# a6f08s - ASSIGNMENT 6 1 Flux V = f n dS = V 2 f dV = f dV =...

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ASSIGNMENT # 6 1. ∫∫ = V V dS f Flux n = = = 5 = = ∫∫∫ V dV f ∫∫∫ V dV f 2 ∫∫∫ V dV π ∫∫∫ 1 2 22 0 2 x x z dydzdx 72 Where the Divergence Theorem was used and the assumption that . 5 2 = f 2. Close that is the part of the paraboloid by the disk and use the 1 S 1 , 1 : 2 2 2 = + z y x S Divergence Theorem to S V S S = = 2 1 : = = ∫∫ ∫∫∫ SV dV dS 1 n F ∫∫ ∫ = θ 2 0 1 0 2 1 2 2 r dzdrd r so we have + ∫∫ 1 S dS n F 2 2 = ∫∫ S dS n F . But , hence ∫∫ ∫∫ ∫∫ = = = 2 ) ( SS S dS dS dS k F n F 2 3 1 = ∫∫ S dS n F 3. The curve C is an intersection of the cylinder and the surface , 1 2 2 = + y x xy z 2 = and is oriented clockwise. Using the Stokes Th. ( with S being the part of the surface bounded by C , n directed downward) we get xy z 2 = = dt t z x y z x y C ) ( ) arctan , arctan , arctan ( 3 2 g + + + = = = dS x y x z S ) 1 , 2 , 2 ( ) 1 , 3 , 2 ( 2 ∫∫ dA x y x xy y x ) 1 , 2 , 2 ( ) 1 , 3 , 4 ( 1 2 2 2 ∫∫ + = =

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## This note was uploaded on 10/21/2010 for the course MATHEMATIC MAT237Y1 taught by Professor Romauldstanczak during the Fall '09 term at University of Toronto.

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a6f08s - ASSIGNMENT 6 1 Flux V = f n dS = V 2 f dV = f dV =...

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