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hwsol_27

# hwsol_27 - Homework Assignment 27(14.4 Solutions Page 1142...

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Homework Assignment 27 - (14.4) - Solutions Page 1142: 5*, 7*, 11*, 13*, 15*, 17*, 38*, 39(extra points) 5. ! C ! x 2 " y " dx ! y 2 dy , C : x 2 ! y 2 " 1, counterclockwise a. r # ! t " "\$ cos t , sin t % , 0 # t # 2 ! , r # & ! t " "\$ " sin t , cos t % ! C ! x 2 " y " dx ! y 2 dy " \$ 0 2 ! \$ cos 2 t " sin t , sin 2 t % # \$ " sin t , cos t % dt " \$ 0 2 ! " sin t cos 2 t ! sin 2 t ! cos t sin 2 t dt " ! b. F 1 ! x , y " " x 2 " y , F 2 ! x , y " " y 2 , R : x 2 ! y 2 # 1 % F 1 % y " " 1, % F 2 % x " 0 ! C ! x 2 " y " dx ! y 2 dy " \$ 0 2 ! \$ 0 1 ! 0 " ! " 1 "" rdrd " " ! 7. ! C x 2 dx " x 3 dy , C : a square from ! 0,0 " to ! 0,2 " to ! 2,2 " to ! 2,0 " to ! 0,0 " a. C " C 1 & C 2 & C 3 & C 4 C 1 : from ! 0,0 " to ! 0,2 " , x " 0, dx " 0, 0 # y # 2 C 2 : from ! 0,2 " to ! 2,2 " , y " 2, dy " 0, 0 # x # 2 C 3 : from ! 2,2 " to ! 2,0 " , x " 2, dx " 0, y is from 2 to 0 C 4 : from ! 2,0 " to ! 0,0 " , y " 0, dy " 0, x is from 2 to 0 F # ! x , y " "\$ x 2 , " x 3 % ! C x 2 dx " x 3 dy " \$ C 1 x 2 dx " x 3 dy ! \$ C 2 x 2 dx " x 3 dy ! \$ C 3 x 2 dx " x 3 dy ! \$ C 4 x 2 dx " x 3 dy " \$ 0 2 0 ! \$ 0 2 x 2 dx " 0 ! \$ 2 0 ! 0 " 8 " dy ! \$ 2 0 x 2 dx " 0 " 1 3 x 3 | 0 2 " 8 y | 2 0 ! 1 3 x 3 | 2 0 " 16 b. F 1 ! x , y " " x 2 , F 2 ! x , y " " " x 3 , R : 0 # x # 1, 0 # y # 1 % F 1 % y " 0, % F 2 % x " " 3 x 2 ! C x 2 dx " x 3 dy " " \$ 0 2 \$ 0 2 ! " 3 x 2 " 0 " dxdy " 16 11. ! C x x 2 ! 1 " y dx ! ! 3 x " 4tan y " dy , C : y " x 2 from ! " 1,1 " to ! 1,1 " , y " 2 " x 2 from ! 1,1 " to ! " 1,1 " R : x 2 # y # 2 " x 2 , " 1 # x # 1 1

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0 0.5 1 1.5 -1 -0.5 0.5 1 x C F # ! x , y " "\$ x x 2 ! 1 " y , 3 x " 4tan y % , % F 1 % y " " 1, % F 2 % x " 3 Check conditions required for the Green°s Theorem: a. C is piecewise-smooth, and simple closed with a positive orientation; b. F 1 , F 2 , % F 1 % y , and % F
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hwsol_27 - Homework Assignment 27(14.4 Solutions Page 1142...

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