a constant quantity; and, therefore, all the other differential coefficients vanish, (by Prop. 19.) Hence, the solid due to QPNn=4axh+4тa. h2 1.2' which being true, whatever be the value of (x), suppose x=0, then h=An, and the solid generated by the area APQn=2πah', and writing (x) for (h), the solid generated by the area, which corresponds to the abscissa (x) = 2 πα x2 = π as before. , 2 Ex. 2. To find the content of a sphere, by differentiation. Here, y2=2ax-x2; hence, f¥' (x) =π . (2ax-x2); ·· ƒ2. 4' (x) = π (2a - 2x), and f3. \' (x) = − 2π. Hence, solid due to QPNn Now, let x=0, then (h) becomes = An, and the area QPNn becomes the area APQn. Hence, the content due to the area APQ n and making h=2a, we have the whole content of Now, if a cylinder be described about this sphere, the radius of this cylinder will = a, and its length = 2a ; and, therefore, its content = 2π α3. Hence, content of sphere content of circumscribing cylinder 2 3. Ex. 3. To find the content of a cone by differentiation. Let (c) be the altitude of the cone, and (a) the radius of its base, then the equation to it is hence, while (x) increases by (h), the quantity, by which '(x), or the content of the cone increases, is Now, to find the whole content, suppose (x) to increase from 0, till it=c, then making r=0, and h=c, we have Hence, the content of a cone is equal to one-third of the content of its circumscribing cylinder. PROP. XLVII. If a curve revolves about its abscissa ANn, as an axis, and if the arc APs, and the surface generated by the revolution of it = S, there results the equation d. S=2. PN.ds. Let Qn (See Fig. Prop. 46.) be an ordinate parallel to PN, and let Nn=h. Then the surface generated by the revolution of the arc PQ, will equal the series The curved surface of a cylinder, whose length equals the arc PQ, and whose radius = PN=y, is, (by Prop. 37. Cor. 4.) equal to 2yx arc PQ. Hence, the surface of this cylinder is equal to the Also the curved surface of a cylinder, whose length equals the arc PQ, and whose radius equals Qn, is equal to 2. Qn. arc PQ. Hence, the curved surface of this latter cylinder is equal to 2x the product of the two series (M) and (N), or h+ yd2 h2 (ads dy+yd's). 1, +&c.). (C). dx dx 1.2 Now, since (by the third Axiom), the surface of the solid lies between these two cylinders, the series (A) must lie between the series (B), and the series (C); hence, dS = dx 2 Tyds ; and . d. S=2πyds. COR. 1. Since ds = √dy+dx2, we have dS= 2ry.dy' +dx. Ex. 1. To find the surface of a sphere. .. d. S (which 2πyds) = 2πadx=d. (2πax); hence, S and 2ax are two functions of (x) which have the same differential coefficient; and, therefore, their difference is constant; but when x=0, S=0; and, therefore, their difference then=0; hence, it always=0; and, consequently, S=2 πax, and when x=2a, the whole surface becomes=4a2; hence, the whole surface of a sphere equal four times the area of one of its great circles. SECT. VI. ON SPIRALS. THERE HERE are many curves, the form and properties of which are not determined by an equation between two co-ordinates, referred to rectangular axes, but by an equation expressing the relation which subsists between the length of a straight line, drawn from a given point, and the angle, which this straight line makes with another given straight line, drawn from the same point. Such curves are called Spirals. Let S be a given point, SL a given straight line, then, if other straight lines, as SP, be drawn from S, whose lengths are functions of the angles LSP, which L 6 they make with SL, the locus of the points P will be a curve, and this curve is, by the definition, a spiral. Since SP may be supposed to revolve round S, we may imagine it to complete a revolution, and to be again brought into the position SP; then producing SP, if necessary, and taking SP', the same function of the angle (LSP + 2), that SP is of |