FL2009 Final Solutions - Math 3210 final exam Friday 11...

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Unformatted text preview: Math 3210 final exam Friday 11 December 2009 2:00-4:30 pm Solutions 1. (a) M is the portion of the paraboloid z = x 2 + y 2 sitting over the disc D . (b) Let f : D R be the function f ( x , y ) = x 2 + y 2 . By a theorem in the notes, ( x , y ) = ( x , y , f ( x , y )) is then an embedding. Therefore M = graph f = ( D ) is a manifold with boundary, of dimension 2, since dim D = 2. (c) Use * = p det ( D T D ) dx dy . From D = 1 1 f x f y we get D T D = 1 + ( f x ) 2 f x f y f x f y 1 + ( f y ) 2 , so det ( D T D ) = 1 + grad f 2 = 1 + 4 ( x 2 + y 2 ) , so * = p 1 + 4 ( x 2 + y 2 ) dx dy . (d) area ( M ) = R M = R D * = R D p 1 + 4 ( x 2 + y 2 ) dx dy . Substitute x = r cos , y = r sin ; then dx dy = r dr d , so area ( M ) = Z 2 Z 1 p 1 + 4 r 2 r dr d = 2 Z 1 r p 1 + 4 r 2 dr = 4 Z 4 1 + s ds = 4 2 3 ( 1 + s...
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FL2009 Final Solutions - Math 3210 final exam Friday 11...

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