1 2 0 2 r 2 x y z x y z f i j k and is oriented up

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1 2, 0 2 r ( , , ) 2 x y z x y z F i j k , and is oriented up and 2 2 ( , ) z g x y x y 2 2 2 2 2 2 2 2 , The flux is then 2 R R z x z y x y x y x y z z dS dA x y x y xi y z dA x y x y              F n F i j k j k i j k 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 1 2 2 14 3 R R R x y z dA x y x y x y x y dA x y x y x y dA r drd     
15) We have ( , , ) x y z x F k , and is the portion of the paraboloid 2 2 z x y below the plane z y , oriented by downward unit normal. 𝑅 is the circular region enclosed by 2 2 0 x y y . ( , , ) x y z x F k , and is oriented up and 2 2 ( , ) z g x y x y   2 , 2 The flux is then 2 2 0 since the region is symmetric across the -axis. R R R z z x y x y z z dS dA x y x x y dA x dA R y       F n F i j k k i j k 16) 2 ( , , ) x y z x yx zx F i j k , and is the portion of the plane 6 3 2 6 x y z in the first octant, oriented by unit normals with positive components. .
9 Contact: [email protected] MA211 Week 13 Tutorial Solution 1/2017
10 Contact: [email protected] MA211 Week 13 Tutorial Solution 1/2017 and 3 3, 2 z z x y 2 2 2 1 2 2 0 0 The flux is then 3 3 2 3 3 2 3 1 3 (6 6 3 ) 2 2 3 1 R R R R x z z dS dA x y x yx zx dA x yx zx dA x yx x y x dA xdydx          F n F i j k i j k i j k 17) ( , , ) x y z x y F i j k , and is the portion of the paraboloid 2 ( , ) cos sin (1 ) u v u v u u u r i j k with 1 2, 0 2 u v . 2 2 We need to determine . cos sin 2 , sin cos Now take the cross product cos sin 2 2 cos 2 sin sin cos 0 r r u v r r v v u u v u v u v r r v v u u v u v u u v u v u v   i j k i j i j k i j k 2 2 3 cos sin then cos sin sin cos sin sin cos sin 2 x y u v u u r r u v u u u u F i j k i j k F i j k i j k 2 2 3 0 1 The flux is then 2 18 R dS dA u u dudv       r r F n F

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