RULE. Multiply the transverse diameter by the conjugate, and multiply that product by ,7854. This last product is the area of the oval.

Let it be required to find the area of an ellipsis, whose transverse diameter is 61,6 feet, and conjugate diameter 44,4? Ans. 2148,100416 area.

16th. OF A PARABOLA.

A parabola is a curvilineal figure, made by the section of a cone, being cut by a plane parallel to one of its sides. Every parabola is two-thirds of its circumscribing parallelogram; therefore, to find the area thereof, this is the

RULE. Multiply the base, or greatest ordinate, by the perpendicular height, and multiply that product by 2, and divide the last product by 3, and the quotient will be the area of the parabola.

Let it be required to find the area of a parabola, whose base is 53,75 feet, and the perpendicular 39,25 ? Ans. 1406,4583 area.

MENSURATION OF SOLIDS.

Solid bodies are such as do consist of length, breadth and thickness, as stone, timber, globes, bullets, &c.

1st. OF A CUBE.

A cube is a square solid, comprehended under six geometrical squares, being in the form of a dye. To find the solid content, this is the

RULE. Multiply the side of the cube into itself, and that product again by the side: the last product will be the solidity, or solid content of the cube.

Let it be required to find the solid contents of a cubic stone, each side thereof measuring 17,5 inches ?

Ans. 5359,375 inches = 3,175.

2nd. OF A PARALLELOPIPEDON.

A parallelopipedon, or square prism, representing a square piece of timber or stone, each side of its square base being 21 inches, and its length 15 feet; required the solid conten thereof?

RULE. Find the area of the base, and multiply said area by the length in inches, and divide that product by 1728; or multiply the area of the base by length in feet, and divide by 144, and you will have the solid content in feet.

Ans. 45,9+ feet, the solidity..

3d. OF A TRIANGULAR PRISM.

A prism is a solid contained under several planes, and having its bases like, equal, and parallel.

The solid content of a prism, (whether triangular or multangular,) is found by multiplying the area of base into the length or height, and the product is the solid content.

Let it be required to find the solid content of a triangular prism, whose base is 15,6 inches, each side and the perpendicular 13,51 inches, and the length of the solid 19,5 feet? · Ans. 14,27 feet.

4th. OF A PYRAMID.

A pyramid is a solid figure, whose base is a polygon, and whose sides are plain triangles, their several tops meeting together in one point. To find the solid content thereof,

RULE. Multiply the area of the base by a third part of the altitude, or length, and the product is the solid content of the pyramid.

Let each side of the base of a square pyramid be 18,5 inches, and the perpendicular height 15 feet; required the solidity of that pyramid ? Ans. 11,88 feet.

NOTE. Every pyramid is a third part of the prism, that hath the same base and height, (by Euclid, 12, 7.)

5th. OF A CYLINDER.

A cylinder is a round solid, having its bases circular, equal, and parallel, in form of a rollingstone. To find the solid content thereof, this is the

RULE. Multiply the area of the base by the length, and the product is the solid content.

Let it be required to find the solid content of a rollingstone, whose diameter is 21,5 inches, and the length is 16 feet? Ans. 40,34 feet.

6th. OF A CONE.

A cone is a solid, having a circular base, and growing smaller and smaller, till it ends in a point, which is called

the vertex, and may be nearly represented by a sugar-loaf. To find the solidity thereof, this is the

RULE.

Multiply the area of the base by a third part of the perpendicular height, the product is the solid content.

Let it be required to find the solid content of a cone, whose diameter of the base is 26,5 inches, and the height of the cone 16,5 inches ? Ans. 21,06 feet.

7th. OF THE FRUSTUM OF A PYRAMID.

A frustum of a pyramid is the remaining part, when the top is cut off by a plane parallel to the base. To find the solid content thereof, this is the

RULE. To the rectangle (or product) of the sides of the two bases, add the sum of their squares, and that sum being multiplied into one-third part of the frustum's height, will give its solidity, if the base be square.

Or multiply the areas of the two bases together, and to the square root thereof add the two areas, and that sum multiplied by one-third of the height, gives the solidity of any frustum, square, or multangular.

Let it be required to find the solid content of the frustum of a square pyramid, whose side of the greater base is 18 inches, and the side of the lesser 12 inches, and the height 18 feet? Ans. 28,5 feet.

8th. OF THE FRUSTUM OF A CONE.

A frustum of a cone is that part which remains when the top-end is cut off by a plane parallel to the base. To find the solid content, the rules are the same in effect as for the frustum of a pyramid.

RULE. To the rectangle of the diameter of the two bases, add the squares of the said diameters, and multiply the sum of ,7854; the product will be the triple of a mean area, which multiplied by one-third of the perpendicular height, the product will be the solid content.

Let the solid content of the frustum of a cone be required, whose greater diameter is 18 inches, and the lesser diameter 9 inches, and the length 14,25 feet? Ans. 14,6775 feet.

9th. TO MEASURE THE FRUSTUM OF A RECTANGLED PYRAMID, CALLED A PRISMOID,

Whose bases are parallel one to another, but dispropor tional, this is the

RULE. To the greater length add half the lesser length, and multiply the sum by the breadth of the greater base, and reserve the product; then to the lesser length add half the greater length, and multiply the sum by the breadth, or the lesser base, and add this product to the reserved, and multiply that sum by a third part of the height, and the product is the solid content.

Let the length of the greater base of a prismoid be 38 inches, and its breadth 16 inc. and the length of the lesser base 30 inches, and its breadth 12 inches, and the height 6 feet; required the solid content ?

Ans. 19,94 feet, the content.

10th. TO MEASURE A CYLINDROID,

That is, a frustum of a cone, having its bases parallel to each other, but unlike.

RULE. To the longest diameter of the greater base, add half the longest diameter of the lesser base, and multiply the sum by the shortest diameter of the greater base, and reserve the product. Then to the longest diameter of the lesser base, add half the longest diameter of the greater base, and multiply the sum by the shortest diameter of the lesser base, and add the product of the former reserved sum, and that sum will be the triple square of a mean diameter, which multiplied by ,7854, and the product multiplied by a third part of the height, the product is the solid content.

Let it be required to find the solid content of a cylindroid, whose bottom base is an oval, the transverse diameter being 44 inches, and the conjugate diameter 14 inches; and the upper base is a circle, whose diameter is 26 inches, and the height of the frustum 9 feet?

Ans. 33,47 feet, the content.

11th. OF A SPHERE OR GLOBE.

A sphere or globe is a round solid body, every part of whose surface is equally from a point within it, called the centre; and it may be conceived to be formed by the revolution of a semicircle round its diameter. To find its solidity, this is the

RULE. Multiply the axis or diameter into the circumference, and the product is the superficial content, which multiplied by a sixth part of the axis, the product is the solidity.

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Or, as 21 is to 11, so is the cube of the axis to the solid content. Or, as 1 is to ,5236, so is the cube of the axis to the solid content.

Let the solid content of a globe be required, whose axis is 20 inches? Ans. 4188,8 inches,

NOTE. If the axis of a globe be 1, the solidity will be,5236; and if the circumference be 1, the solidity will be ,016887. The superficies of every globe is equal to four times the area of its greatest circle. The solidity of every globe is two-thirds of its circumscribing cylinder.

12th. OF A SPHEROID.

A spheroid is a solid, resembling an egg. To find the solid content thereof, this is the

RULE. Multiply the square of the diameter of the greatest circle by the length, and the product multiply again by ,5236. This last product will be the solidity of the spheroid.

Let the solidity of a spheroid be required, the diameter of whose greatest circle is 33 inches, and length 55 inches? Ans. 31361,0220 inches, the solidity.

13th. OF A PARABOLIC CONOID.

A parabolic conoid is something like a half spheroid, having its sides somewhat straiter. It is generated by supposing a semiparabola turned about its axis. To find the solid content thereof, this is the

RULE. Multiply the square of the diameter of its base by ,7854, and multiply that product by half the height, and the last product shall be the solid content.

Let the solidity of a parabolic conoid be required, the diameter of whose base is 36 inches, and its height 33 inc.? Ans. 9,71 feet, the content.

14th. OF A PARABOLIC SPINDLE.

If an acute parabola be supposed to be moved about its greatest ordinate, it will form a solid, called a parabolic spindle. To find the solid content, this is the

RULE. Multiply the square of the diameter of its greatest circle, 41888, (being three-fifths of ,7854,) and that product by its length, and the last product is the solid content.

Let it be required to find the solid content of a parabolic spindle, whose greatest diameter is 36 inches, and its length 99 inches ? Ans. 31,10184 feet.