Editing Spherical cap (section)

==Volume and surface area==
The [[volume]] of the spherical cap and the area of the curved surface may be calculated using combinations of
* The radius <math>r</math> of the sphere
* The radius <math>a</math> of the base of the cap
* The height <math>h</math> of the cap
* The [[Spherical coordinate system|polar angle]] <math>\theta</math> between the rays from the center of the sphere to the apex of the cap (the pole) and the edge of the [[disk (mathematics)|disk]] forming the base of the cap.

These variables are inter-related through the formulas
<math> a = r \sin \theta</math>, <math>h = r ( 1 - \cos \theta )</math>, <math>2hr = a^2 + h^2</math>,
and <math>2 h a = (a^2 + h^2)\sin \theta</math>.

{| class="wikitable"
!
! Using <math>r</math> and <math>h</math>
! Using <math>a</math> and <math>h</math>
! Using <math>r</math> and <math>\theta</math>
|-
! Volume
| <math>V = \frac {\pi h^2}{3} (3r-h)</math> <ref name="handbook">{{citation|title=Handbook of Mathematics for Engineers and Scientists|first1=Andrei D|last1=Polyanin|first2=Alexander V.|last2=Manzhirov|publisher=CRC Press|year=2006|isbn=9781584885023|page=69|url=https://books.google.com/books?id=ge6nk9W0BCcC&pg=PA69}}.</ref>
| <math>V = \frac{1}{6}\pi h (3a^2 + h^2)</math>
| <math>V = \frac{\pi}{3} r^3 (2+\cos\theta) (1-\cos\theta)^2 </math>
|-
! Area
| <math>A = 2 \pi r h</math><ref name="handbook"/>
| <math>A =\pi (a^2 + h^2)</math>
| <math>A=2 \pi r^2 (1-\cos \theta)</math>
|-
! Constraints
| <math> 0 \leq h \leq 2 r </math>
| <math> 0 \leq a, \; 0 \leq h </math>
| <math> 0 \leq \theta \leq \pi, \; 0 \leq r</math>
|}
If <math>\phi</math> denotes the [[latitude]] in [[geographic coordinates]], then <math>\theta+\phi = \pi/2 = 90^\circ\,</math>, and <math>\cos \theta = \sin \phi</math>.

=== Deriving the surface area intuitively from the spherical sector volume ===

Note that aside from the calculus based argument below, the area of the spherical cap may be derived from [[Spherical sector#Volume|the volume <math>V_{sec}</math> of the spherical sector]], by an intuitive argument,<ref>{{cite web |last1=Shekhtman |first1=Zor |title=Unizor - Geometry3D - Spherical Sectors |url=https://www.youtube.com/watch?v=ts3J5onzvQg&t=8m54s  |archive-url=https://ghostarchive.org/varchive/youtube/20211222/ts3J5onzvQg |archive-date=2021-12-22 |url-status=live|website=YouTube |publisher=Zor Shekhtman |access-date=31 Dec 2018}}{{cbignore}}</ref> as
:<math>A = \frac{3}{r}V_{sec} = \frac{3}{r} \frac{2\pi r^2h}{3} = 2\pi rh\,.</math>
The intuitive argument is based upon summing the total sector volume from that of infinitesimal [[Tetrahedron|triangular pyramids]]. Utilizing the [[Pyramid (geometry)#Volume|pyramid (or cone) volume]] formula of <math>V = \frac{1}{3} bh'</math>, where <math>b</math> is the infinitesimal [[area]] of each pyramidal base (located on the surface of the sphere) and <math>h'</math> is the height of each pyramid from its base to its apex (at the center of the sphere). Since each <math>h'</math>, in the limit, is constant and equivalent to the radius <math>r</math> of the sphere, the sum of the infinitesimal pyramidal bases would equal the area of the spherical sector, and: 
:<math>V_{sec} = \sum{V} = \sum\frac{1}{3} bh' = \sum\frac{1}{3} br = \frac{r}{3} \sum b = \frac{r}{3} A</math>

===Deriving the volume and surface area using calculus ===
[[File:Spherical cap from rotation.svg|thumb|Rotating the green area creates a spherical cap with height <math>h</math> and sphere radius <math>r</math>.]]
The volume and area formulas may be derived by examining the rotation of the function 
:<math>f(x)=\sqrt{r^2-(x-r)^2}=\sqrt{2rx-x^2}</math> 
for <math>x \in [0,h]</math>, using the formulas the [[Surface of revolution|surface of the rotation]] for the area and the [[Solid of revolution|solid of the revolution]] for the volume.
The area is
:<math>A = 2\pi\int_0^h f(x) \sqrt{1+f'(x)^2} \,dx </math>
The derivative of <math>f</math> is
:<math>f'(x) = \frac{r-x}{\sqrt{2rx-x^2}} </math>
and hence
:<math>1+f'(x)^2 = \frac{r^2}{2rx-x^2} </math>
The formula for the area is therefore
:<math>A = 2\pi\int_0^h \sqrt{2rx-x^2} \sqrt{\frac{r^2}{2rx-x^2}} \,dx
         = 2\pi \int_0^h r\,dx
         = 2\pi r \left[x\right]_0^h
         = 2 \pi r h </math>
The volume is
:<math>V = \pi \int_0^h f(x)^2 \,dx
         = \pi \int_0^h (2rx-x^2) \,dx
         = \pi \left[rx^2-\frac13x^3\right]_0^h
         = \frac{\pi h^2}{3} (3r - h)</math>