2220hw10sol

2220hw10sol - Math 2220 Problem Set 10 Solutions Spring...

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Unformatted text preview: Math 2220 Problem Set 10 Solutions Spring 2010 Section 6.1 : 2. Calculate the integral of the function f ( x, y, z ) = xyz over the curve c ( t ) = ( t, 2 t, 3 t ), ≤ t ≤ 2. Solution. The integral can be computed as integraldisplay 2 f ( r ( t )) | r ′ ( t ) | dt , where r ( t ) denotes the parametrization of the curve. After substituting the given function f ( x, y, z ) = xyz we get integraldisplay C f ds = integraldisplay 2 ( t )(2 t )(3 t ) | (1 , 2 , 3) | dt = integraldisplay 2 6 √ 14 t 3 dt = 3 2 √ 14 t 4 vextendsingle vextendsingle vextendsingle 2 = 24 √ 14 4. Calculate the integral of f ( x, y, z ) = 3 x + xy + z 3 over the curve c ( t ) = (cos 4 t, sin4 t, 3 t ), ≤ t ≤ 2 π . Solution. The derivative of the parametrization is c ′ ( t ) = (- 4 sin4 t, 4 cos4 t, 3) and its norm is | c ′ ( t ) | = 5. Thus, the integral is integraldisplay C f ds = integraldisplay 2 π (3 cos4 t + cos 4 t sin4 t + 27 t 3 )(5) dt = 5 parenleftbigg 3 4 sin 4 t + 1 8 sin 2 4 t + 27 4 t 4 parenrightbigg vextendsingle vextendsingle vextendsingle 2 π = 540 π 4 8. Calculate integraltext C F · d s for the vector field F ( x, y, z ) = ( x, y,- z ) where C is the curve c ( t ) = ( t, 3 t 2 , 2 t 3 ),- 1 ≤ t ≤ 1. Solution. The derivative of the parametrization is c ′ ( t ) = (1 , 6 t, 6 t 2 ) and the vector field along the curve is F ( c ( t )) = ( t, 3 t 2 ,- 2 t 3 ). Therefore the line integral over the curve can be computed as integraldisplay 1 − 1 ( t, 3 t 2 ,- 2 t 3 ) · (1 , 6 t, 6 t 2 ) dt = integraldisplay 1 − 1 ( t + 18 t 3- 12 t 5 ) dt = t 2 2 + 9 2 t 4- 2 t 6 vextendsingle vextendsingle vextendsingle 1 − 1 = 0 Alternatively, you can start with the line integral integraldisplay C F 1 dx + F 2 dy + F 3 dz = integraldisplay C x dx + y dy- z dz and compute this using ( x, y, z ) = ( t, 3 t 2 , 2 t 3 ), which leads to the same thing....
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2220hw10sol - Math 2220 Problem Set 10 Solutions Spring...

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