« PreviousContinue »
the elemental slice LPKJR contained between two planes parallel to the coordinate plane of (y, z), and at an infinitesimal distance dx apart. In the case therefore of the volume being such as that of the figure, the limits of y are mk and 0, MK being the y to the trace of the surface on the plane of (x,y) and which may therefore be found in terms of x by putting z = 0 in the equation of the surface. Or if the volume is contained between two planes parallel to that of (x,z) and at distances Yn, Yo from it, y, and yo being constants, they are in that case the limits of y; and similarly must the limits be determined if the bounding surface is a cylinder whose generating lines are parallel to the axis of z; in all these cases the result of the y-integration is the volume of a slice contained between two planes at an infinitesimal distance apart, the length of which, viz. that parallel to the axis of y, is a function of its distance, from the plane of (y,z); and therefore if the length is expressed by F(x), v = f (x) dx,
(15) and the volume is the sum of all such differential slices taken between the assigned limits. Thus suppose in fig. 36 the volume contained in the octant to be required, and on = a, then the limits of x are a and 0: or suppose the required volume to be contained between two planes at distances X, and x, from the plane of (y,z), then X, and X, are the limits of x. The following examples how. ever of such cubatures will serve best to clear up any difficulties which arise from inadequate explanations of general principles.
258.] Examples of cubature.
Ex. 1. Find the content of a rectangular parallelepipedon, three of whose edges meeting at a point are a, b, c.
Let the point at which the edges a, b, c meet be the origin, and let the axes of x, y, z severally lie along them; then if v is the volume,
Ex. 2. Determine the volume in the first octant of a right elliptical cylinder whose axis is the x-axis, and whose altitude = a: and the equation to the generating ellipse is
c* y2 +62 m2 = 62 c.
Here the superior limit of z is î (62 — y2)), which I will call 2, and the inferior limit is 0; the limits of y are b and 0; and of x, a and 0.
To find the volume of the ellipsoid whose equation is
Hence the limits of integration for the volume in the first octant
fore for the volume in the first octant the limits are which I will call z, and 0); of y, c and 0; of x, a and 0. Therefore the volume required = 4 699*dz dy dx
= 4 [/0%. (a» — 22)dy dx
= 48*(2* —c*)dr = Tae c Ex. 5. To determine the volume contained within the surface of an elliptic paraboloid whose equation is
2 + 3 = 48, and a plane parallel to that of (y, z), and at a distance c from it.
Hence for the volume in the first octant the limits of integration are, for 2, (46x - y2), which I shall call Z, and 0; for y, 2(ax)}, which I shall call y, and 0; for x', c and 0 ; : the volume required = 4 [T['dz dy dx
= 467" (43)*x2–18yb dy de
Ex. 6. Find the volume of the solid cut from the cylinder x2 + y2 = a2 by the planes 2 = 0, and 2 = x tan a.
Here the limits of integration are, for 2, x tan a, which I will call z, and 0; for y, (a? — 22), which I will call y, and -Y; and for x, a and 0; so that, if v is the volume,
Ex. 7. If the general equation of the second degree represents a closed surface, explain the process by which the several limits of integration as required-in Art. 214 are determined." • Corresponding to given values of x and y there are generally two values of 2, say 2, and Zo, which are the limits of the z-integration, so that the z-integration gives the volume of a right prism which is perpendicular to the plane of (x, y), whose altitude iş 24-2, and whose base is dy dx. The subsequent y-integration gives the sum of all those which lie in a thin slice parallel to the plane of (y, z), the limits of this integration being fixed by those points in that slice at which the tangent plane of the closed surface is parallel to the z-axis, and at which consequently i del 6)=0; and thus if z is eliminated between F(x, y, z) = 0, and
6)=0, we shall have a quadratic in y, which will give two roots, say y, and y,; and these are the limits of the y-integration. The subsequent x-integration will give the sum of all these slices, and thus the volume included within the closed surface; and the limits of the x-integration are determined by the values of the x-ordinate at which these slices parallel to the plane of (y,z) vanish, and at which consequently the surface is touched by planes parallel to that of (y, z). At these points then p = 0, 6-) = 0, () = 0, and the elimination of y and z by means of these three equations will give a quadratic in terms of x, the two roots of which are the limits of the x-integration.
259.] If the infinitesimal element-function of a triple integral is of the form FC ), and the inequality which assigns the limits of integration is also of the form for ) < 1, so that, if i is the integral,
=N/ 63% ) dz dy da ; (16)
Jxo Jyo 20 then, if x = ať, y = bn, z =cs, 1 = abc F(8,9, 8)d5 dn dg,
(17) Jx, NY, Izo and the inequality which assigns the limits becomes f (5, 7, 8)<l; consequently if the definite integral given in (17) is known, that of (16) is also known.
Thus, in Ex. 3 of the preceding article, if x, y, z are replaced severally by aề, bn, cs, v = abc|| | dě dn df, and the equation to the ellipsoid, which assigns the limits of integration, becomes f? + m2 +62 = 1, which is the equation to a sphere whose
radius = 1. Now || a5dn dę for these limits expresses the volume of a sphere whose radius=1; that is, 47; and therefore
v = 4mabc.
F 3 Again, since the value of a right cone of altitude = c, de
Tac scribed on a circular base of radius = a, = " , that of a right cone of the same altitude on an elliptical base = "
260.] It is frequently convenient to refer the element of a volume to a mixed system of coordinates; as, for instance, to the z-axis, and to polar coordinates in the plane of (x,y); so that in the expression for the volume-element given in (12), dy dx must be replaced by r dr dd; and consequently v = |||dz r dr do ;
(18) in this case the base of the elemental prismatic column, which results from the z-integration, is the plane area-element whose area = rdrdo; and the volume of that column = zr dr do. Now if the r-integration is next effected, the sum of all these columns, o being constant, will be a sectorial slice, the limits of r being given by the trace of the surface on the plane of (x, y); and if the inferior limit of r is zero, the edge of the sectorial slice will coincide with the z-axis; and the final 6-integration will give the sum of all such sectorial slices, and consequently the volume required by the conditions of the problem. The following examples illustrate the process.
Ex. 1. The axis of a right circular cylinder of radius 6 passes through the centre of a sphere of radius a, b being less than a; find the volume of the solid bounded by the surfaces.
Let the plane passing through the centre of the sphere and perpendicular to the axis of the cylinder be that of (x,y); then the equations of the surfaces are x2 + y2 + z2 = a’, and x2 + y =6%; or, in terms of polar coordinates, the equation to the cylinder is, p= b.
For the volume therefore contained in the first octant the limits are, of z, (a? — 22 - y2); or (ao —2)}, which I will call 2, and 0; of r, 6 and 0); and 0; therefore