We found before, supposing d = the side of the square base, and a equal the altitude, that the solidity of the solid 4as dome was expressed by d' a , then to find the solidity 3 of the shell, it is only finding the solidity of two solid butdomes, of the same altitude, but of different dimensions at the base, and deducting the greater from the less.
Now let B be the side of the base to the external surface, and d the side of the base corresponding to the internal sur face ; then the solidity of the solid comprehended within the external surface, and the two parallel planes forming the end, is a 3 In like manner, the solidity of 4a2 the solid which would fill d" a 3 ' the cavity, is then, by subtracting the latter of these expressions from a a = a the former, we obtain for the solidity of the shell.
This rule may be thus expressed in words: Alnitiply the difference of the areas of the bases by the altitude of the dome, and the product will give the solidity of the shell.
Erample.Suppose the side of the base, between the external convex surface, to be 20 feet, and the side of the base of the internal cavity, or bason, to be 18 feet, the solidity of the shell is required.
18 20 18 'surface.
144 400 area of the base contained between the convex 324 area of the base contained between the concave surface.
324 76 difference of the areas of the bases.
6 456 solidity of the shell, as required.
To find the convex surface of a dome.
Let the diameter B 0 = d B A = X and c A = y DC =2 we have, by similar triangles, Aocand CED, C A: C 0 :: CE:CD that is, y : : : : z= : 2 but since the fluxion of the surthce whose sections are circu lar is denoted by 2 p y i, where p is equal to 3.1416; therefore, we have 2 and the fluent of 2pyz=pdx: therefore, the superfieies of the segment of a hemispheric dome is equal to the convex surface of a cylinder of the same altitude, and of a diameter equal to the diameter of a great circle of the sphere ; now, when the segment becomes a hemisphere, then pdg p d x but since p = 4 X .7854, we shall have = 2 X that is, the convex area of the hemi spheric dome is double the area of its base ; and since the area of any segmental dome is the same as that of a cylinder of the same altitude, and of a diameter equal to that of its great circle ; it follows also, that the convex surthee of any truncated dome is equal to the surface of a cylinder of the same altitude, and of a diameter equal to the great circle of the sphere.
To find the convex surface of the segment of a dome, inde pendent of the diameter of the great circle.
Let D = the diameter of the great circle, a = the altitude of the dome, d = the diameter of the base of the dome; then, by the property of the circle, D a = therefore D = a + 4a But p D a is equal to the convex surface; therefore p a a = p ag = the area of the convex surface = ). Therefore, To ,find the area of the segment of a dome : Multiply the sum of the square of the altitude and the fourth part of the square of the diameter of the base, by 3.1416, and the product will be the superficies of the dome.
Example.What is the superficies of the segment of a dome, the diameter of the base being 17.25 feet, and the height 4.5 feet ? 17.25 4.5 17.25 4.5 8625 225 3450 150 12075 1725 20.25 4 )297.5625 diameter. 74.3906 the fourth part of the square of the 20.25 square of the altitude.
94 6106 the sum.
3.1416 5678436 946406 3785624 946406 2539218 297.32290S96 feet, the surface required.
To show that the superficies of any portion of a sphere, contained between any two parallel planes, is equal to the product of a circumference of a great circle into the dis tance of the parallel planes ; that is, equal to the surface of a cylinder, the ba-:e of which is equal to the great circle of the sphere, and the altitude equal to the distance of the parallel planes: Let A equal the altitude of the segment of the sphere, including both the altitude of the segment wanting, and the distance of the parallel planes ; also, let a equal the altitude of the segment wanting ; and let c be the circumference of the great circle.
Then, whether the segment include the part contained be tween the parallel planes, or be the segment cut off, the area of the curved surface will still be expressed by the product under the circumference of the great circle and the height of the segment ; therefore, c A is equal to the superfieies of the segment, including that of the solid contained between the two parallel planes ; also, c a is the superficies of the segment cut of but the difference between the areas of the curved surfaces of these two segments is equal to the curved surface of the solid contained between the two parallel planes; therefore, c A ca= c(A a) is the surface of the sphere contained between the parallel planes.