Explanation Refer figure P5.39P and figure given below. Write the equation of curved boundary of shaded region. x 2 a 2 + y 2 b 2 = 1 Here x is x-coordinate of point on the curve, y is the y co-ordinate of the point on curve, a is the length of rectangle on P5.39, and b is the breadth of rectangle on P5.39. Rewrite the above equation in terms of y . y = b a a 2 − x 2 The shaded area in the above figure is a differential area element. Write the expression for the x-coordinate of center of mass of rectangular area element. x ¯ E L = x Here, x ¯ E L is the center of mass of differential area element d A . Write the expression for the y-coordinate of center of mass of d A . y ¯ E L = 1 2 ( y + b ) Here, y ¯ E L is the y-coordinate of center of mass of element d A . Rewrite the above expression by substituting b a a 2 − x 2 for y . y ¯ E L = 1 2 ( b a a 2 − x 2 + b ) = b 2 a ( a + a 2 − x 2 ) Write the expression for area of differential element. d A = ( b − y ) d x Here, d A is the area of rectangular area element and d x denotes a small change in x . Rewrite the above expression by substituting b a a 2 − x 2 for y . d A = ( b − b a a 2 − x 2 ) d x = b a ( a − a 2 − x 2 ) d x To find the area of shaded region in P5.39, integrate the right hand side of above equation from x = 0 to x = a . A = ∫ 0 a b a ( a − a 2 − x 2 ) d x (I) Here, A is the area of shaded region in figure P5.39. Assume the value a sin θ for x . x = a sin θ Differentiate the above relation with respect to x . 1 = a cos θ ( d θ d x ) Rewrite the above relation in terms of d x . d x = ( a cos θ ) d θ Find a 2 − x 2 by substituting a sin θ for x . a 2 − x 2 = a 2 − ( a sin θ ) 2 = a cos θ Now, change the integrating variable from x to θ . Then, the range of integration x = 0 to x = a changes to θ = 0 to θ = π 2 . Rewrite equation (I) by replacing limits x = 0 to x = a with θ = 0 to θ = π 2 , substituting ( a cos θ ) d θ for d x , and a cos θ for a 2 − x 2 . A = ∫ 0 π / 2 b a ( a − a cos θ ) ( a cos θ ) d θ = b a [ a 2 sin θ − a 2 ( θ 2 + sin 2 θ 4 ) ] 0 π / 2 = a b ( 1 − π 4 ) To calculate ∫ x ¯ E L d A , substitute b a ( a − a 2 − x 2 ) d x for d A and x for x ¯ E L and integrate it from x = 0 t0 x = a

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259638091

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 5.2, Problem 5.39P

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The centrod of shaded area in Fig. P5.39.

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Vector Mechanics for Engineers: Statics and Dynamics

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The centrod of shaded area in Fig. P5.39.

Solution Summary: The author illustrates the centroid of shaded area in Fig. P5.39.

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The centrod of shaded area in Fig. P5.39.

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