# SpSolHw8 - MATH-4600 Homework VIII Solutions to Graded...

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Unformatted text preview: MATH-4600 Homework VIII Solutions to Graded Problems 1. 6.3 #1 Z C z 2 dx + 2 ydy + xzdz = Z b a ( z 2 ( t ) x ( t ) + 2 y ( t ) y ( t ) + x ( t ) z ( t ) z ( t ) ) dt (a)-→ x ( t ) = ( t, t, t ) dt = (1 , 1 , 1) R 1 ( t 2 + 2 t + t 2 ) dt = 2 t 2 2 + 2 t 3 3 1 = 1 + 2 3 = 5 3 (b)-→ x ( t ) = ( t, t 2 , t 3 ) 0 ≤ t ≤ 1 dt = (1 , 2 t, 3 t 2 ) R 1 ( t 6 + 4 t 3 + 3 t 6 ) dt = R 1 ( 4 t 6 + 4 t 3 ) dt = 4 t 7 7 + 4 t 4 4 1 = 4 7 + 1 = 11 7 (c) Since the line integral from a to b is not path independent ( 5 3 6 = 11 7 ) , it follows that the vector field is not conservative. 2. 6.3 #4 F = 2 x sin( y ) i + x 2 cos( y ) j N = x 2 cos( y ) , M = 2 x sin( y ) δN δx = 2 x cos( y ) , δM δy = 2 x cos( y ) Since δN δx = δM δy , it follows that F is conservative. ⇒ δf δx = 2 x sin( y ) , δf δy = x 2 cos( y ) f = Z (2 x sin( y )) dx = x 2 sin( y )+ g ( y ) , f = Z ( x 2 cos( y ) ) dx = x 2 sin( y )+ C g ( y ) = C It follows then that f = x 2 sin( y ) + C for C ∈ R ....
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SpSolHw8 - MATH-4600 Homework VIII Solutions to Graded...

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