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1 . A ACB =

x A 6Cc.

sin. A But for the same prism, the section I axis 6Cc is evidently constant.

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This proof which only applies to triangular prisms, may be extended to those of whose ends are any polygons whatever, (and :: to cylinders), by dividing them into their component triangular prisms.

158. Let the radius of the circle = 1. Then, by the question y = vers. = 1

cos I, the equation to the curve. :: Sydx = 1 - sin. x, the area required. Again,

dy

= sin. 2
dix
dPy

dx?
which = 0, when x =

(2n

2

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2

or when y = 1, which values of the co-ordinates give the positions of the several points of contrary flexure.

159. By Problem 132, Vol. II., the area (T) of a plane

S A expressed in terms of its sides a, b, c and semi-perimeter is

2

T= »

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S

S
2
But when the right-angled (c being the hypothenusė),
S

(c + a - b). (c-a-)

-a)(9-0).(-o) (-0

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THE CUBATURE OF SOLIDS.

0000

160. Every Triangular Prism is decomposable into three equal Pyramids, of bases and altitudes, equal to those of the prism, and the prism = its base x its altitude, :, any triangular

its base x its altitude pyramid =

3 If the base of the pyramid be a polygon, it may be decomposed into triangular pyramids of the same altitude, the sum of the bases being = that of the whole pyramid. Hence the volume of any pyramid whatever = (base x altitude).

3

)

161.

Let a, 6 be the axes of the ellipse, then

6%
y= .. (2ax - )

al

y? = (262 )

62 are equations to the ellipse, in the former of which x is measured along a from its origin in the curve ; in the latter x is measured along 6 from its extremity.

Hence the volumes of the solids (A, B, generated about 2a, 2b, are

62 A = S nyodx =

x3a

? (30 - x)
22 3

a2 t
= Swy’dx = x? (3b - x)

62 3 respectively. Let x = 2a in the former, and 2b in the latter, then the whole volumes are

4ab2x

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A=

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and B =

Now when b = a, the spheroid will be a sphere S=

403

3
4ab? 423.
.: A XS=

x
3

3 162*622

= B2 9

3 :. A:B :: B : S. Q. E. D.

162. Let the tangent parallel to the diameter of the semicircle be the line abscissæ, and let the abscissæ x originate at the point of contact, then by property of the o, we have

(r - y) = m - x2
:: y=;- p2 - x
:: y = 2y* — 2r pie — xa

• s ro — x2 - x2 :: the volume V, generated between the convex part of the O, and the tangent, is expressed by

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Let x =r; then the whole exterior volume 2V =

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(10 — 37),

And subtracting this from the cylinder described by the diameter about the tangent which = 772 X 2r = 2r37, we get the

volume V', generated by the 0, expressed by

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Otherwise,
By Guldin's Theorem, the volume V' = area of the

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circumference of the circle described in the revolution by its cen

3п

tre of gravity. But the radius of this circle =

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3

163. Letr be the radius of the circle, then by prob. 161,

4713 the volume of the spheroid V, = and that described by the A V' will evidently consist of two cones of the same base are and r

Hence 3

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:: V : V :: 4 : 2 :: 2 : 1.
Again, to compare their surfaces S, S', we have

y = x 2rx
. S = 2* f ydt V17

dy?

= 477x

d.ro
= 47m2 for the whole sphere.

1 Now the surface of a right cone = ? circumference of its base X slant side (as is known) :: S' = ar XrV2 + xr xr

ana
2 9 cro
::S:S :: N2 : 1.

2

164. Let Aa = 2a' (Fig. 68,) be the given diameter, and Bb = 26' its conjugate, then PM being any ordinate, if through it a plane I plane ABC be made to pass, cutting from the spheroid the segment Ap PP', it is required to find the volume of that segment.

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