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Calculus II, Section 8.3, #28 Applications to Physics and Engineering Sketch the region bounded by the curves, and visually estimate the location of the centroid. Then find the exact coordinates of the centroid.1 y = sin (x), y = 0, 0≤x≤π Here’s a sketch of the region: y 1 x π 2 π From the symmetry about the line x = π2 , it seems that the x-coordinate of the centroid will be visually estimate the y-coordinate to be ≈ 0.35. π 2. I’d The relevant formulas for the centroid (center of mass of the region) are Z Z o 1 b 1 b1n 2 2 dx [f (x)] − [g(x)] x̄ = x [f (x) − g(x)] dx, ȳ = A a A a 2 For this problem, a = 0, b = π, f (x) = sin(x), and g(x) = 0. The area, A, is given by Z π A= sin (x) − 0 dx 0 π = [− cos (x)]0 = − cos (π) − − cos (0) = − (−1) − − (1) =2 So x̄ = = = = = = Z 1 π x [sin (x) − 0] dx 2 0 Z 1 π x sin (x) dx 2 0 1 [−x cos (x) + sin (x)]π0 2 1 [(−π · cos (π) + sin (π)) − (−0 · cos (0) + sin (0))] 2 1 [π] 2 π 2 and 1 ȳ = 2 1 Stewart, Z 0 π 1 2 sin (x) − 02 dx 2 Calculus, Early Transcendentals, p. 567, #28. Integration by parts: Z x sin (x) dx Let u = x, so dv = sin (x)dx. Then du = dx and v = − cos (x). Z = x · − cos (x) − − cos (x) dx Z = −x cos (x) + cos (x) dx = −x cos (x) + sin (x) Calculus II Applications to Physics and Engineering 1 = 2 Z 0 π 1 sin2 (x) dx 2 Using the trigonometric identity sin2 (x) = 1−cos (2x) 2 = 1 2 (1 − cos (2x)), Z 1 π1 1 = · (1 − cos (2x)) dx 2 0 2 2 Z π 1 = 1 − cos (2x) dx 8 0 π sin (2x) 1 x− = 8 2 0 1 sin (2π) sin (0) = π− − 0− 8 2 2 1 = [π − 0 − 0 + 0] 8 π = 8 = 0.3927 Thus, the exact coordinates of the centroid are π π 2,8 or about (1.5708,0.3927). y 1 b x π 2 π