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If the radius of the sphere is denoted by r and the height of the cap by h, the volume of the spherical sector is =. This may also be written as V = 2 π r 3 3 ( 1 − cos φ ) , {\displaystyle V={\frac {2\pi r^{3}}{3}}(1-\cos \varphi )\,,} where φ is half the cone angle, i.e., φ is the angle between the rim of the cap and the direction ...
Utilizing the pyramid (or cone) volume formula of = ′, where is the infinitesimal area of each pyramidal base (located on the surface of the sphere) and ′ is the height of each pyramid from its base to its apex (at the center of the sphere).
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":
Thus, the segment volume equals the sum of three volumes: two right circular cylinders one of radius a and the second of radius b (both of height /) and a sphere of radius /. The curved surface area of the spherical zone—which excludes the top and bottom bases—is given by =.
The pyramid height is defined as the length of the line segment between the apex and its orthogonal projection on the base. Given that is the base's area and is the height of a pyramid, the volume of a pyramid is: [25] =. The volume of a pyramid was recorded back in ancient Egypt, where they calculated the volume of a square frustum, suggesting ...
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 ...
A cone and a cylinder have radius r and height h. 2. The volume ratio is maintained when the height is scaled to h' = r √ π. 3. Decompose it into thin slices. 4. Using Cavalieri's principle, reshape each slice into a square of the same area. 5. The pyramid is replicated twice. 6. Combining them into a cube shows that the volume ratio is 1:3.
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]