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The generation of a bicylinder Calculating the volume of a bicylinder. A bicylinder generated by two cylinders with radius r has the volume =, and the surface area [1] [6] =.. The upper half of a bicylinder is the square case of a domical vault, a dome-shaped solid based on any convex polygon whose cross-sections are similar copies of the polygon, and analogous formulas calculating the volume ...
The fourteenth problem of the Moscow Mathematical calculates the volume of a frustum. Problem 14 states that a pyramid has been truncated in such a way that the top area is a square of length 2 units, the bottom a square of length 4 units, and the height 6 units, as shown. The volume is found to be 56 cubic units, which is correct. [1]
Cylindrical (cylinder): Several problems compute the volume of cylindrical granaries (RMP 41–43), while problem 60 RMP seems to concern a pillar or a cone instead of a pyramid. It is rather small and steep, with a seked (reciprocal of slope) of four palms (per cubit). [ 8 ]
The Egyptians knew the correct formula for the volume of such a truncated square pyramid, but no proof of this equation is given in the Moscow papyrus. 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":
The problem includes a diagram indicating the dimensions of the truncated pyramid. Several problems compute the volume of cylindrical granaries (41, 42, and 43 of the RMP), while problem 60 RMP seems to concern a pillar or a cone instead of a pyramid. It is rather small and steep, with a seked (slope) of four palms (per cubit). [10]
Reuse the volume formula and unit information given in 41 to calculate the volume of a cylindrical grain silo with a diameter of 10 cubits and a height of 10 cubits. Give the answer in terms of cubic cubits, khar, and hundreds of quadruple heqats, where 400 heqats = 100 quadruple heqats = 1 hundred-quadruple heqat, all as Egyptian fractions.
The volume of a cylinder was taken as the product of the base and the height, however, the volume of the frustum of a cone or a square pyramid was incorrectly taken as the product of the height and half the sum of the bases. The Pythagorean rule was also known to the Babylonians. [21] [22] [23]
Within the cylinder is the cone whose apex is at the center of one base of the cylinder and whose base is the other base of the cylinder. By the Pythagorean theorem , the plane located y {\displaystyle y} units above the "equator" intersects the sphere in a circle of radius r 2 − y 2 {\textstyle {\sqrt {r^{2}-y^{2}}}} and area π ( r 2 − y ...