Begin2.DVI

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

d
r =

∂
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)