# There fore integraldisplay γ 2 xyz e x 2 dx z e x 2

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dant of the choice of path and we need only evaluate at the endpoints. There- fore, integraldisplay γ 2 xyz e x 2 dx + z e x 2 dy + y e x 2 dz = f (0 , 1 , 2) f (2 , 0 , 1) = (1)(2)( e 0 ) (0)(1)( e 4 ) = 2. 6. (a) (i) We parametrize γ by γ ( t ) = ( 1 , 1) + t (2 , 2) = ( 1 + 2 t, 1 + 2 t ), 0 t 1. Now integraldisplay γ ( x 2 + x ) dx + xy dy = integraldisplay 1 0 ( (( 1 + 2 t ) 2 + ( 1 + 2 t ))(2) + ( 1 + 2 t )( 1 + 2 t )(2) ) dt = integraldisplay 1 0 ( 16 t 2 12 t + 2 ) dt = bracketleftbigg 16 3 t 3 6 t 2 + 2 t bracketrightbigg 1 0 = 4 3 . (ii) We parametrize γ by γ ( t ) = ( t, t 3 ), 1 t 1. Now integraldisplay γ ( x 2 + x ) dx + xy dy = integraldisplay 1 1 ( ( t 2 + t )(1)+( t )( t 3 )(3 t 2 ) ) dt = integraldisplay 1 1 ( 3 t 6 + t 2 + t ) dt = bracketleftbigg 3 7 t 7 + 1 3 t 3 + 1 2 t 2 bracketrightbigg 1 1 = 32 21 .

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MATB42H Solutions # 4 page 4 (b) If the vector field F ( x, y ) = ( x 2 + x, xy ) were conservative, the answers in part (a) would have been the same. Since they are different, F ( x, y ) can not be con- servative. 7. (a) We wish to parametrize the curve of intersection of x 2 + y 2 + z 2 = 3 and z = 1 y . Substituting for z in the first we have x 2 + y 2 + 1 2 y + y 2 = 3. Completing the square we have x 2 + 2 parenleftbigg y 1 2 parenrightbigg 2 = 5 2 , an ellipse in the xy –plane, which can be parametrized by x = radicalbigg 5 2 cos t , y = 1 2 + 5 2 sin t . Now z = 1 y = 1 2 5 2 sin t , so we can parametrize the curve of intersection, in the counterclockwise direction, when viewed from above, by γ ( t ) = parenleftbigg radicalbigg 5 2 cos t, 1 2 + 5 2 sin t, 1 2 5 2 sin t parenrightbigg , 0 t 2 π . (b) By inspection we note that F ( x, y, z ) = (2 x, 2 yz, y 2 ) is conservative and the general potential function is g ( x, y, z ) = x 2 + y 2 z + C . Since c , the curve of intersection, is a closed curve, we have integraldisplay γ F · d s = 0.
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