Note. For a cube, take the cube of its side by multiplying this twice by itself; and for a parallelopipedon, multiply the length, breadth and depth all together, for the content.

Ex. 1. To find the solid content of a cube, whose side is 24 inches. Ans. 13824.

Ex. 2. How many cubic feet are in a block of marble, its length being 3 feet 2 inches, breadth 2 feet 8 inches, and thickness 2 feet 6 inches? Ans. 213.

Ex. 3. How many gallons of water will the cistern contain, whose dimensions are the same as in the last example, when 282 cubic inches are contained in one gallon?

Ans. 12947.

Ex. 4. Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of its triangular end or base are 3, 4, 5 feet.

Ans. 60.

Ex. 5. Required the content of a round pillar, or cylinder, whose length is 20 feet, and circumference 5 feet 6 inches. Ans. 48:1459 feet.

rectangular parallelopipedon be the solid to be measured, and the cube Pthe solid measuring unit, its side being 1 inch, or 1 foot, &c; also, let the length and breadth of the base ABCD, and also the height AH, be each divided into spaces equal to the length of the base of the cube P, namely, here 3 in the length and 2 in the breadth, making 3 times 2 or 6 squares in the base AC, each equal to the base of the cube P. Hence it

is manifest that the parallelopipedon will contain the cube P, as many times as the base AC contains the base of the cube, repeated as often as the height AH contains the height of the cube. That is, the content of any parallelopipedon is found, by multiplying the area of the base by the altitude of that solid.

And, because all prisms and cylinders are equal to parallelopipedons of equal bases and altitudes, by Geom. theor. 108, it follows that the rule is general for all such solids, whatever the ofigure of the base may be.

PROBLEM

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 content*.

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 6 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 Frustum 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 frustum, will give its contentt.

Note.

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

+ Let ABCD be any pyramid, of which BCUGFE is a frustum. And put «2 for the area of the base BCD, b2 the area of the top

EFG,

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 two 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.

EFG, À the height IH of the frustum, and c the height AI of the top part above it. Then c+AH is the height of the whole pyramid.

B

EV

H

G

Hence, by the last prob. a (c + h) is the content of the whole pyramid ABCD, and c the content of the top part AEFG; therefore the difference 3a2 (c + h) — 3bc is the content of the frustum BCDG FE. But the quantity c 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 : ba :: (c + h)2 : c2, or a:b: : : c + h : c, hence (Geom. th. 69) a bb::h: c, and ah andc+h====

a-ba::b: c+h; hence therefore c

bh

a

then these values of c and c + h being substituted for them in the expression for the content of the frustum, gives that con

ah

bh

a

a

ab + b2); which is the rule above given; ab being the mean be

tween a2 and 6o.

Ex. 3. To find the content of a conic frustum, the altitade being 18, the greatest diameter 8, and the least dia

meter 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 head 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 I. 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.

E

D

R E
M

Let ABCD be a cylinder, circumscribing A the sphere EFGH; 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, OQP, being equiangular, it is, as OP: 0Q or LN:: 10 or KL: KO:: circumference described by KL: circumf. described by Ko; therefore the rectangle OP X circumf. of KO is equal to the rectangle LN x circumf. of KL; that is, the ring described by op on the sphere, is equal to the ring described by LN on the cylinder. VOL. II.

E

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 3.1416, 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 79574 miles, and its circumference 25000 miles. Ans. 198943750 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, described by the whole semicircle FGH, 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 Fo, 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, multiplied 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