Part 3: Area Formulas   

If we let M_y = 1, then M( x,y) = y and thus (1) becomes

óó õõ   R 1dA = - ¶R  ydx

Likewise, if N_x = 1, then N( x,y) = x and (2) becomes

óó õõ   R  1 dA =  ¶R  xdy

and the average of the two results yields

óó õõ   R dA =   1 2 ¶R  xdy-ydx

That is, since A = òò _R dA, Green's theorem for the special case of N_x and M_y constant yields the following:

Theorem 4: The area A of a simply-connected region R in the xy-plane is given by each of the following:

A = - ¶R  ydx,    A = ¶R  xdy,   or  A =  1 2 ¶R  xdy-ydx

In general, we use one of the first two formulas, but in the development of theory or when trigonometric functions are involved, we often use the last formula.      

EXAMPLE 4    Find the area of the region R enclosed by the curve r( t) = át⁴-t²,t⁶-t² ñ , t in [ 0,1] . Solution: Since x = t⁴-t² and y = t⁶-t², the area of R is thus given by

A = ¶R  xdy = ó õ 1 0  x  dy dt dt = ó õ 1 0  ( t4-t2) ( 6t5-2t) dt =  1 60

Notice that we obtain the same result with

A = - ¶R  ydx = - ó õ 1 0  y  dx dt dt = - ó õ 1 0  ( t6-t2) ( 4t3-2t) dt =  1 60

EXAMPLE 5    Find the area of an ellipse R with semi-major axis a and semi-minor axis b. Solution: We can parametrize the ellipse by r(t) = á acos( t) ,bsin( t) ñ since x = acos( t) and y = bsin(t) satisfies

x2 a2 +  y2 b2 =  a2cos2( t) a2 +  b2sin2( t) b2 = cos2(t) +sin2( t) = 1

Since trigonometric functions are involved, we use the last formula in theorem 2 to obtain

A =  1 2 ¶R  xdy-ydx =  1 2 ó õ 2p 0  acos( t) [ bcos( t) ] -bsin(t) [ -asin( t) ] dt

However, this simplifies to

A =  1 2 ó õ 2p 0  ( abcos2( t) +absin2( t) ) dt =  1 2 ó õ 2p 0  abdt =  ab 2 q  2p 0 = pab