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where H is the hypervolume of a 3-sphere and r is the radius. S V = 2 π 2 r 3 {\displaystyle SV=2\pi ^{2}r^{3}} where SV is the surface volume of a 3-sphere and r is the radius.
On the sphere, he showed that the surface area is four times the area of its great circle. In modern terms, this means that the surface area is equal to: =. The result for the volume of the contained ball stated that it is two-thirds the volume of a circumscribed cylinder, meaning that the volume is
For most practical purposes, the volume inside a sphere inscribed in a cube can be approximated as 52.4% of the volume of the cube, since V = π / 6 d 3, where d is the diameter of the sphere and also the length of a side of the cube and π / 6 ≈ 0.5236.
The volume can be computed without use of the Gamma function. As is proved below using a vector-calculus double integral in polar coordinates, the volume V of an n-ball of radius R can be expressed recursively in terms of the volume of an (n − 2)-ball, via the interleaved recurrence relation:
The volume of a sphere with radius r is 4 / 3 πr 3. The surface area of a sphere with radius r is 4π r 2 . Some of the formulae above are special cases of the volume of the n -dimensional ball and the surface area of its boundary, the ( n −1)-dimensional sphere , given below .
Lines, L. (1965), Solid geometry: With Chapters on Space-lattices, Sphere-packs and Crystals, Dover. 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.
A sphere (top), rotational ellipsoid (left) and triaxial ellipsoid (right) The volume of a sphere of radius R is . Given the volume of a non-spherical object V, one can calculate its volume-equivalent radius by setting = or, alternatively:
As can be seen, the area of the circle defined by the intersection with the sphere of a horizontal plane located at any height equals the area of the intersection of that plane with the part of the cylinder that is "outside" of the cone; thus, applying Cavalieri's principle, it could be said that the volume of the half sphere equals the volume ...