University of TexasM 427LHW11-Solutions. Homework Solutions

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University of TexasM 427LHW11-Solutions. Homework Solutions University of TexasM 427LHW11-Solutions. Homework Solutions University of TexasM 427LHW11-Solutions. Homework Solutions University of TexasM 427LHW11-Solutions. Homework Solutions University of TexasM 427LHW11-Solutions. Homework Solutions This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the integral I = ZC F · ds when F(x, y) = 2xy i + 3y j and C is given parametrically by Explanation: When C is given parametrically by c(t) = t i + t2 j , 0 ≤ t ≤ 1 , then c0(t) = i + 2t j , while F(x, y) = F(c(t)) = 2t3 i + 3t2 j on C. Thus F(c(t)) · c0(t) = 2t3 + 6t3 = 8t3 , and so I = ZC F · s = Z01 F(c(t)) · c0(t) dt = 8 Z01 t3 dt = 8h 1 4t4 i1 0 . Consequently, I = 2 . 002 10.0 points Evaluate the integral I = ZC 2z ds when the path C is parametrized by c(t) = (2et, e−t, 2t) for 0 ≤ t ≤ 1. Explanation: A scalar line integral over c(t) = (2et, e−t, 2t), 0 ≤ t ≤ 1, is given by I = Z01 f(c(t))kc0(t)k dt . Now kc0(t)k = q(2et)2 + (−e−t)2 + (2)2 ,silveyra (as54582) – HW11 – goddard – (57110) 2 which after simplification becomes kc0(t)k = p4e2t + e−2t + 4 = q(2et + e−t)2 = 2et + e−t . On the other hand, f(c(t)) = 2z c(t) = 4t . Thus I = Z01 4t(2et + e−t) dt = Z01 (8tet + 4te−t) dt . To evaluate this last integral it’s simpler to use Integration by Parts separately on the two terms. For the first one, Z01 8tet dt = 8h tet i1 0 − 8 Z01 et dt = 8e − 8h et i1 0 = 8 . Consequently, I = 43 − 2e . 003 10.0 points Evaluate the integral I = ZC F · ds when and C is the quarter circle x2 + y2 = 1 , x, y ≥ 0 , oriented counter-clockwise. Explanation: The quarter circle x2 + y2 = 1 , x, y ≥ 0 is given parametrically by c(t) = cos t i + sin t j , 0 ≤ t ≤ π 2 . In this case c0(t) = − sin t i + cos t j , while F(x, y) = F(c(t)) = sin t i − 3 cos t j on C, and so F(c(t)) · c0(t) = − sin2 t − 3 cos2 t . On the other hand, since C is oriented counter-clockwise while the parametrization c(t) traces out C counter-clockwise as t increases, we shall need to integrate from t = 0 to π/2. Thus

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