be taken from o to- x, instead of from o to x. A being a function of r, x, and u, may be represented by 24 (r, x,u). To correct the fluent, let sm (fig. 23) = X, sm' x, then, the attraction of the solid, whose base is the quadrilateral figure prr'p', will be 24 (r, x, u) — 24 (r,X, u).

In figure 24, call ps, u; sm, X; sm', x. The action of the solid, whose base is prr'p', is expressed by 24 (r,-x, u) — 2¢ (r,-X, u).

In fig. 25, put psu, ps'u', sm = s'm = x, tang. of rsmr: the attraction of the solid, whose base is the rhombus srs'p, on a point p in the produced diameter of the section, is 2¢ (1, x,u) — 24 (r,o,u) + 24 (r,— x, u') — 24 (r, — 0, u'′).

Prop. C.

Let fig. 26 represent the base or section of an infinitely long prism, and let this base be any right lined figure whatever, regular or irregular: from p, a point in the same plane, draw any line pq, cutting the base at s and m"". It is required to find the action of the solid on the point p, in the direction pq.

From the angles r, r', r", r'", &c. of the base, let fall the perpendiculars rm, r'm', r"m", r""m"", &c. on the line pq. Prolong the sides of the polygon till they meet pq at the points s, s', s", s"", &c.

Put ups, u' = ps', u" = ps", u"" =ps"", &c.; and [ x' = s′m' x"=s"m"] x''' = s"m"

x = {m}, {x" == "m"}, {x

s'm X"=s"m' X""=s"m" Also, let r = tang. rsm, r'— : tang. r's'm', r" = tang. r"s"m", r""'= tang. r"''s"m", &c. then it appears from the last proposition, that the attraction, of the upper half of the solid, is

t=sm,

},

&c.

expressed by

x,u)
u ) — o (r

,

& (r
+(r
+¢ (r", x",u" ) -
+ ¢ ( r'"',—x''', u'""') —

+

&c.

&c.

And in the same manner is found the attraction of the lower portion. If any part of the polygon, as pp', is parallel to pq, the attraction of that portion of the solid may be found by Prop. A.

o, u )

X', u' )
X",u")

$ (r",

q ( r'"',—X"", u"")

Scholium to Prop. 25, page 273.

The following expression includes the attraction (on a point at the pole or vertex) of all this class of solids, where the generating plane is a regular polygon, and guiding curve a conic section: or where ya (ßx + yx2).

A = 2n (x + arc (tang. = ——√μx+ vxa ) — { (n − 2)

x+ } = + Q

ηβαι ι 1+ya2

in which μ = ßa3 (1 + ro), v = 1 + yao ( 1 + ro), and

2nrBa2

Φ

2nrBa2 √T (1+ya2)

L(+

+1), or =

√= (1+ya2)

arc

(sine ====**) accordingly as is positive or negative.

V

μ

Ex. 1. Let y=0, a = 1, y = ßx; in which case the solid is the polygonal parabolic conoid treated of in the proposition; and we have μ = μ = ß ( 1 + r2), v = 1, whence

A = 2n (x+ß) arc (tang.=;✔ß (1+r) x+x°) — { (n − 2)

~ I √R (1+r2) the same as was found before.

x +nß }≈ + 2nrßL (

+

Er. 2. Let a2 = B =a, y =

b2
a2,

b2

=

1, y' — — ( ax — x') : here the curve pr, fig. 15, is an ellipsis whose diameters are a and b, a being that which coincides with the axis pin. We

b2

have, in this case, μ = ·

a

the attraction of a polygonal A = 2n (x + b) arc (tang. =

b2

(1 + r ), v = 1 − 2/2 (1+r), and

spheroid, on a point at its pole, is

b2

b

√1⁄2 (1+r) x+ (1 − 2 = (1+r'))x′′)

✓

or

nb21

+ a2 —b2

x

B (1 + r2) + 1),

L

{ (1 + r2) b2 —a2 } x

b2a (1+r2)

2nra2b2

arc (sine

a2 b2

(a2—b2) √ (1+r2) b2—a2 accordingly as is greater or less than 1+r2, or as — is greater or less than the secant of half the angle formed at the centre of the generating polygon by one of its sides.

When xa, the first arc in the above expression becomes

2n

simply arc (tang. =) = (-2), and we have for the action ,& representing & after a

2ab2

of the whole solid, A = & has been put for x.

Φ

In like manner, may the action of the solid be found when

+

the guiding curve is an hyperbola; the only difference between that case, and the one we have just considered, being in the value of which must be taken + 1 instead of — 1.

7,

Scholium to Cor. 3, Prop. 27, page 278.

If the variable rectangle is given in species, and the touching curves are conic sections; that is, if

ya = x2 (ßx + yx2), y12 = a12 (ßx + yx2), we shall have, for the action of the generated solid, on a point at its vertex by Prop. 4,

A = 4x arc (tang. = ́x2 + (1 + r2) œa (Bx+yx2)) +

2

Tx

12

4√x arc (tang. =—= √x2 + (1+112) aˆa2 (ßx + yxa)) — 2′′X, r'; and by actually taking the fluent,

where r =
A = 4(x+

Baz
1+ya2

arc (tang. =—-√μx+vx2)

26a2
1+ya2

Ba12
1+ya's

2Ba'z

:)

arc (tang.✔í'x+vx3)

12

+4(x+ 1+yazπ+02 -2x, where μ= = Bæ2 (1 + r2), v = 1+ya2 (1+r2), μ'=Bxˆa (1+r22), v'=1+ ya22 (1 + y22),

LV + √ +1

Φ

4rBa2
~T (1+ya2)

~ + 1), or =

(sine

according as is positive or negative,

4r'Ba'z

L

~T (1+ yα22)

~===

+ √ * + 1), or =
v

4rBa2
√ = (1+ya2)

arc

arc

(sine =

as is positive or negative.

If, in the preceding expression, we make r and a' infinite,

and ro, it is reduced to

arc (tang.==✔Bx+yx3)

✓

4Ba

LV7) + √ + 1), or = √=y(1+7α2)

arc (sine

as v or y is positive or negative.

B

This is the action of an infinitely long cylinder on a point at the vertex of its transverse section, the equation of the said section being y2 = ao (Bx + yx2).

Ex. If the base, or transverse section, is an ellipsis, or if

--

:)

arc (tang.

A=

b

✓ ax = x*) + 4a arc

4a2b a2-b2

ax

When xa, this expression is

zab
a+bπ.

b2

1, ßa, y = — 1; and

Tx

Scholium to Cor. 2, Prop. 30, page 281.

If we would have a general expression for the attraction of such solids as the one we considered in the proposition, when the guiding curve is any conic section, or when

y2 = x2 (ßx+yxr), there arises at first (from the formula for the action of a rhombus)

2

2πx,

A=4fx
4x arc (tang.= √ x2 + ( 1 + r° ) a2 (ßx + yx3)) +
4fx arc (tang. ; ; √ x2 + ( 1 + ~22 ) xa (ßx + yx*))
r3 e* — 27
and by actually taking the fluent
A = 4(x+ = =) { }
arc (tang. = μx + vx3) + arc
(tang. = — √μ'x + v'x') } − ( 422 + 2x) π + 0 + 0,

Baz
1+ya2

D

2

1+ya2