Surprisingly, it often turns out to be simpler to solve problems involving spheres by working in cylindrical coordinates. We indicate here one of the reasons for this.

The equation of a sphere of radius $a$ in cylindrical coordinates is 1) \begin{equation} r^2 + z^2 = a^2 \end{equation} so that \begin{equation} 2r\,dr + 2z\,dz = 0 \label{Spherecyl} \end{equation} Proceeding as for the paraboloid, we take \begin{eqnarray} d\rr_1 &=& r\,d\phi\,\phat \\ d\rr_2 &=& dr\,\rhat + dz\,\zhat  =  \left( -\frac{z}{r}\,\rhat + \zhat \right) dz \label{drdz} \end{eqnarray} where we now view $r$ as a function of $z$. We therefore have \begin{equation} d\SS = d\rr_1 \times d\rr_2 = (r\,\rhat + z\,\zhat) \,d\phi\,dz \end{equation} which results finally in \begin{equation} \dS = |d\SS| = a\,d\phi\,dz \end{equation} The surface element of a sphere is therefore the same as that of the cylinder of the same radius! Among other things, this means that projecting the Earth outward onto a cylinder is an equal-area projection, which is useful for cartographers.