EXAMPLES.

1. What is the solidity of a rectangular prismoid, the length and breadth of one end being 14 and 12 inches, and the corresponding sides of the other 6 and 4 inches, and the perpendicular 30 feet.

Here 14× 12+6x4=168+24=192=sum of the areas

of the two ends.

Whence 10x 8×4=80 × 4=320=4 times the area of the middle rectangle.

366

Or (320+192) ×

=512 × 61=31232 solid inches.

6

And ‘31232÷1728=18.074 solid feet, the content.

2. What is the solid content of a prismoid, whose greater end measures 12 inches by 8, the less end 8 inches by 6, and the length, or height, 60 inches? Ans. 2.453 feet.

3. What is the capacity of a coal wagon, whose inside dimensions are as follow: at the top, the length is 81, and breadth 55 inches; at the bottom the length is 41, and the breadth 29 inches; and the perpendicular depth is 474 inches? Ans. 126340.59375 cubic inches.

PROBLEM XII.

To find the convex surface of a sphere.

RULE.*

Multiply the diameter of the sphere by its circumference, and the product will be the convex superficies required.

Note.-The curve surface of any zone or segment will also be found by multiplying its height by the whole circumference of the sphere.

* Demon. Put the diameter BG=d, BA=x, AC=y, BC=z; and 3.1416=p.

Then, since the triangles AOC and CED are similar, we shall

d

dx

2y

have CA (y) co 2 :: CE (x): CD(x)=" But 2pyz is the general expression for the fluxion of any surface; and therefore, by

dx

substituting 2y for its equal z, the fluxion will become pdx; and consequently pdx=surface of any segment of a sphere whose height is x, and pdd that of the whole sphere. Q. E. D.

Cor. 1. The surface of a sphere is also equal to the curve surface of its circumscribing cylinder.

Cor. 2. The surface of a sphere is also equal to four times the area of a great circle of it.

1. To find the lunar surface included between two great circles of the sphere.

RULE. Multiply the diameter into the breadth of the surface in the middle, and the product will be the superficies required. Or,

As one right angle is to a great circle of the sphere;

So is the angle made by the two great circles,

To the surface included by them.

2. To find the area of a spherical triangle, or the surface included by the intersecting arcs of three great circles of the sphere.

RULE. As two right angles, or 180°,

Is to a great circle of the sphere;

So is the excess of the three angles above two right angles,
To the area of the triangle.

EXAMPLES.

1. What is the convex superficies of a globe BCG whose diameter BG is 17 inches?

D

G

Here 3.1416 x 17 x 17 53.4072 x 17=907.9224 square inches.

And 907.9224-144-6.305 square feet, the answer. 2. What is the convex superficies of a sphere whose diameter is 1 feet, and the circumference 4.1888 feet? Ans. 5.58506 feet.

3. If the diameter, or axis of the earth be 7957 miles, what is the whole surface, supposing it to be a perfect sphere? Ans. 198944286.35235 sq. miles. 4. The diameter of a sphere is 21 inches; what is the convex superficies of that segment of it whose height is 4 inches? Ans. 296.8812 inches.

5. What is the convex surface of a spherical zone, whose breadth is 4 inches, and the diameter of the sphere, from which it was cut, 25 inches? Ans. 314.16 inches.

PROBLEM XIII.

RULE.*

Multiply the cube of the diameter by .5236, and the product will be the solidity.

*Demon. Put AD=x, DC=y, the diameter AB=d, and p=3.146.

EXAMPLES.

1. What is the solidity of the sphere AEBC, whose diameter AB is 17 inches?

Here 173 x 5236 = 17 x 17 x 17 x .5236 = 289 x 17 x 5236-4913x.5236-2572.4468 solid inches. And 2572.4468÷1728-1.48868 solid feet, the answer. 2. What is the solidity of a sphere whose diameter is 1 feet? Ans. 1.2411 feet. 3. What is the solidity of the earth, supposing it to be perfectly spherical, and its diameter 7957 miles? Ans. 263858149120 miles.

Then, by the property of the circle, dx-x-y2. But the general expression for the fluxion of any solid is py'x; and therefore by writing de-2 for its equal y2, we shall have px × dx—x2=pdxx-pxx. The fluent of which is pdx2 px 3pdx-2px3x

6

=content of the segment CAE. 3pd3-2pd3

2 3 And if d be substituted for x, it will become

_pd®

6

6

dx.5236, or .5236d; which is the same as the rule.

Coroll. A sphere, or globe, is equal to two-thirds of its circumscribing cylinder.

PROBLEM XIV.

To find the solidity of the segment of a sphere.

RULE.*

To three times the square of the radius of its base add the square of its height; and this sum multiplied by the height, and the product again by .5236, will give the solidity. Or,

From three times the diameter of the sphere subtract twice the height of the segment, multiply by the square of the height, and that product by .5236; the last product will be the solidity.

EXAMPLES.

1. The radius Cn of the base of the segment CAD is 7 inches, and the height An 4 inches; what is the solidity?

* Demon. Let r-radius of the base of the segment, h =height of the segment, and the other letters as before. Then will (3dh3—2h3)× 2=solidity of the segment, as is shown in the last problem.

go2+h2
h

But since =d, by the property of the circle we shall

N