APM3713 Assignment 6 2025 (Accurate Solutions) Year 2025

APM3713
Assignment 6

Due Year 2025

, Solution

Question 1

Consider the surface parametrized by

√ √

r(u, v) = u2 + 1 cos v, u2 + 1 sin v, u .


We take x1 = u, x2 = v.


(a) Line Element

Tangent vectors:

   √ √ 
ru = √ u cos v, √ u sin v, 1 , rv = − u2 + 1 sin v, u2 + 1 cos v, 0 .
u2 +1 u2 +1



Metric coefficients:

2u2 + 1
guu = , guv = 0, gvv = u2 + 1.
u2 + 1

Thus the line element is

2u2 + 1 2
ds2 = du + (u2 + 1) dv 2 .
u2 + 1

(b) Metric and Inverse Metric
u2 + 1
 2   
2u + 1
0 0
 2
g ij =  2u + 1
 2
gij =  u + 1 , .
 
1
0 u2 + 1 0 2
u +1

(c) Christoffel Symbols

Using
Γkij = 21 g kℓ (∂i gℓj + ∂j gℓi − ∂ℓ gij ),

we compute

u u(u2 + 1) u
Γ111 = 2 , Γ122 =− , Γ212 = Γ221 = .
(u + 1)(2u2 + 1) 2u2 + 1 u2 + 1


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