APM3713
Assignment 6
Due Year 2025
, Question 1
Consider the surface parametrized by
√ √
r(u, v) = x(u, v), y(u, v), z(u, v) = u2 + 1 cos v, u2 + 1 sin v, u ,
with coordinates x1 = u, x2 = v.
(a) Line element
√
Define a(u) = u2 + 1. Then
u u
ru = a
cos v, a
sin v, 1 , rv = − a sin v, a cos v, 0 .
The metric components:
1 + 2u2
g11 = , g22 = 1 + u2 , g12 = 0.
1 + u2
Hence the line element is
1 + 2u2 2
ds2 = du + (1 + u2 ) dv 2 .
1 + u2
(b) Metric and inverse metric
1 + u2
1 + 2u2
0 1 + 2u2 0
2
gij = 1 + u
, g ij = .
2
1
0 1+u 0
1 + u2
(c) Christoffel symbols
1 u 1 1 + u2 u
Γ 11 = , Γ 22 = −u , Γ2 12 = Γ2 21 = .
(1 + u2 )(1 + 2u2 ) 1 + 2u2 1 + u2
All other Christoffel symbols vanish.
1
Assignment 6
Due Year 2025
, Question 1
Consider the surface parametrized by
√ √
r(u, v) = x(u, v), y(u, v), z(u, v) = u2 + 1 cos v, u2 + 1 sin v, u ,
with coordinates x1 = u, x2 = v.
(a) Line element
√
Define a(u) = u2 + 1. Then
u u
ru = a
cos v, a
sin v, 1 , rv = − a sin v, a cos v, 0 .
The metric components:
1 + 2u2
g11 = , g22 = 1 + u2 , g12 = 0.
1 + u2
Hence the line element is
1 + 2u2 2
ds2 = du + (1 + u2 ) dv 2 .
1 + u2
(b) Metric and inverse metric
1 + u2
1 + 2u2
0 1 + 2u2 0
2
gij = 1 + u
, g ij = .
2
1
0 1+u 0
1 + u2
(c) Christoffel symbols
1 u 1 1 + u2 u
Γ 11 = , Γ 22 = −u , Γ2 12 = Γ2 21 = .
(1 + u2 )(1 + 2u2 ) 1 + 2u2 1 + u2
All other Christoffel symbols vanish.
1
