Editing Sphere (section)

===Surface area{{anchor|Area}}===
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The [[surface area]] of a sphere of radius {{mvar|r}} is:
:<math>A = 4\pi r^2.</math>

[[Archimedes]] first derived this formula<ref name=MathWorld_Sphere>{{MathWorld |title=Sphere |id=Sphere}}</ref> from the fact that the projection to the lateral surface of a [[circumscribe]]d cylinder is area-preserving.<ref>{{harvnb|Steinhaus|1969|loc=p. 221}}.</ref> Another approach to obtaining the formula comes from the fact that it equals the [[derivative]] of the formula for the volume with respect to {{mvar|r}} because the total volume inside a sphere of radius {{mvar|r}} can be thought of as the summation of the surface area of an infinite number of spherical shells of infinitesimal thickness concentrically stacked inside one another from radius 0 to radius {{mvar|r}}. At infinitesimal thickness the discrepancy between the inner and outer surface area of any given shell is infinitesimal, and the elemental volume at radius {{mvar|r}} is simply the product of the surface area at radius {{mvar|r}} and the infinitesimal thickness.

{{Collapse top|title=Proof of surface area, using calculus}}
At any given radius {{mvar|r}},{{NoteTag |{{mvar|r}} is being considered as a variable in this computation.}} the incremental volume ({{mvar|δV}}) equals the product of the surface area at radius {{mvar|r}} ({{math|''A''(''r'')}}) and the thickness of a shell ({{mvar|δr}}):
:<math>\delta V \approx A(r) \cdot \delta r. </math>

The total volume is the summation of all shell volumes:
:<math>V \approx \sum A(r) \cdot \delta r.</math>

In the limit as {{mvar|δr}} approaches zero<ref name="delta">{{cite book
|author1=E.J. Borowski |author2=J.M. Borwein |title=Collins Dictionary of Mathematics
|year=1989 |isbn=978-0-00-434347-1|pages=141, 149|publisher=Collins }}</ref> this equation becomes:
:<math>V = \int_0^r A(r) \, dr.</math>

Substitute {{mvar|V}}:
:<math>\frac43\pi r^3 = \int_0^r A(r) \, dr.</math>

Differentiating both sides of this equation with respect to {{mvar|r}} yields {{mvar|A}} as a function of {{mvar|r}}:
:<math>4\pi r^2 = A(r).</math>

This is generally abbreviated as:
:<math>A = 4\pi r^2,</math>
where {{mvar|r}} is now considered to be the fixed radius of the sphere.

Alternatively, the [[area element]] on the sphere is given in [[spherical coordinates]] by {{math|1=''dA'' = ''r''<sup>2</sup> sin ''θ dθ dφ''}}. The total area can thus be obtained by [[Integral|integration]]:
:<math>A = \int_0^{2\pi} \int_0^\pi r^2 \sin\theta \, d\theta \, d\varphi = 4\pi r^2.</math>
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The sphere has the smallest surface area of all surfaces that enclose a given volume, and it encloses the largest volume among all closed surfaces with a given surface area.<ref>{{cite journal |last1=Osserman |first1=Robert |journal=Bulletin of the American Mathematical Society |title=The isoperimetric inequality |date=1978 |volume=84 |issue=6 |page=1187 |doi=10.1090/S0002-9904-1978-14553-4 |url=https://www.ams.org/journals/bull/1978-84-06/S0002-9904-1978-14553-4/ |access-date=14 December 2019 |ref=Osserman|doi-access=free }}</ref> The sphere therefore appears in nature: for example, bubbles and small water drops are roughly spherical because the [[surface tension]] locally minimizes surface area.

The surface area relative to the mass of a ball is called the [[specific surface area]] and can be expressed from the above stated equations as
:<math>\mathrm{SSA} = \frac{A}{V\rho} = \frac{3}{r\rho}</math>
where {{mvar|ρ}} is the [[density]] (the ratio of mass to volume).