Editing Sphere (section)

===Enclosed volume{{anchor|Volume}}===
[[File:Sphere and circumscribed cylinder.svg|thumb|upright=1.1|Sphere and circumscribed cylinder]]

In three dimensions, the [[volume]] inside a sphere (that is, the volume of a [[ball (mathematics)|ball]], but classically referred to as the volume of a sphere) is
:<math>V = \frac{4}{3}\pi r^3 = \frac{\pi}{6}\ d^3 \approx 0.5236 \cdot d^3</math>
where {{mvar|r}} is the radius and {{mvar|d}} is the diameter of the sphere. [[Archimedes]] first derived this formula (''[[On the Sphere and Cylinder]]'' c. 225 BCE) by showing that the volume inside a sphere is twice the volume between the sphere and the [[circumscribe]]d [[cylinder (geometry)|cylinder]] of that sphere (having the height and diameter equal to the diameter of the sphere).<ref>{{harvnb|Steinhaus|1969|loc=p. 223}}.</ref> This may be proved by inscribing a cone upside down into 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 [[Cavalieri's principle]].<ref>{{cite web|url=http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html|title=The volume of a sphere – Math Central|website=mathcentral.uregina.ca|access-date=2019-06-10}}</ref> This formula can also be derived using [[integral calculus]] (i.e., [[disk integration]]) to sum the volumes of an [[infinite number]] of [[Circle#Properties|circular]] disks of infinitesimally small thickness stacked side by side and centered along the {{mvar|x}}-axis from {{math|1=''x'' = −''r''}} to {{math|1=''x'' = ''r''}}, assuming the sphere of radius {{mvar|r}} is centered at the origin.
{{Collapse top|title=Proof of sphere volume, using calculus}}
At any given {{mvar|x}}, the incremental volume ({{mvar|δV}}) equals the product of the cross-sectional [[area of a disc#Onion proof|area of the disk]] at {{mvar|x}} and its thickness ({{mvar|δx}}):
:<math>\delta V \approx \pi y^2 \cdot \delta x.</math>

The total volume is the summation of all incremental volumes:
:<math>V \approx \sum \pi y^2 \cdot \delta x.</math>

In the limit as {{mvar|δx}} approaches zero,<ref name="delta"/> this equation becomes:
:<math>V = \int_{-r}^{r} \pi y^2 dx.</math>

At any given {{mvar|x}}, a right-angled triangle connects {{mvar|x}}, {{mvar|y}} and {{mvar|r}} to the origin; hence, applying the [[Pythagorean theorem]] yields:
:<math>y^2 = r^2 - x^2.</math>

Using this substitution gives
:<math>V = \int_{-r}^{r} \pi \left(r^2 - x^2\right)dx,</math>

which can be evaluated to give the result
:<math>V = \pi \left[r^2x - \frac{x^3}{3} \right]_{-r}^{r} = \pi \left(r^3 - \frac{r^3}{3} \right) - \pi \left(-r^3 + \frac{r^3}{3} \right) = \frac43\pi r^3.</math>

An alternative formula is found using [[spherical coordinates]], with [[volume element]]
:<math> dV=r^2\sin\theta\, dr\, d\theta\, d\varphi</math>
so
:<math>V=\int_0^{2\pi} \int_{0}^{\pi} \int_0^r r'^2\sin\theta\, dr'\, d\theta\, d\varphi
= 2\pi \int_{0}^{\pi} \int_0^r r'^2\sin\theta\, dr'\, d\theta
= 4\pi \int_0^r r'^2\, dr'\ =\frac43\pi r^3.</math>
{{Collapse bottom}}

For most practical purposes, the volume inside a sphere [[Inscribed figure|inscribed]] in a cube can be approximated as 52.4% of the volume of the cube, since {{math|1=''V'' = {{sfrac|{{pi}}|6}} ''d''<sup>3</sup>}}, where {{mvar|d}} is the diameter of the sphere and also the length of a side of the cube and {{sfrac|{{pi}}|6}}&nbsp;≈&nbsp;0.5236. For example, a sphere with diameter 1&nbsp;m has 52.4% the volume of a cube with edge length 1{{Spaces}}m, or about 0.524&nbsp;m<sup>3</sup>.