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finf98s - S in the Form oF a graph z-1 = g x y =-² 4-x 2-y...

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Solutions to MATH 20E FINAL EXAM Fall 98, Lindblad, Problems 1-5 only. 1. (a) f = 6 xye x 2 y - 2 i + 3 x 2 e x 2 y - 2 j so it increases most in the direction 12 i + 3 j . b) z - z 0 = f x ( x 0 , y 0 )( x - x 0 ) + f y ( x 0 , y 0 )( y - y 0 ) so z - 3 = 12( x - 1) + 3( y - 2). 2. a) ∇ × F = 0. b ) C F · d R = C xdx + ydy x 2 + y 2 = 2 π 0 r cos t r ( - sin t ) + r sin t r cos t r 2 dt = 0 . c) F = φ , where φ = ln x 2 + y 2 . d) The flow lines are lines from the origin: dx x/ ( x 2 + y 2 ) = dy y/ ( x 2 + y 2 ) so dx x = dy y and hence ln | x | + C = ln | y | which is equivalent to | y | = e C | x | . 3. a) ∇ × F = 0. b ) C F · d R = C ydx - xdy x 2 + y 2 = 2 π 0 r sin t r ( - sin t ) - r cos t r cos t r 2 dt = - 2 π 0 dt = - 2 π c) F is not conservative since the integral around the closed curve in (b) is = 0. d) The flow lines are circles centered at the origin: dx y/ ( x 2 + y 2 ) = dy - x/ ( x 2 + y 2 ) so - xdx = ydy and hence y 2 = - x 2 + C which is equivalent to x 2 + y 2 = C . 4. a) F (0 , 2) = (2 , 0), F (0 , 4) = (0 , 4), F ( π / 2 , 2) = (0 , 2) and F ( π / 2 , 4) = (0 , 4). b) ( x, y ) ( u, v ) = 2 v . c ) Area = S dxdy = R ( x, y ) ( u, v ) dudv = π / 2 0 4 2 2 vdudv = ... = π 2 4 2 - 2 2 5. a) 0
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Unformatted text preview: S in the Form oF a graph z-1 = g ( x, y ) =-² 4-x 2-y 2 where ( x, y ) ∈ D = { ( x, y ); 3 ≤ x 2 + y 2 ≤ 4 } . dS = dxdy/ | n · k | = µ 1 + g 2 x + g 2 y dxdy = ... = 2 dxdy/ ² 4-x 2-y 2 so Area( S ) = ±± S dS = ±± D dxdy ² 4-x 2-y 2 = ± 2 √ 3 ± 2 π rdrdθ √ 4-r 2 =-2 π ² 4-r 2 ¶ ¶ ¶ ¶ 2 √ 3 = 2 π ±±± B x 2 dxdydz = ± 2 ± 2 π ± π r 2 sin 2 φ cos 2 θ r 2 sin φ dφdθdr = ± 2 ± 2 π-cos 3 φ 3 ¶ ¶ ¶ ¶ π cos 2 θ r 4 dθdr = 2 3 r 5 5 ¶ ¶ ¶ ¶ 2 ± 2 π cos 2 θdθ = 2 3 2 5 5 π NOTE: THERE ARE NO SOLUTIONS TO PROBLEMS 6-9!...
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