Example 8-10. Illustrate Stokes theorem using the vector field
F=yz ˆe 1 +xz^2 ˆe 2 +xy ˆe 3 ,
where the surface Sis a portion of a sphere of radius rinside a circle on the sphere.
The surface of the sphere can be described by the para-
metric equations.
x=rsin θcos φ, y =rsin θsin φ, z =rcos θ
for rconstant, 0 ≤θ≤π and 0 ≤φ≤ 2 π. The position vector
to a point on the sphere being represented
r =r (θ, φ ) = rsin θcos φˆe 1 +rsin θsin φˆe 2 +rcos θˆe 3 (ris constant)
From the previous chapter we found an element of surface area on the sphere can
be represented
dS =
∣∣
∣∣∂r
∂θ
×∂r
∂φ
∣∣
∣∣dθ dφ =
√
EG −F^2 dθ dφ =r^2 sin θ dθ dφ (ris constant) (8 .41)
The physical interpretation of dS being that
it is the area of the parallelogram having the
sides ∂r∂θ dθ and ∂r∂φ dφ with diagonal vector
dr =
∂r
∂θ dθ +
∂r
∂φ dφ
If one holds θ=θ 0 constant, one obtains a circle Con the sphere described by
r =r (φ) = rsin θ 0 cos φˆe 1 +rsin θ 0 sin φˆe 2 +rcos θ 0 eˆ 3 , 0 ≤φ≤ 2 π (8 .42)
A unit outward normal to the sphere and inside the circle Cis given by
ˆen= sin θcos φˆe 1 + sin θsin φˆe 2 + cos θˆe 3 ,^0 ≤φ≤^2 π
0 ≤θ≤θ 0
(8 .43)
The vector curlF is calculated from the determinant
curlF =∇× F=
∣∣
∣∣
∣∣
ˆe 1 ˆe 2 ˆe 3
∂
∂x
∂
∂y
∂
∂z
yz xz^2 xy
∣∣
∣∣
∣∣
=ˆe 1 (x− 2 xz) + ˆe 2 (0) + ˆe 3 (z^2 −z)
