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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]
Surface areas of hyperspheres in dimensions 0 through 25. Let A n − 1 (R) denote the hypervolume of the (n − 1)-sphere of radius R. The (n − 1)-sphere is the (n − 1)-dimensional boundary (surface) of the n-dimensional ball of radius R, and the sphere's hypervolume and the ball's hypervolume are related by:
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 surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m −1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus
All of the curves are circles: the curves that intersect 0,0,0,1 have an infinite radius (= straight line). In mathematics , an n -sphere or hypersphere is an n {\displaystyle n} - dimensional generalization of the 1 {\displaystyle 1} -dimensional circle and 2 {\displaystyle 2} -dimensional sphere to any non-negative ...
An approximation for the volume of a thin spherical shell is the surface area of the inner sphere multiplied by the thickness t of the shell: [2] V ≈ 4 π r 2 t , {\displaystyle V\approx 4\pi r^{2}t,}
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 =.