As outlined in the worksheet, expressions can be derived for $d\rr$ in other coordinate systems. When completing the worksheet, make sure you think about what the correct lengths are in each coordinate direction. Angles do not measure lengths. Also make sure you know where the center of a circle of constant latitude is. Upon completing the worksheet, you should have obtained the expressions \begin{eqnarray} d\rr &=& dx\,\xhat + dy\,\yhat + dz\,\zhat \\ &=& dr\,\rhat + r\,d\phi\,\phat + dz\,\zhat \\ &=& dr\,\rhat + r\,d\theta\,\that + r\,\sin\theta\,d\phi\,\phat \end{eqnarray} in rectangular, cylindrical, and spherical coordinates, respectively.

The same geometric reasoning can be used to determine the area of surfaces adapted to these coordinates. For instance, on a ($z=\hbox{constant}$) plane, one has \begin{equation} dA = dx\,dy = r\,dr\,d\phi \end{equation} and on the surface of a cylinder one has \begin{equation} dA = r\,d\phi\,dz \end{equation} and on the surface of a sphere one has \begin{equation} dA = r^2\, \sin\theta\,d\theta\,d\phi \end{equation} Similarly, the volume elements in rectangular, cylindrical, and spherical coordinates are given by \begin{equation} d\tau = dx\,dy\,dz = r\,dr\,d\phi\,dz = r^2\, \sin\theta\,dr\,d\theta\,d\phi \end{equation} where we use $d\tau$ rather than $dV$ to avoid confusion with the electric potential.

The geometry behind these formulas is already contained in the derivation of $d\rr$, so that each of these formulas can be quickly recovered if you know the expression for $d\rr$ in the relevant coordinate system.