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Right-rectangular pyramid: a, b = the sides of the base h = the distance is from base to the apex General triangular prism: b = the base side of the prism's triangular base, h = the height of the prism's triangular base
Perimeter#Formulas – Path that surrounds an area; List of second moments of area; List of surface-area-to-volume ratios – Surface area per unit volume; List of surface area formulas – Measure of a two-dimensional surface; List of trigonometric identities; List of volume formulas – Quantity of three-dimensional space
This formula holds whether or not the cylinder is a right cylinder. [7] This formula may be established by using Cavalieri's principle. A solid elliptic right cylinder with the semi-axes a and b for the base ellipse and height h. In more generality, by the same principle, the volume of any cylinder is the product of the area of a base and the ...
The ratio of the volume of a sphere to the volume of its circumscribed cylinder is 2:3, as was determined by Archimedes. The principal formulae derived in On the Sphere and Cylinder are those mentioned above: the surface area of the sphere, the volume of the contained ball, and surface area and volume of the cylinder.
The Pythagoreans dealt with the regular solids, but the pyramid, prism, cone and cylinder were not studied until the Platonists. Eudoxus established their measurement, proving the pyramid and cone to have one-third the volume of a prism and cylinder on the same base and of the same height.
The volume of a conical or pyramidal frustum is the volume of the solid before slicing its "apex" off, minus the volume of this "apex": =, where B 1 and B 2 are the base and top areas, and h 1 and h 2 are the perpendicular heights from the apex to the base and top planes. Considering that
The condition of balance ensures that the volume of the cone plus the volume of the sphere is equal to the volume of the cylinder. The volume of the cylinder is the cross section area, times the height, which is 2, or . Archimedes could also find the volume of the cone using the mechanical method, since, in modern terms, the integral involved ...
Reprint of 1935 edition. A problem on page 101 describes the shape formed by a sphere with a cylinder removed as a "napkin ring" and asks for a proof that the volume is the same as that of a sphere with diameter equal to the length of the hole. Pólya, George (1990), Mathematics and Plausible Reasoning, Vol.