Page 395 - Advanced engineering mathematics

12.2 Green’s Theorem 375

C is positively oriented then, as we walk around C in the positive direction, the interior is over
our left shoulder.
We will use the term path for a piecewise smooth curve. And we often denote a line integral

over a closed path C as , with a small oval on the integral sign. This is not obligatory and does
C
not affect the meaning of the line integral or the way it is evaluated.

THEOREM 12.1 Green’s Theorem

Let C be a simple closed positively oriented path in the plane. Let D consist of all points on C
and in its interior. Let f, g, ∂ f/∂y and ∂g/∂x be continuous on D. Then

∂g ∂ f
f (x, y)dx + g(x, y)dy = − dA.
∂x ∂y
C D
A proof under special conditions on D is sketched in Problem 14.

EXAMPLE 12.8

Sometimes Green’s theorem simplifies an integration. Suppose we want to compute the work
done by F(x, y)=(y − x e )i+(cos(2y )− x)j in moving a particle counterclockwise about the
2
2 x
rectangular path C having vertices (0,1),(1,1), (1,3) and (0,3).

If we attempt to evaluate F · dR we encounter integrals that cannot be done in elementary
C
form. However, by Green’s theorem, with D the solid rectangle bounded by C,
∂ 2 ∂ 2 x

work = F · dR = (cos(2y ) − x) − (y − x e ) dA
C D ∂x ∂y

= −2dA = (−2)[area ofD]=−4.
D

EXAMPLE 12.9
Another use of Green’s theorem is in deriving general results. Suppose we want to evaluate

2
2x cos(2y)dx − 2x sin(2y)dy
C
for every positively oriented simple closed path C in the plane.
There are infinitely many such paths. However, f (x, y) and g(x, y) have the special property
that

∂
∂ 2
−2x sin(2y) − (2x cos(2y))
∂x ∂y
=−4x sin(2y) + 4x sin(2y) = 0.
By Green’s theorem, for any such closed path C in the plane,

2 0dA = 0.
2x cos(2y)dx − 2x sin(2y)dy =
C D

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October 14, 2010 14:53 THM/NEIL Page-375 27410_12_ch12_p367-424