The document provides 4 examples of calculating triple integrals over different regions. Example 1 calculates a triple integral over a region bounded by a paraboloid and plane in rectangular coordinates, then evaluates it in polar coordinates. Example 2 calculates a triple integral over a hemisphere in spherical coordinates. Example 3 finds the volume under a paraboloid and above a rectangle using a double integral. Example 4 calculates a triple integral over a tetrahedron bounded by 4 planes.

1. TRIPLE INTEGRALS
Example 1.Calculate the value of multiple integral E
y 2 z 2 dV where
2 2
E is bounded by the paraboloid x = 1 − y − z and plane x = 0.
Solution. In the rectangular coordinate system we can describe our region
E as
E = {(x, y, z)|(y, z) ∈ D, 0 ≤ x ≤ 1 − y 2 − z 2 }
where
D = {(x, y)|y 2 + z 2 ≤ 1}.
Therefore
2
−z 2
y 2 z 2 dV = y 2 z 2 x|1−y
0 dA = y 2 z 2 (1 − y 2 − z 2 )dA.
E D D
In order to evaluate the double integral over D it is better to use the polar
coordinate system. We set
z = rcosθ, y = rsinθ.
In the polar coordinate system we can describe the region D as
D = {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}.
So we have
2π 1 2π 1
2 2 2 2 2 2 2 sin2 (2θ)
y z dV = r cos θr sin θ(1−r )rdrdθ = (r5 −r7 ) drdθ =
E 0 0 0 0 4
2π 2π
r6 r8 sin2 (2θ) 1 1
( − ) |0 dθ = sin2 (2θ)dθ.
0 6 8 4 86 0
Note that
1 1
sin2 (2θ) = θ − sin(4θ).
2 8
Typeset by AMS-TEX
1

2. 1 1 1 π
y 2 z 2 dV = ( θ − sin(4θ))|2π =
0 .
E 86 2 8 86
Example 2.Calculate the value of multiple integral H
z 3 x2 + y 2 + z 2 dV
where H is the solid hemisphere with the center at the origin, radius 1, that
lies above xy-plane.
Solution.
Using the spherical coordinates we can describe the hemisphere H as
π
H = {(x, y, z)|0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ }.
2
Then we have
π
2 2π 1
3
z x2 + y 2 + z 2 dV = ρ3 cos3 φρ3 sinφdρdθdφ
H 0 0 0
π π
2π 1 2π
2
6 3
2 ρ7
= ρ cos φsinφdρdθdφ = cos3 φsinφ|1 dθdφ
0
0 0 0 0 0 7
π π
2π
1 2 2π 2 2π π
= cos3 φsinφdθdφ = cos3 φsinφdφ = − cos4 φ|02
7 0 0 7 0 28
2π
= .
28
Example 3. Find the volume of the given solid. Under the paraboloid
z = x2 + 4y 2 and above the rectangle R = [0, 2] × [1, 4].
Solution.
2 4 2
4
volume = (x2 +4y 2 )dA = (x2 +4y 2 )dydx = (x2 y+ y 3 )|4 dx =
1
R 0 1 0 3
2
(3x2 + 84)dx = (x3 + 84x)|2 = 8 + 168 = 176.
0
0
Example 4. Calculate the multiple integral T
ydV , where T is the
tetrahedron bounded by the planes x = 0, y = 0, z = 0 and 2x + y + z = 2.
Solution.
2

3. 1 2−2x 2−2x−y 1 2−2x
2−2x−y
ydV = ydzdydx = zy|0 dydx =
T 0 0 0 0 0
1 2−2x
(2y − 2xy − y 2 )dydx =
0 0
1 3 1
y 2−2x (2 − 2x)3
(y 2 − xy 2 − )| dx = ((2 − 2x)2 − x(2 − 2x)2 − dx =
0 3 0 0 3
1
4 1 1
(1 − x)3 dx = − (1 − x)4 |1 = .
0
3 0 3 3
3