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EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = 3y² and the planes x = Z, z = 0, and x = 3.

EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = 3y² and the planes x = Z, z = 0, and x = 3.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Mxy = = = Р = = -1 3 = ²/27y-27y7 2p 25 (27-27,6 /0 31x = 3 3 108p 7 JACE Z 3 -LLLE = 1² P P (x, y, z) = = X = = 3y² 1 3 P -LLx² 54p 7 Z Therefore the center of mass is Myz, Mxz m m Z 2 dy x² dx dy (27-271,6 Mxy m 15 5 7' 14" 1,0 ). ) dy 1. Jp dv )p dz dx dy X 1²=* dx dy z 0 )dy
Transcribed Image Text:Mxy = = = Р = = -1 3 = ²/27y-27y7 2p 25 (27-27,6 /0 31x = 3 3 108p 7 JACE Z 3 -LLLE = 1² P P (x, y, z) = = X = = 3y² 1 3 P -LLx² 54p 7 Z Therefore the center of mass is Myz, Mxz m m Z 2 dy x² dx dy (27-271,6 Mxy m 15 5 7' 14" 1,0 ). ) dy 1. Jp dv )p dz dx dy X 1²=* dx dy z 0 )dy
E 0 Z = X 0 EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = 3y² and the planes X = Z, Z = 0, and x = 3. SOLUTION The solid E and its projection onto the xy-plane are shown in the figure. The lower and upper surfaces of E are the planes z = 0 and z = x, so we describe E as a type 1 region: ≤ y ≤ 1 ✔, 3y² ≤ x ≤ 3,0 ≤ z ≤x}. E = = {(x, y, z)-1 Then, if the density is p(x, y, z) =p, the mass is 1 3 X 16₂0 •LL F m = 1 3 - LLE = P 2 = 0/² || ²/²/² 2₁9-91¹ 1 = of ² ( 9 - 9x²) dy 0 = - By 905 = 9y 5 p dv = Myz 36p 5 = SSLC EX - ILLE = 1 3 -LL(²² P 3y2 = P Because of the symmetry of E and p about the xz-plane, we can immediately say that Mxz = 0 and therefore y = 0. The other moments are Jp dv 31x = 3 3 p dz dx dy dy It ) dx dy 7x = 3 x = 3y² ) dy dy dz dx dy ) dx dy
Transcribed Image Text:E 0 Z = X 0 EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = 3y² and the planes X = Z, Z = 0, and x = 3. SOLUTION The solid E and its projection onto the xy-plane are shown in the figure. The lower and upper surfaces of E are the planes z = 0 and z = x, so we describe E as a type 1 region: ≤ y ≤ 1 ✔, 3y² ≤ x ≤ 3,0 ≤ z ≤x}. E = = {(x, y, z)-1 Then, if the density is p(x, y, z) =p, the mass is 1 3 X 16₂0 •LL F m = 1 3 - LLE = P 2 = 0/² || ²/²/² 2₁9-91¹ 1 = of ² ( 9 - 9x²) dy 0 = - By 905 = 9y 5 p dv = Myz 36p 5 = SSLC EX - ILLE = 1 3 -LL(²² P 3y2 = P Because of the symmetry of E and p about the xz-plane, we can immediately say that Mxz = 0 and therefore y = 0. The other moments are Jp dv 31x = 3 3 p dz dx dy dy It ) dx dy 7x = 3 x = 3y² ) dy dy dz dx dy ) dx dy
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