx

dt+f

y

dy

dt+f

z

dz

dt

)dt

=

ba

d

dtf (r(t))dt = f (r(b)) f (r(a))

THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS

Recall from Section 5.3 that Part 2 of the Fundamental Theorem of Calculus can be writ-ten as

where is continuous on . We also called Equation 1 the Net Change Theorem: Theintegral of a rate of change is the net change.

If we think of the gradient vector of a function of two or three variables as a sortof derivative of , then the following theorem can be regarded as a version of the Funda-mental Theorem for line integrals.

THEOREM Let be a smooth curve given by the vector function , .Let be a differentiable function of two or three variables whose gradient vector

is continuous on . Then

Theorem 2 says that we can evaluate the line integral of a conservative vectorfield (the gradient vector field of the potential function ) simply by knowing the value of

at the endpoints of . In fact, Theorem 2 says that the line integral of is the netchange in f. If is a function of two variables and is a plane curve with initial point

and terminal point , as in Figure 1, then Theorem 2 becomes

If is a function of three variables and is a space curve joining the point to the point , then we have

Lets prove Theorem 2 for this case.

FIGURE 1

0

A(x,y,z)B(x,y,z)

C

0

A(x,y) B(x,y)

C

y

z

x

x

y

yC ! f ! dr ! f !x2, y2, z2 " " f !x1, y1, z1 "

B!x2, y2, z2 "A!x1, y1, z1 "Cf

yC ! f ! dr ! f !x2, y2 " " f !x1, y1 "

B!x2, y2 "A!x1, y1 "Cf

fCff

NOTE

yC ! f ! dr ! f !r!b"" " f !r!a""

C ff

a # t # br!t"C2

ff f

#a, b$F$

yba F$!x" dx ! F!b" " F!a"1

16.3




V d

TnhC y

2dx + xdy vi

a) C = C1 l on thng t (5,3) n (0, 2)b)C = C2 l ng cong parabol x = 4 y2 t (5,3) n (0, 2).

a) Phng trnh tham s ca on thng

x = 5 + 5t, y = 3 + 5t, 0 t 1

C1

y2dx + xdy =

10

(5t 3)2(5dt) + (5t 5)(5dt)

= 5

10

(25t2 25t + 4)dt

= 5

[25t3

3 25t

2

2+ 4t

]= 5

6

resentation of the line segment that starts at and ends at is given by

(See Equation 12.5.4.)

EXAMPLE 4 Evaluate , where (a) is the line segment fromto and (b) is the arc of the parabola from

to . (See Figure 7.)SOLUTION(a) A parametric representation for the line segment is

(Use Equation 8 with and .) Then , , andFormulas 7 give

(b) Since the parabola is given as a function of , lets take as the parameter and writeas

Then and by Formulas 7 we have

M

Notice that we got different answers in parts (a) and (b) of Example 4 even though thetwo curves had the same endpoints. Thus, in general, the value of a line integral dependsnot just on the endpoints of the curve but also on the path. (But see Section 16.3 for con-ditions under which the integral is independent of the path.)

Notice also that the answers in Example 4 depend on the direction, or orientation, of thecurve. If denotes the line segment from to , you can verify, using theparametrization

that y !C1

y 2 dx " x dy ! 56

0 # t # 1y ! 2 ! 5tx ! !5t

!!5, !3"!0, 2"!C1

! #! y 42 ! y 33 " 4y$!32

! 40 56

! y2!3

!!2y 3 ! y 2 " 4" dy

y C2

y 2 dx " x dy ! y2!3

y 2!!2y" dy " !4 ! y 2 " dy

dx ! !2y dy

!3 # y # 2y ! yx ! 4 ! y 2C2

yy

! 5#25t 33 ! 25t 22 " 4t$01

! !56

! 5 y10 !25t 2 ! 25t " 4" dt

yC1

y 2 dx " x dy ! y10 !5t ! 3"2!5 dt" " !5t ! 5"!5 dt"

dy ! 5 dtdx ! 5 dtr1 ! %0, 2 &r0 ! %!5, !3 &

0 # t # 1y ! 5t ! 3x ! 5t ! 5

!0, 2"!!5, !3"x ! 4 ! y 2C ! C2!0, 2"!!5, !3"

C ! C1xC y 2 dx " x dyV

0 # t # 1r!t" ! !1 ! t"r0 " t r18

r1r0


FIGURE 7

0 4

(_5,_3)

(0,2)

C C

x=4-

x

y



b) Parabol C2 l hm theo bin y , ta xem yl tham s

x = 4 y2, y = y ,3 y 2

C2

y2dx + xdy =

23

y2(2ydy) + (4 y2)dy

=

23

(2y3 y2 + 4)dy

=

[y

4

2 y

3

3+ 4y

]= 40

5

6

resentation of the line segment that starts at and ends at is given by

(See Equation 12.5.4.)

EXAMPLE 4 Evaluate , where (a) is the line segment fromto and (b) is the arc of the parabola from

to . (See Figure 7.)SOLUTION(a) A parametric representation for the line segment is

(Use Equation 8 with and .) Then , , andFormulas 7 give

(b) Since the parabola is given as a function of , lets take as the parameter and writeas

Then and by Formulas 7 we have

M

Notice that we got different answers in parts (a) and (b) of Example 4 even though thetwo curves had the same endpoints. Thus, in general, the value of a line integral dependsnot just on the endpoints of the curve but also on the path. (But see Section 16.3 for con-ditions under which the integral is independent of the path.)

Notice also that the answers in Example 4 depend on the direction, or orientation, of thecurve. If denotes the line segment from to , you can verify, using theparametrization

that y !C1

y 2 dx " x dy ! 56

0 # t # 1y ! 2 ! 5tx ! !5t

!!5, !3"!0, 2"!C1

! #! y 42 ! y 33 " 4y$!32

! 40 56

! y2!3

!!2y 3 ! y 2 " 4" dy

y C2

y 2 dx " x dy ! y2!3

y 2!!2y" dy " !4 ! y 2 " dy

dx ! !2y dy

!3 # y # 2y ! yx ! 4 ! y 2C2

yy

! 5#25t 33 ! 25t 22 " 4t$01

! !56

! 5 y10 !25t 2 ! 25t " 4" dt

yC1

y 2 dx " x dy ! y10 !5t ! 3"2!5 dt" " !5t ! 5"!5 dt"

dy ! 5 dtdx ! 5 dtr1 ! %0, 2 &r0 ! %!5, !3 &

0 # t # 1y ! 5t ! 3x ! 5t ! 5

!0, 2"!!5, !3"x ! 4 ! y 2C ! C2!0, 2"!!5, !3"

C ! C1xC y 2 dx " x dyV

0 # t # 1r!t" ! !1 ! t"r0 " t r18

r1r0


FIGURE 7

0 4

(_5,_3)

(0,2)

C C

x=4-

x

y



Khng ph thuc vo ng i

Gi s C1 v C2 l hai ng cong trn tng khc vi im u l A vim cui l B.Tng qut, (nh v d trn) ta thy

C1F .dr 6= C2 F .dr . Tuy nhin

C1

f dr =C2

f dr

vi f l trng vector lin tc.Cho F l trng vector lin tc trn min D, ta ni tch phn ngC F dr khng ph thuc vo ng i nu

C1F dr = C2 F dr vi

mi ng cong C1, C2 bt k trong D c cng im u v im cui.



Mt ng cong c gi l kn nu im u v im cui ca n trngnhau (r(b) = r(a)).

that have the same initial and terminal points. With this terminology we can say that lineintegrals of conservative vector fields are independent of path.

A curve is called closed if its terminal point coincides with its initial point, that is,. (See Figure 2.) If is independent of path in and is any closed

path in , we can choose any two points and on and regard as being composedof the path from to followed by the path from to . (See Figure 3.) Then

since and have the same initial and terminal points.Conversely, if it is true that whenever is a closed path in , then we

demonstrate independence of path as follows. Take any two paths and from to in and define to be the curve consisting of followed by . Then

and so . Thus we have proved the following theorem.

THEOREM is independent of path in if and only if forevery closed path in .

Since we know that the line integral of any conservative vector field is independentof path, it follows that for any closed path. The physical interpretation is thatthe work done by a conservative force field (such as the gravitational or electric field inSection 16.1) as it moves an object around a closed path is 0.

The following theorem says that the only vector fields that are independent of path areconservative. It is stated and proved for plane curves, but there is a similar version forspace curves. We assume that is open, which means that for every point in there isa disk with center that lies entirely in . (So doesnt contain any of its boundarypoints.) In addition, we assume that is connected. This means that any two points in can be joined by a path that lies in .

THEOREM Suppose is a vector field that is continuous on an open connectedregion . If is independent of path in , then is a conservative vectorfield on ; that is, there exists a function such that .

PROOF Let be a fixed point in . We construct the desired potential function bydefining

for any point in . Since is independent of path, it does not matter which path from to is used to evaluate . Since is open, thereexists a disk contained in with center . Choose any point in the disk with

and let consist of any path from to followed by the horizontalline segment from to . (See Figure 4.) Then

f !x, y" ! y C1

F ! dr ! y C2

F ! dr ! y!x1, y"!a, b"

F ! dr ! y C2

F ! dr

!x, y"!x1, y"C2!x1, y"!a, b"C1Cx1 " x

!x1, y"!x, y"DDf !x, y"!x, y"!a, b"C

xC F ! drD!x, y"

f !x, y" ! y!x, y"!a, b"

F ! dr

fDA!a, b"

f ! FfDFDxC F ! drD

F4

DDD

DDPDPD

xC F ! dr ! 0F

DCxC F ! dr ! 0DxC F ! dr3

x C1 F ! dr ! x C2 F ! dr

0 ! yC F ! dr ! y

C1 F ! dr ! y

#C2 F ! dr ! y

C1 F ! dr # y

C2 F ! dr

#C2C1CDBAC2C1

DCxC F ! dr ! 0#C2C1

yC F ! dr ! y

C1 F ! dr ! y

C2 F ! dr ! y

C1 F ! dr # y

#C2 F ! dr ! 0

ABC2BAC1CCBAD

CDxC F ! drr!b" ! r!a"

D


FIGURE 2A closed curve

C

FIGURE 3

C

CB

A

FIGURE 4

(a,b)

x0

y

D

(x,y)

C

C

(x,y)

NuC F dr khng ph thuc ng i trong D th vi mi ng cong

kn C trong D, chn hai im bt k A,B trn C v chia C thnh haing cong C1 t A n B v C2 t B n A. Ta c

CF dr =

C1

F dr +C2

F dr =C1

F dr C2

F dr = 0

NuC F dr = 0 vi C l ng cong kn th

0 =

CF dr =

C1

F dr + intC2F dr =C1

F dr C2

F dr



nh l

Tch phnC F dr khng ph thuc vo ng i trong D nu v ch nu

C F dr = 0 vi mi ng cong kn C trong D

Ta s ch ra rng

nh l

Gi s F l trng vector lin tc trn min D. Nu tchC F dr khng

ph thuc vo ng i th tn ti hm s f sao cho f = F .



Chng minh

Cho A(a, b) l im c nh trong D. Ta xy dng hm f bi

f (x , y) =(x ,y)(a,b)

F dr .

DoC F dr khng ph thuc ng i, ta

c th chn ng C cha: C1 t (a, b) n(x1, y) vi x1 < x v C2 l on thng nini (x , y). Khi

f (x , y) =

C1

F dr +C2

F dr

=

(x1,y)(a,b)

F dr +C2

F dr

that have the same initial and terminal points. With this terminology we can say that lineintegrals of conservative vector fields are independent of path.

A curve is called closed if its terminal point coincides with its initial point, that is,. (See Figure 2.) If is independent of path in and is any closed

path in , we can choose any two points and on and regard as being composedof the path from to followed by the path from to . (See Figure 3.) Then

since and have the same initial and terminal points.Conversely, if it is true that whenever is a closed path in , then we

demonstrate independence of path as follows. Take any two paths and from to in and define to be the curve consisting of followed by . Then

and so . Thus we have proved the following theorem.

THEOREM is independent of path in if and only if forevery closed path in .

Since we know that the line integral of any conservative vector field is independentof path, it follows that for any closed path. The physical interpretation is thatthe work done by a conservative force field (such as the gravitational or electric field inSection 16.1) as it moves an object around a closed path is 0.

The following theorem says that the only vector fields that are independent of path areconservative. It is stated and proved for plane curves, but there is a similar version forspace curves. We assume that is open, which means that for every point in there isa disk with center that lies entirely in . (So doesnt contain any of its boundarypoints.) In addition, we assume that is connected. This means that any two points in can be joined by a path that lies in .

THEOREM Suppose is a vector field that is continuous on an open connectedregion . If is independent of path in , then is a conservative vectorfield on ; that is, there exists a function such that .

PROOF Let be a fixed point in . We construct the desired potential function bydefining

for any point in . Since is independent of path, it does not matter which path from to is used to evaluate . Since is open, thereexists a disk contained in with center . Choose any point in the disk with

and let consist of any path from to followed by the horizontalline segment from to . (See Figure 4.) Then

f !x, y" ! y C1

F ! dr ! y C2

F ! dr ! y!x1, y"!a, b"

F ! dr ! y C2

F ! dr

!x, y"!x1, y"C2!x1, y"!a, b"C1Cx1 " x

!x1, y"!x, y"DDf !x, y"!x, y"!a, b"C

xC F ! drD!x, y"

f !x, y" ! y!x, y"!a, b"

F ! dr

fDA!a, b"

f ! FfDFDxC F ! drD

F4

DDD

DDPDPD

xC F ! dr ! 0F

DCxC F ! dr ! 0DxC F ! dr3

x C1 F ! dr ! x C2 F ! dr

0 ! yC F ! dr ! y

C1 F ! dr ! y

#C2 F ! dr ! y

C1 F ! dr # y

C2 F ! dr

#C2C1CDBAC2C1

DCxC F ! dr ! 0#C2C1

yC F ! dr ! y

C1 F ! dr ! y

C2 F ! dr ! y

C1 F ! dr # y

#C2 F ! dr ! 0

ABC2BAC1CCBAD

CDxC F ! drr!b" ! r!a"

D


FIGURE 2A closed curve

C

FIGURE 3

C

CB

A

FIGURE 4

(a,b)

x0

y

D

(x,y)

C

C

(x,y)



xf (x , y) = 0 +

x

C2

F dr

Nu F = P~i + Q~j thC2F dr = C2 Pdx + Qdy .

Trn C2, y l hng s do dy = 0, xt tham s t vi x1 t x

xf (x , y) =

x

C2

Pdx + Qdy =

x

xx1

P(t, y) = P(x , y)

Tng t ta chng minhc

yf (x , y) = Q(x , y)

Notice that the first of these integrals does not depend on , so

If we write , then

On , is constant, so . Using as the parameter, where , we have

by Part 1 of the Fundamental Theorem of Calculus (see Section 5.3). A similar argu-ment, using a vertical line segment (see Figure 5), shows that

Thus

which says that is conservative. M

The question remains: How is it possible to determine whether or not a vector field is conservative? Suppose it is known that is conservative, where and have continuous first-order partial derivatives. Then there is a function such that

, that is,

Therefore, by Clairauts Theorem,

THEOREM If is a conservative vector field,where and have continuous first-order partial derivatives on a domain , thenthroughout we have

The converse of Theorem 5 is true only for a special type of region. To explain this, wefirst need the concept of a simple curve, which is a curve that doesnt intersect itself any-where between its endpoints. [See Figure 6; for a simple closed curve, but

when .]In Theorem 4 we needed an open connected region. For the next theorem we need a

stronger condition. A simply-connected region in the plane is a connected region such D

a ! t1 ! t2 ! br!t1 " ! r!t2 "r!a" " r!b"

"P"y""Q"x

DDQP

F!x, y" " P!x, y" i # Q!x, y" j5

"P"y"

"2 f"y "x

""2 f"x "y

""Q"x

Q " "f"y

andP ""f"x

F " ffQ

PF " P i # Q jF

F

F " P i # Q j " "f"x

i #"f"y

j " f

"

"y f !x, y" " "

"y y

C2 P dx # Q dy " "

"y yy

y1 Q!x, t" dt " Q!x, y"

""

"x y x

x1

P!t, y" dt " P!x, y" "

"x f !x, y" " "

"x y

C2 P dx # Q dy

x1 $ t $ xtdy " 0yC2

y C2

F ! dr " y C2

P dx # Q dy

F " P i # Q j

"

"x f !x, y" " 0 # "

"x y

C2 F ! dr

x

SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS | | | | 1049

FIGURE 5

(a,b)

x0

y

D

(x,y)

C

C(x,y)

FIGURE 6Types of curves

simple,not closed

not simple,closed

not simple,not closed

not simple,closed

simple,closed

Vy F = P~i + Q~j = fx~i + fy

~j = f .



Ta thy: Nu tch phnC Pdx + Qdy khng ph thuc vo ng i, v

gi s P,Q lin tc v c cc o hm ring bc nht. Khi , tn tihm f sao cho (P,Q) = f

P =f

xQ =

f

y.

Ngoi ra ta cP

y=

2f

yx=

2f

xy=Q

x

nh l

Cho F = P~j + Q~j l trng vector trn min lin thng n D. Gi sP,Q c cc o hm ring cp mt lin tc v

P

y=Q

xtrn D

th tch phnC Pdx + Qdy khng ph thuc vo ng i trn D.



V d

Tnh tch phn I =C ydx + xdy .

Ta thyQ

x=P

y= 1, tch phn trn

khng ph thuc vo ng i.Cch 1.Ta chn ng i khc t O n B l nggp khc OAB. Khi

I =

OA

+

AB

=

10

0dy +

30

1dx = 3

-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5

-2.5

-2

-1.5

-1

-0.5

0.5

1

1.5

(3,1)

C

O

A B



Ta thyQ

x=P

y= 1, tch phn trn

khng ph thuc vo ng i.Cch 2.Tn ti hm kh vi U(x , y) sao cho vi phndU = Pdx + Qdy{

U x = PU y = Q

U(x , y) = xy

I =

(1,3)0,0

ydx + xdy = U(x , y)|(1,3)(0,0) = 3

-1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5

-2.5

-2

-1.5

-1

-0.5

0.5

1

1.5

(3,1)

C

O

A B



V d

Tnh I =C

xdx + ydy

x2 + y2vi C l mt ng cong tu t A(1, 0) n

B(2, 0).a) Khng bao quanh gc to ;b) Bao quanh gc to .

a) Ta kim tra rngQ

x=P

y, do tch phn khng ph thuc ng

i t A n B. Ta chn l on thng ni AB.

I =

21

dx

x= ln |x ||21 = ln 2



b) Tch phn khng ph thuc vo ng i, tuy nhin ta khng th tnhtch phn theo ng t A n B. Ta thy khng tn ti min D cha ccng cong kn bao quanh gc O sao cho P v Q l cc o hm ringcp 1 lin tc trn DTa tm hm U(x , y) sao cho vi phn dU(x , y) = Pdx + Qdy

U x = P =x

x2 + y2 U(x , y) = ln(x

2 + y2)

2+ g(y)

U y = Q =y

x2 + y2 g(y) = C

U(x , y) =ln(x2 + y2)

2+ C

I = U(x , y)|(2,0)(1,0) =ln 4 ln 1

2= ln 2


nh l Green

nh l Green


nh l Green

Cho min D c gii hn bi ng cong n lin CGREENS THEOREM

Greens Theorem gives the relationship between a line integral around a simple closedcurve and a double integral over the plane region bounded by . (See Figure 1. Weassume that consists of all points inside as well as all points on .) In stating GreensTheorem we use the convention that the positive orientation of a simple closed curve refers to a single counterclockwise traversal of . Thus if is given by the vector func-tion , , then the region is always on the left as the point traverses .(See Figure 2.)

GREENS THEOREM Let be a positively oriented, piecewise-smooth, simpleclosed curve in the plane and let be the region bounded by . If and havecontinuous partial derivatives on an open region that contains , then

The notation

gC

is sometimes used to indicate that the line integral is calculated using the positive orienta-tion of the closed curve . Another notation for the positively oriented boundary curve of

is , so the equation in Greens Theorem can be written as

Greens Theorem should be regarded as the counterpart of the Fundamental Theorem ofCalculus for double integrals. Compare Equation 1 with the statement of the FundamentalTheorem of Calculus, Part 2, in the following equation:

In both cases there is an integral involving derivatives ( , , and ) on the leftside of the equation. And in both cases the right side involves the values of the originalfunctions ( , , and ) only on the boundary of the domain. (In the one-dimensional case,the domain is an interval whose boundary consists of just two points, and .)ba!a, b"

PQF

!P#!y!Q#!xF"

yba F"$x% dx ! F$b% # F$a%

yyD

&!Q!x # !P!y ' dA ! y!D P dx $ Q dy1!DD

C

P dx $ Q dyor!yC P dx $ Q dy

NOTE

yC P dx $ Q dy ! yy

D

&!Q!x # !P!y ' dAD

QPCDC

FIGURE 2 (a) Positive orientation

y

x0

D

C

(b) Negative orientation

y

x0

D

C

Cr$t%Da % t % br$t%CC

CCCDCDC

16.4

SECTION 16.4 GREENS THEOREM | | | | 1055

FIGURE 1

y

x0

D

C

N Recall that the left side of this equation is another way of writing , where

.F ! P i $ Q jxC F ! dr

Ta nh ngha chiu dng ca ng cong l chiu ngc chiu kim ngh. Do nu C cho bi phng trnh tham s r(t), a t b th min Dlun nm bn tri ca im r(t) khi chy trn C . Tng t ta c chiu m

GREENS THEOREM

Greens Theorem gives the relationship between a line integral around a simple closedcurve and a double integral over the plane region bounded by . (See Figure 1. Weassume that consists of all points inside as well as all points on .) In stating GreensTheorem we use the convention that the positive orientation of a simple closed curve refers to a single counterclockwise traversal of . Thus if is given by the vector func-tion , , then the region is always on the left as the point traverses .(See Figure 2.)

GREENS THEOREM Let be a positively oriented, piecewise-smooth, simpleclosed curve in the plane and let be the region bounded by . If and havecontinuous partial derivatives on an open region that contains , then

The notation

gC

is sometimes used to indicate that the line integral is calculated using the positive orienta-tion of the closed curve . Another notation for the positively oriented boundary curve of

is , so the equation in Greens Theorem can be written as

Greens Theorem should be regarded as the counterpart of the Fundamental Theorem ofCalculus for double integrals. Compare Equation 1 with the statement of the FundamentalTheorem of Calculus, Part 2, in the following equation:

In both cases there is an integral involving derivatives ( , , and ) on the leftside of the equation. And in both cases the right side involves the values of the originalfunctions ( , , and ) only on the boundary of the domain. (In the one-dimensional case,the domain is an interval whose boundary consists of just two points, and .)ba!a, b"

PQF

!P#!y!Q#!xF"

yba F"$x% dx ! F$b% # F$a%

yyD

&!Q!x # !P!y ' dA ! y!D P dx $ Q dy1!DD

C

P dx $ Q dyor!yC P dx $ Q dy

NOTE

yC P dx $ Q dy ! yy

D

&!Q!x # !P!y ' dAD

QPCDC

FIGURE 2 (a) Positive orientation

y

x0

D

C

(b) Negative orientation

y

x0

D

C

Cr$t%Da % t % br$t%CC

CCCDCDC

16.4


FIGURE 1

y

x0

D

C

N Recall that the left side of this equation is another way of writing , where

.F ! P i $ Q jxC F ! dr


nh l Green

nh l Green

nh l

Cho l ng cong ng C n lin theo hng dng, trn tng khc.Nu P,Q l cc hm s c cc o hm ring lin tc trn min D th

C

Pdx + Qdy =

D

(Q

x Py

)dA


nh l Green

Chng minh

Ta chng minhCPdx =

D

P

ydA v

CQdy =

D

Q

xdA

Ta biu din min D di dng

D = {(x , y)|a x b, g1(x) y g2(x)}

Greens Theorem is not easy to prove in general, but we can give a proof for the specialcase where the region is both of type I and of type II (see Section 15.3). Lets call suchregions simple regions.

PROOF OF GREENS THEOREM FOR THE CASE IN WHICH IS A SIMPLE REGION Notice that GreensTheorem will be proved if we can show that

and

We prove Equation 2 by expressing as a type I region:

where and are continuous functions. This enables us to compute the double integralon the right side of Equation 2 as follows:

where the last step follows from the Fundamental Theorem of Calculus.Now we compute the left side of Equation 2 by breaking up as the union of the

four curves , , , and shown in Figure 3. On we take as the parameter andwrite the parametric equations as , , . Thus

Observe that goes from right to left but goes from left to right, so we can writethe parametric equations of as , , . Therefore

On or (either of which might reduce to just a single point), is constant, soand

Hence

! yba P!x, t1!x"" dx ! yb

a P!x, t2!x"" dx

yC P!x, y" dx ! y

C1 P!x, y" dx " y

C2 P!x, y" dx " y

C3 P!x, y" dx " y

C4 P!x, y" dx

y C2

P!x, y" dx ! 0 ! y C4

P!x, y" dx

dx ! 0xC4C2

y C3

P!x, y" dx ! !y !C3

P!x, y" dx ! !yba P!x, t2!x"" dx

a # x # by ! t2!x"x ! x!C3!C3C3

y C1

P!x, y" dx ! yba P!x, t1!x"" dx

a # x # by ! t1!x"x ! xxC1C4C3C2C1

C

! yba #P!x, t2!x"" ! P!x, t1!x""$ dx yy

D

$P$y

dA ! yba yt2!x"t1!x"

$P$y

!x, y" dy dx4

t2t1

D ! %!x, y" & a # x # b, t1!x" # y # t2!x"'D

yC Q dy ! yy

D

$Q$x

dA3

yC P dx ! !yy

D

$P$y

dA2

D


N Greens Theorem is named after the self-taught English scientist George Green(17931841). He worked full-time in his fathersbakery from the age of nine and taught himselfmathematics from library books. In 1828 he published privately An Essay on the Applicationof Mathematical Analysis to the Theories ofElectricity and Magnetism, but only 100 copieswere printed and most of those went to hisfriends. This pamphlet contained a theorem that is equivalent to what we know as GreensTheorem, but it didnt become widely known at that time. Finally, at age 40, Green entered Cambridge University as an undergraduate but died four years after graduation. In 1846 William Thomson (Lord Kelvin) located a copy of Greens essay, realized its significance, andhad it reprinted. Green was the first person to try to formulate a mathematical theory of elec-tricity and magnetism. His work was the basisfor the subsequent electromagnetic theories ofThomson, Stokes, Rayleigh, and Maxwell.

FIGURE 3

y

x0 a b

D

C

y=g(x)

y=g(x)

C

C

C


nh l Green

Ta tnh tch phn trn cc ng cong C1,C2,C3,C4.Trn C1, ta xem x l tham s: x = x , y = g1(x), a x b.

C1

P(x , y) =

ba

P(x , g1(x))dx

Trn C3, x l tham s v i t b n a, do C3

P(x , y) = ba

P(x , g2(x))dx

Trn C2,C4, x l hng s do dx = 0, ta cC2

P(x , y)dx = 0 =

C4

P(x , y)dx = 0


nh l Green

Vy CP(x , y)dx =

C1

+

C2

+

C3

C4

P(x , y)dx

=

ba

P(x , g1(x))dx ba

P(x , g2(x))dx

Ta li cD

P

ydA =

ba

dx

g2(x)g1(x)

dy =

ba

[P(x , g2(x)) P(x , g1(x))]dx

Do CP(x , y)dx =

D

P

ydA

Tng t ta c CP(x , y)dy =

D

Q

xdA


nh l Green

There are several possibilities:

Then Greens Theorem gives the following formulas for the area of :

EXAMPLE 3 Find the area enclosed by the ellipse .

SOLUTION The ellipse has parametric equations and , where. Using the third formula in Equation 5, we have

M

Although we have proved Greens Theorem only for the case where is simple, we cannow extend it to the case where is a finite union of simple regions. For example, if isthe region shown in Figure 5, then we can write , where and are bothsimple. The boundary of is and the boundary of is so, apply-ing Greens Theorem to and separately, we get

If we add these two equations, the line integrals along and cancel, so we get

which is Greens Theorem for , since its boundary is .The same sort of argument allows us to establish Greens Theorem for any finite union

of nonoverlapping simple regions (see Figure 6).

EXAMPLE 4 Evaluate , where is the boundary of the semiannularregion in the upper half-plane between the circles and .

SOLUTION Notice that although is not simple, the -axis divides it into two simpleregions (see Figure 7). In polar coordinates we can write

D ! !"r, !# $ 1 " r " 2, 0 " ! " #%

yD

x 2 $ y 2 ! 4x 2 $ y 2 ! 1DC!xC y 2 dx $ 3xy dyV

C ! C1 ! C2D ! D1 ! D2

y C1!C2

P dx $ Q dy ! yyD

&%Q%x & %P%y ' dA&C3C3

yC2!"&C3 #

P dx $ Q dy ! yyD2

&%Q%x & %P%y ' dA y

C1!C3 P dx $ Q dy ! yy

D1

&%Q%x & %P%y ' dAD2D1

C2 ! "&C3#D2C1 ! C3D1D2D1D ! D1 ! D2

DDD

!ab2

y2#0

dt ! #ab

! 12 y2#0 "a cos t#"b cos t# dt & "b sin t#"&a sin t# dt A ! 12 yC x dy & y dx

0 " t " 2#y ! b sin tx ! a cos t

x 2

a 2$

y 2

b 2! 1

A ! !yC x dy ! &!y

C y dx ! 12 !yC x dy & y dx5

D

Q"x, y# ! 12 x Q"x, y# ! 0 Q"x, y# ! xP"x, y# ! &12 yP"x, y# ! &yP"x, y# ! 0


FIGURE 5

C

_CCCD D

FIGURE 6

C

There are several possibilities:

Then Greens Theorem gives the following formulas for the area of :

EXAMPLE 3 Find the area enclosed by the ellipse .

SOLUTION The ellipse has parametric equations and , where. Using the third formula in Equation 5, we have

M

Although we have proved Greens Theorem only for the case where is simple, we cannow extend it to the case where is a finite union of simple regions. For example, if isthe region shown in Figure 5, then we can write , where and are bothsimple. The boundary of is and the boundary of is so, apply-ing Greens Theorem to and separately, we get

If we add these two equations, the line integrals along and cancel, so we get

which is Greens Theorem for , since its boundary is .The same sort of argument allows us to establish Greens Theorem for any finite union

of nonoverlapping simple regions (see Figure 6).

EXAMPLE 4 Evaluate , where is the boundary of the semiannularregion in the upper half-plane between the circles and .

SOLUTION Notice that although is not simple, the -axis divides it into two simpleregions (see Figure 7). In polar coordinates we can write

D ! !"r, !# $ 1 " r " 2, 0 " ! " #%

yD

x 2 $ y 2 ! 4x 2 $ y 2 ! 1DC!xC y 2 dx $ 3xy dyV

C ! C1 ! C2D ! D1 ! D2

y C1!C2

P dx $ Q dy ! yyD

&%Q%x & %P%y ' dA&C3C3

yC2!"&C3 #

P dx $ Q dy ! yyD2

&%Q%x & %P%y ' dA y

C1!C3 P dx $ Q dy ! yy

D1

&%Q%x & %P%y ' dAD2D1

C2 ! "&C3#D2C1 ! C3D1D2D1D ! D1 ! D2

DDD

!ab2

y2#0

dt ! #ab

! 12 y2#0 "a cos t#"b cos t# dt & "b sin t#"&a sin t# dt A ! 12 yC x dy & y dx

0 " t " 2#y ! b sin tx ! a cos t

x 2

a 2$

y 2

b 2! 1

A ! !yC x dy ! &!y

C y dx ! 12 !yC x dy & y dx5

D

Q"x, y# ! 12 x Q"x, y# ! 0 Q"x, y# ! xP"x, y# ! &12 yP"x, y# ! &yP"x, y# ! 0


FIGURE 5

C

_CCCD D

FIGURE 6

C

Therefore Greens Theorem gives

M

Greens Theorem can be extended to apply to regions with holes, that is, regions thatare not simply-connected. Observe that the boundary of the region in Figure 8 con-sists of two simple closed curves and . We assume that these boundary curves are oriented so that the region is always on the left as the curve is traversed. Thus the positive direction is counterclockwise for the outer curve but clockwise for the innercurve . If we divide into two regions and by means of the lines shown in Figure 9 and then apply Greens Theorem to each of and we get

Since the line integrals along the common boundary lines are in opposite directions, theycancel and we get

which is Greens Theorem for the region .

EXAMPLE 5 If , show that for everypositively oriented simple closed path that encloses the origin.

SOLUTION Since is an arbitrary closed path that encloses the origin, its difficult to compute the given integral directly. So lets consider a counterclockwise-oriented circle

with center the origin and radius , where is chosen to be small enough that liesinside . (See Figure 10.) Let be the region bounded by and . Then its positivelyoriented boundary is and so the general version of Greens Theorem gives

Therefore

that is, yC F ! dr ! y

C! F ! dr

yC P dx " Q dy ! y

C! P dx " Q dy

! yyD

! y 2 # x 2"x 2 " y 2 #2 # y 2 # x 2"x 2 " y 2 #2$ dA ! 0 y

C P dx " Q dy " y

#C! P dx " Q dy ! yy

D

%$Q$x # $P$y & dAC " "#C!#

C!CDCC!aaC!

C

xC F ! dr ! 2%F"x, y# ! "#y i " x j#'"x 2 " y 2 #VD

yyD

%$Q$x # $P$y & dA ! y C1 P dx " Q dy " y C2 P dx " Q dy ! yC P dx " Q dy

! y$D!

P dx " Q dy " y$D&

P dx " Q dy

yyD

%$Q$x # $P$y & dA ! yyD!

%$Q$x # $P$y & dA " yyD&

%$Q$x # $P$y & dAD&,D!

D&D!DC2C1

CDC2C1

DC

! y%0

sin ' d' y21 r 2 dr ! [#cos ']0% [ 13 r 3 ]12 ! 143

! yyD

y dA ! y%0

y21 "r sin '# r dr d'

!yC

y 2 dx " 3xy dy ! yyD

! $$x "3xy# # $$y "y 2 #$ dA


FIGURE 7

0

y

x

C

+=4

+=1

D

FIGURE 8

DC

C

FIGURE 9

D

D

FIGURE 10

y

xD

C

C


nh l Green

V d

Tnh tch phnC x

4dx + xydy vi C l cc cnh tam gic vi cc nh

(0, 0), (1, 0), (0, 1) theo chiu dng.

p dng nh l Green ta cCx4dx + xydy =

D

(Q

x Py

)dA

=

10

1x0

(y 0)dydx

=

10

y2

2

y=1xy=0

dx

=1

2

10

(1 x2)2dx = 16

Comparing this expression with the one in Equation 4, we see that

Equation 3 can be proved in much the same way by expressing as a type II region (seeExercise 28). Then, by adding Equations 2 and 3, we obtain Greens Theorem. M

EXAMPLE 1 Evaluate , where is the triangular curve consisting of theline segments from to , from to , and from to .

SOLUTION Although the given line integral could be evaluated as usual by the methods ofSection 16.2, that would involve setting up three separate integrals along the three sidesof the triangle, so lets use Greens Theorem instead. Notice that the region enclosedby is simple and has positive orientation (see Figure 4). If we let and

, then we have

M

EXAMPLE 2 Evaluate , where is the circle .

SOLUTION The region bounded by is the disk , so lets change to polarcoordinates after applying Greens Theorem:

M

In Examples 1 and 2 we found that the double integral was easier to evaluate than theline integral. (Try setting up the line integral in Example 2 and youll soon be convinced!)But sometimes its easier to evaluate the line integral, and Greens Theorem is used in thereverse direction. For instance, if it is known that on the curve ,then Greens Theorem gives

no matter what values and assume in the region .Another application of the reverse direction of Greens Theorem is in computing areas.

Since the area of is , we wish to choose and so that

!Q!x

"!P!y! 1

QPxxD 1 dADDQP

yyD

!!Q!x " !P!y " dA ! yC P dx # Q dy ! 0CP#x, y$ ! Q#x, y$ ! 0

! 4 y2$0

d% y30 r dr ! 36$! y2$

0 y3

0 #7 " 3$ r dr d%

! yyD

% !!x (7x # sy 4 # 1) " !!y #3y " e sin x$& dA!y

C #3y " e sin x $ dx # (7x # sy 4 # 1) dy

x 2 # y 2 & 9CD

x 2 # y 2 ! 9C!xC #3y " e sin x $ dx # (7x # sy 4 # 1) dyV

! " 16 #1 " x$3 ]01 ! 16 ! y1

0 [ 12 y 2 ]y!0y!1"x dx ! 12 y10 #1 " x$2 dx

yC x 4 dx # xy dy ! yy

D

!!Q!x " !P!y " dA ! y10 y1"x0 #y " 0$ dy dxQ#x, y$ ! xy

P#x, y$ ! x 4CCD

#0, 0$#0, 1$#0, 1$#1, 0$#1, 0$#0, 0$CxC x 4 dx # xy dy

D

yC P#x, y$ dx ! "yy

D

!P!y

dA


N Instead of using polar coordinates, we couldsimply use the fact that is a disk of radius 3and write

yyD

4 dA ! 4 ! $#3$2 ! 36$

D

FIGURE 4

y

x

C

(1,0)(0,0)

(0,1) y=1-x

D


nh l Green

V d

Tnh I =C (x y)2dx + (x + y)2dy trong C l na trn ng trn

x2 + y2 = 2x theo hng cng chiu kim ng h.

Cung C khng kn, ta thm vo on AO c min D l na hnh trn.

I =

C

=

CAO

AO

CAO=

D

(Q

x Py

)dxdy

=

D2 ((x + y) + 2(x y)) dxdy

= pi/20

d

2 cos0

4r cosrdr = 2pi

-0.5 0 0.5 1 1.5 2 2.5

-0.5

0.5

1


nh l Green

Trn cung AO ta c phng trnh tham s

x = 2 t y = 0 vi 0 t 2AO

= 20

(2 t)2dt = (2 t)3

3

20

= 83

Vy ta c

I =

CAO

AO

= 2pi + 83


nh l Green

Bi tp 1.

Tnh tch phnC F dr trn ng cong C vi

1 F (x , y) = x2~i + y2~j , C l ng parabol y = 2x2 t (1, 2) n(2, 8).

2 F = xy2~i + x2y~j vi C : r(t) =t + sin 12pit, t + cos

12pit

0 t 1.3 F (x , y) =

y2

1 + x2~i + 2y arctan x~j vi C : r(t) = t2~i + 2t~j , 0 t 1.

4 TnhC (1 yex)dx + exdy vi C l ng cong t (0, 1) n

(1, 2).

5 Tnh cng sinh ra khi tc dng trng lc F khi di chuyn vt t Pn Q via) F (x , y) = 2y3/2~i + 3x

y~j ; P(1, 1) v Q(2, 4);

b) F (x , y) = ey~i xey~j ; P(0, 1) v Q(2, 0).


nh l Green

Bi tp 2.

1 TnhC (x y)dx + (x + y)dy vi C l ng trn tm ti gc to

v bn knh 2 theo chiu m.

2 TnhC xydx + x

2y3dy vi C l tam gic vi nh (0, 0), (1, 0), (1, 2)

theo chiu dng.

3 TnhC xy

2dx + 2x2ydy vi C l tam gic vi cc nh (0, 0), (2, 2

v (2, 4) theo chiu dng.

4C (y + e

x)dx + (2x + cos y2)dy vi C l bin ca min gii hn vi

hai parabol y = x2 v x = y2 ngc chiu kim ng h.

5C sin ydx + x cos ydy vi C l ellipse x

2 + xy + y2 = 1 ngc chiu

kim ng h.


Tch phn ng loai mtTch phn ng loai haiMt s tnh cht cua tch phn nginh l Green

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