derivation of surface area measure on sphere

The sphere of radius $r$ can be described parametrically by spherical coordinates (what else ;) ) as follows:

x=r\\sin u\\sin v

y=r\\sin u\\cos v

z=r\\cos u

Then, using trigonometric identities to simplify expressions we find that

\\frac{\\partial(x,y)}{\\partial(u,v)}=\\left|\\begin{matrix}r\\cos u\\sin v&r\\sin u%\\cos v\\\\r\\cos u\\cos v&-r\\sin u\\sin v\\end{matrix}\\right|=-r^{2}\\cos u\\sin u

\\frac{\\partial(y,z)}{\\partial(u,v)}=\\left|\\begin{matrix}r\\cos u\\cos v&-r\\sin u%\\sin v\\\\-r\\sin u&0\\end{matrix}\\right|=-r^{2}\\sin^{2}u\\sin v

\\frac{\\partial(z,x)}{\\partial(u,v)}=\\left|\\begin{matrix}-r\\sin u&0\\\\r\\cos u\\sin v&r\\sin u\\cos v\\end{matrix}\\right|=r^{2}\\sin^{2}u\\cos v

and hence, using more trigonometric identities, we find that

\\sqrt{\\left(\\frac{\\partial(x,y)}{\\partial(u,v)}\\right)^{2}+\\left(\\frac{%\\partial(y,z)}{\\partial(u,v)}\\right)^{2}+\\left(\\frac{\\partial(z,x)}{\\partial(u%,v)}\\right)^{2}}=

\\sqrt{r^{4}\\cos^{2}u\\sin^{2}u+r^{4}\\sin^{4}u\\sin^{2}v+r^{4}\\sin^{4}u\\cos^{2}v}%=r^{2}\\sin u.

This means that, on a sphere

d^{2}A=r^{2}\\sin u\\>du\\,dv.

Note that in the case of a unit sphere, ( $r=1$ ) this agrees with the formula presented in the second paragraph of subsection 2 of the main entry.

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