# Volume integrals

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Volume integrals
```LINE, SURFACE AND VOLUME INTEGRALS
In particular, when the surface is closed Ω = 0 if O is outside S and Ω = 4π if O
is an interior point.
Surface integrals resulting in vectors occur less frequently. An example is
aﬀorded, however, by the total resultant force experienced by a body immersed in
a stationary ﬂuid in which the hydrostatic pressure is given by p(r)./The pressure
is everywhere inwardly directed and the resultant force is F = − S p dS, taken
over the whole surface.
11.6 Volume integrals
Volume integrals are deﬁned in an obvious way and are generally simpler than
line or surface integrals since the element of volume dV is a scalar quantity. We
may encounter volume integrals of the forms
φ dV ,
a dV .
(11.12)
V
V
Clearly, the ﬁrst form results in a scalar, whereas the second form yields a vector.
Two closely related physical examples, one of each kind, are provided by the total
mass of a ﬂuid contained in a volume V , given by V ρ(r) dV , and the total linear
momentum of that same ﬂuid, given by V ρ(r)v(r) dV where v(r) is the velocity
ﬁeld in the ﬂuid. As a slightly more complicated example of a volume integral we
may consider the following.
Find an expression for the angular momentum of a solid body rotating with angular
velocity ω about an axis through the origin.
Consider a small volume element dV situated at position r; its linear momentum is ρ dVṙ,
where ρ = ρ(r) is the density distribution, and its angular momentum about O is r × ρṙ dV .
Thus for the whole body the angular momentum L is
L = (r × ṙ)ρ dV .
V
Putting ṙ = ω × r yields
[r × (ω × r)] ρ dV =
ωr2 ρ dV − (r · ω)rρ dV . L=
V
V
V
The evaluation of the ﬁrst type of volume integral in (11.12) has already been
considered in our discussion of multiple integrals in chapter 6. The evaluation of
the second type of volume integral follows directly since we can write
a dV = i ax dV + j ay dV + k az dV ,
(11.13)
V
V
V
V
where ax , ay , az are the Cartesian components of a. Of course, we could have
written a in terms of the basis vectors of some other coordinate system (e.g.
spherical polars) but, since such basis vectors are not, in general, constant, they
396
11.6 VOLUME INTEGRALS
dS
S
r
V
O
Figure 11.9 A general volume V containing the origin and bounded by the
closed surface S.
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 region V is simply
V = V dV , which may be evaluated directly once the limits of integration have
been found. However, the volume of the region obviously depends only on the
surface S that bounds it. We should therefore be able to express the volume V
in terms of a surface integral over S. This is indeed possible, and the appropriate
expression may derived as follows. Referring to ﬁgure 11.9, let us suppose that
the origin O is contained within V . The volume of the small shaded cone is
dV = 13 r · dS; the total volume of the region is thus given by
0
1
r · dS.
V =
3 S
It may be shown that this expression is valid even when O is not contained in V .
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 radius a centred on the origin and a circular
cone of half-angle α with its vertex at the origin.
The element of vector area dS on the surface of the sphere is given in spherical polar
coordinates by a2 sin θ dθ dφ r̂. Now taking the axis of the cone to lie along the z-axis (from
which θ is measured) the required volume is given by
α
0
1 2π
1
r · dS =
dφ
a2 sin θ r · r̂ dθ
V =
3 S
3 0
0
α
2πa3
1 2π
dφ
a3 sin θ dθ =
=
(1 − cos α). 3 0
3
0
397
```
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