11.3 GREEN’S THEOREM IN A PLANE
y=y 2 (x) be the equations of the curvesSTUandSVUrespectively. We then
write
∫∫
R
∂P
∂y
dx dy=
∫b
a
dx
∫y 2 (x)
y 1 (x)
dy
∂P
∂y
=
∫b
a
dx
[
P(x, y)
]y=y 2 (x)
y=y 1 (x)
=
∫b
a
[
P(x, y 2 (x))−P(x, y 1 (x))
]
dx
=−
∫b
a
P(x, y 1 (x))dx−
∫a
b
P(x, y 2 (x))dx=−
∮
C
Pdx.
If we now letx=x 1 (y)andx=x 2 (y) be the equations of the curvesTSVand
TUVrespectively, we can similarly show that
∫∫
R
∂Q
∂x
dx dy=
∫d
c
dy
∫x 2 (y)
x 1 (y)
dx
∂Q
∂x
=
∫d
c
dy
[
Q(x, y)
]x=x 2 (y)
x=x 1 (y)
=
∫d
c
[
Q(x 2 (y),y)−Q(x 1 (y),y)
]
dy
=
∫c
d
Q(x 1 ,y)dy+
∫d
c
Q(x 2 ,y)dy=
∮
C
Qdy.
Subtracting these two results gives Green’s theorem in a plane.
Show that the area of a regionRenclosed by a simple closed curveCis given byA=
1
2
∮
C(xdy−ydx)=
∮
Cxdy=−
∮
Cydx. Hence calculate the area of the ellipsex=acosφ,
y=bsinφ.
In Green’s theorem (11.4) putP=−yandQ=x;then
∮
C
(xdy−ydx)=
∫∫
R
(1+1)dx dy=2
∫∫
R
dx dy=2A.
Therefore the area of the region isA=^12
∮
C(xdy−ydx). Alternatively, we could putP=0
andQ=xand obtainA=
∮
Cxdy, or putP=−yandQ= 0, which givesA=−
∮
Cydx.
The area of the ellipsex=acosφ,y=bsinφis given by
A=
1
2
∮
C
(xdy−ydx)=
1
2
∫ 2 π
0
ab(cos^2 φ+sin^2 φ)dφ
=
ab
2
∫ 2 π
0
dφ=πab.
It may further be shown that Green’s theorem in a plane is also valid for
multiply connected regions. In this case, the line integral must be taken over
all the distinct boundaries of the region. Furthermore, each boundary must be
traversed in the positive direction, so that a person travelling along it in this
direction always has the regionRon their left. In order to apply Green’s theorem