Search results
Results from the WOW.Com Content Network
The volume of a n-ball is the Lebesgue measure of this ball, which generalizes to any dimension the usual volume of a ball in 3-dimensional space. The volume of a n -ball of radius R is R n V n , {\displaystyle R^{n}V_{n},} where V n {\displaystyle V_{n}} is the volume of the unit n -ball , the n -ball of radius 1 .
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 aperture angle, i.e., φ is the angle between the rim of the cap and the ...
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 3-sphere is the boundary of a -ball in four-dimensional space. The -sphere is the boundary of an -ball. Given a Cartesian coordinate system, the unit -sphere of radius can be defined as:
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.
For example, assuming the Earth is a sphere of radius 6371 km, the surface area of the arctic (north of the Arctic Circle, at latitude 66.56° as of August 2016 [7]) is 2π ⋅ 6371 2 | sin 90° − sin 66.56° | = 21.04 million km 2 (8.12 million sq mi), or 0.5 ⋅ | sin 90° − sin 66.56° | = 4.125% of the total surface area of the Earth.
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 =.
First we need to write surface area of the sphere, in terms of the volume of the object being measured, A s 3 = ( 4 π r 2 ) 3 = 4 3 π 3 r 6 = 4 π ( 4 2 π 2 r 6 ) = 4 π ⋅ 3 2 ( 4 2 π 2 3 2 r 6 ) = 36 π ( 4 π 3 r 3 ) 2 = 36 π V p 2 {\displaystyle A_{s}^{3}=\left(4\pi r^{2}\right)^{3}=4^{3}\pi ^{3}r^{6}=4\pi \left(4^{2}\pi ^{2}r^{6 ...