1. Calculate the volume of the solid defined by
2. Find the volume cut off from the paraboloid
by the plane z = h.
3. Find the volume cut off from the ellipsoid
by the plane lx+ my + nz = p.
4. (a) Show that if any closed curve q = f(j) is drawn on the surface
(r, q, j being polar
co-ordinates in space), the area of the surface
so enclosed is equal to the area enclosed by the projection of
the curve onto the sphere
r = a, the origin of co-ordinates being the
vertex of projection.
(b) Express the area by a simple integral.
(c) Find the area of the whole surface.
5. Find the area of the surface of the spheroid formed by rotating an ellipse about its major axis and show that, if the fourth and higher powers of the eccentricity e may be neglected, this area is equal to that of the sphere with volume equal to that of the spheroid.
6. Find the volume and surface area of the solid, generated by rotating the triangle ABO about the side AB.
7*. A tube-surface is generated by the spheres of unit radius the centres of which form the closed plane curve L. Prove that the area A of the surface is 2p times the length of L.
8*. (a) Calculate the volume of the region defined by
(b) Calculate the area of the spherical part of the boundary of this region, i.e. the area of the surface
9. Calculate the area of that part of the screw surface
for which
10. Calculate the area of the surface
In 4.2.7, we have already seen how the concept of mass is linked to that of a multiple integral. We shall now study some of the other concepts of Mechanics. We begin with a more detailed study of moment and of moment of inertia than was possible in Volume I, 10.6.6
4.7.1 Moments and Centre of Mass: The moment with respect to the xy-plane of a particle with mass m is defined as the product mz of the mass and the z-co-ordinate. Similarly, the moment with respect to the yz-plane is mx and that with respect to the zx-plane is my. The moments of several particles are combined by addition, i.e., the three moments of a system of particles with masses m1, m2, ··· , mn and co-ordinates (x1, y1, z1), ··· , (xn, yn, zn) are given by
If instead of with a finite number of particles we are dealing with a mass distributed continuously with density m=m(x,y,z) through a region in space or over a surface or curve, we define the moment of the mass-distribution by a limiting process, as in Volume I, A10.6.6, and thus express the moments by integrals. For example, with a distribution in space, we subdivide the region R into n sub-regions, imagine the total mass of each sub-region concentrated at any one of its points and then form the moment of the system of these n particles. We see at once that as n ® ¥ and at the same time the largest diameter of the sub-regions tends to zero, the sums tend to the limits
which we call the moments of the volume-distribution.
Similarly, if the mass is distributed over a surface S given by the equations x = f(u, v), y = y(u, v), z = c(u, v) with surface density m(u, v), we define the moments of the surface distribution by
Finally, the moments of a curve x(s), y(s), z(s) in space with mass density m(s) are
where s denotes the length of arc.
The centroid (centre of mass) of a mass of total amount M, distributed through a region R, is defined as the point with the co-ordinates
Hence, for a distribution in space, the co-ordinates of the centre of mass are given by
As a first example, we consider the uniform hemispherical region H with mass density 1:
The first two moments
are zero, since the integrations with respect to x and y yield the value zero. For the third
we introduce cylindrical co-ordinates (r, z, q) by means of the equations
and obtain
Since the total mass is 2p/3, the co-ordinates of the centre of mass are x = 0, y = 0, z = 3/8.
Next, calculate the centre of mass of a hemispherical surface of unit radius over which a mass of unit density is distributed uniformly. We calculate for the parametric representation
the surface element and find that
whence the three moments are
Since obviously the total mass is 2p, we see that the centre of mass lies at the point with co-ordinates x=0, y=0, z=1/2.
The generalization of the concept of moment of inertia is equally obvious. The moment of inertia of a particle with respect to the x-axis is the product of its mass and r² = y² + z², i.e., the square of the distance of the point from the x-axis. In the same way, we define the moment of inertia about the x-axis of a mass distributed with density m(x, y, z) through a region R by
The moments of inertia about the other axes are represented by similar expressions. Occasionally, the moment of inertia with respect to a point, say the origin, is defined by
and the moment of inertia with respect to a plane, say the yz-plane, by
Similarly, the moment of inertia of a surface distribution with respect to the x-axis. is given by
where m(u,v) is a continuous function of the two parameters u and v.
The moment of inertia of a mass, distributed with density m(x, y, z) through a region R , with respect to an axis parallel to the x-axis and passing through the point (x, h, z) is given by
If, in particular, we let (x, h, z) be the centre of mass and recall the relations for the co-ordinates of the centre of mass, we obtain directly the equation
Since any arbitrary axis of rotation of a body can be chosen as the x-axis, the meaning of this equation can be expressed as follows: T!ie moment of inertia of a rigid body with respect to an arbitrary axis of rotation is equal to the moment of inertia of the. body about a parallel axis through its centre of mass plus the product of the total mass and the square of the distance between the centre of mass and the axis of rotation (Steiner's theorem).
The physical meaning of the moment of inertia for regions in several dimensions is exactly the same as that already stated in Volume I, 5.2.9:
The kinetic energy of a body rotating uniformly about an axis is equal to half the product of the square of the angular velocity and the moment of inertia.
The following examples will serve to illustrate the concept and the actual calculation of the moment of inertia in simple cases. For the sphere V with centre at the origin, unit radius and unit density, we see that, by symmetry, the moment of inertia with respect to any axis through the origin is
If we add the three integrals, we obtain
or, if we introduce polar co-ordinates,
For a beam with edges a, b, c, parallel to the x-axis, the y-axis and the z-axis, respectively, with unit density and centre of mass at the origin, we find that the moment of inertia with respect to the xy-plane is
4.7.3 The Compound Pendulum: The above ideas find an application in the mathematical treatment of the compound pendulum, i.e., of a rigid body which oscillates about a fixed axis under the influence of gravity.
We consider a plane through G, the centre of
mass of the rigid body, perpendicular to the axis of rotation;
let this plane intersect the axis in the point O (Fig.
14). Then the motion of the body is obviously given, if we state
the angle j = j(t) which OG makes
at time t with the downward vertical line through O.
In order to determine this function j(t)
and also the period of oscillation of the pendulum, we must
assume a knowledge of certain physical facts. We make use of the law
of conservation of energy, which states that
during the motion of the body the sum of its kinetic and potential energies
remain constant. Here V, the potential energy of the body,
is the product Mgh, where M is the total mass, g
the gravitational acceleration
and. h the height of the
centre of mass above an arbitrary horizontal
line, for example, above the horizontal line through the lowest
position reached, by the centre of mass during the motion. If we
denote OG, the distance of the centre of mass from the
axis, by s, then V = Mgs(1 - cos j).The kinetic energy is given by 
where I is the moment of inertia of the body with respect
to the axis of rotation and we have written 
for
dj/dt.
Hence, the law of
conservation of energy yields the
equation
If we introduce the constant l = I/Ms, this is exactly the same as the equation previously found for the simple pendulum; accordingly, I is known as the length of the equivalent simple pendulum.
We can now apply directly the formulae found equation previously . The period of oscillation is given by
where j0 corresponds to the largest displacement of the centre of mass; for small angles, this is approximately
Naturally, the formula for the simple pendulum is included in this as a special case. In fact, if the whole mass M is concentrated at the centre of mass, then I = Ms² so that l = s.
Investigating further, we recall that I, the moment of inertia about the axis of rotation, is connected with I0, the moment of inertia about a parallel axis through the centre of mass, by the relation
whence
or, if we introduce the constant a = I0/M,
We see at once that in a compound pendulum l always exceeds s, so that the period of a compound pendulum is always longer than that of the simple pendulum obtained by concentrating the mass M at the centre of mass. Moreover, we note that the period is the same for all parallel axes at the same distance a from the centre of mass. In fact, the length of the equivalent simple pendulum depends only on the two quantities s and a = I0/M, whence it remains the same provided neither the direction of the axis of rotation nor its distance from the centre of mass is altered.
If we replace in the formula l = s + a/s the quantity s by as, i.e., if the axis is moved from the distance s to the distance a/s from the centre of mass, then l remains unchanged. This means that a compound pendulum has the same period of oscillation for all parallel axes which have the distance s or a/s from the centre of mass.
The formula 
shows at once that the period T
increases beyond all bounds as s tends to zero or to
infinity, whence it must have a minimum for some value s0.
Differentiating, we obtain
A pendulum with axis at a distance 
from the centre of
mass will be relatively insensitive to small displacements of the
axis. In fact, in this case dT/ds vanishes, so that
first-order changes in s produce only second-order changes
in T. This fact has been applied by Prof. Schuler
of Göttingen in the construction of very accurate
clocks.
4.7.4 Potential of Attracting Masses:. We have seen already that, according to Newton's law of gravitation, the force which a fixed particle Q with co-ordinates (x, h, z) and mass m exerts on a second particle P with co-ordinates (x, y, z) and unit mass is given, apart from the gravitational constant g, by
where 
is the distance between the points P and Q.
The direction of the force is along the line joining the two
particles and its magnitude is inversely proportional to the
square of the distance. If we now consider the force exerted on P
by a number of points Q1, Q2,
··· , Qn with respective
masses m1, m2,
··· , mn, we can express the
total force as the gradient of the quantity
where rn denotes the distance of the point Qn from the point P. If a fore can be expressed as a gradient of a function, it is customary to call this function the potential of the force; accordingly, define the gravitational potential of the system of particles Q1, Q2, ··· , Qn at the point P as the expression
We now assume that, instead of being concentrated at a finite number of points, the gravitating masses are distributed with continuous density m over a portion R of space or a surface S or a curve C. Then the potential of this mass-distribution at a point with co-ordinates (x, y, z) outside the system of masses is defined as
In the first case, the integration is taken throughout the region R with rectangular co-ordinates (x, h, z), in the second case over the surface S with the surface element ds and in the third case along the curve with arc length s. In all three formulae, r denotes the distance of the point P from the point (x, h, z) of the region of integration and m the mass density at the point (x,h,z).
Thus, for example, the potential at a point P with co-ordinates (x, y, z), due to a sphere K of constant density equal to unity, with unit radius and centre at the origin, is given by the integral
In all these expressions, the co-ordinates (x, y, z) of the point P do not appear as variables of integration, but as parameters, and the potentials are functions of these parameters.
In order to obtain the components of the force from the potential, we must differentiate the integral with respect to the parameters. The rules for differentiation with respect to a parameter extend directly to multiple integrals and, by 4.1.2, the differentiation can be performed under the integral sign, provided that the point P does not belong to the region of integration, i.e., provided that we are certain that there is no point of the closed region of integration for which the distance r has the value zero. Thus, for example, we find that the components of the gravitational force on unit mass due to a mass distributed with unit density throughout the region R in space are given by
Finally, we point out that the expressions for the potential and its first derivatives continue to have a meaning if the point P lies inside the region of integration. The integrals are then improper integrals and, as is easily shown, their convergence follows from the criteria of 4.6.2.
As an example, we shall calculate the potentials at an internal and at an external point, due to a spherical surface S with radius a and unit surface density. If we take the centre of the sphere as origin and make the x-axis pass through the point P (inside or outside the sphere), the point P will have the co-ordinates (x, 0, 0), and the potential will be
If we introduce polar co-ordinates on the sphere by means of the equations
then
If we now set put x² + a² - 2ax cos q = r², so that ax sin q dq = rdr, then, provided that x ¹ 0, the integral becomes
Hence we find for |x| > a
and for |x| < a
Thus, the potential at an external point is the same as if the entire mass 4pa² were concentrated at the centre of the sphere. On the other hand, throughout the interior of the sphere, the potential is constant. At the surface of the sphere, the potential is continuous; the expression for U is still defined (as an improper integral) and has the value 4pa. However, the force component Fx in the x-direction has a jump of magnitude -4p at the surface of the sphere, because, if |x| > a, we have
while Fx = 0 if |x| < a.
The potential of a solid sphere of unit density is found from the above by multiplying by da and then integrating with respect to a so that
is the value of the potential at an external point, which is again the same as if the total mass 4pa³/3 were concentrated at the centre.
1. Find the position of the centre of mass of the curved surface of a right cone.
2. Find the co-ordinates of the centre of mass of the portion of the paraboloid
cut off by the plane x = x0.
3*. A tube surface is generated by a family of spheres of unit radius with their centres in the xy-plane. Let S be a portion of the surface lying above the xy-plane and P the area of the projection of S onto the xy-plane. Prove that the z-co-ordinate of the centre of mass of S is equal to P /S.
4. Calculate the moment of inertia of the solid, enclosed between the two cylinders
and the two planes z = h and z = —h, with respect to (a) the z-axis, (b) the x-axis.
5. If A, B, C denote the moments of inertia of an arbitrary solid of positive density with respect to the x-, y-, z-axes, then the triangular inequalities
are satisfied.
6. Find the moment of inertia of the ellipsoid
with respect to the (a) z-axis, (b) an arbitrary axis through the origin, given by
7*. Find the envelopes of the planes with respect to which the ellipsoid
has the same moment of inertia h.
8. Let O be an arbitrary point and 8S an
arbitrary body. On every ray from O, we take the
point at the distance 
from O, where I denotes
the moment of inertia of S with respect to the straight
line coinciding with the ray. Prove that these points form an ellipsoid (the so called momental ellipsoid).
9. Find the momental ellipsoid of the ellipsoid
at the point (x, h, z).
10. Find the co-ordinates of the centre of mass of the surface of the sphere x² + y² + z² = 1, the density being given by
11. Find the x-co-ordinate of the centre of mass of the octant of the ellipsoid
12. A system of masses S consists of two parts S1 and S2; I1, I2 and I are the respective moments of inertia of S1, S2 and S about three parallel axes passing through the respective centres of mass. Prove that
where m1 and m2 are the masses of S1 and S2 and d is the distance between the axes passing through their centres of mass.
13. Calculate the potential of the ellipsoid of revolution
at its centre (b > a).
14. Calculate the potential of a solid of revolution
at the origin.
A4.1 THE EXISTENCE OF THE MULTIPLE INTEGRAL
A4.1.1 The Content of Plane Regions and Regions of Higher Dimensions: In order to obtain the analytical proof of the existence of the multiple integral of a continuous function, we must begin with a study of the idea of content.
In Volume I, 5.2.2, we saw how, in general, the content of a plane region can be expressed by an integral. Without employing that fact and considering the existence of area as guaranteed by intuition, we shall now proceed to give a general definition of the idea of content and investigate under what conditions this concept has a meaning.
We begin with a rectangle with sides parallel to the x- and y-axes and define the area of such a rectangle as the product of its base and height. If the given rectangle is subdivided into smaller rectangles by a number of parallels to the sides, it follows from this definition that the area of the rectangle is equal to the sum of the areas of all the sub-rectangles. The area of a region, which is composed of a finite number of rectangles, can now be defined as the sum of the areas of these rectangles.
Throughout this section, the word rectangle will always be understood to mean a rectangle with sides parallel to the axes.
The area thus defined does not depend on the way in which the region is subdivided (or resolved) into rectangles. In fact, if we are given two different resolutions, we can find a third resolution which is a finer subdivision of the two original ones. We do this by extending throughout the region all the lines which occur in either of the resolutions. These lines subdivide the two subdivisions into still smaller rectangles. The sum of the areas of these smaller rectangles is equal to the sum of the areas of the rectangles both of the first and the second resolution.
Now, in order to define the area of an arbitrary bounded region B, we form an inner and an outer approximation to the region, i.e., we find two regions Bi and Be , each consisting of rectangles, the region Bi lying entirely within B and the region Be containing B. For this purpose, we first enclose the region B in a large square. Then, we subdivide this square into small rectangles by drawing parallels to the axes. Those rectangles, having points in common with B, form together a region Be which encloses B; those rectangles, which lie wholly within B, form a region Bi which is contained in B.
We now wish to define the area C(B) of B in such a way that for every choice of Bi and Be the area of B lies between that of Bi and that of Be:
If we make the subdivisions finer, so that the diameters of the rectangles tend to zero, then the areas C(Bi) form a monotonic increasing sequence and the areas C(Be) a monotonic decreasing sequence. In fact, one can only add to the regions Bi rectangles and only remove rectangles from B, whence C(Bi) has a limit and so has C(Be). If these two limits are equal, we call this common limit the area of the region B.
Under what conditions are the two limits C(Bi) and C(Be) equal? Of course, the answer is: When the difference [C(Be)-C(Bi)] tends to zero as the fineness of the subdivisions increases. The region Be - Bi consists of those rectangles which have points in common with the boundary of B. Hence, if the area of this region Be - Bi tends to zero, it follows that the boundary of B can be enclosed in a region composed of rectangles and having as small an area as we please, namely in Be-Bi. Conversely, if the boundary of B can be enclosed in the interior of a region S, consisting of rectangles with a total area as small as we please and the subdivision is sufficiently fine, the rectangles Be-Bi will all lie in S; the area of Be-Bi will then be less than that of S, so that it tends to zero.
The result of these considerations follows: The limits of C(Bi) and C(Be) are equal if, and only if, the boundary of B can be enclosed in a region consisting of rectangles of total area as small as we please. In this case, our definition actually does assign a content to B.
From a geometrical point of view, it is somewhat unsatisfactory that in defining content we have singled out a particular co-ordinate system. However, as a matter of fact, there is no difficulty in showing that content is independent of the co-ordinate system, not only for two dimensions, but also for n dimensions. However, we shall omit this discussion here. In fact, on the one hand, it is not necessary for our particular purpose, which is the proof of the existence of the double integral; and, on the other hand, the fact that content is independent of the co-ordinate system follows immediately when we represent it by a multiple integral and recall that the transformation formula shows that the value of this integral is unchanged when new rectangular co-ordinates are introduced.
In 4.1.2, we shall prove the intuitively plausible fact that every sectionally smooth continuous curve (i.e., every continuous curve which has a continuously turning tangent except at a finite number of points) can be enclosed in a region formed from rectangles, the areas of which are as small as we please. The condition is therefore satisfied whenever the region B consists of a finite number of parts, each bounded by a finite number of sectionally smooth curves. Such regions have a unique area; others do not arise in practical applications.
We shall show in 4.1.2 that, if a
region B is subdivided by sectionally smooth curves, the
sum of the contents of the
sub-regions is equal to the content
of the entire region B. Hence, we shall
merely show that the present definition of area
agrees with the integral formulae obtained previously.
We begin by considering a region B, bounded by the x-axis,
the lines x = a, x = b and a
curve y=f(x). For the regions Be
and Bi, contained in and containing B,
respectively , we can take the regions composed
of-rectangles shown in Fig. 15 (Be by
dotted, Bi by continuous lines).
According to the definition of a simple integral in Volume
I, 2.1.2., the areas Bi and Be
are an upper sum 
and a lower sum 
, respectively, for
the integral 
. In addition to our formula
we have accordingly, by the definition of integral, the further inequality
Since lim C(Bi) = lim C(Be),
it follows, in agreement with what we have said already, that C(B)
= In
the case of an arbitrary region B, subdivision of a region
by lines parallel to the axes shows that our definition of content agrees with the
expression for area:
The present definition of area can immediately be extended to three-dimensional regions, and, in fact, to regions in n dimensional space. The content of a parallelepiped with sides parallel to the axes is defined as the product of the lengths of the three sides. We then extend the definition to regions composed of a finite number of such parallelepipeds. For an arbitrary region B, we then find regions Bi, composed of parallelepipeds lying inside B, and similar regions Be, containing B. The definition of the content of the region B as the common limit of the content of Be and that of Bi again has a meaning, provided that the boundary of the region B can be enclosed in a set of parallelepipeds of arbitrarily small total content. In A4.1.3, we shall show that this can always be done for regions bounded by surfaces having sectionally continuous tangent planes. As before, we shall in the sequel restrict ourselves to such regions. The word region is always to mean a bounded closed region with a boundary consisting of a finite number of surfaces with sectionally continuous derivatives.
The volume of a cylinder with its axis in the direction of the z-axis and base in the xy-plane is the product of the area of the base and the height. This is at once clear when the base consists of rectangles with sides parallel to the axes. In the general case, the cylinder can be enclosed between two cylinders the bases of which are regions composed of rectangles and the volumes of which differ from that of the given cylinder by arbitrarily small amounts. The theorem therefore holds for cylinders with any base. Hence, as before, the double integral
gives the volume of a portion of space bounded above by the surface z = f(x, y), below by the plane region B and at the sides by the vertical lines by which the edge of the surface is projected onto the boundary of B. Moreover, we see that the definition of volume for a general region in space R agrees with the integral
A4.1.2 A Theorem on Smooth Arcs: In discussing area, we have used the theorem: A continuous curve with a continuously turning tangent at all but a finite number of points can always be enclosed in a region composed of rectangles with sides parallel to the axes and arbi trarily small total content. It is obviously sufficient to prove the theorem for individual arcs with continuous tangents. Let such an arc be given by the equations
where the parameter s is the arc length and f(s) and y(s) are continuously differentiable functions. Then,
By the Mean value theorem of differential calculus, for any two values s and s1of s in the interval a £ s £ b, we have
Hence, if we subdivide the curve into n arcs of length e = (b— a)/n and denote the initial point of then-th arc by (xn, yn,) and an arbitrary point of that arc by (x, y), we have
Hence, all the points of the n-th arc lie in a square with side 2e and area 4e ². The entire curve is included in n such squares with at most total area
This quantity can be made as small as we please by taking e sufficiently small.
There is no difficulty in proving the corresponding theorem for surfaces in space defined by the equations
where the functions f, y, c have sectional continuous derivatives. It is found that every such surface can be enclosed in a region of arbitrarily small volume, consisting of a number of parallelepipeds.
A consequence of this theorem is that, if a plane region R, bounded by a sectional smooth curve, is subdivided into two sub-regions R', R", which are separated by sectionally smooth arcs, the area of R is equal to the sum of the areas of R' and R". In fact, we can subdivide the plane by straight lines parallel to the coordinate axes and so close together that all the rectangles, which have points in common with the boundary of B or with the arcs separating R' and R", have an arbitrarily small total area. As before, we define Re as the region consisting of all rectangles having points in common with R and Ri as the region consisting of all rectangles entirely within R; the regions Re', Ri', Re", Ri" are similarly defined. The regions Re" and Ri" together cover Re, some rectangles being counted twice; hence C(Re') + C(Re") ³ C(Re) ³ C(R). Again, Ri' and Ri" are contained in Ri and are completely separate; hence C(R) ³ C(Ri) ³ C(Ri') + C(Ri''). Since C(Ri') and C(Ri")can be made to approximate as closely as we desire to C(R') and C(R") by making the subdivision fine enough, the first of these inequalities yields C(R') + C(R") ³ C(R); similarly, the second C(R') + C(R") £ C(R). Taken together, these inequalities prove our statement.
It is clear that this addition theorem still holds when the region R is subdivided into any finite number of regions R(1),R(2),···,R(n). The extension to more than two dimensions follows the same lines and offers no difficulty.
A4.1.3 The Existence of the Multiple Integral of a Continuous Function:. Let the function f(x, y) be continuous in the interior and on the boundary of a region R. We wish. to show that, as the diameters of the sub-regions Rn tend to zero, the upper and lower sums
tend to a common limit which is independent of the mode of subdivision. The proof is essentially the same as the corresponding proof in Volume I, A2.1 and can therefore be given quite briefly here.
We first assume that the subdivision of R into sub-regions Rn is effected by polygonal paths. We choose the maximum diameter d of the sub-regions Rn so small that for every two points, the distance of which is less than d, the values of the function differ by less than e. Then, we have in each of these regions
Thus, we have for the difference between the upper and lower sums
Obviously, every subdivision, obtained by further subdividing the given subdivision, has a lower sum which lies between the upper and lower sums of the original subdivision.
The proof is complete once we have shown that, for every two subdivisions of R into sub-regions with diameters less than d, the corresponding upper and lower sums of the two subdivisions differ by as little as we please, provided only that d is chosen sufficiently small
If we are given a second subdivision into sub-regions Rn' with diameters less than d, then, in this subdivision, also the upper and lower sums will differ by less than eC(R):
The two subdivisions together define a new subdivision which is a further subdivision of each of the two and which is obtained by collecting the common points of each pair of regions Rn and Rm' (if such points exist) into a region Rnm ". By the previous remark, the lower sum of this third subdivision is not smaller than the lower sum of the two original subdivisions and differs from each of them by less than eC(R). Hence, the lower sums
differ by less than 2e C(R). If we now let e tend to zero, it follows from Cauchy's Convergence test that the lower sums have a limit independent of the mode of subdivision. Since we have already seen that the upper sums differ from the lower sums by as little as we please, the upper sums have the same limit. This proves the existence of the double integral
for polygonal subdivisions of R.
We have made this assumption in order to be sure that a common subdivision into a finite number of regions Rnm" really exists. For example, if the boundaries of the sub-regions are curves and a portion of a boundary curve in one subdivision consists of the line x = 0 and a portion of a boundary in the other one consists of the curve x² sin 1/x = y, then the common subdivision will have an infinite number of cells in the neighbourhood of x = 0. However, we can easily get rid of this assumption of polygonal subdivision. In fact, by A4.1.2, we can replace every curvilinear subdivision by a polygonal subdivision such that the total difference of the areas and hence the difference of the corresponding lower sums is arbitrarily small. This obviously reduces the case of sub-regions with arbitrary boundaries to the special case already discussed. Clearly, the proof is independent of the number of dimensions.
The corollaries on the existence of the double integral stated in 4.2.2 follow immediately from the approximation formula developed there and do not require a further proof.
The transformations in A4.1 enable us to give a simple proof of the following theorems:
If a straight-line segment S of constant or variable length l is in motion in a plane and t represents the time, then the area swept out by the moving segment is
where t0 and t1 are the initial and final positions of the segment S and dn/dt the component of the velocity of the mean centre of S in the direction perpendicular to S.
Again, the volume V swept out by a moving
plane area P of area A is
where dn/dt is the component velocity of the mean centre of the area A perpendicular to the plane of P.
In both these formulae and in the proofs, we assume, to begin with, that the moving segment S or plane area A passes once and only once through each point of the swept out region (Fig. 16).
We first give the proof in the case of a segment moving in a plane. The generating segment must be represented by an equation of the form
or else in the form obtained by solving this equation for
We first carry out the transformation of 
by means of the formula in A4.3.1 for
the special case
Denoting by ds the line element taken along the segment S, we obtain
for the area. It is easy to see, by substituting t = f(x, y) in (a) above and differentiating with respect to x and y, that
Hence the area is given by
Here a ', b', g' denote the derivatives of a, b, g with respect to t. The integration with respect to s is to be taken along the segment S.
The single integral with respect to s is equal to
where (X, Y) are the co-ordinates of the mean centre of S. However, X and Y satisfy the equation aX + bY + g = 0. Differentiation of this equation with respect to t yields
Thus,
Here a, b are
the components of the unit vector perpendicular to the segment S
and X', Y' the velocity components of the mean
centre at the time t. Thus, the expression a'X+ b'Y
+ g' is equal
to the velocity of the mean centre perpendicular to S.
This proves our formula.
This result can be shown to be intuitively plausible by the following argument. Consider two neighbouring positions of the segment S, say PQ and P'Q' (Fig. 17). These two segments determine an area which is given approximately by the product of the length PQ of S and the distance M'M of the mean centre of one segment from that of the other one. The error in this approximation is of higher order than that of the increment of time d t corresponding to the displacement. It would be an instructive example for the reader to try to fill in the details of this geometrical argument and provide a strict proof.
The corresponding theorem in three dimensions can be proved in the same way by the use of the transformation formulae for volume integrals in A4.3.2. There is no need to go through the proof here.
In the special case of a plane region, which is rotated about an axis while retaining its original size and shape, we have the problem already considered in Volume I, 5.2.8, where Guldin's rule for the volume of a solid of revolution was given.
Our formulae associate a definite sign with the area of the region swept out. In the two-dimensional case, the sign depends on which of the two directions normal to S is regarded as positive. (The same is true in three dimensions.) The area obtained is positive, if the segment S, as it passes through any point, moves in the direction of the positive normal; otherwise it is negative.
These observations allow us to extend our results to cases in which the segment or plane area does not always move in the same sense or covers part of the plane (or space) more than once. The integrals given above will then express the algebraic sum of the areas (or volumes) of the parts of the region described, each taken with the appropriate sign. We leave it to the reader to work out how this may be taken account in practice.
As an example, let a segment of constant length move
so as to have its end-points always on two fixed curves C
and C' in a plane as in Fig. 18. We can determine from the
arrows showing the positive direction of the normal the sign with
which each area appears in the integral and discover that the
integral gives the difference between the areas enclosed by C
and C'. If C' contains zero area, as it does when
it degenerates into a single segment of a curve,
multiply-described, the integral gives the area enclosed by C.
This principle is used in the construction of the well-known polar planimeter of Amsler, a mechanical device for measuring plane areas. It consists of a rigid rod at the centre of which is a measuring-wheel which can roll on the drawing paper. The plane of the wheel is perpendicular to the rod. When the instrument is to be used to measure the area enclosed by a curve C drawn on the paper, one end of the rod is moved around the curve, while the other is connected to a fixed point O, the pole, by means of a rigid member joined to it. Hence, this end of the rod describes (multiply) an arc of a circle - a closed curve containing zero area. It follows that the normal motion of the mean centre of the rod gives the area enclosed by C, apart from the constant multiplier l. But this normal component is proportional to the angle through which the measuring-wheel turns, provided the circumference of the wheel moves on the paper as the rod moves, in which case the position of the wheel is only affected by the motion normal to the rod.
In the standard instrument, the wheel is not exactly at the centre of the rod, but this only alters the factor of proportionality in the result and this factor can be determined directly by calibration of the instrument.
Let S be a Tube-surface, generated by a family of unit spheres with their centres on a closed curve C in the xy-plane. Prove that the volume enclosed by S is p times the length of C.
A4.3 VOLUMES AND AREAS IN SPACE OF ANY NUMBER OF DIMENSIONS
A4.3.1 Resolution of Multiple Integrals: If the region R of the xy-plane is covered by a family of curves f(x, y) = const in such a way that there passes through every point of R one, and only one, curve of the family, we can take the quantity f(x,y) = x as a new independent variable, i.e., we can take the curves represented by f(x,y) = const as a family of parametric curves.
For the second independent variable, we can take h = y, provided that we restrict ourselves to a region R in which the curves f(x,y) = const and y = const determine uniquely the points .
If we introduce these new variables, a double integral is transformed as follows:
If we keep x constant and integrate the right-hand side with respect to h, the integral with respect to h can be written in the form
Since
this integral may be regarded as an integral along the curve f(x, y) = x , the arc length s being the integration variable. Thus, we obtain the resolution
for our double integral. The intuitive meaning of this resolution is very easily recognized, if we assume that there is, corresponding to the curves f(x, y) = const, a family of orthogonal curves which intersect each separate curve f = const at right angles in the direction of the vector grad f. If the orthogonal curves are represented by the functions x(s) and y(s), where s is the arc length on them, then
Since
we obtain
We now consider the quadrilateral mesh bounded by the two curves f(x, y) = x, f(x, y) = x + Dx and two orthogonal curves which cut off a portion of length Ds from f(x, y) = x. The area of this mesh is given approximately by the product DsDs, and this, in turn, is approximately equal to
The transformation of the double integral
simply means: Instead of calculating the double integral by subdividing the region into small squares, we may use a subdivision determined by the curves f(x, y) = const. and their orthogonal curves.
A similar resolution can be effected in three-dimensional space. If the region R is covered by a family of surfaces f(x,y,z)=const in such a way that there pass through every point one, and only one, surface, then we can take x=f(x, y, z) as a variable of integration. In this way, we resolve a triple integral
into an integration
over the surface f = x and a subsequent integration with respect to x :
A4.3.2 Areas of Surfaces and Integration over Surfaces in more than Three Dimensions: In n-dimensional space, that is, in the region of sets of values with n co-ordinates, an (n — l)-dimensional surface is defined by
Let a portion of this surface correspond to a certain region B of the variables x1, x2, . . . , xn-1, where xn is to be calculated from the equation f( x1, x2, . . . , xn) = const.
We now define the area of this portion of surface as the absolute value of the integral
In the first instance, this definition is only a formal generalization of the formulae for the area obtained by intuition in the case of three dimensions. Nevertheless, it has a certain justification in the fact that the quantity A is independent of the choice of the co-ordinate xn. This may be proved in the same way as for the three-dimensional case in 4.6.3.
We define the integral of a function f(x1, x2, . . . , xn) over this (n - 1)-dimensional surface by
where, as before, we assume that xn is expressed in terms of x1, x2, . . . , xn-1 by means of the equation f(x1,x2,...,xn)=const. We again find that the expression is independent of the choice of the variable xn.
As in the case of two or three dimensions, a multiple integral over an n-dimensional region R
can be resolved as above. We assume that the region R is covered by a familiy of surfaces
is covered by a family of surfaces
in such a way that there passes through each point (x1, x2, . . . , xn) of R one, and only one, surface. If we introduce instead of x1, x2, . . . , xn-1, xn
as independent variables, the multiple integral becomes
A4.3.3 Area and Volume of the n-Dimensional Unit Sphere: As an example, calculate the area and volume of the sphere in n-dimensional space, i.e., the area of the (n - 1)-dimensional surface determined by the equation
and the volume inside the (n — l)-dimensional surface, which is the volume given by the inequality
Let there be given a continuous function f(r)
of 
inside the sphere. We shall first find the multiple
integral 
over the sphere 
We introduce the new
variable
and, by virtue of the relations
obtain the resolution
where Wn(r) is the area of the sphere
According to our general definition, the area of a hemisphere of radius r is given by the integral
where the integration is extended throughout the interior of the (n - l)-dimensional sphere
If we introduce instead of the variables xn the quantities
we obtain
where we denote the area of the unit sphere 
by
It then follows that
We can now calculate wn conveniently from this formula; we extend the integration on the left hand side throughout the entire x1x2 . . .xn-space (i.e., we let R increase beyond all bounds) and choose for f(r) a function for which both the n-tuple integral on the left hand side and the single integral on the right hand side can be evaluated explicitly . Such a function is
With this function, the equation becomes
,
we obtain
Here G
(n/2) denotes
the elementary expression 
, if n is even,
and 
if n is odd. By the general definition of
the G- function as well as by Volume I,
4.8.4, in order to find the volume of the n-dirnensional
unit sphere, we now set f(r) = 1 and obtain 
whence
???A4.3.4 Generalizations. Parametric Representations:. In n-dimensional space, we can consider an r-dimensional manifold for any r £ n and seek to define its content. For this purpose, a parametric representation is advantageous. Let the r-dimensional manifold be given by the equations
where the functions fn possess continuous derivatives in a region B of the variables u1, ··· , ur. As these range over this region, the point (x1, ··· , xn) describes an r-dimensional surface.
We now form deom the rectangular array
all possible r-rowed determinants
the first of which, for example, is the determinant
The content of the r-dimensional surface is then given by the integral
By means of the theorem on the transformation of multiple integrals and simple calculations with determinants, which we omit here, we can prove that this expression for the content remains unchanged if we replace the parameters u1, ··· , ur by other parameters. We likewise see that in the case r = 1 this reduces to the usual formula for the length of arc, in the case r = 2, in a space of three dimensions, it becomes the formula for the area.
We shall give a proof for the case r = n - 1, where n is arbitrary, i.e., we shall prove the theorem: If f(x1, ··· ,xn) = 0 is an arbitrary (n — l)-dimensional portion of a surface in n-dimensional space and this portion can also be represented parametrically by the equations
then its area is given by
where Di is a Jacobian with (n—1) rows:
As always, we assume here the existence and continuity of all the derivatives involved.
Without loss of generality, we may assume that 
.Then,
since, by A4.3.1,
A is given by
we must only show that
Now, by the properties of Jacobians, we have
This last Jacobian corresponds to the introduction of (x1, ··· ,xn-1, xn+1···
,xn) instead of (x1, ··· ,xn-1)
as independent variables. However, since the partial derivatives 
are
obtained from the equations
we have
whence
which proves the formula for A.
Note that the expression 
may be represented as a
determinant of (n— 1) rows
whence
The elements of the determinant here are the inner products of the vectors
i.e., the expressions
1. Calculate the volume of the n-dimensional ellipsoid
2. Express the integral I of a function of x1,
depending only on x1, over the unit sphere 
in
n-dimensional space as a single integral.
A4.4 IMPROPER INTEGRALS AS FUNCTIONS OF A PARAMETER
A4.4.1 Uniform Convergence. Continuous Dependence on the Parameter: Improper integrals frequently appear as functions of a parameter; thus, for example, the integral of the general power
in the interval -1 < x < 0 is an improper integral.
We have seen that an integral over a finite interval is continuous when regarded as a function of a parameter, provided that the integrand is continuous. However, in the case of an infinite interval, the situation is not so simple. For example, consider the integral
As x > 0 or x < 0, this is transformed by the substitution xy = z into
The integral 
converges, as we have seen. in Volume I, 4.8.5,
A8.1.2
and, in fact, it has the value p/2
(Volume
I, 9.5.1 and A4.4.3.
Thus, in spite of the fact that the function (sin xy)/y,
regarded as a function of x and y, is continuous everywhere and
its integral converges for every value of x, the function F(x)
is discontinuous;
it is equal top/2 for
positive values of x, -p/2 for
negative values of x and 0 for x= 0.
By itself, this fact is not at all surprising, for it is analogous to the situation which we have already encountered in the case of infinite series (Volume I, 8.4.4 and we must remember that the process of integration is a generalized summation. In the case of an infinite series of continuous functions, we required, if we were to be sure that the series represented a continuous function, that the convergence should be uniform. Here, in the case of convergent integrals depending on a parameter, we shall again have to introduce the concept of uniform convergence.
We say that the convergent integral
converges uniformly (in x) in the interval a £ x £ b, provided the remainder of the integral can be made arbitrarily small simultaneously for all values of x in the interval under consideration, and, more precisely, provided that for a given positive number e there is a positive number A = A(e), which does not depend on x and is such that whenever B £ A
As a useful test, we mention the fact that the integral
converges uniformly {and absolutely) if from a point y = y0 onward there holds the relation
where M is a positive constant and a > 1. In fact, in this case
the right-hand side can. be made as small as we please by choosing A sufficiently large, and it is independent of x. This is a straightforward analogue of the test for the uniform convergence of series.
We readily see that a uniformly convergent integral of a continuous function is itself a continuous function. In fact, if we choose a number A such that
for all values of x in the interval under consideration, we have
By virtue of the continuity of the function f(x, y), we can choose k so small that the finite integral on the right hand side is less than e, which proves the continuity of the integral.
A similar result holds when the region of integration is finite, but the integrand has a point of infinite discontinuity. For example, let the function f(x, y) tend to infinity as y ® a. We then say that the convergent integral
converges uniformly in a £ x £ b if we can find for every positive number e a number k such that
p
provided h £ k, where k does not depend on x. Uniform convergence in this sense occurs if there holds in the neighbourhood of the point y = a the relation
where, as before, M is a positive constant and n <. 1. Just as above, we show that in this case of uniform convergence F(x) is a continuous function.
If the convergence is uniform, the improper integrals F(x) are continuous in a certain interval, say in a £ x £ b. We can then integrate them over this interval and thus form the corresponding improper repeated integral
Naturally, we can consider instead of the finite interval a £ x £ b an infinite integration interval.