[size=100][size=150]In three dimensions, the [url=https://en.wikipedia.org/wiki/Volume]volume[/url] inside a sphere (that is, the volume of a [url=https://en.wikipedia.org/wiki/Ball_(mathematics)]ball[/url], but classically referred to as the volume of a sphere) is:[br][br][img]https://wikimedia.org/api/rest_v1/media/math/render/svg/42de7dbcc7c51105e0f10e71f53c61987454ced6[/img][br][br]where r is the radius and d is the diameter of the sphere. [url=https://en.wikipedia.org/wiki/Archimedes]Archimedes[/url] first derived this formula by showing that the volume inside a sphere is twice the volume between the sphere and the [url=https://en.wikipedia.org/wiki/Circumscribe]circumscribed[/url] [url=https://en.wikipedia.org/wiki/Cylinder_(geometry)]cylinder[/url] of that sphere (having the height and diameter equal to the diameter of the sphere).[sup][url=https://en.wikipedia.org/wiki/Sphere#cite_note-6][6][/url][/sup] This may be proved by inscribing a cone upside down into a semi-sphere, noting that the area of a cross-section of the cone plus the area of a cross-section of the sphere is the same as the area of the cross-section of the circumscribing cylinder, and applying [url=https://en.wikipedia.org/wiki/Cavalieri%27s_principle]Cavalieri's principle[/url].[sup][url=https://en.wikipedia.org/wiki/Sphere#cite_note-7][7][/url][/sup] This formula can also be derived using [url=https://en.wikipedia.org/wiki/Integral_calculus]integral calculus[/url], i.e. [url=https://en.wikipedia.org/wiki/Disk_integration]disk integration[/url] to sum the volumes of an [url=https://en.wikipedia.org/wiki/Infinite_number]infinite number[/url] of [url=https://en.wikipedia.org/wiki/Circle#Properties]circular[/url] disks of infinitesimally small thickness stacked side by side and centered along the x-axis from [i]x[/i] = −[i]r[/i] to [i]x[/i] = [i]r[/i], assuming the sphere of radius r is centered at the origin. At any given x, the incremental volume (δV) equals the product of the cross-sectional [url=https://en.wikipedia.org/wiki/Area_of_a_disk#Onion_proof]area of the disk[/url] at x and its thickness (δx):[br][br][img]https://wikimedia.org/api/rest_v1/media/math/render/svg/b3f5294b86738de4a6370477df266e31adea58e6[/img]The total volume is the summation of all incremental volume:[br][br][img]https://wikimedia.org/api/rest_v1/media/math/render/svg/9d25825c42ffeebfbf63349381f1bb020398910a[/img]In the limit as δx approaches zero, this equation becomes:[br][br][img]https://wikimedia.org/api/rest_v1/media/math/render/svg/89d1ea760eaab5ec19ef3372ea4c9ee24addee22[/img][br][br]At any given x, a right-angled triangle connects x, y, and r to the origin; hence, applying the [url=https://en.wikipedia.org/wiki/Pythagorean_theorem]Pythagorean theorem[/url] yields: [img]https://wikimedia.org/api/rest_v1/media/math/render/svg/61fa971d8c5410f0cbbbb2f948b29db288e26550[/img]Using this substitution gives:[br]which can be evaluated to give the result[br][img]https://wikimedia.org/api/rest_v1/media/math/render/svg/4c081de9760153a5ab7e59be1b9de1aa97d08dec[/img][br]An alternative formula is found using [url=https://en.wikipedia.org/wiki/Spherical_coordinates]spherical coordinates[/url], with [url=https://en.wikipedia.org/wiki/Volume_element]volume element[/url]{\displaystyle dV=r^{2}[img]https://wikimedia.org/api/rest_v1/media/math/render/svg/e7baab55bb4d5559e61d50df77cca1d7f6befc27[/img]so:[br][br][img]https://wikimedia.org/api/rest_v1/media/math/render/svg/16b51d4bd953c2d8ddb0b746770be5d790eb6e01[/img][br][br]For most practical purposes, the volume inside a sphere [url=https://en.wikipedia.org/wiki/Inscribed_figure]inscribed[/url] in a cube can be approximated as 52.4% of the volume of the cube, since [i]V[/i] = π/6 [i]d[/i][sup]3[/sup], where d is the diameter of the sphere and also the length of a side of the cube and π/6 ≈ 0.5236. For example, a sphere with a diameter of 1 m has 52.4% of the volume of a cube with edge length 1 m, or about 0.524 m[sup]3[/sup][/size][sup][/sup].[/size]