MATH103 April 2012
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q2 (e) • Q2 (f) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q4 (e) • Q4 (f) • Q5 (a) • Q5 (b) • Q5 (c) •
Question 02 (c)
Short answer question. Simplify your answer to the best possible. Show in detail how you arrive at your answer (work will be considered for this problem).
Find the center of mass of the distribution
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The centre of mass of the shape delimited by the x-axis, the vertical lines x = a, x = b and the curve y = ƒ(x) is given by
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The centre of mass is given by
We start with the simpler integral M and use the substitution w = x/2. Then dw = dx/2, or equivalently, dx = 2dw. Further, when x = π we have w = π/2 and when x = 0 we have w = 0. The integral M therefore becomes
Next we calculate I. This is done using integration by parts: Choose u = x and . Then du = dx and from the integral we just solve, we know that . Hence
For the second integral we again used our favourite substitution w = x/2. The center of mass is thus
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MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Center of mass, MER Tag Integration by parts, MER Tag Substitution