PROBLEM 5.48 Determine by direct integration the centroid of the area shown. SOLUTION y = a sin πx L 1 y, dA = y dx 2 L /2 L /2 πx sin A= y dx = a dx 0 0 L xEL = x, yEL = L /2 L π x A = a − cos L 0 π Setting u = πx L , we have x = Integrating by parts, L /2 xEL dA = L u , dx = L π 0 π xydx = L /2 aL = π xa sin 0 πx L du, π /2 L L L π u a sin u π du = a π xEL dA = L xEL dA = a [−u cos u ]π0 /2 + π yEL dA = 0 2 = dx L /2 0 a2 L 2π 2 1 2 1 y dx = a 2 2 2 π /2 1 0 2 L /2 0 π /2 0 sin 2 2 π /2 0 u sin x du 2 aL cos u du = 2 π πx L dx = a2 L 2π π /2 0 sin 2 u du π /2 (1 − cos 2u )du = 2 aL aL xA = xEL dA: x = 2 π π aL 1 2 yA = yEL dA: y = a L π 8 a2 L 1 u − sin 2u 4π 2 0 1 = a2 L 8 x= y= π 8 L π a PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 609 PROBLEM 5.53 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.2 about (a) the line y = 72 mm, (b) the x-axis. SOLUTION From the solution of Problem 5.2, we have A = 2808 mm 2 x = 36 mm y = 48 mm Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the line y = 72 mm: Volume = 2π (72 − y ) A = 2π (72 − 48)(2808) Volume = 423 × 103 mm3 Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y3 L3 ) where y1 and y3 are measured with respect to line y = 72 mm. Area = 2π (36) ( ) 482 + 722 + (36) ( ) 302 + 722 Area = 37.2 × 103 mm 2 (b) Rotation about the x-axis: Volume = 2π yarea A = 2π (48)(2808) Volume = 847 × 103 mm3 Area = 2π yline L = 2π Σ( yline ) L = 2π ( y1 L1 + y2 L2 + y3 L3 ) = 2π (36) ( ) 482 + 722 + (72)(78) + (36) ( ) 302 + 722 Area = 72.5 × 103 mm 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 616 PROBLEM 5.62 A 34 - in.-diameter hole is drilled in a piece of 1-in.-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process. SOLUTION The required volume can be generated by rotating the area shown about the y-axis. Applying the second theorem of Pappus-Guldinus, we have V = 2π x A 3 1 1 1 1 1 = 2π + in. × × in. × in. 8 3 4 2 4 4 V = 0.0900 in 3 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 626 PROBLEM 5.64 Determine the capacity, in liters, of the punch bowl shown if R = 250 mm. SOLUTION The volume can be generated by rotating the triangle and circular sector shown about the y-axis. Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have V = 2π xA = 2π Σ xA = 2π ( x1 A1 + x2 A2 ) 1 1 1 1 3 2 R sin 30° π 2 = 2π × R × R × R + R 2 3 × π6 6 3 2 2 2 R3 R3 = 2π + 16 3 2 3 3 3 π R3 8 3 3 = π (0.25 m)3 8 = 0.031883 m3 = Since 103 l = 1 m3 V = 0.031883 m3 × 103 l 1 m3 V = 31.9 l PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 628

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