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MATH 101 HOMEWORK 9 – SOLUTIONS 1. Find the mass of a circular cone of base radius 10 cm and height 20 cm, if its density at a point P is 1 + r/100, where r is the distance from P to the axis of the cone. The density is constant on cylindrical shells. The area of a shell at distance r from the axis is 2πrh, where h/(10 − r) = 20/10 = 2, i.e. h = 2(10 − r). Thus Z 10 m= 0 r )dr = 4π 2πr · 2(10 − r) · (1 + 100 = 4π(5r 2 − Z 10 (10r − 0 r3 9r 2 − )dr 10 100 r 4 10 3r 3 − ) = 700π ≈ 2199.11 10 400 0 2. Find the centroids of the following planar regions. (a) The region enclosed between the parabola y = 3x − x2 and the line y = 2x − 6. We first find the intersection points: 3x − x2 = 2x − 6, x2 − x − 6 = 0, x = −2, 3. Hence the centroid has coordinates x̄ = Mx /m, ȳ = My /m, where Z Z x2 125 x3 3 m= (3x − x − (2x − 6))dx = (6 + x − x )dx = 6x + − = , 2 3 −2 6 −2 −2 3 3 2 2 1 Z Mx = Z 3 (3x − x − (2x − 6))xdx = 3 2 −2 1 My = 2 Z −2 (6x + x2 − x3 )dx = 3x + x3 125 x4 3 − = , 3 4 −2 12 Z 1 3 ((3x − x ) − (2x − 6) )xdx = (5x2 − 6x3 + x4 + 24x − 36)dx 2 −2 −2 3 1 5x3 625 3x4 x5 = − + + 12x2 − 36x = − , 2 3 2 5 12 −2 3 2 2 2 so that x̄ = 5 1 , ȳ = − . 2 2 (b) The region in Figure 7.41(d) in the textbook (page 442). By symmetry, x̄ = 0. To find ȳ, we write ȳ = m1 ȳ1 + m2 ȳ2 + m3 ȳ3 , m1 + m2 + m3 where m1 , ȳ1 denote the area of the large half-disc and the y-coordinate of its centroid, and m2 , m3 , ȳ2 , ȳ3 are the areas and y-coordinates of the centroids of the two small half-discs. We have m1 = π/2, m2 = m + 3 = π/8, and from the example in class, y1 = 4/3π, y2 = y3 = −2/3π. Hence π 2 · 4 + 2 · π8 · 3π 2 ȳ = 2 3ππ = . π 3π 2 +2· 8 3. A random variable X has probability density f (x) = Cx2 on [0, 1], for some value of C. Find this C. For this C, find the mean value of X. Z 1 1 x3 C = 1, so that C = 3, f (x) = 3x2 . Then Cx dx = C = 30 3 2 We want 0 Z 1 x · 3x2 dx = E(X) = 0 2 3x4 1 3 = . 4 0 4