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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 and Cylinder (Greek: Περὶ σφαίρας καὶ κυλίνδρου) is a treatise that was published by Archimedes in two volumes c. 225 BCE. [1] It most notably details how to find the surface area of a sphere and the volume of the contained ball and the analogous values for a cylinder, and was the first to do so. [2]
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.
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 ...
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.
The volume of a sphere with radius r is 4 / 3 πr 3. The surface area of a sphere with radius r is 4πr 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. Apart from circles, there are other curves of constant width.
Plot of the surface-area:volume ratio (SA:V) for a 3-dimensional ball, showing the ratio decline inversely as the radius of the ball increases. A solid sphere or ball is a three-dimensional object, being the solid figure bounded by a sphere. (In geometry, the term sphere properly refers only to the surface, so a sphere thus lacks volume in this ...
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.