PROBLEM V.

To find the Content of any Pyramid or Cone.

FIND the area of the base, and multiply that area by the perpendicular height; then take of the product for the con13 tent.*

Ex. 1. Required the solidity of the square pyramid, each side of its base being 30, and its perpendicular height 25. Ans. 7500.

Ex. 2. To find the content of a triangular pyramid, whose perpendicular height is 30, and each side of the base 3. Ans. 38 97117.

Ex. 3. To find the content of a triangular pyramid. its height being 14 feet 6 inches, and the three sides of its base 5, 6, 7 feet. Ans. 71.0352.

Ex. 4. What is the content of a pentagonal pyramid its height being 12 feet, and each side of its base 2 feet?

Ans. 27.5276.

Ex. 5. What is the content of the hexagonal pyramid, whose height is 6.4 feet, and each side of its base & inches ? Ans. 1.38564 feet.

Ex. 6. Required the content of a cone, its height being 10 feet and the circumference of its base 9 feet.

Ans. 22.56093.

PROBLEM VI.

To find the Solidity of the Frustrum of a Cone or Pyramid.

ADD into one sum, the areas of the two ends, and the mean proportional between them: and take of that sum for a mean area; which being multiplied by the perpendicular height or length of the frustrum will give its content t

Note.

* This rule follows from that of the prism, because any pyramid is of prism of equal base and altitude; by Geom. theor. 115, cor. 1 and 2.

+ Let ABCD be any pyramid, of which BCDGFE is a frustum. And put a for the area of the base BCD, 6 the area of the top gre, h the height 1 of the frus

tuna,

Note. This general rule may be otherwise expressed, as follows, when the ends of the frustum are circles or regular polygons. In this latter case, square one side of each polygon, and also multiply the one side by the other; add all these three products together; then multiply their sum by the tabular area proper to the polygon, and take one-third of the product for the mean area, to be multiplied by the length, to give the solid content. And in the case of the frustum of a cone, the ends. being circles, square the diameter or the circumference of each end, and also multiply the same twe dimensions together; then take the sum of the three products and multiply it by the proper tabular number, viz. by 7854 when the diameters are used, or by 07958 in using the circumferences; then taking one-third of the product to multiply by the length, for the content.

Ex. 1. To find the number of solid feet in a piece of timber. whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches; also, the length or perpendicular altitude 24 feet.

Ans. 19.

Ex. 2. Required the content of a pentagonal frustum, whose height is 5 feet, each side of the base 18 inches, and each side of the top or less end 6 inches. Ans. 9.31925 feet.

E

H

B

G

Hence, by the last prob. a (c + h) is the content of the whole pyramid ABCD, and be the content of the top part AEFG; therefore the difference a2 (c + h) — be is the content of the frustum BCDGFE. But the quantity e being no dimension of the frustum, it must be expelled from this formula, by substituting its value, found in the following manner. By Geom theor. 112, a2 : 6a : : (c + h)a: c2, or ab:: c+h: c, hence (Geom. th. 69) a-bbh e, and a - b: a:: ::b:ch; hence therefore c=

c+h=

ah
a—b

a

bh

and

; then these values of c and ch being substituted for them

in the expression for the content of the frustum, gives that contenta X

is the rule above given; ab being the mean between «a and ¿a.

Ex. 3. To find the content of a conic frustum, the altitude being 18, the greatest diameter 8, and the least diameter 4. Ans. 527-7888.

Ex. 4. What is the solidity of the frustum of a cone, the altitude being 25, also the circumference at the greater end being 20, and at the less end 10? Ans. 464.216.

Ex. 5. If a cask, which is two equal conic frustums joined together at the bases, have its bung diameter 28 inches, the bead diameter 20 inches, and length 40 inches; how many gallons of wine will it hold. Ans. 79.0613.

PROBLEM VII.

To find the Surface of a Sphere, or any Segment.

RULE 1.

MULTIPLY the circumference of the sphere by its diameter, and the product will be the whole surface of it.*

RULE II.

These rules come from the following theorems, for the surface of a sphere, viz. That the said surface is equal to the curve surface of its circumscribing cylinder; or that it is equal to 4 great circles of the same sphere, or of the same diameter; which are thus proved.

B

R

-I

M

QP

E

G

1

Let ABCD be a cylinder, circumscribing the sphere BFGH; the former generated by the rotation of the rectangle FBCH about the axis or diameter FH; and the latter by the rotation of the semicircle FGH about the same diameter FH. Draw two lines KL, MN, perpendicular to the axis intercepting the parts LN, OP, of the cylinder and sphere; then will the ring or cylindric surface generated by the rotation of LN, be equal to the ring or spherical surface generated by the arc OP. For first, suppose the parallels KL and MN to be indefinitely near together; drawing 10, and also og parallel to LN. Then, the two triangles IKO, OOP, being equiangular, it is, as oP: 09 or LN:: 10 or KL KO:: circumference described by KL: circumf. described by ko; therefore the rectangle or X circumf. of Ko is equal to the rectangle LN circumf. of KL; that is, the ring described by or on the sphere, is equal to the ring described by LN on the cylinder.

D

And

RULE II. Square the diameter and multiply that square by 3.1416, for the surface.

Rule III. Square the circumference; then either multiply that square by the decimal 3183, or divide it by S∙1416, for the surface.

Note. For the surface of a segment or frustum, multiply the whole circumference of the sphere by the height of the part required.

Ex. 1. Required the convex superficies of a sphere, whose diameter is 7, and circumference 22 Ans. 154.

Ex. 2. Required the superficies of a globe, whose diameter is 24 inches. Ans. 1809-5616.

Ex. 3. Required the area of the whole surface of the earth, its diameter being 7918-7 miles.

Ans. 196994111 sq. miles.

Ex 4. The axis of a sphere being 42 inches, what is the convex superficies of the segment whose height is 9 inches? Ans. 1187-5248 inches.

Ex. 5. Required the convex surface of a spherical zone, whose breadth or height is 2 feet, and cut from a sphere of 12 feet diameter. Ans. 78 54 feet.

And as this is every where the case, therefore the sums of any corresponding number of these are also equal; that is, the whole surface of the sphere, describ ed by the whole semicircle BGH, is equal to the whole curve surface of the cylinder, described by the height BC; as well as the surface of any segment described by so, equal to the surface of the corresponding segment described by BL.

Corol. 1. Hence the surface of the sphere is equal to 4 of its great circles, or equal to the circumference EFGH, or of DC, muitiplied by the height BC, or by the diameter FH.

Corol 2. Hence also, the surface of any such part as a segment or frustum, or zone, is equal to the same circumference of the sphere, multiplied by the height of the said part. And consequently such spherical curve surfaces are to one another in the same proportion as their altitudes.

PROBLEM

PROBLEM VIII.

To find the Solidity of a Sphere or Globe.

RULE I. Multiply the surface by the diameter, and take Or, which is the same of the product for the content.*

thing, multiply the square of the diameter, by the circumference, and take of the product.

RULE II. Take the cube of the diameter, and multiply it by the decimal 5236, for the content.

RULE III. Cube the circumference, and multiply by

⚫01688.

Ex. 1. To find the content of a sphere whose axis is 12. Ans. 904-7808. Ex. 2. To find the solid content of the globe of the earth supposing its diameter to be 7918-7, and consequently its circumference 24877-4 miles.

Ans. 260002677535 miles.

PROBLEM IX.

To find the Solid Content of a Spherical Segment."

RULE I. From 3 times the diameter of the sphere take

double

the surface of *For, put d the diameter, c➡ the circumference, and s = the sphere, or of its circumscribing cylinder; also, a the number 3·1416. Then, s is the base of the cylinder, or one great circle of the sphere; and d is the height of the cylinder; therefore ds is the content of the cylinder. But of the cylinder is the sphere, by th. 117, Geom. that is, of ds, or ds is the sphere; which is the first rule.

Again, because the surface s is: ad2; therefore &ds = fad3 = ·5236d3, is the content, as in the 2d rule. Also d being =ca therefore & ad3 = → a2 = 01688, the 3d rule for the content.

By corol. 3, of theor. 117, Geom. it appears that the spheric segment PFN, is equal to the difference between the cylinder ABLO, and the conic frustum ABMQ.

But, putting d = AB or FH the diameter of the sphere or cylinder, h=FK the height of the segment, PK the radius of its base, and a 3.1416; then the content of the cone ABI is = ad1=2ad3; and by the similar cones ABI, QMI, as F13: KI

3

A

B

K

MN

L

E

G

I

D