--

Then (aacc × 11 — — 2ac) × p =

(100 X 900 X

20-3
2500

tt

-18 × 27.968765) × 3*14159

= (36 × 50 - 18 x 27.968765) × 3.14159 = 18

* 5*364568 x 3.14159 303 35917 the content of the fegment LAF required.

PROBLEM XX.

To find the Content of an Univerfal Spheroid, or a
Solid conceived to be Generated by the Revolution
of a Semi-Ellipfe about its Diameter, whether
that Diameter be one of the Axes of the
Ellipfe or not.

RULE 1.

Divide the fquare of the product of the axes of the ellipfe, by the axe of the folid, or the diameter about which the femi-ellipfe is conceived to revolve; multiply the quotient by 5236, and the product will be the content required.

That is, 12 × ·5236

d

the content; T and c

being the tranfverfe and conjugate axes of the ellipfe, and d the axe of the folid.`

RULE II.

The continual product of 5236, the diameter about which the revolution is made, the fquare of its conjugate diameter, and the fquare of the fine of the angle made by thofe diameters, the radius being 1, will be the content.

That

That is, deess X5236 the content; c being the conjugate diameter to d, and s the fine of the angle made by the diameters.*

of the DCB or FEG, and its cofine, x = AE, and p =

the part generated by AFG. And when x=d, it becomes

Corol. I. If d T, the rule becomes prc2 for the oblong fpheroid. And if d = c, it will be pcr for the oblate fpheroid. Alfo if T, C, and d, be all equal, the rule will be pds for the fphere. Which are the fame with the rules before found for the fame bodies.

Corol

If the axes of an ellipfe be 50 and 30, and it be cut in two by a diameter whofe length is 40; required the content of the folid generated by one of the halves about that diameter.

5

The diameter of an ellipfe about which it revolves being 40, its conjugate diameter 30/2, and the fine of the angle made by thofe diameters 2; required the content of the folid formed by the revolution of the ellipfe.

By rule 2, we have 40 X 30 X 30 X2 X3 X5 X2 X *5236 = 225000 X 130929452.5 the content, the fame as before.

SEC

Carol. 2. Draw CH perpendicular, and DH parallel to AB, and about the axes AB and 2CH defcribe the femi-ellipfe AHB; then the fpheroid generated by the revolution of the femi-ellipfe AH, about AB, will be equal to the folid generated by the femi-ellipfe AFDE about the fame axc AB.

For 2CH ac, and therefore the folid AFBI, or pda2 c2, is = pdx (2CH)2; the fpheroid whofe axes are dand 2CH.

And fo the folids generated by all femi-cllipfes upon the fame bafe and between the fame parallels, are all equal to

each other.

SECTION VI.

OF PARABOLIC LINES, AREAS, SURFACES,

AND SOLIDITIES.

PROBLEM I.

To Construct a Parabola; having given any Ordinate Po to the Axe, and its Abfciffa vp.

2. In the axe, produced without the vertex v, take vc VF; draw feveral double ordinates SRS; then with the radii ck, and center F, defcribe arcs. cutting the correfponding ordinates in the points s.

Draw a curve through all the points of interfection, and it will be the parabola required.

Of any Abfcifs x, its Ordinate y, and Parameter p; having Two given, to find the Third.

CASE I.

To find the Parameter.

Divide the fquare of the ordinate by its abfcifs, and the quotient will be the parameter.

Or, take a third proportional to the abfciffa and ordinate, for the parameter.

That is, pyy÷x.

EXAMPLE.

If the abfcifs be 9, and its ordinate 6; required the parameter.

Here 6×69369=4= the parameter.

CASE II.

To find the Abfcifs.

Divide the fquare of the ordinate by the parameter, and the quotient will be the abfcifs.

That is, xyy ÷ p•

EXAMPLE.

If the ordinate be 6, and the parameter 4; required the abfcifs.

Here 6 x64 = 36 ÷ 4 = 9 = the abfcifs.

CASE III.

To find the Ordinate.

Multiply the parameter by the abfcifs, and the fquare root of the product will be the ordinate.

That is, y√px.