Differential Geometric Aspects in Image Processing

Differential Geometric Aspects in Image Processing

Dr. Marcelo C´ ardenas

Classroom exercises: December 19, 2019

Problem C6.1

a) Letu= (u, v).We have

Dσ(u, v) =





−sinucosv −cosucosv

−sinusinv cosusinv

cosu 0



,

thus

I_(u,v)=

1 0 0 cos²u

.

Therefore A=

Z

S²

dS= Z 2π

0

Z ^π₂

−^π₂

pdet(Iu)dudv

= Z 2π

0

Z ^π₂

−^π₂

cos(u)dudv= 2π Z ^π₂

−^π₂

cos(u)du= 2π[sin(u)]

π 2

−^π₂ = 4π b)

(U◦σ) (u, v) = sin(u)

∇_S2U =a1

∂σ

∂u +a2

∂σ

∂v and

g⁻¹=I⁻¹_(u,v)=

1 0 0 cos⁻²(u)

.

In general

a_i=

N

X

j=1

g^ij∂_jU,

thus for our surface

a1=g¹¹∂1U+g¹²∂2U = cos(u)

1

and

a2=g²¹∂1U+g²²∂2U = 0.

Therefore

∇σU = cos(u)∂σ

∂u =





−sinucosvcosu sinusinvcosu

cos²u





c) We have

pdet(g) = cosu and

g⁻¹=

1 0 0 cos⁻²(u)

. Therefore

∆_S2f = 1 cos(u)

∂

∂u

cos(u)∂f

∂u

+ 1

cos²(u)

∂²f

∂v

2