<<

10.4 Green's theorem in the plane

Double 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 . The transformation can be done by the following theorem.

Theorem 1: Green's theorem in the plane (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 (see Sec. 10.1). Let Fxy1(, ) and Fxy2 (, ) be

functions that are continuous and have continuous partial Fy1 / and

F2 / x everywhere in some domain containing R . Then,

æö 綶FF21÷ ç -=+÷dx dy( F12 dx F dy). (10.4.1) òò ç ¶¶xy÷ òC R èø

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. 232).

Setting FiFF12j and using eqn (9.9.1), the vectorial form of eqn (10.4.1) is,

(curlFk ) ··dx dy= F d r. (10.4.1*) òò òC R

The proof follows after Example 1.

Example 1: Verification of Green's theorem in the plane Before proving Green's theorem in the plane, we get used to the theorem by verifying it for Fi(7)(22)y2 yxyx  j and C the circle xy22 1.

Solution: In eqn (10.4.1) on the left we get,

æö¶¶FF ç 21-=+--=÷dx dy[(2 y 2) (2 y 7)] dx dy 9 dx dy = 9 , òò ç ¶¶xy÷ òò òò RRèø R

since the circular disk R has an of  .

June 25, 2011 10.4-1/6

We now show that the line integral in eqn (10.4.1) on the right gives the same value of 9 . We must orient C counterclockwise, say, ri()tt cos sintj . Then ri()ttsin  costj, and on C ,

22 Fy127 y sin t  7sin t , F2 xyx  2 2cos t sin t  2cos t .

Hence the line integral in eqn (10.4.1) becomes,

()F12 dx+= F dy() F 12 x¢¢ + F y dt, òòCC 2 =--++[(sin2 ttt 7sin )( sin ) 2(cos tttt sin cos )(cos )]dt , ò0 2 =-+( sin32tt 7sin + 2cos 2 tttdt sin + 2cos 2 ) , ò0 =+07 -+ 02 = 9,

verifying Green's theorem.

Proof: Green's theorem in the plane is first proved for a special region R that can be represented in both forms,

axb,( uxyvx)(), (see Fig. 233),

and

cyd,( pyxqy)(), (see Fig. 234).

Using eqn (10.3.3), we obtain for the second term on the left side of eqn (10.4.1) taken without the minus sign (see Fig. 233),

¶¶FFbvxéù() 11dx dy= êú dy dx . (10.4.2) òò¶¶yy òauxêú ò () R ëû

Integrate the inner integral,

vx() yvx= () ¶F1 dy==- F111(, x y ) F [,()] x v x F [,()] x u x . òux() ¶y yux= ()

By inserting the above expression into eqn (10.4.2) we find,

June 25, 2011 10.4-2/6

bb ¶F1 dx dy=- F11[,()] x v x dx F [,()], x u x dx òò¶y òaa ò R ab =-Fxvx11[,()] dx - Fxux [,()]. dx òòba

Since y  vx() represents the C** (Fig. 233) and y  ux() represents C* , the last two integrals may be written as line integrals over C** and C* (oriented as in Fig. 233); therefore,

¶F1 dx dy=- F11(, x y ) dx - F (, x y ) dx , òò¶y òCC** ò * R (10.4.3)

=- Fxydx1(, ) . òC

This proves eqn (10.4.1) in Green's theorem if F2  0 .

 The result remains valid if C has portions parallel to the y -axis (such as C and C in Fig. 235). Indeed, the integrals over these portions are zero because in eqn (10.4.3) on the right we integrate with respect to x . Hence these integrals may be added to the integrals over C* and C** to obtain the integral over the whole boundary C in eqn (10.4.3).

We now treat the first term in eqn (10.4.1) on the left in the same way. Instead of eqn (10.3.3) we use eqn (10.3.4), and the second representation of the special region (see Fig. 234). Then,

¶¶FFdqyéù() 22dx dy= êú dx dy, òò¶¶ òcpyêú ò () R xxëû dd =-F22[(),] q y y dy F [(),] p y y dy , òòcc dc =+F22[(),] q y y dy F [(),] p y y dy , òòcd

= Fxydy2 (, ) . òC

Together with eqn (10.4.3) this gives eqn (10.4.1) and proves Green's theorem for special regions.

June 25, 2011 10.4-3/6

We now prove the theorem for a region R that itself is not a special region but can be subdivided into finitely many special regions (Fig. 236). 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 R while the right-hand members add up to the line integral over C plus integrals over the curves introduced for subdividing R . The simple key observation now 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 .

10.4.1 Some applications of Green's theorem

Example 2: Area of a plane region as a line integral over the boundary

In eqn (10.4.1) we first choose FF12 0,  x and then FyF12 ,0. This gives,

dx dy= x dy and dx dy=- y dx , òò òC òòòC R R

respectively. The double integral is the area A of R . By addition we have,

1 Axdyy=-(dx), (10.4.4) 2 òC

where we integrate as indicated in Green's theorem. This interesting formula expresses the area of R in terms of a line integral over the boundary. It is used, for instance, in the theory of certain planimeters (mechanical instruments for measuring area).

For an ellipse xa22// yb 221 or x  atybcos , sin t we get xatybsin ,  cost; thus from eqn (10.4.4) we obtain the familiar formula for the area of the region bounded by an ellipse,

22 11éù22 Axyyxdtabtabtdt=-=()¢¢ êúcos --= ( sin ) ab . 22òò00ëû

Example 3: Area of a plane region in polar coordinates Let r and  be polar coordinates defined by xr cos , yr sin . Then,

dxcos dr r sin d , dy  sin dr r cos  d ,

and eqn (10.4.4) becomes a formula that is well known from , namely,

1 A = rd2  . (10.4.5) 2 òC

As an application of eqn (10.4.5), consider the cardioid ra(1 cos ) , where 02  (Fig. 237). Here,

a2 2 3 Ad=-(1 cos )22 =a . 22ò0

June 25, 2011 10.4-4/6

Example 4: Transformation of a double integral of the Laplacian of a into a line integral of its normal Use Green's theorem for deriving a basic integral formula involving the Laplacian.

Take a function wxy(,) that is continuous and has continuous first and second partial derivatives in a domain of the xy -plane containing a region R of the type indicated in

Green's theorem. We set F1  w/ y and Fw2  / .x Then Fy1 / and Fx2 /  are continuous in R , and in eqn (10.4.1) on the left we obtain,

FF22ww 21  2w, (10.4.6) xyx22  y

the Laplacian of w . Furthermore, using those expressions for F1 and F2 , we get in (10.4.1) on the right,

æöæö ç dx dy÷ ç ¶¶w dx w dy÷ ()F12 dx+= F dyç F 1 + F 2÷ ds =-+ç ÷ds , (10.4.7) òòCCèøç ds ds ÷ ò  Cèøç dy ds dx ds ÷

where s is the of C , and C is oriented as shown in Fig. 238. The integrand of the last integral may be written as the ,

w w dy dx   w dy  w dx  (gradw ) nijij    . (10.4.8) dx dy ds ds  dx ds dy ds 

The vector n is a unit normal vector to C, because the vector rr()s d / dsdx / ds i  dy / ds j is the unit vector of C , and rn   0, so that n is perpendicular to r . Also, n is directed to the exterior of C because in Fig. 238 the positive x -component dx / ds of r is the negative y -component of n , and similarly at other points. From this and eqn (9.7.4) we see that the left side of eqn (10.4.8) is the derivative of w in the direction of the outward normal of C . This derivative is called the normal derivative of w and is denoted by wn/ ; that is,

June 25, 2011 10.4-5/6

wn/(grad) w n . Because of eqns (10.4.6)-(10.4.8), Green's theorem gives the desired formula relating the Laplacian to the normal derivative,

¶w =2wdxdy ds. (10.4.9) òò òC ¶n R

For instance, wx2y2 satisfies Laplace's equation 2w  0. Hence, its normal derivative integrated over a closed curve must give 0. This can be verified directly by integration, say for the square 0  xy1, 0 1.

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.

June 25, 2011 10.4-6/6