Tuesday April 16 Solutions Surface integrals of vector fields and related theorems 1 Consider the region D in R3 bounded by the x y plane and the surface x 2 y 2 z 1 a Make a sketch of D Solution The sketch of D is shown below b The boundary of D denoted D has two parts the curved top S 1 and the flat bottom S 2 Parameterize S 1 and calculate the flux of F 0 0 z through S 1 with respect to the upward pointing unit normal vector field Check you answer with the instructor Solution To parametrize S 1 one has r u v u cos v u sin v 1 u 2 0 u 1 0 v 2 In order to calculate the flux first we have ru cos v sin v 2u rv u sin v u cos v 0 and so ru rv 2u 2 cos v 2u 2 sin v u Therefore the flux of F 0 0 z through S 1 with respect to the upward pointing unit normal vector field is Z 2 Z 1 F n dS F u v ru rv d ud v S1 0 Z 0 0 2 Z 1 Z 0 2 Z 1 0 0 0 0 1 u 2 2u 2 cos v 2u 2 sin v u d ud v 1 u 2 u d ud v 2 c Without doing the full calculation determine the flux of F through S 2 with the downward pointing normals Solution Since F 0 on S 2 we know the flux of F through S 2 is F n dS 0 n d S 0 S2 S2 d Determine the flux of F through D with the outward pointing normals Solution By adding up the result from a and b one gets F n dS F n dS F n dS 2 D S1 S2 e Apply the Divergence Theorem and your answer in d to find the volume of D Check your answer with the instructor Solution By the Divergence Theorem one has F n dS divF dV D D Since divF 1 one gets Volume D D 1 dV D divF dV D F n dS 2 Consider the vector field F y x z a Compute curl F Solution i j k curl F x y z 0 0 2 y x z b For the surface S 1 above evaluate S1 curl F n dA Solution By applying the parametrization of S 1 in 1 b one gets Z 2 Z 1 curl F n dA curl F ru rv d ud v S1 0 Z 0 0 2 Z 1 0 2u d ud v 2 2 3 If time remains a Check your answer in 1 e by directly calculating the volume of D Solution One can use the polar coordinate to calculate the volume of D in 1 e Let x r cos y r sin 0 r 1 0 2 Then Volume D 2 x 2 y 2 1 Z 2 1 x y d A 0 2 Z 1 0 1 r 2 r d r d 2 b Repeat 2 b for the surface S 2 and also for the surface D Solution The normal vector of S 2 pointing downward is n k Thus curl F n dA 0 0 2 0 0 1 d A 2 d A 2 S2 S2 S2 For the surface D we get our answer by adding up the two integrals curl F n dA curl F n dA curl F n dA 2 2 0 D S1 S2 c For the vector field F y x z from the second problem compute div curl F Now suppose F F 1 F 2 F 3 is an arbitrary vector field Can you say anything about the function div curl F Solution We already know in 2 a that curl F 0 0 2 so div curl F 0 Generally suppose suppose F F 1 F 2 F 3 is an arbitrary vector field Then i j k curl F det x y z F1 F2 F3 F 3 F 2 F 1 F 3 F 2 F 1 y z z x x y and div curl F x F 3 F 2 F 1 F 3 F 2 F 1 y z y z z x x y 2 F 3 2 F 2 2 F 1 2 F 3 2 F 2 2 F 1 0 x y x z y z y x z x z y
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