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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 = 2y' and the planes x = z, z = 0, and x = 2. 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: E = {(x, y, z) | 1 ✔ sys1 ✔,2y² ≤x≤ 2,0 sz sx} Then, if the density is p(x, y, z) = p, the mass is m = / / / pov = [ 1 ² √² pozo =PLLE 30 -44 =p -24-494 =Pf' (4-6 P =pl| 4y (4-4y¹)dy Myz= M₁2 = // /\x -LOVE sp EP 3.20 Because of the symmetry of E and p about the xz-plane, we can immediately say that Myz = 0 and therefore y = 0. The other moments are Mxy= -1 3 2p 30 2p 3 D 2² -L₁ -1 24p = /// E LAVE 2² 2 X (x, y, z)=( Myz pdy x² dx dy √24² Therefore the center of mass is Mxz Mxy m m m ✔rdy ✓dy p dz dx dy dx dy x l dy 0, ✓ pov X dy ✔p dz dx dy ✔ dx dy ✓ pov p dz dx dy 1dx dy [+] √₁ n'

EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = 2y' and the planes x = z, z = 0, and x = 2. 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: E = {(x, y, z) | 1 ✔ sys1 ✔,2y² ≤x≤ 2,0 sz sx} Then, if the density is p(x, y, z) = p, the mass is m = / / / pov = [ 1 ² √² pozo =PLLE 30 -44 =p -24-494 =Pf' (4-6 P =pl| 4y (4-4y¹)dy Myz= M₁2 = // /\x -LOVE sp EP 3.20 Because of the symmetry of E and p about the xz-plane, we can immediately say that Myz = 0 and therefore y = 0. The other moments are Mxy= -1 3 2p 30 2p 3 D 2² -L₁ -1 24p = /// E LAVE 2² 2 X (x, y, z)=( Myz pdy x² dx dy √24² Therefore the center of mass is Mxz Mxy m m m ✔rdy ✓dy p dz dx dy dx dy x l dy 0, ✓ pov X dy ✔p dz dx dy ✔ dx dy ✓ pov p dz dx dy 1dx dy [+] √₁ n'

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 12T
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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 = 2y² and the planes x = z, z = 0, and x = 2. 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: E = {(x, y, z) | Then, if the density is p(x, y, z)=p, the mass is m = / / / pov = [1 ² [P -/1 =p 4-414 (4-4y¹) dy 4y -4³ Myz = /// [x -LAVE 20 My = ///E -LAVE 4L** L 24p 3.2p Because of the symmetry of E and p about the xz-plane, we can immediately say that Mz = 0 and therefore y = 0. The other moments are ✓ pov x² dx dy (x, y, z)=( =( ✔ sys1✔✔, 2y² ≤x≤ 2,0 sz sx} Therefore the center of mass is Myz Mxz Mxy mm m o dz dx dy ✓dx dy ✓ Bardy ✓dy x l 0, X dy •p dz dx dy ✔ dx dy ✓ pov p dz dx dy 1dx dy + x +² §² 0!
Transcribed Image Text:EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by the parabolic cylinder x = 2y² and the planes x = z, z = 0, and x = 2. 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: E = {(x, y, z) | Then, if the density is p(x, y, z)=p, the mass is m = / / / pov = [1 ² [P -/1 =p 4-414 (4-4y¹) dy 4y -4³ Myz = /// [x -LAVE 20 My = ///E -LAVE 4L** L 24p 3.2p Because of the symmetry of E and p about the xz-plane, we can immediately say that Mz = 0 and therefore y = 0. The other moments are ✓ pov x² dx dy (x, y, z)=( =( ✔ sys1✔✔, 2y² ≤x≤ 2,0 sz sx} Therefore the center of mass is Myz Mxz Mxy mm m o dz dx dy ✓dx dy ✓ Bardy ✓dy x l 0, X dy •p dz dx dy ✔ dx dy ✓ pov p dz dx dy 1dx dy + x +² §² 0!
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