INTEGRATION IN THREE DIMENSIONS
In elementary calculus courses one learns how to integrate but most of the
discussion is confined to one and two dimensions or three dimensional problems
where only one or two variables are present. Let us here look in more detail at 3D
integration problems involving the independent variables x, y, and z. We start with
the volume V bounded by the ellipsoid-
x2
a2
+
y2
b2
z2
+
c2
= 1 with a, b, c
positive cons tan ts
and the z plane z=0. Mathematically one has-
a
b 1−
V = ∫∫∫ dxdydz = 4 ∫ dx
x =0
∫
x2
a2
c 1−
dy
y =0
x2
y2
a
b2
−
2
∫ dz
z =0
Note how the limits are constructed with the variables in the dz limit being x and y,
while those in the dy integral being only a function of x and those in the dx integral
having constant limits. Carrying out the integration one finds after one integration-
a
b 1−
V = 4c ∫ dx
0
∫
0
x2
a2

x 2 y 2 

−
dy 1 −
2

a
b 2 

Integrating two more times, we finda
x2
2π
V = bcπ ∫ (1 − )dx =
abc
2
3
a
x =0
Note that when a=b=c=R one just obtains the volume of a hemisphere of radius R.
A bit more complicated triple integration involves finding the mean density of a
hemisphere of radius R=1 whose variable density goes as ρ=1-x2-y2 . Here one has-
1
ρ = ∫∫∫
2
∫ dx
2
(1 − x − y )dxdydz x = −1
=
∫∫∫ dxdydz
1− x 2
(1 − x 2 − y 2 )dy
∫
y = − 1− x
1− x 2 − y 2
∫ dz
z =0
2
2
π
3
which integrates to-
5 1 1
3
ρ = B( , ) ∫ (1 − x 2 ) 2 dx =
π 2 2 x =0
5
3
The mass of the hemisphere will be 2π/5. Whenever the surfaces of a volume have
the form of a cylinder, sphere, ellipsoid, or a hyperboloid, it is often more
convenient to use a coordinate system suited for the surface. This will generally
reduce the effort required to determine the volume. In the above example one can
employ a cylindrical set of coordinates and write-
1
1− z 2
3 2π
3 1
3
2
4
(
1
)
ρ=
d
θ
dz
−
r
rdr
=
∫
∫
∫
∫ (1 − z )dz =
2π θ = 0 z = 0 r = 0
4 z =0
5
This is a little easier calculation.
Another 3D calculation involves the volume of the funnel shown-
It can be defined as r=sqrt(x2 + y2 )=z4 with 0.5<z<1. Its volume expressed in
Cartesian and in cylindrical coordinates reads-
1
z4
V = 4 ∫ dz ∫ dx
1/ 2 0
z8 − x2
and V =
∫ dy
0
1
2π
z4
∫ dz ∫ dθ ∫ rdr
z = 0.5 θ = 0
r =0
Clearly the simpler calculation is found with the second integral and reads1
V = π ∫ z 8 dz =
1/ 2
π
1  511π
1
−
=
9  29  4608
Next consider the volume of a torus. This doughnut shaped solid has its major
radius equal to ‘a’ and its minor radius equal to ‘b’. A cross-section of a vertical cut
through the torus looks as shown-
This time we use a cylindrical coordinate description with the ring of cross-section
dA having a volume-
dV = 2π (a + r cos(θ )rdrdθ
On summing up all these rings, one findsb
2π
r =0
θ =0
V = 2π ∫ rdr ∫ [a + r cos(θ )]dθ = 2π 2 ab 2
Notice that this same result may also be obtained via one of the Pappus Theorems
encountered in solid mechanics which states that the volume of a rotationally
symmetric solid equals the product of the circumferential path length of its centroid
(2πa) and its cross-sectional area (πb2).
Next let us examine the volume of the two parts of a sphere x2+y2+z2=a2 when cut by
the plane z=b with a>b. A computer generated view of the problem follows-
This time it will be most convenient to use spherical coordinates. We have for the
cap showing above the plane a volume-
2π
arccos(b / a )
ϕ =0
θ =0
V = ∫ dϕ
∫
a
1
sin(θ )dθ ∫ r 2 dr − πb(a 2 − b 2 )
3
r =0
where φ is the azimuthal angle and θ the polar angle. A simple integration yields-
V=
π
3
( a − b ) 2 ( 2 a + b)
Thus the cap has zero volume when b=a and a volume of a half-sphere when b=0.
One could also solve this problem using Cartesian coordinates but it would be more
complicated.
As a final problem consider the volume between the two slanted planes x+y+z=1 and
x+y+2z=1 and the planes x=0 and y=0. We are here dealing with a four-sided solid
with six edges. The simplest way to find its volume is to subtract two tetrahedrons
from each other. The first tetrahedron has volume1
1− x 1− x − y
V1 = ∫ dx ∫ dy
0
0
∫ dz =
0
1 1
1
2
∫ (1 − x) dx =
2 x =0
6
while the second volume is-
1
1
(1− x − y )
1− x 2
V2 = ∫ dx ∫ dy
0
0
∫ dz =
0
1
11
2
∫ (1 − x) dx =
12
40
Hence the volume of the solid becomes V=V1-V2=1/12. By recalling that the volume
of a pyramid equals one-third its base area times its height, one can also deduce this
result without any integration.