Calculus- Early Transcendentals, 2021a

506 Multiple Integration
∫ 1
= 1 x − 3x 2 + 3x 3 − x 4 dx = 1
6 0
120 .
We compute moments as before, except now there is a third moment:
M xy =
M xz =
M yz =
∫ 1 ∫ 1 ∫ y−x
0 x 0
∫ 1 ∫ 1 ∫ y−x
0 x 0
∫ 1 ∫ 1 ∫ y−x
0
x
0
xz 2 dzdydx = 1
360 ,
xyzdzdydx= 1
144 ,
x 2 zdzdydx= 1
360 .
Finally, the coordinates of the center of mass are ¯x = M yz /M = 1/3, ȳ = M xz /M = 5/6, and ¯z = M xy /M =
1/3. ♣
Exercises for 14.5
Exercise 14.5.1 Evaluate
Exercise 14.5.2 Evaluate
Exercise 14.5.3 Evaluate
Exercise 14.5.4 Evaluate
Exercise 14.5.5 Evaluate
Exercise 14.5.6 Evaluate
Exercise 14.5.7 Evaluate
Exercise 14.5.8 Compute
∫ 1 ∫ x ∫ x+y
0
0
0
∫ 2 ∫ x 2 ∫ y
0 −1 1
∫ 1 ∫ x ∫ lny
2x + y − 1dzdydx.
xyzdzdydx.
0 0 0
∫ π/2 ∫ sinθ ∫ r cosθ
0
0
e x+y+z dzdydx.
0
∫ π ∫ sinθ ∫ r sinθ
0
0
0
∫ 1 ∫ y 2 ∫ x+y
0
0
0
∫ 2 ∫ y 2 ∫ ln(y+z)
1
y
0
∫ π ∫ π/2 ∫ 1
0
0
0
r 2 dzdrdθ.
r cos 2 θ dzdrdθ.
xdzdxdy.
e x dxdzdy.
zsinx + zcosydzdydx.
Exercise 14.5.9 For each of the integrals in the previous exercises, give a description of the volume (both
algebraic and geometric) that is the domain of integration.
∫ ∫ ∫
Exercise 14.5.10 Compute x + y + zdV over the region inside x 2 + y 2 + z 2 ≤ 1 in the first octant.