This document verifies Stokes's theorem for a given vector field F and surface S. It shows that the line integral of F around the boundary of S equals the surface integral of the curl of F over S. Specifically: - F = zi + xj + yk and the surface S is the cone z = x^2 + y^2 from z = 0 to z = 4 - It calculates the line integral of F around the boundary curve C as 2π - It calculates the surface integral of curl(F) over S as 2π - Since these two integrals are equal, Stokes's theorem is verified for this case.

1. MATHEMATICS METHODS FOR ENGINEERING
QUESTION 6:
Verify Stokes’s theorem for the vector F = zi +xj + yk and the surface S represented by the come
z = x2 +y 2 where 0 £ z £ 4.
Let us proof that òò =
R
ò .
(curlF).ndA F dr
c
r =[cos s; sin s;0] r¢ =[-sin s;cos s;0] F =zi 1 F = xj 2 F = yk 3
F(r(s)) =[0,cos s,sin s]
2
0 ò
F dr F r s r s ds
c
. ( ( ). ( )
ò
P
= ¢
2
P
= + ´ +
ò
0
[0 cos s cos s 0]ds
P
ò sds
=
2
0
cos2
P P
2
òds ò sds
= +
0
2
0
cos 2
0 2
0 = + P s
= 2P
We now consider the surface integral, we have F =zi 1 F = xj 2 F = yk 3 so can now calculate the
curlF .
dz - dy
+ ( )N
3 dy - dx
+ ( )N
2 curlF = 1 ( )N
dz
dy
dx
dz
dz
dy
dx -
dx
=curl z x y
( , , )
=
(1,1,1)
A normal vector of S is N =grad(z - f (x, y)) =[2x,2y,2z]. Hence
(curlF) · N = 2x + 2y + 1 .
16
Now ndA =Ndxdy ,using polar coordinates r,q defined by x =r cosq, y =r sinq and denoting
the projection of S into the xy -plane by R, we thus obtain:

2. òò =òò
R R
(curlF).ndA (curlF).Ndxdy
( 1
=òò + +
R
x y dxdy
16
2
P
= + +
ò ò
0
4
0
(2r(cosq sinq ) 1 rdrdq
)
16
P
= + +
32(cosq sinq ) 1dq
2
0 ò
P = + 2
0 0 q
= 2P