Page images
[ocr errors][ocr errors][ocr errors]

Ex. 2. Through the centre of a sphere of radius a passes the surface of a right circular cylinder the radius of whose directorcircle is half that of the sphere: find the volume of the solid bounded by the surfaces.

If the centre of the sphere is the origin, and the plane through it and perpendicular to the axis of the cylinder is that of (x, y), then the equations of the sphere and the cylinder are

202 + y2 + z2 = a?,. x2 + y2 = ax ; and if the angular point of the volume-element nearest to the origin is (z, r, 0), the equations to these surfaces may be put into the following forms, za = a? —p, r = a cos 0; so that if v is the required volume,

facoso (a? ---2)

dz r dr do

[ocr errors]
[ocr errors]


[ocr errors]
[ocr errors][ocr errors][merged small][ocr errors]

Ex. 3. To determine the volume contained within a surface of revolution, and two planes perpendicular to the axis of revolution.

Let the axis of revolution be the z-axis; then, by Vol. I, Art. 369, it is evident that the general equation of such surfaces is x2 + y2 = f(z); and if r is the distance of (x,y,z) from the z-axis, the equation to these surfaces is p2 =f(z). Now if the angular point of the volume-element nearest to the origin is (z,r, 0), and if r2 = f(2), and if z, and z, are the distances from the origin of the two bounding planes,

par er

[ocr errors]


which is the same result as that obtained in Art. 251. PRICE, VOL. II.

3 A

261.] The value of the volume-element given in (18), and the ensuing triple integration is readily applicable to those problems in which the trace of the bounding surface on the plane of (x,y) is a circle or other curve which is expressed in polar coordinates more simply than in rectangular coordinates. For other traces other systems of reference may be more convenient, and these must be left to the ingenuity of the student, the preceding principles being of sufficient breadth for all. I will however take another example, that of the cubature of the ellipsoid. Let v be the volume of the octant contained within the positive rectangular coordinate planes. Then, if z is the height of an elemental prism whose base is dy dx, z dy dx is the volume of that prism, and the double integral of this volume-element is the volume required. Now if the equation to the ellipsoid is

22 22 22

a2 T 72 + 2 = l, we may as in (38), Art. 238, replace as follows; let x=a sin a cos B, y= b sin a sin ß, z= c cosa; then dy dx = ab sin a cos a da ; so that

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

= abc :17 (cos a) sin a da dB

6 Ex. 4. The volume of the solid contained between the paraboloid x2 + y2 = 2 az, the cylinder x2 + y2 – 2 ax = 0, and the plane

3 77a3 of (x, y), = 7.

Ex. 5. The volume of the solid contained between the surface z = ae and the plane of (x,y), = n ac?.

262.] If the system of reference is that of polar coordinates in space, which has been explained in Art. 165, the volume-element will be expressed in the following form. Let E, fig. 42, be the point (r, 0, 0), and let u be the point (r+dr, 0+do, o+do); this latter point having been determined from the former by the following process. Let the meridian plane, in which E is, be drawn, and let oe be increased by the infinitesimal element EJ = dr; let the radius oe revolve in the meridian-plane through an infinitesimal angle EoF = do, so that EF =rdo, and the area

EFGJ = r dr do. Also let the meridian-plane revolve about oz through an infinitesimal angle nog=do, whereby the rectangle EFGJ will come into the position idhk, and each one of the angular points will have passed through a distance equal to EI or nq, neglecting infinitesimals of the highest orders; but NQ = On do = pdp = r sin o dd; and therefore the volume of the elemental parallelepipedon, of which E and h are opposite angular points, is på sin o drdo do ; and consequently by integration between assigned limits,

the volume = || | sin 0 dr do do. (20) The order in which the integrations are to be effected is scarcely arbitrary ; for the form of the equations to surfaces in most cases renders it necessary to integrate first with respect to r; and as the infinitesimal expression represents an element of a pyramid whose vertex is at o, and whose base is an element of the bounding surface, so does the r-integration produce the volume of the pyramid, complete or truncated, according as the inferior limit of r is 0 or a given finite quantity; and every solid admits of being resolved into such pyramids; but the order in which the 0- and ointegrations are effected is arbitrary; and if the 6-integration is effected next after that of r, it produces a sectorial or cuneiform slice of the solid, complete or truncated, according to the limits of r; and the subsequent p-integration will produce the aggregate of all such slices : or if the -integration is effected next after r, it produces a complete or truncated cuneiform conical slice, the aggregate of all which is found by the subsequent 6-integration.

The particular values of the limits of course may be different for every case, and are assigned by the geometrical conditions of the problem. If the origin is in the interior of a closed surface, the volume contained within which is to be determined, then, if the equation to the bounding surface is r =f(0, 0), this value, as the superior limit of the r-integration, it is convenient to express by r; whereby if v is the volume,


[ocr errors]

{f(0, 0)}3 sin 0 do do.

(22) The preceding expression for the volume-element in terms of polar coordinates, viz. på sin dr do do, may also be deduced by transformation from the expression of the volume-element, viz., dz dy dx, which is in terms of rectangular coordinates. The equations which connect rectangular and polar coordinates have already been given in (70), Art. 165, and the transformation has been made in (2), Ex. 2, Art. 213, the result being given in (38); so that it is necessary here only to call the attention of the reader to those equivalents.

263.] Examples illustrating the preceding process. Ex. 1. Determine the volume of a sphere.

Let the radius of the sphere = a; then the volume = /" [" /^r2 sin 0 dr do do

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

Ex. 2. Determine the volume of that part of a sphere which abuts at the centre, and is bounded by the ring contained between two parallels of latitude, for which the values of 0 are on and

[ocr errors][ocr errors][ocr errors][ocr errors][merged small]

evidently the volume of the solid common to the sphere and to a right circular cone whose vertical angle is 2 a, and whose centre is at the centre of the sphere.

Ex. 3. Determine the general expression of the volume included within a conical surface and a given base.

Let the origin be at the vertex of the cone, and let the axis of z fall within the conical surface; so that the equation to the surface is

= f(); then this equation, in terms of polar coordinates given by (70), Art. 165, becomes

tan 0 sin p = f(tan 6 cos 0);

(23) (24)

and let us suppose this to be put into the form

tan 0 = F (sin ). Now suppose the required volume to be contained between the conical surface, and a plane perpendicular to the z-axis at a distance c from the origin; then the limits of the r-integration are c sec 0 and 0; of the 6-integration, tan-F (sin o), which I shall call e, and 0; and of the o-integration 27 and 0; so that

(21 porc sece v=

g2 sin o dr do do

[ocr errors]
[ocr errors][ocr errors][ocr errors]

but by (24), CF (sin 0) = c tan 0 = the radius vector of the plane base of the cone drawn from the point where it is pierced by the axis of z; therefore by equation (12), Art. 226,

["{F (sin )}2 = the area of the base of the cone;

:: v= the area of the base of the cone ; consequently the volume of a cone is one-third of the right cylinder of the same base and of the same altitude.

« PreviousContinue »