Then f 1 u v h g is a dieomorphism note that

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Unformatted text preview: ˆ Therefore, � � �Tφ × Tθ � = a2 sin φ(sin2 φ cos2 θ + sin2 φ sin2 θ + cos2 φ)1/2 = a2 sin φ. As 0 < φ < π , note that a2 sin φ > 0. Hence area(M ) = area(M ∩ W ) �� = dS � �M ∩W � � = �Tφ × Tθ � dφ dθ U � 2π � π = a2 sin φ dφ dθ 0 0 � 2π = 2a2 dt 0 2 = 4πa . Notice that this is the surface area of a sphere of radius a. Let’s now suppose that there are two different ways to parametrise the same piece M ∩ W of the manifold M : � : U −→ M ∩ W g and � : V −→ M ∩ W. h Let use (u, v ) coordinates for U ⊂ R2 and (s, t) coordinates for V ⊂ R2 . Then � f = (� )−1 ◦ � : U −→ V , h g is a diffeomorphism. Note that � = � ◦ f . We then have gh� ∂� g ∂ (� ◦ f ) h� (u, v ) = (u, v ) ∂u ∂u h ∂s ∂� h ∂t ∂� = (s, t) (u, v ) + (s, t) (u, v ). ∂s ∂u ∂t ∂u 3 Similarly ∂� g ∂ (� ◦...
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This note was uploaded on 01/27/2014 for the course MATH 324 taught by Professor Kopp during the Winter '08 term at University of Washington.

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