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If a solid be referred to rectangular co-ordinates its volume (V) is found by integrating the triple integral Sffdx dydz. If we integrate first with respect to , and suppose the integral to begin when z = 0,

V = ${xdvdy is the volume, z being given in terms of x and y by the equation to the surface, and the integrals being taken between the proper limits, which differ according to the nature of the surfaces which bound the surface laterally. The most simple case is when the solid is bounded laterally by four planes, two parallel to the plane of wz, and two to that of yz. The limits of x and y being then constant are independent of each other, and the integration may be easily effected. But if the surface be terminated laterally by the curve surface, the extreme values of the variables are connected together by a relation derived from the equation to the surface by making % = 0. If we integrate first with respect to y, and if the

equation to the surface give us, on making x = 0, a relation
f(x,y) = 0 between Q and y, we have to take the values of
y in terms of x derived from this equation as the limits of
the integral with respect to y. There remains now only to
integrate a function of x, and to take it between the limits
of the value of x derived from the equation to the surface
by making x= 0 and y = 0.
. If we wish to find the volume of the solid terminated
laterally by a cylinder perpendicular to the plane of xy,
having for its base any curve as LL'NN' (fig. 54), we take
the integral with respect to y from y = MN to y = MN',
which are given in terms of x by the equation to the cylinder ;
and then we integrate with respect to v between the limits
of that variable corresponding to the extreme points of the
curve which is the base of the cylinder, such as HK and H'K'
in the figure. It is to be observed that in getting the limit-
ing values of y in terms of xwe introduce into the integral
new functions which may often render the formula unin-
tegrable.

(1) To find the volume of the octant of the ellipsoid

We have V = SSfdæ dy dx = $fx dx dy, the limits of being 0 and its value given in terms of x and y by the equation to the surface, that is,

*=0(1-. - ? Therefore V= c ssdæ dy (1- )?

Integrating with respect to y, we have v=sdo {v (1-) +(1-) sin-'olor 2 cm)

Now fody represents the area of a section parallel to the plane yx, and at a distance equal to x: the integral must

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therefore be taken between the limits given by that section, or from y = 0 to y = b (1-): This gives

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The limits of x are 0 and a, so that we have

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and the volume of the whole ellipsoid, being eight times this quantity, is 4 a abc. When a = b = c, the ellipsoid becomes a sphere, the volume of which consequently is a'.

(2) The equation to the elliptic paraboloid being

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On integrating with respect to y from y=0 to y=(26x)}, this gives

V = (ab): fx dx = *** (ab)}; which taken from x = 0 to X = c, and multiplied by 4, gives a (ab) as the volume of the paraboloid intercepted between the vertex and a plane parallel to the plane of yz at a distance c. (3) The equation to the Cono-Cuneus of Wallis is

cx = y* (a– xo); the whole volume is

V = mac. (4) The general equation to conical surfaces is (the origin being at the vertex and the axis of x being the axis of the cone)

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Therefore V = ssdx dy xo (*), or putting y = ax and therefore dy = x da, we have

V = Ssdæ da xod(a) = Sda (a). If the base be a plane curve parallel to yx, the limits of a are 0 and a, so that

Vasda o(a). Now the equation to the base is found by making w=a in the equation to the surface, which then becomes z=a0 (2),

and, as in this case, y = a a, dy = ada,

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Now as z is the ordinate of the base, sædy is its area, so that the volume of the cone is one third of its base multiplied into its altitude.

(5) The axes of two equal right circular cylinders intersect at right angles; find the volume of the solid common to both.

Taking the intersection of the axes of the cylinders as the origin, and their axes as the axes of y and x, their equations are

«? + z = a’, m? + y = a,

V = Sdx dyz = S/dx dy (a – m?)!.
Integrating with respect to y from y = 0 to y = (a’ – ?),

V = sdx (a’ – 2”);
and integrating with respect to a from x = 0 to w = a,

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This is the eighth part of the whole intercepted solid, which is therefore

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(6) The axis of a right circular cylinder passes through the centre of a sphere; find the volume of the solid which is common to both surfaces.

Taking the centre of the sphere as origin, and the axis of the cylinder as the axis of , the equations to the surfaces are

? + y2 + m2 = a, x2 + y = 6?. The direct integration of sfxdxdy in terms of x and y, leads to operations of considerable complexity which may be avoided by transforming x and y into polar co-ordinates r and : in which case by Chap. III. Sect. 2. of the Diff. Calc., we have

dx dy = rdr do, and

V = fzrdr d0 = Sdrdor (a’ – 52)}; the limits of o being 0 and 27 ; and those of r being O and b. llence V = 27 f.dr r (a’ – po?)?

=**{«® – (a – b)}}: and the whole volume of the included solid being double this quantity is **{a– (abo)!}.

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The volume of the sphere is – a', and therefore the volume of the solid intercepted between the concave surface of the sphere and the convex surface of the cylinder, is

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(7) A sphere is cut by a cylinder, the radius of whose base is half of that of the sphere, and whose axis bisects the radius of the sphere at right angles ; find the volume of the solid common to both surfaces. The equations to the surfaces in this case are

x? + y + m2 = a’, and x2 + y = ax. Transforming w and y into polar co-ordinates as in the last example, we have

V = S/drdor x = Ssdr dor (a’ – j?)};

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