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Section 10.4 General Surface Elements

Since surfaces are two-dimensional, chopping up a surface is usually done by drawing two families of curves on the surface. Then you can compute \(d\rr\) on each family and take the cross product, see Section 1.20, to get the vector surface element in the form

\begin{equation} d\AA = d\rr_1 \times d\rr_2 .\tag{10.4.1} \end{equation}

In order to determine the area of the vector surface element, we need the magnitude of this expression, which is

\begin{equation} \dA = |d\AA| = |d\rr_1\times d\rr_2|\tag{10.4.2} \end{equation}

and which is called the (scalar) surface element. This usage should remind you of the corresponding relation for line integrals, namely \(ds=|d\rr|\text{.}\)

Figure 10.2. Chopping up a paraboloid in rectangular coordinates.

We illustrate this technique by computing the surface element for the paraboloid given by \(z=x^2+y^2\text{,}\) as shown in Figure 10.2, with the two families of curves corresponding to \(\{x=\hbox{const}\}\) and \(\{y=\hbox{const}\}\text{.}\) We start with the basic formula for the vector differential \(d\rr\text{,}\) namely

\begin{equation} d\rr = dx\,\xhat + dy\,\yhat + dz\,\zhat .\tag{10.4.3} \end{equation}

What do you know? The expression for \(z\) leads to

\begin{equation} dz = 2x\,dx + 2y\,dy .\tag{10.4.4} \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 \(dy=0\text{,}\) and we obtain

\begin{equation} d\rr_1 = dx\,\xhat + 2x\,dx\,\zhat = (\xhat + 2x\,\zhat)\, dx .\tag{10.4.5} \end{equation}

Similarly, in the \(y\) direction, \(dx=0\text{,}\) which leads to

\begin{equation} d\rr_2 = dy\,\yhat + 2y\,dy\,\zhat = (\yhat + 2y\,\zhat)\, dy .\tag{10.4.6} \end{equation}

Putting this together, we obtain

\begin{align} d\AA = d\rr_1 \times d\rr_2 \amp= (\xhat + 2x\,\zhat) \times (\yhat + 2y\,\zhat) \,dx\,dy \nonumber\tag{10.4.7}\\ \amp= (-2x\,\xhat - 2y\,\yhat + \zhat) \,dx\,dy\tag{10.4.8} \end{align}

for the vector surface element, and

\begin{equation} \dA = \left|{-}2x\,\xhat - 2y\,\yhat + \zhat\right| \,dx\,dy = \sqrt{1+4x^2+4y^2}\,dx\,dy\tag{10.4.9} \end{equation}

for the scalar surface element. Can you repeat this computation in cylindrical coordinates? (See Section 11.4 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 (10.4.1) and (10.4.2), 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 \(d\rr_2\text{.}\) Rather than worrying too much about getting the “right” orientation from the beginning, it is usually simpler to check after you've calculated \(dSS\) whether the orientations 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 you must write everything, including the limits of integration, in terms of exactly two parameters before starting to integrate.

Finally, a word about notation. You will often see \(\dS\) instead of \(\dA\text{,}\) and \(d\SS\) instead of \(d\AA\text{;}\) most authors use \(dA\) in the \(xy\)-plane.