Chopping up the surface is usually done by drawing two families of curves on the surface; surfaces are 2-dimensional. So compute $d\rr$ on each family, that is, take the cross product, and get the vector surface element in the form \begin{equation} d\SS = d\rr_1 \times d\rr_2 \label{Surface} \end{equation} In order to determine the area of the vector surface element, we need the magnitude of this expression, which is \begin{equation} \dS = |d\SS| = |d\rr_1\times d\rr_2| \label{Scalar} \end{equation} and which is called the (scalar) surface element. This should remind you of the corresponding expression for line integrals, namely $ds=|d\rr|$.

Figure 1: Chopping up a paraboloid in rectangular coordinates.

We illustrate this technique by computing the surface element for the paraboloid given by $z=x^2+y^2$, as shown in Figure 1, with the two families of curves corresponding to $\{x=\hbox{const}\}$ and $\{y=\hbox{const}\}$. We start with the basic formula for the vector differential $d\rr$, namely \begin{equation} d\rr = dx\,\ii + dy\,\jj + dz\,\kk \end{equation} What do you know? The expression for $z$ leads to \begin{equation} dz = 2x\,dx + 2y\,dy \end{equation} In rectangular coordinates, it is natural to consider infinitesimal displacements in the $x$ and $y$ directions. In the $x$ direction, $y$ is constant, so $dy=0$, and we obtain \begin{equation} d\rr_1 = dx\,\ii + 2x\,dx\,\kk \end{equation} Similarly, in the $y$ direction, $dx=0$, which leads to \begin{equation} d\rr_2 = dy\,\jj + 2y\,dy\,\kk \end{equation} Putting this together, we obtain \begin{eqnarray} d\SS = d\rr_1 \times d\rr_2 &=& (\ii + 2x\,\kk) \times (\jj + 2y\,\kk) \,dx\,dy \nonumber\\ &=& (-2x\,\ii - 2y\,\jj + \kk) \,dx\,dy \label{parabolaR} \end{eqnarray} for the vector surface element, and \begin{equation} \dS = \left|{-}2x\,\ii - 2y\,\jj + \kk\right| \,dx\,dy = \sqrt{1+4x^2+4y^2}\,dx\,dy \end{equation} for the scalar surface element. Can you repeat this computation in cylindrical coordinates? (See § {Less Symmetric Surfaces} for the answer.)

This construction emphasizes that “area” is really a vector, whose direction is perpendicular to the surface, and whose magnitude is the area. Note that there are always two choices for the direction; choosing one determines the orientation of the surface.

When using ($\ref{Surface}$) and ($\ref{Scalar}$), it doesn't matter how you chop up the surface. It is of course possible to get the opposite orientation, for instance by interchanging the roles of $d\rr_1$ and $d\rr_2$. Rather than worrying too much about getting the “right” orientation from the beginning, it is usually simpler to check at the end whether the one you've got agrees with the requirements of the problem. If not, insert a minus sign.

Just as a curve is a 1-dimensional set of points, a surface is 2-dimensional. When computing line integrals, it was necessary to write everything in terms of a single parameter before integrating. Similarly, for surface integrals one must write everything in terms of two parameters before starting to integrate.

Finally, a word about notation. You will often see $dS$ instead of $dA$, and $d\vec S$ instead of $d\SS$; most authors use $dA$ in the $xy$-plane.