fegment; from the fum take the product of the faid length and altitude; and call the difference P.

Multiply the cube of the altitude of the fegment by the fquare of the greatest diameter of the fpindle; divide the product by the fourth power of the length of the femi-spindle; and call the quotient o

Then the product of P and Q multiplied by 785398, will give the content of the fegment.

That is,

4AC2 -ACX AG + AG2

AC+

X AG3 X DE X

785398, is the fegment AFH, by corollary 2 of the Jaft problem.

EXAMPLE.

If the diameter of the bafe of the fegment of a parabolic fpindle be 24, and its altitude 20; what will be the content, fuppofing the length of the whole fpindle to be 80?

Here

Corol. 3. If from the value of the femi-fpindle ADE, be taken

that of the fegment AFH, there will remain 4bb

b x + xx Cz

ca2 % or

pp

8cbs

15PP

z

2

× (64 - 362x2 + 4) for

the value of the fruftum HDEF; putting z instead of b

x or GC.

And if instead of z, in the two laft terms of this expreffion, be fubftituted its value b, the value of the faid fruftum

Here AC 40, AG 20, and GH = 12; then 20; and, by the parabola, AC2 GC2: Ac2:: GH: CD = 16; the double of

GC = 40

20 =

202
5

3

× 785398 (- ÷ +20) × 5 × 322 × 785398 = 45223 × 785398 = 3552'094093659 is the content required.

PROBLEM XVII.

To find the Content of the Fruftum of a Parabolic Spindle, one of the Ends of the Fruftum paffing through the Center of the Spindle.

Add into one fum, 8 times the fquare of the diameter of the greater end, 3 times the fquare of the diameter of the lefs end, and 4 times the product of

the

be added, or fubtracted, the fruft. DIKE = ca X

8d2 + 4d♪+3d2 60

8D2 (Aa) +da (4D + 3d) ±da (4D +3d)

60

there will result c X for the value of the fruftum HIKF, neither of whofe ends pafs through the center of the fpindle; the upper or under figns being ufed, according as the faid center is within or without the fruftum, and in which D reprefents the diameter through the center of the spindle, d the diameter of the lefs end of the fruftum, and that of the greater; alfo A the distance of d from the center of the fpindle, and a that of from the fame. Or in a cafk of this form, itanding upon one end, D will be the bung diameter, d the head diameter, 2A the length of the cafk, Aaw the wet inches of 2A, and the diameter at the furface of the liquor, the value of which diameter is as determined in the following corollary.

2

Corol. 5. Since, by the property of the parabola, ▲2 : a2 :: DA2 a2 (D − d) p-d; D, the value of ♪ will be

AA

the diameters; multiply the fum by the length; and the product multiplied by 05236, or

will be the content.

of 785398,

That is, (3DE+4DE × FH + 3FH2) × CGX05236 is the fruftum DEFH, by corollary 3 to prob. 15.

EXAMPLE.

Required the content of a cafk in the form of the middle zone of a parabolic fpindle, the bung diameter being 32 inches, the head diameter 24, and length 40 inches.

The cafk is evidently two equal fruftums, whofe greater diameters are 32, and leaft 24, and their lengths each 20; therefore

2

(8 × 322 + 4× 32 × 24 + 3 × 243) × 40 × ·05236 83 × (8 × 42 + 3 × 42 + 3 × 32) × 5 × •05236 = (11 × 16 + 27) × 83 × 5 × 05236 = 519680 × 05236 = 27210.38117 inches = 96-4907 ale gallons, the content required.

PROBLEM XVIII.

To find the Content of the Fruftum of a Parabolic
Spindle, neither of whofe Ends pass through
the Center of the Spindle.

Multiply the diameter of each end, by its diftance from the diameter in the middle of the fpindle; and multiply each product by the fum of 3 times the faid diameter of the end, and 4 times the faid middle diameter; to or from the product belonging to the lefs diameter, add or fubtract that which belongs to the greater, according as the center of the fpindle is within or without the fruftum; to the fum or difference, add 8 times the product arifing from the multiplication of the length of the fruftum by the

fquare

fquare of the diameter in the middle of the spindle; and this fum multiplied by 05236, or of 785398, will produce the content of the fruftum.

That is, [8DEX GL + (4DE + 3FH) FH X GC ± (4DE +3IK) IK X LC] × 05236 is the fruftum HIKF; using the upper or under fign, according as GL is greater or lefs than GC; as is proved in corollary 4 to prob. 15.

Note. The value of the greater diameter IK is DE X GC2 — LC2 X (DE-FH); as in corollary 5 to the

GC2

fame problem.

EXAMPLE I.

If the liquor in the cafk, in the example to the last problem, when standing upon its end, with its axe perpendicular to the horizon, rife to the height of 30 inches; what quantity of liquor will there be?

Then, by the rule,[8 × 32 × 30+(4×32+3 × 24) 24 × 20+ (4 × 32 + 3 × 30)30 × 10] × 05236 = [8 × 162 + (128 + 72)4 + (64 + 45) 5] X 120 X 05236=407160 x 0523621318.84775 cubic inches 75'59875 ale gallons, the quantity required.

EXAMPLE II.

If in the fame cafk, placed as before, the liquor rife but to the height of 10 inches; how much will be in it?

Here all the dimenfions are the fame quantities as in the laft example, excepting GL, which here is only 10, instead of 30; and the cafk is less than half full. Wherefore,

Wherefore, by the rule, [8 x 32 x 10 + (4 × 32 + 3×24) 24 X 20 (4× 32 + 3 × 30) 30 X 10] X052368 × 162+(128+72)12—(64+45)15] X40X05236 = 112520 X 05236 = 5891*533423 cubic inches 20.89196 ale gallons, the quantity required.

PROBLEM XIX.

To find the Content of the Univerfal Parabolic Spindle,
or Solid generated by the Revolution of a Parabolic
Segment about its Bafe, being an Ordinate
to Any Diameter.

Multiply the area of the greateft fection by the length, and of the product will be the content; as in the common fpindle.

8

E

That is, DE X AB X = the folid ADBEA, n T generated from the revolution of the parabolic fegment ADB about its bafe AB, which is a double ordinate to the diameter DP, and DCE being perpendicular to AB.

*

E X

*

DEMONSTRATION.

and put

Let GH be parallel to CD, and HF parallel to DP; a = DP, 1⁄2 b = AP = AB, C = CD = {DE, z = AF, and 4314159. Then, by the property of the parabola, AP2 AF X FB: PD: FH; by fim. triangles, CD: GH

=

= CD