Explanation Refer the figure P5.50 and figure given below. The area in P5.50 is divided into two sections for the purpose of integration. Rectangular region 0 ≤ x ≤ 1 a and region 1 a ≤ x ≤ a . Consider the region 0 ≤ x ≤ 1 a in P5.50. The region 0 ≤ x ≤ 1 a is a rectangle with breadth d x along x-axis and length a along y-axis. Write the expression for the x-coordinate of center of mass of differential area element in this region. x ¯ E L 1 = x (I) Here, x ¯ E L 1 is the center of mass of differential area element in region 0 ≤ x ≤ 1 a . Write the expression for the y-coordinate of center of mass of differential area element in region 0 ≤ x ≤ 1 a in P5.50. y ¯ E L 1 = a 2 (II) Here, y ¯ E L 1 is the y-coordinate of differential area element Write the expression to calculate the differential area element in 0 ≤ x ≤ 1 a . d A 1 = a d x (III) Here, d A 1 is the differential area element in 0 ≤ x ≤ 1 a and d x denotes a small change in x . Consider the region 1 a ≤ x ≤ a in in P5.50. Write the expression for the x-coordinate of center of mass of differential area element in region 1 a ≤ x ≤ a . x ¯ E L 2 = x (IV) Here, x ¯ E L 2 is the center of mass of differential area element in region 1 a ≤ x ≤ a Write the expression for the y-coordinate of center of mass of differential area element in region 1 a ≤ x ≤ a . y ¯ E L 2 = y 2 Rewrite the above relation by substituting 1 x for y . y ¯ E L 2 = 1 2 x (V) Here, y ¯ E L 2 is the y-coordinate of center of mass of differential area element in region 1 a ≤ x ≤ a and y is a curve in region. Write the expression for the differential area element in 1 a ≤ x ≤ a . d A 2 = 1 x d x (VI) Here, d A 2 is the differential area element in 1 a ≤ x ≤ a . Write the equation to calculate the total area of shaded region in P5.44. A = ∫ 0 1 / A d A 1 + ∫ 1 / A a d A 2 Here, A is the area shaded region in P5.50. Rewrite the above equation by substituting equations (III) and (VI). A = ∫ 0 1 / a a d x + ∫ 1 / a a 1 x d x = a [ x ] 0 1 / a + [ ln x ] 1 / a a = a ( 1 a ) + [ ln a − ln ( 1 a ) ] = 1 + ln a − ln a − 1 = 1 + ln a − ( − ln a ) = 1 + 2 ln a Write the expression for ∫ x ¯ E L d A . ∫ x ¯ E L d A = ∫ 0 1 / a x ¯ E L 1 d A 1 + ∫ 1 / a a x ¯ E L 2 d A 2 = ∫ 0 1 / a x d A 1 + ∫ 1 / a a x d A 2 Rewrite the above equation by substituting equations (I), (III), (IV), and (VI). ∫ x ¯ E L d A = ∫ 0 1 / a x ( a d x ) + ∫ 1 / a a x ( 1 x d x ) = a [ x 2 2 ] 0 1 / a + [ x ] 1 / a a = 1 2 a + a − 1 a = 2 a 2 − 1 2 a Write the expression for ∫ y ¯ E L d A

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ISBN: 9781259633133

Author: BEER

Publisher: MCG

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Chapter 5.2, Problem 5.50P

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The centroid of area in Fig. P5.450 by method of direct integration.

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Chapter 5.2, Problem 5.49P

Chapter 5.2, Problem 5.51P

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The centroid of area in Fig. P5.450 by method of direct integration.

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The centroid of area in Fig. P5.450 by method of direct integration.

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