Multivariate Calculus - Bruce E. Shapiro

162 LECTURE 20. TRIPLE INTEGRALS
Solution We integrate first over z, then over y, then over x, because that is the order
of the “d”s inside the integral.
∫ 2 ∫ 3 ∫ 2x−y
∫ 2 ∫ 3
(∫ 2x−y
)
dzdydx =
dz dydx
0
−2
Example 20.2 Find
0
∫ π/2 ∫ z ∫ y
0
0
0
=
=
=
=
=
0 −2
∫ 2 ∫ 3
0 −2
∫ 2 ∫ 3
0
∫ 2
0
∫ 2
0
∫ 2
0
−2
(
0
z| 2x−y
0
)
dydx
(2x − y)dydx
(2xy − 1 2 y2 )∣ ∣∣∣
3
dx
−2
[6x − 9/2 + 4x + 2]dx
(10x − 5/2)dx
= [ 5x 2 − (5/2)x ] 2
0
= 5(4) − (5/2)(2) = 15
sin(x + y + z)dxdydz
Solution We integrate first over x, then over y, then over z. Since ∫ sin udu =
− cos u,
∫ π/2 ∫ z ∫ y
∫ π/2
[∫ z
(∫ y
) ]
sin(x + y + z)dxdydz =
sin(x + y + z)dx dy dz
0
0
0
=
=
0
∫ π/2
0
∫ π/2
0
0
0
[∫ z
]
(− cos(x + y + z)| y 0 ) dy dz
0
[∫ z
Next we integrate over y, using the fact that ∫ cos udu = sin u
∫ π/2 ∫ z ∫ y
0
0
0
sin(x + y + z)dxdydz =
=
=
=
∫ π/2
0
∫ π/2
0
∫ π/2
0
∫ π/2
0
[∫ z
0
0
]
(− cos(2y + z) + cos(y + z)) dy dz
]
(− cos(2y + z) + cos(y + z)) dy dz
[− 1 2 sin(2y + z) + sin(y + z) ]∣ ∣∣∣
z
[(− 1 2 sin 3z + sin 2z )
−
[− 1 2 sin 3z + sin 2z − 1 2 sin z ]
dz
dz
0
(− 1 2 sin z + sin z )]
dz
Revised December 6, 2006. Math 250, Fall 2006