# 23ss216-HW5-solutions - MATH 23 SS II 2016 HOMEWORK...

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MATH 23, SS II, 2016 HOMEWORK ASSIGNMENT #5 SOLUTIONS (1) Find the mass of the region in space bounded below by the xy -plane and bounded above by the hemisphere z = p 4 - x 2 - y 2 , if the density ρ = z .
(2) Find the center of mass of the region in space bounded below by the xy -plane, above by the plane z = x + 2 , and by the cylinder ( x - 1) 2 + y 2 = 1 , if the density ρ = x 2 + y 2 .
MATH 23, SS II, 2016 2 the region, and the density, so x = R R R V x ( x 2 + y 2 ) dV R R R V ( x 2 + y 2 ) dV = R π 2 - π 2 R 2 cos( θ ) 0 R r cos( θ )+2 0 r cos( θ ) ( r 2 ) rdzdrdθ R π 2 - π 2 R 2 cos( θ ) 0 R r cos( θ )+2 0 ( r 2 ) rdzdrdθ = R π 2 - π 2 R 2 cos( θ ) 0 r 4 cos( θ ) ( r cos( θ ) + 2) drdθ R π 2 - π 2 R 2 cos( θ ) 0 r 3 ( r cos( θ ) + 2) drdθ = R π 2 - π 2 1 6 r 6 cos 2 ( θ ) + 2 5 r 5 cos( θ ) 2 cos( θ ) 0 R π 2 - π 2 1 5 r 5 cos( θ ) + 1 2 r 4 2 cos( θ ) 0 = R π 2 - π 2 2 5 3 cos 8 ( θ ) + 2 6 5 cos 6 ( θ ) R π 2 - π 2 2 4 3 cos 6 ( θ ) + 2 5 5 cos 4 ( θ ) = R π 2 - π 2 2 5 3 ( 1 2 (1 + cos(2 θ )) ) 4 + 2 6 5 ( 1 2 (1 + cos(2 θ )) ) 3 R π 2 - π 2 2 4 3 ( 1 2 (1 + cos(2 θ )) ) 3 + 2 5 5 ( 1 2 (1 + cos(2 θ )) ) 2 = R π 2 - π 2 2 3 ( 1 + 4 cos(2 θ ) + 6 cos 2 (2 θ ) + 4 cos 3 (2 θ ) + cos 4 (2 θ ) ) + 2 3 5 ( 1 + 3 cos(2 θ ) + 3 cos 2 (2 θ ) + cos 3 (2 θ ) ) R π 2 - π 2 2 3 (1 + 3 cos(2 θ ) + 3 cos 2 (2 θ ) + cos 3 (2 θ )) + 2 3 5 (1 + 2 cos(2 θ ) + cos 2 (2 θ )) = R π 2 - π 2 2 3 ( 1 + 4 cos(2 θ ) + 6 cos 2 (2 θ ) + 4 cos 3 (2 θ ) + 1 4 ( 1 + 2 cos(4 θ ) + cos 2 (4 θ ) )) + 2 3 5 ( 1 + 3 cos(2 θ ) + 3 cos 2 (2 θ ) + cos 3 R π 2 - π 2 2 3 (1 + 3 cos(2 θ ) + 3 cos 2 (2 θ ) + cos 3 (2 θ )) + 2 3 5 (1 + 2 cos(2 θ ) + cos 2 (2 θ )) = 2 3 ( π + 3 π + 1 4 ( π + π 2 )) + 2 3 5 ( π + 3 2 π ) 2 3 ( π + 3 2 π ) + 2 3 5 ( π + π 2 ) = 2 3 ( 1 + 3 + 3 8 ) + 2 3 5 ( 5 2 ) 2 3 ( 1 + 3 2 ) + 2 3 5 ( 3 2 ) = 35 12 + 4 5 3 + 12 5 = 415 244 , which seems a little large, since x 2 everywhere on the region. However, the density does increase as we move out, as does the height of the region, so maybe this is reasonable. Well, that is, the answer is plausible.
MATH 23, SS II, 2016 3 Now, for z : z = R R R V z ( x 2 + y 2 ) dV R R R V ( x 2 + y 2 ) dV = R π 2 - π 2 R 2 cos( θ ) 0 R r cos( θ )+2 0 z ( r 2 ) rdzdrdθ R π 2 - π 2 R 2 cos( θ ) 0 R r cos( θ )+2 0 ( r 2 ) rdzdrdθ = R π 2 - π 2 R 2 cos( θ ) 0 1 2 r 3 ( r cos( θ ) + 2) 2 drdθ 61 15 π = R π 2 - π 2 R 2 cos( θ ) 0 ( 1 2 r 5 cos 2 ( θ ) + 2 r 4 cos( θ ) + 2 r 3 ) drdθ 61 15 π = R π 2 - π 2 ( 1 12 r 6 cos 2 ( θ ) + 2 5 r 5 cos( θ ) + 1 2 r 4 ) 2 cos( θ ) 0 61 15 π = R π 2 - π 2 2 4 3 cos 8 ( θ ) + 2 6 5 cos 6 ( θ ) + 8 cos 4 ( θ ) 61 15 π = R π 2 - π 2 2 4 3 ( 1 2 (1 + cos(2 θ )) ) 4 + 2 6 5 ( 1 2 (1 + cos(2 θ )) ) 3 + 8 ( 1 2 (1 + cos(2 θ )) ) 2 61 15 π = R π 2 - π 2 1 3 ( 1 + 4 cos(2 θ ) + 6 cos 2 (2