The tetrahedron is e x y z 0 x 1 0 y 2 2x 0

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Unformatted text preview: then E = {(r, θ, z ) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1, 3r2 ≤ z ≤ 4 − r2 } and V= 4−r 2 1 2π dV = E r dz dr dθ 3r 2 0 0 = 2π 7. The tetrahedron is E = {(x, y, z ) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 2 − 2x, 0 ≤ z ≤ 3(1 − x − y/2)} So the mass is 3(1−x−y/2) 2−2x 1 mass = (x2 + y 2 + z 2 ) dz dy dx = 7/5 ρ dV = 0 E 0 0 and the center of the mass is 1 mass 1 y= ¯ mass 1 z= ¯ mass xρ dV = 4/21, x= ¯ 8. E yρ dV = 11/21, E zρ dV = 8/7. E (a) x+y x 3 xy dz dy dx = 81/2 xy dV = E 0 0 4 0 (b) Use cylindrical coordinates: π 2 r sin θ yz dV = E (r sin θ)zr dz dr dθ 0 0 0 2 π = 0 0 14 3 r sin θ dr dθ 2 16 π 3 = sin θ dθ 50 16 π (1 − cos2 θ) sin θ dθ = 50 1...
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This document was uploaded on 03/08/2014.

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