Derivation FOR THE SURFACE AREA OF A SPHERE

[tex] 4 \pi r^2 = SA [/tex] for a sphere. everybody knows that.

but, how do you derive this by integrating infinitesimal amounts of area over the curvature of the sphere????
 

lanedance

Homework Helper
3,305
2
think of a curve you could rotate around an axis to get a sphere

an surface area element is dA = 2.pi.ds, with ds an infinitesimal length along curve... imagine unrotating a tiny strip
 

lurflurf

Homework Helper
2,410
116
There are many ways
-integrate over the area
note r^2=x^2+y^2+z^2
or use parameters like
x=cos(s)cos(t)
y=cos(s)sin(t)
z=sin(s)
-project a circle
the projection will be 4 circles
4*pi*r^2
-consider the solid of revolution of a semi circle
-differentiate the volume
[(4/3)pi*r^3]'=4pi*r^2
 

HallsofIvy

Science Advisor
41,626
821
The "standard" way of finding surface area (although I like orthovector's method, treating it as a semicircle rotated around the y-axis better):

In spherical coordinates, [itex]x= \rho cos(\theta)sin(\phi)[/itex], [itex]y= \rho sin(\theta)sin(\phi)[/itex], [itex]z= \rho cos(\phi)[/itex].

Taking [itex]\rho[/itex] constant, say [itex]\rho= R[/itex], gives the surface of the sphere of radius R in terms of parameters [itex]\theta[/itex] and [itex]\rho[/itex] and we can write the vector equation as
[tex]\vec{r}= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)\sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}[/tex]
Differentiating with respect to [itex]\theta[/itex] and [itex]\phi[/itex] gives to tangent vectors to that surface
[tex]\vec{r}_\theta= -Rsin(\theta)sin(\phi}\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}[/tex]
[tex]\vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rcos(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}[/tex]

The "fundamental vector product" for the surface is the cross product of those two tangent vectors:
[tex]R^2cos(\theta)sin^2(\phi)\vec{i}- R^2sin(\theta)sin^2(\phi)\vec{j}- R^2sin(\phi)cos(\phi)\vec{k}[/tex]
is perpendicular to the surface and its length, [itex]R^2 sin(\phi)[/itex] gives the "differential of surface area", [itex]R^2 sin(\phi) d\theta d\phi[/itex]. To cover the entire surface of the sphere, [itex]\theta[/itex] must range from 0 to [itex]2\pi[/itex] and [itex]\phi[/itex] must range from 0 to [itex]\pi[/itex]. The surface area is given by
[tex]\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^\pi R^2 sin(\phi)d\phi d\theta[/tex]

Integrating with respect to [itex]\theta[/itex] immediately gives
[tex]2\pi R^2 \int_{\phi= 0}^\pi sin(\phi)d\phi= -2\pi R^2 cos(\phi)\right|_0^\pi[/tex]
[tex]= -2\pi R^2 \left(-1- 1)= 4\pi R^2[/tex]
 

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