Part 2: Green's Theorem   

Suppose R is a simply-connected region which is both type I and type II and has piecewise differentiable boundaries. Then it can be shown that given N( x,y) , we can use the fact that R is also type II to obtain

óó õõ   R NxdA =  ¶R  N( x,y) dx

(2)

Combining (2) with (1) then yields a special case of the following theorem:      

Green's Theorem: If R is simply connected, and if M(x,y) and N( x,y) are differentiable over R, then

¶R  Mdx+Ndy =   óó õõ   R   æ è  ¶N ¶x -  ¶M ¶y ö ø dA

       

In particular, while Green's theorem holds for any simply connected region R, we have shown it only for those simply connected regions that are both type I and type II.       

EXAMPLE 2    Given that D is the unit disk, evaluate

¶D  ( xy) dx+( x+y) dy

Solution: We first identify M( x,y) = xy and N(x,y) = x+y, so that

¶D  ( xy) dx+( x+y)dy =   óó õõ   D æ è  ¶( x+y) ¶x -  ¶( xy) ¶y ö ø dA =  óó õõ   D ( 1-x)dA

We then convert the double integral into polar coordinates to obtain

óó õõ   D ( 1-x) dA =  ó õ 2p 0  ó õ 1 0  ( 1-rcos( q) ) rdrdq = ó õ 2p 0  ó õ 1 0  (r-r2cos( q) ) drdq

As a result, evaluation of the integrals leads to

óó õõ   D ( 1-x) dA = ó õ 2p 0   r2 2 -  r3 3 cos( q) ê ê 1 0  dq = ó õ 2p 0  æ è  1 2 -  1 3 cos( q) ö ø dq =  q 2 -  1 3 sin( q) ê ê 2p 0  = p

Thus, Green's theorem yields

¶D  ( xy) dx+( x+y) dy = p

       

EXAMPLE 3    Given that R is the unit square, evaluate

¶R  x2y dx+x2y3 dy

Solution: In this case, evaluating the line integral directly would require dividing the line integral into 4 separate integrals over the 4 separate sides of the square. However, M( x,y) = x²y and N( x,y) = x²y³ implies that M_y = x² and N_x = 2xy³, so that

¶R  x2y dx+x2y3 dy =   óó õõ   R (2xy3-x2) dA

As a result, we have

óó õõ   R ( 2xy3-x2) dA = ó õ 1 0  ó õ 1 0  (2xy3-x2) dydx =  -1 12

       

Check your Reading: What is the value of

¶R  x2y dx+x2y3 dy