Part 1: Double Integrals and Boundary Curves 

When a vector field F( x,y) = á M(x,y) ,N( x,y) ñ is conservative, then it has a potential and line integrals of the form ò_A^BF·dr are independent of the path C which connects A to B. When a vector field is not conservative, then another form of the fundamental theorem called Green's theorem is used to reveal properties of the field.

To begin with, a region R is simply connected if its boundary is a simple closed curve. Moreover, we denote its boundary curve by ¶R and we define the orientation of ¶R to be such that the region R is always to the left of a tangent vector to ¶R.

For example, if f( x) > g( x) for all x in (a,b) and if f( a) = g( a) and f(b) = g( b) , then y = f( x) and y = g(x) enclose a simply connected region R whose boundary ¶R is made up of two curves, C₁ and C₂.  The curve C₁ is parametrized by r( t) = á -t,f( -t) ñ for t in [ -b,-a] so that it starts at ( b,f( b) ) and ends at ( a,f( a)).

The double integral of the derivative M_y with respect to y of a function M(x,y) over R is given by

R  MydA = ó õ b a  ó õ f( x) g( x)   My( x,y) dydx

Since M_y is a derivative with respect to y, we must have

R MydA = ó õ b a  M( x,y)     y = f( x) y = g( x)   dx = ó õ b a  M( x,f( x) ) -M( x,g(x) ) dx = ó õ b a  M( x,f( x) ) dx- ó õ b a  M(x,g( x) ) dx

If we let x = -t, dx = -dt, t( a) = -a, t(b) = -b in the first integral, then

R MydA = - ó õ -b -a  M( -t,f( -t) )dt - ó õ b a  M( x,g( x) ) dx

However, if x = -t, then dx/dt= -1. Likewise, in the second integral we let x = t, so that dx/dt = 1 and

R MydA = - ó õ -a -b  M( -t,f( -t) )  dx dt dt - ó õ b a  M( t,g( t) )  dx dt dt = - ó õ C1  M( x,y) dx- ó õ C2  M( x,y) dx = - é ë ó õ C1  M( x,y) dx+ ó õ C2  M( x,y)dx ù û = - ¶R  M(x,y) dx

where denotes a line integral around a closed curve with the natural orientation. That is, we have shown that

R MydA = - ¶R  M( x,y) dx

(1)

Thus, a line integral on the boundary of R can be used to evaluate a double integral over R.       

EXAMPLE 1    Use (1) to evaluate

ó õ 1 -1  ó õ Ö 1-x2 3xy2dydx -Ö 1-x2

Solution: First we convert into a double integral over the unit disk D:

ó õ 1 -1  ó õ Ö 1-x2 3xy2dydx =   óó õõ   D 3xy2 dA -Ö 1-x2

Identifying M_y = 3xy² implies that M = òM_ydy = ò3xy²dy = xy³. Thus, (1) says that

óó õõ   D 3xy2dA = - ¶D  xy3dx

Moreover, r( t) = á cos( t),sin( t) ñ , t in [ 0,2p] , is a parametrization of ¶D, so that

¶D  xy3dx = ó õ 2p 0  xy3  dx dt  dt = ó õ 2p 0  cos( t) sin3( t) [-sin( t) ] dt

If u = sin( t) , then du = cos( t) dt and u(0) = u( 2p) = 0. Thus,

óó õõ   D 3xy2dA = - ¶D  xy3dx = ó õ 2p 0  sin4( t) cos( t) dt = - ó õ 0 0  u4du = 0