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.

To return to the main entry click here