Applications of multiple integrals

6.3 APPLICATIONS OF MULTIPLE INTEGRALS
and let N → ∞ as each of the volumes ∆Vp → 0. If the sum S tends to a unique
limit, I, then this is called the triple integral of f(x, y, z) over the region R and is
written
f(x, y, z) dV ,
(6.5)
I=
R
where dV stands for the element of volume. By choosing the subregions to be
small cuboids, each of volume ∆V = ∆x∆y∆z, and proceeding to the limit, we
can also write the integral as
I=
f(x, y, z) dx dy dz,
(6.6)
R
where we have written out the element of volume explicitly as the product of the
three coordinate differentials. Extending the notation used for double integrals,
we may write triple integrals as three iterated integrals, for example,
x2 y2 (x) z2 (x,y)
dx
dy
dz f(x, y, z),
I=
x1
y1 (x)
z1 (x,y)
where the limits on each of the integrals describe the values that x, y and z take
on the boundary of the region R. As for double integrals, in most cases the order
of integration does not affect the value of the integral.
We can extend these ideas to define multiple integrals of higher dimensionality
in a similar way.
6.3 Applications of multiple integrals
Multiple integrals have many uses in the physical sciences, since there are numerous physical quantities which can be written in terms of them. We now discuss a
few of the more common examples.
6.3.1 Areas and volumes
Multiple integrals are often used in finding areas and volumes. For example, the
integral
dA =
dx dy
A=
R
R
is simply equal to the area of the region R. Similarly, if we consider the surface
z = f(x, y) in three-dimensional Cartesian coordinates then the volume under this
surface that stands vertically above the region R is given by the integral
V =
z dA =
f(x, y) dx dy,
R
R
where volumes above the xy-plane are counted as positive, and those below as
negative.
191
MULTIPLE INTEGRALS
z
c
dV = dx dy dz
dz
b
dx
y
dy
a
x
Figure 6.3 The tetrahedron bounded by the coordinate surfaces and the
plane x/a + y/b + z/c = 1 is divided up into vertical slabs, the slabs into
columns and the columns into small boxes.
Find the volume of the tetrahedron bounded by the three coordinate surfaces x = 0, y = 0
and z = 0 and the plane x/a + y/b + z/c = 1.
Referring to figure 6.3, the elemental volume of the shaded region is given by dV = z dx dy,
and we must integrate over the triangular region R in the xy-plane whose sides are x = 0,
y = 0 and y = b − bx/a. The total volume of the tetrahedron is therefore given by
V =
y
x
dy c 1 − −
b
a
0
0
y=b−bx/a
a 2
y
xy
=c
dx y −
−
2b
a y=0
0
a 2
abc
bx b
bx
=
=c
dx
−
+
.
2a2
a
2
6
0
z dx dy =
R
a
b−bx/a
dx
Alternatively, we can write the volume of a three-dimensional region R as
V =
dV =
R
dx dy dz,
(6.7)
R
where the only difficulty occurs in setting the correct limits on each of the
integrals. For the above example, writing the volume in this way corresponds to
dividing the tetrahedron into elemental boxes of volume dx dy dz (as shown in
figure 6.3); integration over z then adds up the boxes to form
the shaded column
in the figure. The limits of integration are z = 0 to z = c 1 − y/b − x/a , and
192
6.3 APPLICATIONS OF MULTIPLE INTEGRALS
the total volume of the tetrahedron is given by
a b−bx/a c(1−y/b−x/a)
V =
dx
dy
dz,
0
0
(6.8)
0
which clearly gives the same result as above. This method is illustrated further in
the following example.
Find the volume of the region bounded by the paraboloid z = x2 + y 2 and the plane
z = 2y.
The required region is shown in figure 6.4. In order to write the volume of the region in
the form (6.7), we must deduce the limits on each of the integrals. Since the integrations
can be performed in any order, let us first divide the region into vertical slabs of thickness
dy perpendicular to the y-axis, and then as shown in the figure we cut each slab into
horizontal strips of height dz, and each strip into elemental boxes of volume dV = dx dy dz.
Integrating first with respectto x (adding up the elemental boxes to get a horizontal strip),
the limits on x are x = − z − y 2 to x = z − y 2 . Now integrating with respect to z
(adding up the strips to form a vertical slab) the limits on z are z = y 2 to z = 2y. Finally,
integrating with respect to y (adding up the slabs to obtain the required region), the limits
on y are y = 0 and y = 2, the solutions of the simultaneous equations z = 02 + y 2 and
z = 2y. So the volume of the region is
2 2y
2 2y √z−y2
dy
dz √
dx =
dy
dz 2 z − y 2
V =
y2
0
2
dy
=
0
4
3
−
(z − y 2 )
z−y 2
3/2 z=2y
z=y 2
y2
0
2
=
0
dy 43 (2y − y 2 )3/2 .
The integral over y may be evaluated straightforwardly by making the substitution y =
1 + sin u, and gives V = π/2. In general, when calculating the volume (area) of a region, the volume (area)
elements need not be small boxes as in the previous example, but may be of any
convenient shape. The latter is usually chosen to make evaluation of the integral
as simple as possible.
6.3.2 Masses, centres of mass and centroids
It is sometimes necessary to calculate the mass of a given object having a nonuniform density. Symbolically, this mass is given simply by
M = dM,
where dM is the element of mass and the integral is taken over the extent of the
object. For a solid three-dimensional body the element of mass is just dM = ρ dV ,
where dV is an element of volume and ρ is the variable density. For a laminar
body (i.e. a uniform sheet of material) the element of mass is dM = σ dA, where
σ is the mass per unit area of the body and dA is an area element. Finally, for
a body in the form of a thin wire we have dM = λ ds, where λ is the mass per
193
MULTIPLE INTEGRALS
z
z = 2y
z = x2 + y 2
0
2
y
dV = dx dy dz
x
Figure 6.4 The region bounded by the paraboloid z = x2 + y 2 and the plane
z = 2y is divided into vertical slabs, the slabs into horizontal strips and the
strips into boxes.
unit length and ds is an element of arc length along the wire. When evaluating
the required integral, we are free to divide up the body into mass elements in
the most convenient way, provided that over each mass element the density is
approximately constant.
Find the mass of the tetrahedron bounded by the three coordinate surfaces and the plane
x/a + y/b + z/c = 1, if its density is given by ρ(x, y, z) = ρ0 (1 + x/a).
From (6.8), we can immediately write down the mass of the tetrahedron as
a
c(1−y/b−x/a)
x
x b−bx/a
dV =
ρ0 1 +
dx ρ0 1 +
dy
dz,
M=
a
a
0
0
R
0
where we have taken the density outside the integrations with respect to z and y since it
depends only on x. Therefore the integrations with respect to z and y proceed exactly as
they did when finding the volume of the tetrahedron, and we have
a
x bx2
bx b
.
(6.9)
dx 1 +
−
M = cρ0
+
a
2a2
a
2
0
We could have arrived at (6.9) more directly by dividing the tetrahedron into triangular
slabs of thickness dx perpendicular to the x-axis (see figure 6.3), each of which is of
constant density, since ρ depends on x alone. A slab at a position x has volume dV =
1
c(1 − x/a)(b − bx/a) dx and mass dM = ρ dV = ρ0 (1 + x/a) dV . Integrating over x we
2
5
abcρ0 . again obtain (6.9). This integral is easily evaluated and gives M = 24
194
6.3 APPLICATIONS OF MULTIPLE INTEGRALS
The coordinates of the centre of mass of a solid or laminar body may also be
written as multiple integrals. The centre of mass of a body has coordinates x̄, ȳ,
z̄ given by the three equations
x̄ dM = x dM
ȳ dM = y dM
z̄ dM = z dM,
where again dM is an element of mass as described above, x, y, z are the
coordinates of the centre of mass of the element dM and the integrals are taken
over the entire body. Obviously, for any body that lies entirely in, or is symmetrical
about, the xy-plane (say), we immediately have z̄ = 0. For completeness, we note
that the three equations above can be written as the single vector equation (see
chapter 7)
1
r̄ =
r dM,
M
where r̄ is the position vector of the body’s centre of mass with respect to the
origin, r is the position vector of the centre of mass of the element dM and
M = dM is the total mass of the body. As previously, we may divide the body
into the most convenient mass elements for evaluating the necessary integrals,
provided each mass element is of constant density.
We further note that the coordinates of the centroid of a body are defined as
those of its centre of mass if the body had uniform density.
Find the centre of mass of the solid hemisphere bounded by the surfaces x2 + y 2 + z 2 = a2
and the xy-plane, assuming that it has a uniform density ρ.
Referring to figure 6.5, we know from symmetry that the centre of mass must lie on
the z-axis. Let us divide the hemisphere into volume elements that are circular slabs of
thickness dz parallel to the xy-plane. For a slab at a height z, the mass of the element is
dM = ρ dV = ρπ(a2 − z 2 ) dz. Integrating over z, we find that the z-coordinate of the centre
of mass of the hemisphere is given by
a
a
ρπ(a2 − z 2 ) dz =
zρπ(a2 − z 2 ) dz.
z̄
0
0
The integrals are easily evaluated and give z̄ = 3a/8. Since the hemisphere is of uniform
density, this is also the position of its centroid. 6.3.3 Pappus’ theorems
The theorems of Pappus (which are about seventeen centuries old) relate centroids
to volumes of revolution and areas of surfaces, discussed in chapter 2, and may be
useful for finding one quantity given another that can be calculated more easily.
195
MULTIPLE INTEGRALS
z
a
√
a2 − z 2
dz
a
y
a
x
Figure 6.5 The solid hemisphere bounded by the surfaces x2 + y 2 + z 2 = a2
and the xy-plane.
y
A
dA
y
ȳ
x
Figure 6.6 An area A in the xy-plane, which may be rotated about the x-axis
to form a volume of revolution.
If a plane area is rotated about an axis that does not intersect it then the solid
so generated is called a volume of revolution. Pappus’ first theorem states that the
volume of such a solid is given by the plane area A multiplied by the distance
moved by its centroid (see figure 6.6). This may be proved by considering the
definition of the centroid of the plane area as the position of the centre of mass
if the density is uniform, so that
1
y dA.
ȳ =
A
Now the volume generated by rotating the plane area about the x-axis is given by
V = 2πy dA = 2πȳA,
which is the area multiplied by the distance moved by the centroid.
196
6.3 APPLICATIONS OF MULTIPLE INTEGRALS
y
ds
y
ȳ
x
Figure 6.7 A curve in the xy-plane, which may be rotated about the x-axis
to form a surface of revolution.
Pappus’ second theorem states that if a plane curve is rotated about a coplanar
axis that does not intersect it then the area of the surface of revolution so generated
is given by the length of the curve L multiplied by the distance moved by its
centroid (see figure 6.7). This may be proved in a similar manner to the first
theorem by considering the definition of the centroid of a plane curve,
ȳ =
1
L
y ds,
and noting that the surface area generated is given by
S=
2πy ds = 2πȳL,
which is equal to the length of the curve multiplied by the distance moved by its
centroid.
A semicircular uniform lamina is freely suspended from one of its corners. Show that its
straight edge makes an angle of 23.0◦ with the vertical.
Referring to figure 6.8, the suspended lamina will have its centre of gravity C vertically
below the suspension point and its straight edge will make an angle θ = tan−1 (d/a) with
the vertical, where 2a is the diameter of the semicircle and d is the distance of its centre
of mass from the diameter.
Since rotating the lamina about the diameter generates a sphere of volume 43 πa3 , Pappus’
first theorem requires that
4
πa3
3
Hence d =
4a
3π
and θ = tan−1
4
3π
= 2πd × 12 πa2 .
= 23.0◦ . 197
MULTIPLE INTEGRALS
a θ
d
C
Figure 6.8 Suspending a semicircular lamina from one of its corners.
6.3.4 Moments of inertia
For problems in rotational mechanics it is often necessary to calculate the moment
of inertia of a body about a given axis. This is defined by the multiple integral
I=
l 2 dM,
where l is the distance of a mass element dM from the axis. We may again choose
mass elements convenient for evaluating the integral. In this case, however, in
addition to elements of constant density we require all parts of each element to
be at approximately the same distance from the axis about which the moment of
inertia is required.
Find the moment of inertia of a uniform rectangular lamina of mass M with sides a and
b about one of the sides of length b.
Referring to figure 6.9, we wish to calculate the moment of inertia about the y-axis.
We therefore divide the rectangular lamina into elemental strips parallel to the y-axis of
width dx. The mass of such a strip is dM = σb dx, where σ is the mass per unit area of
the lamina. The moment of inertia of a strip at a distance x from the y-axis is simply
dI = x2 dM = σbx2 dx. The total moment of inertia of the lamina about the y-axis is
therefore
a
σba3
σbx2 dx =
I=
.
3
0
Since the total mass of the lamina is M = σab, we can write I = 13 Ma2 . 198