Part 4: Extending Green's Theorem  

Green's theorem can also be extended to regions that are not simply connected, but to do so requires that we carefully define the orientation of the boundary of the region. For example, suppose R is the region shown below: The orientation of the outside boundary is easily defined, but what about the orientation of the inside boundary? To answer that question, we imagine that R is divided into two regions as shown below: If we now imagine the two regions are ``glued'' back together, then the orientation of the boundary inside the region is implied:

EXAMPLE 6    Use the area form of Green's theorem to find the area of the annulus centered at the origin with inner radius 1 and outer radius 3.

Solution: The inner boundary C₂ is a unit circle oriented clockwise, so it is parametrized by r( t) = ásin( t) ,cos( t) ñ , t in [0,2p] . The outer boundary C₁ is a circle centered at the origin with radius 3, so it is parametrized by r( t) = á 3cos( t) ,3sin( t) ñ , t in [ 0,2p] . Thus,

Area =  1 2 ¶R  xdy-ydx =  1 2 ó õ C1  xdy-ydx    +     1 2 ó õ C2  xdy-ydx

However, on C₁ we have

1 2 ó õ C1  xdy-ydx =  1 2 ó õ 2p 0  æ è x  dy dt -y  dx dt ö ø dt =  1 2 ó õ 2p 0  3cos( t) ( 3cos(t) ) -3sin( t) ( -3sin( t)) dt =  1 2 ó õ 2p 0  ( 9cos2( t) +9sin2( t) ) dt = 9p

while on C₂ we have

1 2 ó õ C2  xdy-ydx =  1 2 ó õ 2p 0  æ è x  dy dt -y  dx dt ö ø dt =  1 2 ó õ 2p 0  sin( t) [ -sint]-cos( t) [ cos( t) ] dt =  -1 2 ó õ 2p 0  ( cos2( t) +sin2( t) ) dt = -p

Combining the two yields

Area =  1 2 ó õ C1  xdy-ydx+  1 2 ó õ C2  xdy-ydx = 9p-p = 8p