Tiêu đề MSTE 21 Solutions.pdf ✅

21 Integral Calculus: Centroid and Moment of Inertia Solutions ▣ 1. Find the centroid of the area in the first quadrant under the curve y = 4 − x 2 . [SOLUTION] A = ∫ (4 − x 2 )dx 2 0 = 16 3 x = ∫ x(4 − x 2 )dx 2 0 16 3 = 3 4 y = ∫ 1 2 (4 − x 2 ) 2dx 2 0 16/3 = 8 5 The coordinates are (x, y) = ( 3 4 , 8 5 ) ▣ 2. Find the moment of inertia of the area under the curve y = sin x from x = 0 to x = π 2 , with respect to the x-axis. [SOLUTION] Ix = ∫ 1 3 (sin x) 3 π 2 0 dx = 2 9 units 4 ▣ 3. The curve y = ax 2 + 120 passes through (40,0). Find the radius of gyration of the area bounded by the curve and the x-axis, with respect to the x-axis. [SOLUTION] At (40,0), 0 = a(40) 2 + 120 a = − 3 40 y = − 3 40 x 2 + 120 The x-intercepts: 0 = − 3 40 x 2 + 120 x = ±40

Find the limits of integration x 2 − 4x + y = 5. y = 9 x 2 − 4x + 9 = 5 x 2 − 4x + 4 = 0 (x − 2) 2 = 0 → x = 2 Find the area A = ∫ [9 − (5 + 4x − x 2 )]dx 2 0 A = 8 3 For the centroid: x = ∫ x[9 − (5 + 4x − x 2 )]dx 2 0 8 3 = 1 2 y = ∫ [ 9 + (5 + 4x − x 2 ) 2 ][9 − (5 + 4x − x 2 )] 2 0 dx 8 3 y = 39 5 The centroid is at (x, y) = ( 1 2 , 39 5 )