Page 398 - Advanced engineering mathematics

378 CHAPTER 12 Vector Integral Calculus

If D is the interior of C , then by Green’s theorem,
∗
∗
∂g ∂ f

fdx + gdy = − dA. (12.1)
C ∗ D ∗ ∂x ∂y
Now take a limit in Figure 12.7 as the channels are made narrower. The opposite sides
of each channel merge to single line segments, which are integrated over in both directions in
equation (12.1). The contributions to the sum in this equation from the channel cuts is therefore
zero. Further, as the channels narrow, the small arcs of C and each K j cut out in making the
channels are restored, and the line integrals in equation (12.1) are over all of C and the circles
K j . Recalling that in equation (12.1) the integrations over the K j ’s are clockwise, equation (12.1)
can be written
∂g ∂ f
n

fdx + gdy − fdx + gdy = − dA. (12.2)
∂x ∂y
C K j D ∗
j=1
in which all integrations (over C and each K j ) are now taken in the positive, counterclock-

wise sense. This accounts for the minus sign on each of the integrals fdx + gdy in
K j
equation (12.2). Finally, write equation (12.2) as
n

∂g ∂ f
fdx + gdy = fdx + gdy + − dA. (12.3)
C K j D ∗ ∂x ∂y
j=1
This is the extended form of Green’s theorem. When D contains points at which f, g,∂ f/∂y

and/or ∂g/∂x are not continuous, then fdx + gdy is the sum of the line integrals fdx +
C K j
gdy about small circles centered at the P j ’s, together with
∂g ∂ f

− dA
∂x ∂y
D ∗
∗
over the region D formed by excising from D the disks bounded by the K j ’s.

EXAMPLE 12.10
We will evaluate
−y x

dx + dy
2
2
C x + y 2 x + y 2
in which C is any simple closed positively oriented path in the plane, but not passing through the
origin.
With
−y x
f (x, y) = and g(x, y) =
2
x + y 2 x + y 2
2
we have
2
∂g ∂ f y − x 2
= = .
2
2 2
∂x ∂y (x + y )
This suggests that we consider two cases.

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October 14, 2010 14:53 THM/NEIL Page-378 27410_12_ch12_p367-424