Summary
When a surface is represented in parametric form, the position vector of a point
on the surface can be represented as
r =r (u, v ) = x(u, v )ˆe 1 +y(u, v )ˆe 2 +z(u, v )ˆe 3
where
x=x(u, v ), y =y(u, v ), z =z(u, v )
is the parametric representation of the surface. The differential of the position vector
r =r (u, v )is
dr =
∂r
∂u du +
∂r
∂v dv =
S 1 +S 2 (7 .101)
and this differential can be thought of as a vector addition of the component vectors
S 1 =∂u∂r du and S 2 =∂r∂v dv which make up the sides on an elemental parallelogram
having area dS lying on the surface. The vectors S 1 and S 2 are tangent vectors to the
coordinate curves r (u, v 2 )and r (u 1 , v)where u 1 and v 2 are constants. A representation
of coordinate curves on a surface and an element of surface area are illustrated in
the figure 7-23.
The unit normal to the surface at a point having the surface coordinates (u, v ),
can be found from either of the cross product relations
ˆen=±
∂r
∂u ×
∂r
∣ ∂v
∣∂r∂u ×∂r∂v ∣∣ or ˆen=∓
∂r
∂v ×
∂r
∣ ∂u
∣∂r∂v ×∂∂vu∣∣ (7 .102)
The above results differing in sign. That is, ˆen and −ˆen are both normals to the
surface and selecting one of these vectors gives an orientation to the surface.
Example 7-32. Find the unit normal to the sphere defined by
x=rcos φsin θ, y =rsin φsin θ, z =rcos θ
Solution Here
∂r
∂θ =rcos φcos θˆe^1 +rsin φcos θeˆ^2 −rsin θˆe^3
∂r
∂φ =−rsin φsin θˆe^1 +rcos φsin θˆe^2
and the cross product is
∂r
∂θ ×
∂r
∂φ =
∣∣
∣∣
∣
ˆe 1 ˆe 2 ˆe 3
rcosφcosθr sin φcosθ −rsin θ
−rsin φsin θrcosφsin θ 0
∣∣
∣∣
∣=ˆe^1 (r
(^2) sin (^2) θcosφ) + eˆ 2 (r (^2) sin (^2) θsin φ) + ˆe 3 (r (^2) sin θcosθ)