Explanation Refer Fig. P5.127 and Fig. 1. Because of the symmetry of the diagram, x ¯ = h and z ¯ = 0 . Let the element of volume of the disk be of radius r and thickness dx . Find the equation for radius from the equation of the generating curve. x 2 = h 2 a 2 ( a 2 − y 2 ) r = h − h a ( a − y ) Here, h is the semi major axis, a is the semi minor axis and y is the distance along y axis. When y = a sin θ , then d y = a cos θ d θ (I) Write the expression for the volume of the shaded region. d V = π r 2 d y = π h 2 a 2 ( a − a 2 − y 2 ) 2 d y (II) Write the expression for y ¯ of the volume. y ¯ V = ∫ y ¯ E L d V y ¯ = ∫ y ¯ E L d V V (III) Here, y ¯ is the y coordinate of the centroid and y ¯ E L is the y coordinate of the centroid of the element. Conclusion: Calculate the volume obtained by rotating the shaded area using equations (I) and (II). V = ∫ 0 a π h 2 a 2 ( a − a 2 − y 2 ) 2 d y = ∫ 0 π / 2 π h 2 a 2 ( a − a 2 − a 2 sin 2 θ ) 2 ( a cos θ d θ ) = π h 2 a 2 ∫ 0 π / 2 [ a 2 − 2 a ( a cos θ ) + ( a 2 − a 2 sin 2 θ ) ] ( a cos θ d θ ) = π a h 2 ∫ 0 π / 2 ( 2 cos θ − 2 cos 2 θ − sin 2 θ cos θ ) d θ = π a h 2 [ 2 sin θ − 2 ( θ 2 + sin 2 θ 4 ) − 1 3 sin 3 θ ] 0 π / 2 = π a h 2 [ 2 − 2 ( ( π 2 ) 2 ) − 1 3 ] 0 π / 2 = 0

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ISBN: 8220102809888

Author: BEER

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Chapter 5.4, Problem 5.127P

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To locate the centroid of the volume obtained by rotating the shaded area.

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Chapter 5.4, Problem 5.126P

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To locate the centroid of the volume obtained by rotating the shaded area.

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