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Now

DE is very small, it may be taken as a straight line, and the surface generated by its revolution round AB, as that of a cylinder, with radius Ff, which will be 2π×Fƒ×DE. (284). Also the surface of the small cylinder generated by the revolution of de round mfn=2′′×dm×DG. the triangles DEG, BFf, have each one right angle, and the angles EDG, BFf equal; therefore they are similar, and the sides about their equal angles proportional; hence

BF: Ff:: DE: DG,

.. BFXDG=FfxDE, or dmxDG=Ff×DE;

and therefore the area of the curved surface described by DE is equal to that of the cylinder described by de.

The same may be shewn to be true of all the successive very small portions of the arc of the quadrant ADC. Hence, the whole hemispherical surface generated by the

revolution of ADC round AC

=2π×dm× (sum of all the small quantities like DG), =2×dmxAB=2π×(rad.), since dm=BC;

and the surface of the whole sphere =4x(rad.).

..

OBS. The section of a sphere made by a plane through its centre is called a great circle of the sphere. Hence, the surface of a spherc=4 great circles of the sphere. This is especially worth remembering.

Ex. If the diameter of the earth be 8,000 miles, what is its surface considered as a sphere?

Surface=4x(rad.)'=4x

22

TM×(rad.)2=4×~×16,000,000 sq. miles, 7

=201,142,855 sq. miles.

294. To measure the volume of a sphere, when its radius, or diameter, is given.

A sphere may be considered as composed of a number of very small equal cones, having their common vertex at the centre, and the same altitude, viz. the radius of the sphere, and having as the base of each, a very small portion of the surface of the sphere.

Then the vol. of the sphere

= sum of vol. of all these cones,

= } ht. × sum of the areas of their bases,
=ht. × surface of the sphere,

=

=

rad.x4x(rad.), (since height = rad.),

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COR. The vol. of the cylinder circumscribing the sphere, and of the same height as the diameter of the sphere,

=ht. x base (289),

=2rad.×TM×(rad.)3,
=2π×(rad.)3;

.. vol. of sphere =3×2× (rad.)3= of circumscribing cylinder.

Ex. The inner diameter of a foot-ball is 6 in.; how cubic inches of air does it contain?

many

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295. Given that the volume of a cone is equal to onethird of the cylinder with the same base and height, prove that the volume of a sphere is two-thirds of the circumscribing cylinder.

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Let ABCD be a cylinder,

AEB a hemisphere, O its centre;

GODC a cone, E centre
of its base;

F any point in OE;
GFIKH a straight line
through F parallel to
AOB.

Then FH2 = OK2 = OF2 + FK2,

= FI2 + FK2, ··· OF : FI :: OE : ED

:: OE : OA

:: 1: 1,

.. T×HF2=π× FI2+π×FK2,

i. e. Circular section of cylinder

= Circular section of cone+circular section of sphere, and since F is any point in OE, .. the same holds for all corresponding sections of the cylinder, cone, and hemisphere; and consequently for all corresponding laminæ of very small thickness. Hence

Whole Cylinder = whole cone + whole sphere; But, by supposition, Cone = Cylinder,

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Double the hemisphere, and also the Cylinder, then

=

vol. of Sphere Circumscribing Cylinder.

296. OBS. We now see, that just as the area of a circle was proportional to the square of the radius, so the volume of a sphere is proportional to the cube of the radius. Hence the volumes of spheres are to one another as the cubes of their radii.

Thus, if two spheres have respectively radii of 9 in. and 5 in., the vol. of the larger : vol. of the smaller :: 93 : 53

:: 729 125.

It may also be mentioned, that any similar* solids, however irregular their shape may be, have also, as in the case of spheres, volumes proportional to the cubes of any two corresponding lines in them.

Thus, if AB, ab were corresponding slant sides or radii of two similar cones; or the heights, or radii, of two similar cylinders; and if it be known that

AB-ab, or AB : ab ::: 1 :: 4 : 3,

* DEF. Similar solids are such as have all their solid angles equal, each to each, and are bounded by the same number of similar plane surfaces.

Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals.

PART III.

10

then, vol. of larger solid : vol. of smaller ::. (AB)3 : (ab)3,

:: 43: 33,
:: 64: 27.

Ex. How many spheres of 3 inches diameter can be placed in another of 12 inches diameter, supposing the small spheres made of plastic material, so as to fill the whole interior of the large sphere?

Large Vol. (12 in.)3 _ 1728
Small Vol. (3 in.)

=

=

27

=

64

i.e. the large sphere contains 64 small ones.

The same result may also be obtained thus. Since the large rad. = 4 times the small radius,

.. Large vol. : small vol. :: 43 ; 1 :: 64 : 1, as before.

297. To measure the solid matter of a pipe, or hollow cylinder.

The pipe, or hollow cylinder, has been virtually included in the article on the cylinder, only that it is, as it were, one cylinder within another, and therefore its volume has not actually been measured.

Now the quantity of material employed in its construction may be found in two ways:

1st. By finding the volumes of both the outer and inner cylinders, and taking their difference.

2nd. By finding the surface of a cylinder which is a mean between the inner and outer cylindrical surfaces, i. e. whose radius is a mean of the outer and inner radii, and then multiplying the surface so obtained, by the thickness of the material of which the pipe is composed.

Ex. How many cubic feet of iron are required to make a cylindrical chimney for a marine engine, which shall be 30 ft. high, have its inner radius 12 in., thickness of metal three-fourths of an inch?

By the first method, the volume) of the smaller cylinder

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=π×12×30 cub. ft.,

and

==×(11%)3×80cub.ft.;

.. volume of iron = diff. of vol. of cylinders,

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By the second method, the radius of the cylinder which is a mean between the outer and inner radii, =132 ft.; and therefore the surface of that cylinder

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298. To find the content, or volume, of a haystack, (1) with circular base, (2) with oblong base.

B

A

F

E

I. Let ABCDE be a vertical section through the middle of a stack whose base is circular, and the sides diverging upwards in the usual way; the upper part will be a cone, whose height is 40, and the radius of its base BO; the lower part will be a frustum of a cone inverted, whose height is OP. Let FEG be a vertical line through E, meeting the horizontals AF, and CD produced, in Fand G. Let AO, or FE, be measured, and also EG. Since the radii OE, PD, cannot be directly measured, measure the circumferences of the circular sections through E and D; then if c be the circumference of the mid-section parallel to the base, and C1, C2, those through E and D,

C

P

D

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