quiz4sol - t t α sin t Solution Compute c t =(1 αt α-1...

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TAKE HOME QUIZ 4 20 minutes closed book NAME: 1
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1. Let F ( x, y, z ) = ( - xy 2 z 4 , - yz 2 x 4 , zx 2 y 4 ) be a vector field and let S be the surface given by the equation z = x 2 + y 2 with x, y 0 and z [0 , a ]. Let S have the upwards orientation. Find the flux of F through S . Solution: Parametrize the surface by ( x, y, z ) = ( r cos θ, r sin θ, r ) with r a and θ [0 , π/ 2]. Then T r = (cos θ, sin θ, 1) and T θ = ( - r sin θ, r cos θ, 0) and so T r × T θ = ( - r cos θ, - r sin θ, r ). Therefore ∫∫ S F · dS = a 0 π/ 2 0 F ( r cos θ, r sin θ, r ) · ( - r cos θ, - r sin θ, r ) drdθ = a 0 π/ 2 0 r 8 ( cos 2 θ sin 2 θ + sin 2 θ cos 4 θ + cos 2 θ sin 4 θ ) drdθ = a 0 π/ 2 0 2 r 8 cos 2 θ sin 2 θdrdθ = a 0 π/ 2 0 r 8 2 sin 2 (2 θ ) drdθ = a 0 π/ 2 0 r 8 4 (1 - cos(4 θ )) drdθ = a 0 π 8 r 8 dr = πa 9 72 2
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2. Find a number α 0 such that c F · ds = 0 where F ( x, y, z ) = ( z, 4 y 3 - 1 , x ) and c : [0 , 2 π ] R 3 is the parametrized curve c ( t ) = (
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Unformatted text preview: t, t α , sin t )? Solution: Compute c ( t ) = (1 , αt α-1 , cos t ) and ∫ c F · ds = ∫ 2 π F ( t, t α , sin t ) · (1 , αt α-1 , cos t ) = ∫ 2 π (sin t, 4 t 3 α-1 , t ) · (1 , αt α-1 , cos t ) = ∫ 2 π (sin t + (4 αt 4 α-1-αt α-1 ) + t cos t ) dt But ∫ 2 π sin t = 0, ∫ 2 π (4 αt 4 α-1-αt α-1 ) dt = ( t 4 α-t α ) | 2 π and ∫ 2 π t cos tdt = t sin t | 2 π-∫ 2 π sin tdt = 0. Therefore ± c F · ds = 0 implies (2 π ) 4 α = (2 π ) α so α = 0. 3...
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This note was uploaded on 11/05/2011 for the course MAT 203 at Princeton.

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quiz4sol - t t α sin t Solution Compute c t =(1 αt α-1...

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