# VC5.16

From Example Problems

Here the path of integration C consists of the straight lines say OA,AB,BC and CD where the coordinates of A,B,C are (a,0),(a,b) and (0,b) respectively.Let S be the plane region bounded by closed curve C,then by Stokes' theorm,we must have

--(1)

Since the integration is performed in the xy-plane,we have so that

LHS of (1)=

=

= --(2)

Along OA,y=0,dy=0 and x varies from 0 to a. Along AB,x=a,dx=0 and y varies from 0 to b. Along BC,y=b,dy=0 and x varies from a to 0. Along CO,x=0,dx=0 and y varies from b to 0.

Therefore,from (2),LHS=

=

= --(3)

Now,

=

Also,n=unit vector normal to surface S=k.

Therefore

Now,RHS=

= --(4)

From (3) and (4),Stokes'theorm is verified.