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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 γ 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 γ 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 γ 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 γ . 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 γ 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 γ 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 γ 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 π . Solution. We know, for a given curve parametrized as in this problem, the arc length is given by the equation L = Z Δ s dx 1 dt 2 + dx 2 dt 2 + dx 3 dt 2 dt, where Δ 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 Δ ds dt dt = √ 3 Z 2 π e t dt = √ 3 e t 2 π = √ 3 ( e 2 π 1 ) . So, the required integral is simply √ 3 ( e 2 π 1 ) ....
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This note was uploaded on 05/03/2010 for the course MAT 218 taught by Professor Robertc.gunning during the Fall '09 term at Princeton.
 Fall '09
 RobertC.Gunning
 Math

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