Page 125 - Vector Analysis

§4.3 The Surface Integrals 121

ϕ
R

θ=ϕ θ
D = ψ´1(Σ)
θ+ϕ=π

Since the first fundamental form associate with tR, ψu is the first fundamental form

associated with tR, ψu is

g(u, v) = ››(ψθ ˆ ψϕ)(u, v)››R2 3
= ››(´ sin θ sin ϕ, cos θ sin ϕ, 0) ˆ (cos θ cos ϕ, sin θ cos ϕ, ´ sin ϕ)››2R3
= ››(´ cos θ sin2 ϕ, ´ sin θ sin2 ϕ, ´(sin2 θ + cos2 θ) sin ϕ cos ϕ)››2R3
= (cos2 θ + sin2 θ) sin4 ϕ + sin2 ϕ cos2 ϕ = sin2 ϕ ,

the area of the desired surface can be computed by

żż ?g ż π ż π´ϕ ż π
dS = 2
dA sin ϕ dθdϕ 2 sin
= = ´ 2ϕ) ϕ dϕ
(π

Σ ψ´1(Σ) 0ϕ 0

= ( ´ π cos ϕ + 2ϕ cos ϕ ´ 2 sin ˇϕ= π = π ´2.
ϕ)ˇ 2

ˇϕ=0

Another way to parameterize Σ is to view Σ as the graph of function z = a1 ´ x2 ´ y2

over D, where D is the projection of Σ along z-axis onto xy-plane. We note that the

boundary of D can be parameterized by tP [ π π] .
rr(t) = (cos t sin t, sin t sin t) , ´ ,
2 2

Let (x, y) P B D. Then x2 + y2 = y; thus Σ can also be parameterized by ψ : D Ñ R3, where

ψ(x, y) = ( y, a ´ x2 ´ ) and D = ␣(x, y) ˇ x2 + y2 ď y( .
x, 1 y2 ˇ

Therefore, with f denoting the function f (x, y) = a1 ´ x2 ´ y2, Remark 4.50 implies that

the surface area of Σ can be computed by

?
ż 1 ż y´y2
żb 1

1 + fx2 + fy2 dA = ? a1 x2 y2 dxdy

D 0 ´ y´y2 ´ ´
?
= ż1 arcsin x y2 ˇx= y´y2 dy = 2 ż1 arcsin ?y y dy ;
a1 ´ ˇ? ?
0 ˇx=´ y´y2 0
1+