Mathematical Methods for Physics and Engineering : A Comprehensive Guide

6.3 APPLICATIONS OF MULTIPLE INTEGRALS

the total volume of the tetrahedron is given by

V=

∫a

0

dx

∫b−bx/a

0

dy

∫c( 1 −y/b−x/a)

0

dz, (6.8)

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 paraboloidz=x^2 +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
dyperpendicular to they-axis, and then as shown in the figure we cut each slab into
horizontal strips of heightdz, and each strip into elemental boxes of volumedV=dx dy dz.
Integrating first with respect tox(adding up the elemental boxes to get a horizontal strip),
the limits onxarex=−

√

z−y^2 tox=

√

z−y^2. Now integrating with respect toz
(adding up the strips to form a vertical slab) the limits onzarez=y^2 toz=2y. Finally,
integrating with respect toy(adding up the slabs to obtain the required region), the limits
onyarey=0andy= 2, the solutions of the simultaneous equationsz=0^2 +y^2 and
z=2y. So the volume of the region is

V=

∫ 2

0

dy

∫ 2 y

y^2

dz

∫√z−y 2

−

√

z−y^2

dx=

∫ 2

0

dy

∫ 2 y

y^2

dz 2

√

z−y^2

=

∫ 2

0

dy

[ 4

3 (z−y

(^2) ) 3 / 2 ]z=2y
z=y^2 =

∫ 2

0

dy^43 (2y−y^2 )^3 /^2.

The integral overymay be evaluated straightforwardly by making the substitutiony=
1+sinu, and givesV=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 non-

uniform density. Symbolically, this mass is given simply by

M=

∫

dM,

wheredMis 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 justdM=ρdV,

wheredVis an element of volume andρis the variable density. For a laminar

body (i.e. a uniform sheet of material) the element of mass isdM=σdA,where

σis the mass per unit area of the body anddAis an area element. Finally, for

a body in the form of a thin wire we havedM=λds,whereλis the mass per