tutorial9_sol

tutorial9_sol - MATH1211/10-11(2)/Tu9sol/TNK Department of...

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Unformatted text preview: MATH1211/10-11(2)/Tu9sol/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010-11 2nd semester) Solution to Tutorial 9 Note : Some solutions given below are outline only. You may need to give more details in your solution. 1. For the line integral, we split the path into four line segments: x 1 ( t ) = (2 t, 0) , x 2 ( t ) = (2 ,t ) , x 3 ( t ) = (2- 2 t, 1) , x 4 ( t ) = (0 , 1- t ), with 0 ≤ t ≤ 1 for all four segments. (See figure at right.) Then I ∂D M dx + N dy = I ∂D F · d s = Z x 1 F · d s + Z x 2 F · d s + Z x 3 F · d s + Z x 4 F · d s = Z 1 ( (2 t ) 2- , 2 t + 0 2 ) · (2 , 0) dt + Z 1 (2 2- t, 2 + t 2 ) · (0 , 1) dt + Z 1 ( (2- 2 t ) 2- 1 , (2- 2 t ) + 1 2 ) · (- 2 , 0) dt + Z 1 ( 2- (1- t ) , 0 + (1- t ) 2 ) · (0 ,- 1) dt = Z 1 8 t 2 + ( 2 + t 2 )- 2 ( (2- 2 t ) 2- 1 )- (1- t ) 2 dt = Z 1 (18 t- 5) dt = h 9 t 2- 5 t i 1 = 4 ....
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tutorial9_sol - MATH1211/10-11(2)/Tu9sol/TNK Department of...

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