As an example of the use of triple integrals, consider a solid occupying a set of points U ⊆ ℝ^{3} having
density ρ. Thus ρ is a function of position and the total mass of the solid equals ∫_{U}ρdV . This is just like
the two dimensional case. The mass of an infinitesimal chunk of the solid located at x would be ρ
(x )
dV
and so the total mass is just the sum of all these, ∫_{U}ρ
(x)
dV .
Example 14.4.1Find the volume of R where R is the bounded region formed by the plane
15
x + y +
15
z = 1 and the planes x = 0,y = 0,z = 0.
When z = 0, the plane becomes
1
5
x + y = 1. Thus the intersection of this plane with the xy plane is this
line shown in the following picture.
PICT
Therefore, the bounded region is between the triangle formed in the above picture by the x axis, the y
axis and the above line and the surface given by
15
x + y +
15
z = 1 or z = 5
( ( ))
1− 15x + y
= 5 − x − 5y.
Therefore, an iterated integral which yields the volume is
∫ 5∫ 1− 1x∫ 5−x−5y
5 dzdydx = 25.
0 0 0 6
Example 14.4.2Find the mass of the bounded region R formed by the plane
1
3
x +
1
3
y +
1
5
z = 1
and the planes x = 0,y = 0,z = 0 if the density is ρ
(x,y,z)
= z.
This is done just like the previous example except in this case, there is a function to integrate. Thus the
answer is
∫ 3∫ 3−x∫ 5− 53x− 53y
zdzdydx = 75 .
0 0 0 8
Example 14.4.3Find the total mass of the bounded solid determined by z = 9 − x^{2}− y^{2}andx,y,z ≥ 0 if the mass is given by ρ
(x,y,z)
= z
When z = 0 the surface z = 9 −x^{2}−y^{2} intersects the xy plane in a circle of radius 3 centered at
(0,0)
.
Since x,y ≥ 0, it is only a quarter of a circle of interest, the part where both these variables are
nonnegative. For each
(x,y)
inside this quarter circle, z goes from 0 to 9 −x^{2}−y^{2}. Therefore, the iterated
integral is of the form,