Michael Corral: Vector Calculus

3.6 Application: Center of Mass 127
so
M xy =
=
=
∫ 2π ∫ π/2
0
∫ 2π
0
∫ 2π
0
= πa4
4 ,
0
a 4
8
sin2φdφdθ (since sin2φ=2sinφ cosφ)
(
− a4
16 cos2φ ∣ ∣∣∣ φ=π/2
φ=0
a 4
8 dθ
)
dθ
¯z= M πa 4
xy
M = 4
2πa 3
3
Thus, the center of mass of S is (¯x,ȳ,¯z)= ( )
0,0, 3a
8 .
= 3a 8 .
☛ ✟
✡Exercises
✠
A
For Exercises 1-5, find the center of mass of the region R with the given density functionδ(x,y).
1. R={(x,y) : 0≤ x≤2, 0≤y≤4},δ(x,y)=2y
2. R={(x,y) : 0≤ x≤1, 0≤y≤ x 2 },δ(x,y)= x+y
3. R={(x,y) : y≥0, x 2 +y 2 ≤ a 2 },δ(x,y)=1
4. R={(x,y) : y≥0, x≥0, 1≤ x 2 +y 2 ≤ 4},δ(x,y)= √ x 2 +y 2
5. R={(x,y) : y≥0, x 2 +y 2 ≤ 1},δ(x,y)=y
B
For Exercises 6-10, find the center of mass of the solid S with the given density functionδ(x,y,z).
6. S={(x,y,z) : 0≤ x≤1, 0≤y≤1, 0≤z≤1},δ(x,y,z)= xyz
7. S={(x,y,z) : z≥0, x 2 +y 2 +z 2 ≤ a 2 },δ(x,y,z)= x 2 +y 2 +z 2
8. S={(x,y,z) : x≥0, y≥0, z≥0, x 2 +y 2 +z 2 ≤ a 2 },δ(x,y,z)=1
9. S={(x,y,z) : 0≤ x≤1, 0≤y≤1, 0≤z≤1},δ(x,y,z)= x 2 +y 2 +z 2
10. S={(x,y,z) : 0≤ x≤1, 0≤y≤1, 0≤z≤1− x−y},δ(x,y,z)=1