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Unformatted text preview: Mathematics 218: Analysis in Several Variables Professor Robert Gunning Fall Term 2009 Eighth Problem Set Rik Sengupta December 15, 2009 1 Group 1 1. Evaluate R x 1 dx 1 + x 2 1 dx 2 + x 2 dx 3 along the curve x 1 = x 2 = x 3 from (0 , , 0) to (1 , 1 , 1). Solution. If we parametrize the curve, we simply get x 1 = x 2 = x 3 = t , with t [0 , 1]. Now, the simplest thing about the integral is the fact that we also immediately get dx 1 = dx 2 = dx 3 = dt , and so our integral reduces to an extremely simple integral. In fact, if I denotes the given integral, i.e. I = Z x 1 dx 1 + x 2 1 dx 2 + x 2 dx 3 , then we have I = Z 1 ( tdt + t 2 dt + tdt ) = Z 1 ( t 2 + 2 t ) dt = t 3 3 + t 2 1 = 1 3 + 1 = 4 3 . So we get the value of the required integral as 4 3 . 1 2. Show that the integral Z sin x 2 sin x 3 dx 1 + x 1 cos x 2 sin x 3 dx 2 + x 1 sin x 2 cos x 3 dx 3 along a path from the point (0 , , 0) to the point (1 , 1 , 1) in R 3 is indepen dent of the path . Solution. Consider the function (or more relevantly, 0form) from R 3 to R given by f = x 1 sin x 2 sin x 3 . Note that df = d ( x 1 sin x 2 sin x 3 ) = sin x 2 sin x 3 dx 1 + x 1 sin x 3 d (sin x 2 ) + x 1 sin x 2 d (sin x 3 ) = sin x 2 sin x 3 dx 1 + x 1 cos x 2 sin x 3 dx 2 + x 1 sin x 2 cos x 3 dx 3 . We will use this to simplify our calculations immensely. If we denote the given integral by I , i.e. I = Z sin x 2 sin x 3 dx 1 + x 1 cos x 2 sin x 3 dx 2 + x 1 sin x 2 cos x 3 dx 3 , then our given integral reduces to I = Z df = [ f ] (1 , 1 , 1) (0 , , 0) = [ x 1 sin x 2 sin x 3 ] (1 , 1 , 1) (0 , , 0) = sin(1)sin(1) = sin 2 (1) , which clearly does not depend on the path at all, as long as its endpoints are the given ones. So, the integral is pathindependent. 3. Find the arc length of the curve in R 3 described parametrically by x 1 = e t sin t, x 2 = e t cos t, x 3 = e t 2 for 0 t 2 . Solution. We know, for a given curve parametrized as in this problem, the arc length is given by the equation L = Z s dx 1 dt 2 + dx 2 dt 2 + dx 3 dt 2 dt, where is the set of possible values of t . So, note in particular that dx 1 dt = e t sin t + e t cos t , dx 2 dt = e t cos t e t sin t , and dx 3 dt = e t . Denote dx 1 dt 2 + dx 2 dt 2 + dx 3 dt 2 = ds dt 2 . Hence, we get ds dt 2 = ( e t sin t + e t cos t ) 2 + ( e t cos t e t sin t ) 2 + ( e t ) 2 = e 2 t sin 2 t + e 2 t cos 2 t + 2 e 2 t sin t cos t + e 2 t sin 2 t + e 2 t cos 2 t 2 e 2 t sin t cos t + e 2 t = 2 e 2 t (sin 2 t + cos 2 t ) + e 2 t = 3 e 2 t , so that ds dt = 3 e t . Hence, we get L = Z ds = Z ds dt dt = 3 Z 2 e t dt = 3 e t 2 = 3 ( e 2  1 ) . So, the required integral is simply 3 ( e 2  1 ) ....
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 Fall '09
 RobertC.Gunning
 Math

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