11.3 GREEN’S THEOREM IN A PLANE
y=y 2 (x) be the equations of the curvesSTUandSVUrespectively. We then
write
∫∫R∂P
∂ydx dy=∫badx∫y 2 (x)y 1 (x)dy∂P
∂y=∫badx[
P(x, y)]y=y 2 (x)y=y 1 (x)=∫ba[
P(x, y 2 (x))−P(x, y 1 (x))]
dx=−∫baP(x, y 1 (x))dx−∫abP(x, y 2 (x))dx=−∮CPdx.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
∂xdx dy=∫dcdy∫x 2 (y)x 1 (y)dx∂Q
∂x=∫dcdy[
Q(x, y)]x=x 2 (y)x=x 1 (y)=∫dc[
Q(x 2 (y),y)−Q(x 1 (y),y)]
dy=∫cdQ(x 1 ,y)dy+∫dcQ(x 2 ,y)dy=∮CQdy.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∫∫
Rdx 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 byA=
1
2
∮
C(xdy−ydx)=1
2
∫ 2 π0ab(cos^2 φ+sin^2 φ)dφ=
ab
2∫ 2 π0dφ=πab.It may further be shown that Green’s theorem in a plane is also valid formultiply 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