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Unformatted text preview: 3 ). Solution: Here, the center as well as the area of disk is clear, so we use the formula F = w hA to calculate the force. F = w hA = 62 . 4 (3 + 2) 2 2 = 62 . 4 20 = 1248 lbs. 3. Evaluate the integral Z 8 1 log 8 x x dx Solution: Let u = log x 8 , then du = dx x ln8 . Also, when x = 1, log x 8 = log 1 8 = 0, and when x = 8, log x 8 = log 8 8 = 1 . So, Z 8 1 log 8 x x dx = Z 1 u ln 8 dx = u 2 2 ln 8  u =1 u =0 = ln 8 2 = ln 2 3 2 = 3 2 ln 2 . 1...
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This note was uploaded on 10/08/2010 for the course MATH 21B taught by Professor Vershynin during the Spring '08 term at UC Davis.
 Spring '08
 Vershynin
 Calculus

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