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Ex. 7. The whole surface of a prolate spheroid, the equation to whose generating ellipse is

1. x2 u2
O2 +

... 27 ab in-1 and whose eccentricity is e, is 27 62+ sin-le.

е Ex. 8. The area of a surface generated by the revolution of a logarithmic curve, y = e*, about the axis of x is equal to

{y(1+y?) + log(y+(1+y2))}. Ex. 9. The whole area of the surface generated by the revolution of a cycloid about its axis is 8

233.] If the line about which the generating plane curve revolves is the axis of y, then, see fig. 34, if om = x, MP = y, PQ = ds, the convex surface of the band generated by PQ in one revolution is equal to 21 x ds, and as this is an infinitesimal bandelement of the required surface,

the whole surface = 2* | « dx ; (19) the integral being definite, and the limits being assigned by the problem.

Ex. 1. To determine the surface of an oblate spheroid.
Let the equation to the revolving ellipse be

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Ex. 2. To determine the area of the surface generated by a given length of the catenary revolving about the axis of y, when the equation is yzfé + e-ál.

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234.] If the curve, whose equation is y=f(x), generates a surface by revolving about, not one of its axes of reference, but an axis parallel to, say, its axis of x, at a distance a from it, and in the plane of the curve, then the surface generated


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XOXO X and X, being the abscissæ corresponding to the extremities of the generating curve; and therefore if s is the length of the generating arc, and s' is the area of the surface generated by the revolution of it about its axis of x,

the required surface = 2ta s+s'. If the generating curve is a closed figure, such as that drawn in fig. 39, and capable of being divided into two equal and symmetrical parts by a line EBC which is its axis of x, then, if AB=a, and the equation to EPC is y=f(x), EC being its axis of x, the surface generated by the revolution of EPC about ox

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= 27/ '{a+f(x)}ds ; and that generated by the revolution of Eric about the same line

= 2* | **{af(x)} ds ; therefore the surface generated by the revolution of the closed

figure EPCP

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which is equal to 47 a s; that is, to 271a x length of the generating curve; therefore the area of the surface generated by the revolution about an axis in its own plane of a closed curve which is symmetrical with respect to a line parallel to that about which it revolves, is equal to the product of the length of the curve and the path described by a point on the line of symmetry.

Hence if a circle of radius a revolves about an axis in its own plane at a distance c from its centre, the surface of the generated ring = 47° ac.

235.] If the surface is generated by a curve referred to polar coordinates, the area of it may be determined as follows. Let the axis of revolution be the prime-radius, and from P which is (r, ), see fig. 26, let pm be drawn at right angles to sx. Then PM = r sin 0; and the arc-element PQ, which = ds, will by its revolution about sx describe an infinitesimal band, whose breadth is ds, and whose circumference = 2 r sin 0. Consequently 2ar sin 0 ds is the band-element of the surface, and the surface = 2 ar sin 0 ds ;

(21) this integral is of course definite, and the limits are assigned by the geometrical conditions of the particular problem.

Ex. 1. Find the area of the surface of a spherical sector, 2 a being the vertical angle of the sector, and a being the radius of the sphere. The surface = 1 27 asin 0 do

= 47 a? (sinm)". Consequently, if a = 7, the whole surface of a sphere = 4ta2.

Ex. 2. Find the area of the surface generated by the revolution of a loop of the lemniscata about its axis. Since pe = a* cos 20, r sin 0 ds = asin 0 do ;

the surface = 2 1a*sin o d

= na? {2–25}. Ex. 3. The area of the surface generated by the revolution of

32 na the cardioid, whose equation is r = a (1+cos O), is ~

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Ex. 4. The area of a surface generated by the revolution of a loop of the lemniscata, whose equation is g2 = asin 20, is 27a2.

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SECTION 4.-Quadrature of Curved Surfaces. 236.] We now come to the problem of Quadrature in its most general form, only particular cases having been investigated in the preceding articles. It will be observed also that in the preceding section, the integrals, on which the quadrature depends have been single, whereas the quadrature of an area must involve a double integral: the revolution of the arc-element about the axis is however equivalent to one integration.

Let the equation to the surface, on which the area, whose quadrature is to be determined, lies, be F(x, y, z) = 0;

(22) and employing the same notation as in Art. 332, Vol. I., let : ) = v, (s) = v, (4-1) = w; (23) U2 + 12 + 2 = Q?;

(24) so that if a, b, y are the direction angles of the normal at (x, y, z), cos a = cos B = = cos y = =

(25) Let P, fig. 35, be the point (x, y, z) on the surface. Through p let planes PSLN, PRJN be drawn parallel to the planes of (y, z), (X, 2) respectively; and also let two other planes respectively parallel to them be drawn, and at infinitesimal distances dx, dy; so that NL = dy, NJ = dx, and PsQR is the intercepted infini. tesimal element of the surface; then Q is (x + dx,y+dy, z+d2); and let us imagine the whole surface by a similar process to be resolved into similar infinitesimal elements : then the area of one of these baving been expressed in general terms, the area of the surface will be given by the double integral which expresses the sum of such elements, the integral of course being definite.

Let A represent the required area of the surface, and da the area of the element PRQS. As a tangent plane to a surface at a given point contains not only the point but also an infinity of other points immediately contiguous to it, so da being infinitesimal will be coincident with the tangent plane at P, and therefore the angle between it and any other plane is equal to the angle between the tangent plane and that plane.

Now the projection of da on the plane of (x,y) is the rectangle NK, which = dx dy; s dx dy = da cos y = d.


Similarly, if da is projected on the planes (y, z), and (2,x),

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each of these being a double integral; and either one being em-
ployed according as it is best suited to the equation of the surface
and to the given limits.
Also squaring and adding (26) (27) and (28),

da? = dyż dz2 + dz2 dm2 + dæ2 dy> ;
da = {dyo dz? +dzo dx? + dx? dy?}#; (30)

A= ||{dy? dz2 + dz2 dx? + dxdy2}, (31) which is also the general value of the area ; this form is convenient whenever the equation to the surface is given in the explicit form. Thus if z =f(x, y), then from (31), we have

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The formula (32) may also be deduced from the last of (29), by the theory explained in Art. 50, Vol. I; or as follows; since

F(x, y, z) = f(x,y)—% = 0,

v = (dz), v = (dir.), w=-1; and therefore (29) becomes

^ = S/{1+cm+*+ 1.657)"} *de dy. Now in all these cases, by means of substitution from the equation to the surface, the element-function will become a function of those two variables, whose differentials enter into the element. Thus, let us suppose the element-function to be a function of x and y, and let us consider the effects of the successive integrations.

We will suppose the surface, of which the area is required, to be closed, and to be such as is contained in the octant delineated in fig. 36; then, since PRQs is the element of the surface, the effect of a y-integration, x being constant, will be, the summa

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