Final Exam Solution

8 2012 2 let f r3 r3 be given by f x y z y x

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Unformatted text preview: exist, explain. 1 1 1 1 f (x, y ) dy dx, 0 f (x, y ) dx dy, 0 0 0 f (x, y ) dA. D Solution: Using part (b), 1 1 1−δ dx = lim arctan x|1−δ δ δ →0 δ 1 + x2 δ→0 π = lim (arctan (1 − δ ) arctan (δ )) = δ →0 4 f (x, y ) dy dx = lim 0 0 Similarly, 1 0 D 1 0 π f (x, y ) dx dy = − . 4 f (x, y ) dA does not exist since the two are not equal (Fubini’s theorem). (e) (1 point) State the condition of Fubini’s theorem (for improper integrals) that is not satisfied by f (x, y ) on [0, 1] × [0, 1]. Solution: It does not satisfy f (x, y ) 0. Math 3010 Final Exam - Page 5 of 14 Dec. 8, 2012 2. Let F : R3 → R3 be given by F (x, y, z ) = (−y, x, 0). Let G : R3 \ {z-axis} → R3 be y x given by G (x, y, z ) = − x2 +y2 , x2 +y2 , 0 . (a) (3 points) Evaluate ∇ × F and ∇ × G. Solution: ∇ × F = (0, 0, 2) , ∇ × G = (0, 0, 0) . (b) (2 points) Which (if any) vector fields are conservative fields? Which one of the vector fields is “irrotational”? Solution: Neither is conservative (for it to be conservative, G would need to be defined everywhere in R3 except for a finite number of points). G is an irrotational field. t t (c) (3 points) Let c (t) = (R cos(t), R sin(t), S ) and p (t) = R cos R2 , R sin R2 , S for R > 0 and S ∈ R. Show that c and p are flow lines for F and G respectively. Solution: c′ (t) = (−R sin(t), R cos(t), 0) = F (c(t)) , 1 t 1 t p′ (t) = − sin( 2 ), cos( 2 ), 0 , R RR R t 1 t 1 = − 2 R sin( 2 ), 2 R cos( 2 ), 0 = G (p(t)) ....
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