Mathematical Methods for Physics and Engineering : A Comprehensive Guide

11.6 VOLUME INTEGRALS

V

O

S

r

dS

Figure 11.9 A general volumeVcontaining the origin and bounded by the

closed surfaceS.

cannot be taken out of the integral sign as in (11.13) and must be included as

part of the integrand.

11.6.1 Volumes of three-dimensional regions

As discussed in chapter 6, the volume of a three-dimensional regionVis simply

V=

∫
VdV, which may be evaluated directly once the limits of integration have
been found. However, the volume of the region obviously depends only on the

surfaceSthat bounds it. We should therefore be able to express the volumeV

in terms of a surface integral overS. This is indeed possible, and the appropriate

expression may derived as follows. Referring to figure 11.9, let us suppose that

the originOis contained withinV. The volume of the small shaded cone is

dV=^13 r·dS; the total volume of the region is thus given by

V=

1

3

∮

S

r·dS.

It may be shown that this expression is valid even whenOis not contained inV.

Although this surface integral form is available, in practice it is usually simpler

to evaluate the volume integral directly.

Find the volume enclosed between a sphere of radiusacentred on the origin and a circular

cone of half-angleαwith its vertex at the origin.

The element of vector areadSon the surface of the sphere is given in spherical polar
coordinates bya^2 sinθdθdφˆr. Now taking the axis of the cone to lie along thez-axis (from
whichθis measured) the required volume is given by

V=

1

3

∮

S

r·dS=

1

3

∫ 2 π

0

dφ

∫α

0

a^2 sinθr·ˆrdθ

=

1

3

∫ 2 π

0

dφ

∫α

0

a^3 sinθdθ=

2 πa^3

3

(1−cosα).