Page 123 - Vector Analysis

§4.3 The Surface Integrals 119

Example 4.49. Let Σ Ď R3 be the upper half sphere; that is, Σ = ␣(x, y, z) P R3 ˇ x2 + y2 +
ˇ

z2 = R2, z ą 0(, and tV, ψu be a global parametrization of Σ given by

ψ(u, v) = (u, v, ?R2 ´ u2 ´ v2) , (u, v) V = ␣(u, v) R2 ˇ u2 + v2 R2( .
ˇ
P P ď

To find the surface area using this parametrization, we first compute tψ,1 , ψ,2 u as follows:

ψ,1 (u, v) = ( 0, ? ´u ) and ψ,2 (u, v) = ( 1, ? ´v ) ,
1, R2 ´ u2 0, R2 ´ u2
´ v2 ´ v2

thus the first fundamental form associated with the parametrization tV, ψu is

g(u, v) = }ψ,1 (u, v) ˆ ψ,2 (u, v)}2R3 = ›( u ´ , ? v ´ )›2
›? ´ u2 R2 ´ u2 ,1 ›
› R2 v2 v2 ›R3

= R2 R2 ´ v2 .
´ u2

Therefore, the surface area of Σ is

?
żR ż R2´u2
ż dS = ż ? R ´ v2 dA = ? R ´ v2 dvdu
R2 ´ u2 ´R ? R2 ´ u2
Σ V ´ R2´u2
?
= R żR arcsin v u2 du = R żR π du = 2πR2 .
? ˇv= R2´u2
´R ˇ ´R
R2 ´ ˇ?

v=´ R2´u2

Note the the computation above also shows that the surface area of the sphere in R3 with
radius R is 4πR2 which is the same as what we have conclude in Example 4.48.

Remark 4.50. The example above provides one specific way of evaluating the surface

integrals: if the surface Σ is in fact a subset of the graph of a function f : D Ď R2 Ñ R;

that is, Σ Ď ␣x, y, f (x, y)) ˇ (x, y) P D(, then Σ has a global parametrization
ˇ

() (x, y) P V ,
ψ(x, y) = x, y, f (x, y) ,

where V is the projection of Σ onto the xy-plane along the z-direction. Then the first
fundamental form associated to this parametrization is

g(x, y) = }ψ,1 (x, y) ˆ ψ,2 (x, y)}2R3 = 1 + ˇ B f (x, y)ˇˇ2 + ˇ B f (x, y)ˇˇ2 ;
ˇ ˇ
Bx By

thus the surface area of Σ is

ż żc f f

dS = 1 + ˇ B (x, y)ˇˇ2 + ˇ B (x, y)ˇˇ2 d(x, y) .
ˇ ˇ
Σ V Bx By