Exam_solutions_1_(2)

Exam_solutions_1_(2) - February 9, 2007 MATH 141 TEST 1...

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Math 141, Sections 03**, T Pilachowski, Spring 2007 February 9, 2007 MATH 141 – TEST 1 (6.1 – 6.8) [Pilachowski] SOLUTIONS 1. (15 points) Let R be the region between the graph of () ( ) 4 1 1 3 + = x x x f and the x -axis on the interval [0 , 1]. Find the volume V of the solid obtained by rotating R about the x -axis. (You must show integral-calculus work and state a numeric answer in simplest form to receive full credit.) radius of circular cross-section = f, so ( ) 2 1 1 3 2 2 + = x x r ; ( ) + = 1 0 3 2 2 1 1 dx x x V π ; let 1 3 + = x u, dx x du 2 3 = ; = = = 1 2 9 2 1 2 3 2 3 3 2 3 2 3 2 1 2 1 u du u V or ( ) 1 2 2 9 2 2. a. (15 points) The base of a solid S is a circle with equation 9 2 2 = + y x . The cross-sections are squares. Write the integral you’d need to find the volume of S . DO NOT EVALUATE. domain of x = [–3, 3]; circle (base) is on both sides of x-axis so side of square (cross-section) = 2y = 2 9 2 x ; area of cross-section = (side) 2 ; ( ) = 3 3 2 9 4 dx x V b. (4 extra points) The base of a solid T
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Exam_solutions_1_(2) - February 9, 2007 MATH 141 TEST 1...

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