Explanation Given: The height of the shaded area is a . The width of the shaded area is a . Show the area of the differential element that is parallel to the y -axis as in Figure (1). From Figure 1, Compute the area of the differential element that parallel to the y -axis as shown with shaded area from Figure 1. d A = y d x (I) Here, the area of the differential element is d A . Substitute ( a 1 2 − x 1 2 ) 2 for y in Equation (I). d A = ( a 1 2 − x 1 2 ) 2 d x Compute the coordinate of the centroid of the differential element along the x axis. x ˜ = x Compute the coordinate of the centroid of the differential element along the y axis. y ˜ = y 2 (II) Substitute ( a 1 2 − x 1 2 ) 2 for y in Equation (II). y ˜ = ( a 1 2 − x 1 2 ) 2 2 = 1 2 ( a 1 2 − x 1 2 ) 2 Compute the product of inertia of differential element with respect to the x and y axes. d I x y = d I ¯ x ′ y ′ + d A x ˜ y ˜ (III) Here, the first integral of the moment of inertia of the area about the centroidal x and y axes is I ¯ x ′ y ′ . Substitute 0 for d I ¯ x ′ y ′ , ( a 1 2 − x 1 2 ) 2 d x for d A , x for x ˜ , and 1 2 ( a 1 2 − x 1 2 ) 2 for y ˜ in Equation (III). d I x y = 0 + [ ( a 1 2 − x 1 2 ) 2 d x ] ( x ) [ 1 2 ( a 1 2 − x 1 2 ) 2 ] = [ ( a + x − 2 a 1 2 x 1 2 ) d x ] x [ 1 2 ( a + x − 2 a 1 2 x 1 2 ) ] = 1 2 ( x 3 + a 2 x + 6 a x 2 − 4 a 3 2 x 3 2 − 4 a 1 2 x 5 2 ) d x Conclusion: Compute the product of inertia of the shaded area with respect to x and y axes. I x y = ∫ 0 a d I x y (IV) Substitute 1 2 ( x 3 + a 2 x + 6 a x 2 − 4 a 3 2 x 3 2 − 4 a 1 2 x 5 2 ) d x for d I x y in Equation (IV)

Engineering Mechanics: Statics Plus Mastering Engineering with Pearson eText -- Access Card Package (14th Edition) (Hibbeler, The Engineering Mechanics: Statics & Dynamics Series, 14th Edition)

14th Edition

ISBN: 9780134160689

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 10.7, Problem 59P

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The product of inertia of the shaded area with respect to x and y axes.

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