This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 3210 final exam Friday 11 December 2009 2:004:30 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...
View Full
Document
 '06
 SJAMAAR
 Math

Click to edit the document details