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Unformatted text preview: Numerical Analysis in Engineering ME 140A, Fall 2008 Midterm #1 Solution By: Mohamad M. NasrAzadani mmnasr@engr.ucsb.edu December 3, 2008 Problem 1 a) Since h = h ( r ) (only a function of r ), volume of the given shape can be found as V = Z Z A h ( r ) dA (1) V = Z R h ( r )2 rdr (2) Note that dV = h ( r )2 rdr is the differential volume of a cylinder with radius r , thickness dr and height h ( r ) (see figure 1). b) Total mass of the pollutant in the reservoir can be found by M = Z Z Z V c ( r ) dV (3) M = Z R c ( r ) h ( r )2 rdr (4) Note that in equation (4), dM = c ( r ) h ( r )2 rdr is the differential mass of the pollutant in the differential volume (see figure 1). 1 Figure 1: Left: Side view of the reservoir. Right: Three dimensional view of the element. Problem 2 a) Analytical integral for the given function f ( x ) = cos 2 x is found as Z 1 (cos 2 x ) dx = sin 2 x 2 fl fl fl fl 1 = 0 . (5) b) Using the Trapezoidal rule for n = 1, the given integral is found as...
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This note was uploaded on 01/04/2011 for the course ME 140A taught by Professor Meiburg during the Spring '08 term at UCSB.
 Spring '08
 Meiburg

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