Now, when this normal meets the plane of (xy), or (xz), or (yz),

we evidently have z′ = 0, or y′ = 0, x′ = 0.

determine the points of intersection required.

Again, to find the volume we have (Lac. Trans. p. 311.)
V = f.dy Szdx

y

the symbols S,,, denoting that the integral is taken on the supposition that x or y is constant.

But f,zdx = = (ac—
a (ac— y3) $

the limits 0 and

S,zdx =

√ _—_-_. √ ac-y3 of x, gives, after reduction,

α

for the area of the section of the solid made by a plane parallel to that of (x, z).

Hence the volume is generally expressed by

b

V = Sdy Szdx = 2√2 × (ac – yoa dy, which, by

means of two substitutions

P = (ac — yo)3⁄431y, P′ = (ac — y2)3 y

and proper reductions, becomes

V =

yo)1y

Let yac (or x = 0, z = 0). Then the whole volume comprehended by the planes (x, y), (y, z), (x, z), is expressed by

194.

Let the radius of the base of either cylinder = r, and suppose (x, y,) originating in the intersection of the axes of

the cylinders, to be measured along those axes. Then the equations to the surfaces of the cylinders being

if a section of the part common to both be made by a plane parallel to that of (x, y), we shall have z = z', and .. y' = a. Hence it is evident that the section is a square whose side is 2. Consequently the solid common to both cylinders may be generated by the motion of that square parallel to itself, and its volume will therefore be expressible by

Then the whole volume of the part common to both cylinders will be expressed by

This solid is evidently the common groin.

Mr. Peacock (A Collection of Examples, &c., page 451,) has arrived at the above result, but by a process of reasoning obviously incorrect. This work, however, upon the whole, is worthy our best recommendation.

195. Let be the radius of the base of the cone, and .. —that of the base of the cylinder. Then referring the co-ordinates to the centre of the base of the cone, its equation will be (a = its alt.)

Hence the equations to the curve of their intersection (x = x, y=y), are

But projecting the tangent and curve upon the planes (x, y), (y, z), we have (Lac. Trans. p. 171.)

Hence the equation to the tangent plane. (Luc. Trans. p. 176.)

197.

Let r be the radius of the sphere, y the radius of the base of the cone, and xr its altitude. Then its volume V is

198.

Let APB (Fig. 71,) be the circular arc whose centre is C. Then D being the middle point of the radius CD, describe a circle passing through C, D, and touching the are AB (p. 7, Vol. I.) in the point P. P is the point required.

Now

For the CPD = ≤ CQD > CPD.

the circles touch at P, they have a common tangent, which is their radii, and .. 1 CP. Hence CP is the diameter of CDP, and .. CDP is a right

.

or the max. value of P is 30°.

... sin. P =

199.

2

Let y be the radius of the base of a cylinder inscribed in the sphere, whose radius supposer, and 2x its altitude. Then it easily appears that the volume (V) is expressed by V = #у2 × 2x = max.

200.

3√3

If y and x be the co-ordinates of the O referred to

its centre, and r its rad, then

201.

Let AC, BC, be the given lines, and P the point Detween them. Draw APB CB, and let QPq be any other line passing through P, &c.

Then, putting PA ≈ 6, PB, = ß, CB = a, and ≤ 9QC = 0,