Derivation FOR THE SURFACE AREA OF A SPHERE

[orthovector]

$$4 \pi r^2 = SA$$ 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]

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]

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]

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, $x= \rho cos(\theta)sin(\phi)$, $y= \rho sin(\theta)sin(\phi)$, $z= \rho cos(\phi)$.

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

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

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