Begin2.DVI

One finds

∣∣

∣∂
r∂θ ×∂
r∂φ

∣∣

∣=r^2 sin θso that a unit vector to the surface of the sphere is

ˆen= sin θcos φeˆ 1 + sin θsin φˆe 2 + cos θeˆ 3

For
r a position vector to a point on the surface, the element d
r lies in the

tangent plane to the surface at the point determined by the parameters uand v. The

element of surface area dS is also determined from the differential element d
r and is

given by the magnitude of the cross products of the vectors S
1 and S
2 representing

the sides of the elemental parallelogram which defines the element of surface area.

This element of surface area is calculated from the cross product

dS =

∣∣

∣∣∂
r

∂u

×∂
r

∂v

∣∣

∣∣dudv

Figure 7-23.

Coordinate curves on surface and elemental parallelogram.

Using the dot product relation

(A
×B
)·(C
×D
) = (A
·C
)(B
·D
)−(A
·D
)(B
·C
)

one can readily verify that

∣∣

∣∣∂
r

∂u ×

∂
r

∂v

∣∣

∣∣=

√

EG −F^2 , (7 .103)