Final Exam Solution

A 2 points show that w r x y z dv z d r x

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Unformatted text preview: lit S into two surfaces Sa and Sb . The boundaries ∂Sa and ∂Sb coincide but they are oriented in the opposite directions. S ∇ × F · dS = Sa ∇ × F · dS + Sb ∇ × F · dS ∂ Sa F · ds + ∂ Sb F · ds, ∂ Sa F · ds − ∂ Sa F · ds = = using Stokes’ theorem = 0. (b) (5 points) Prove that S ∇ ×F · dS = 0 with the help of Gauss’ divergence theorem. Solution: Let G = ∇ × F . Then ∇ · G = ∇ · ∇ × F = 0 (true for any C 2 F ). By divergence theorem, S ∇ × F · dS = S G · dS = interior of S = 0 dV interior of S = 0. ∇ · G dV Math 3010 Final Exam - Page 11 of 14 Dec. 8, 2012 8. Proof of Gauss’ divergence theorem Let W = {(x, y, z ) | (x, y ) ∈ D, γ1 (x, y ) z γ2 (x, y )} where D is an elementary region in R2 . Let ∂W be the closed surface bounding W with outward orientation. We can split ∂W into six surfaces, ∂W = S1 + S2 + · · · + S6 where S1 is the bottom surface (which satisﬁes z = γ1 (x, y )), S2 is the top surface (which satisﬁes z = γ2 (x, y )) and Si for i = 3, . . . , 6 are the sides (in any order). Let F (x, y, z ) = (P (x, y, z ) , Q (x, y, z ) , R (x, y, z )) where P , Q, and R be smooth scalarvalued functions on W . (a) (2 points) Show that W ∂ R (x, y, z ) dV = ∂z D [R (x, y, γ2 (x, y )) − R (x, y, γ1 (x, y ))] dA. Solution: Refer to textbook or lecture notes. (b) (2 points) Show that Si ˆ R (x, y, z ) k · dS = 0 for i = 3, 4, 5 and 6. (c) (3 points) Show that S1 ˆ R (x, y, z ) k · dS = − (d) (3 points) Show...
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This note was uploaded on 01/19/2014 for the course MATH 3010 taught by Professor Magpantay during the Winter '13 term at York University.

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