There are many ways to describe a surface. Consider the following following descriptions:

• the unit sphere;
• $x^2+y^2+z^2=1$;
• $r=1$ (where $r$ is the spherical radial coordinate);
• $x=\sin\theta\cos\phi$, $y=\sin\theta\sin\phi$, $z=\cos\theta$, with $\theta\in[0,\pi]$ and $\phi\in[0,2\pi)$;
• $\rr = x\,\ii + y\,\jj + z\,\kk$, with $x$, $y$, $z$ given as above;

all of which describe the same surface. Which representation is best for a given problem depends on the circumstances.

The simplest surfaces are those given by holding one of the coordinates constant. Thus, the $xy$-plane is given by $z=0$. Its (surface) area element is $dA=(dx)(dy)=(dr)(r\,d\phi)$, as can easily be seen by drawing the appropriate small rectangle. The surface of a cylinder is nearly as easy, as it is given by $r=a$ in cylindrical coordinates, and drawing a small “rectangle” yields for the surface element \begin{eqnarray*} \hbox{cylinder}: \qquad && \dS = (a\,d\phi)(dz) = a\, d\phi \, dz \\ \noalign{\noindent while a similar construction for the sphere given by $r=a$ in spherical coordinates yields} \noalign{\smallskip} \hbox{sphere}: \qquad && \dS = (a\,d\theta)(a\sin\theta\,d\phi) = a^2 \sin\theta \, d\theta \, d\phi \end{eqnarray*} The last expression can of course be used to compute the surface area of a sphere, which is 1) $$\Int_{\rm sphere} \!\! \dS = \int_0^{2\pi} \int_0^\pi a^2 \sin\theta \, d\theta \, d\phi = 4 \pi a^2 $$

What about more complicated surfaces?

The basic building block for surface integrals is the infinitesimal area $\dS$, obtained by chopping up the surface into small pieces. If the pieces are small parallelograms, then the area can be determined by taking the cross product of the sides!

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, 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 surface area, 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 $ds=|d\rr|$.

We illustrate this technique by computing the surface element for the paraboloid given by $z=x^2+y^2$. We start with the basic formula for the vector differential $d\rr$, namely $$d\rr = dx\,\ii + dy\,\jj + dz\,\kk$$ What do you know? The expression for $z$ leads to $$dz = 2x\,dx + 2y\,dy$$ 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 $$d\rr_1 = dx\,\ii + 2x\,dx\,\kk$$ Similarly, in the $y$ direction, $dx=0$, which leads to $$d\rr_2 = dy\,\jj + 2y\,dy\,\kk$$ Putting this together, we obtain $$d\SS = d\rr_1 \times d\rr_2 = (\ii + 2x\,\kk) \times (\jj + 2y\,\kk) \,dx\,dy = (-2x\,\ii - 2y\,\jj + \kk) \,dx\,dy $$ for the vector surface element, and $$\dS = \left|{-}2x\,\ii - 2y\,\jj + \kk\right| \,dx\,dy = \sqrt{1+4x^2+4y^2}\,dx\,dy $$ for the scalar surface element. Can you repeat this computation in cylindrical coordinates?

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 a 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.

A common technique, which we now discuss, is to introduce these two parameters at the beginning, thus representing the surface as a parametric surface. However, the expression (\ref{Surface}) for $d\SS$ makes no reference to the parameters! In fact, $d\SS$ can be computed using any two (independent) infinitesimal displacements in the surface; no parameterization is needed! You will have to decide for yourself which approach you prefer. But real-world problems rarely come equipped with parameterizations, unlike those found in calculus texts.

Recall that a parametric curve can be written $$\rr = \rr(u) = x(u)\,\ii + y(u)\,\jj + z(u)\,\kk$$ together with an appropriate domain for the parameter $u$. A parametric surface is similar, except there are now two parameters $u$,$v$ (and an appropriate domain): $$\rr = \rr(u,v) = x(u,v)\,\ii + y(u,v)\,\jj + z(u,v)\,\kk$$

Here are some examples of parametric surfaces (without domains): \begin{eqnarray*} \hbox{sphere}: \qquad && \rr(\theta,\phi) = a\sin\theta\cos\phi\,\ii + a\sin\theta\sin\phi\,\jj + a\cos\theta\,\kk \\ \hbox{cylinder}: \qquad && \rr(\phi,z) = a\cos\phi\,\ii + a\sin\phi\,\jj + z\,\kk \\ z=f(x,y): \qquad && \rr(x,y) = x\,\ii + y\,\jj + f(x,y)\,\kk\cr z=f(r,\phi): \qquad && \rr(r,\phi) = r\cos\phi\,\ii + r\sin\phi\,\jj + f(r,\phi)\,\kk\\ \hbox{surface of revolution}: \qquad && \rr(x,\phi) = x\,\ii + f(x)\cos\phi\,\jj + f(x)\sin\phi\,\kk \\ \hbox{(2-d) change of variables}: \qquad && \rr(u,v) = x(u,v)\,\ii + y(u,v)\,\jj \end{eqnarray*}

How do you find the surface element of a parametric surface? Use curves with $u$ or $v$ constant to chop up the surface. The sides of a small parallelogram are then $$ d\rr_1 = \Partial{\rr}{u}\,du \qquad d\rr_2 = \Partial{\rr}{v}\,dv $$ so that $$d\SS = d\rr_1 \times d\rr_2 = \left( \Partial{\rr}{u} \times \Partial{\rr}{v} \right) \, du \, dv $$ and $$\dS = |d\SS| = \left| \Partial{\rr}{u} \times \Partial{\rr}{v} \right| \> du \, dv $$ We reiterate, however, that it is not always necessary to have an explicit parameterization in order to determine the surface element, as illustrated by the above computation for the paraboloid.

Here are some common surface elements: \begin{eqnarray*} \hbox{(plane)}\quad\hbox{$z={\rm const}$}: \qquad && d\SS=\pm \kk\, dx\,dy \\ \hbox{(plane)}\quad\hbox{$z={\rm const}$}: \qquad && d\SS=\pm \kk\, r\,dr\,d\phi \\ \hbox{(cylinder)}\quad\hbox{$r={\rm const}$}: \qquad && d\SS=\pm \,\rhat\, r\,d\phi\,dz \\ \hbox{(sphere)}\quad\hbox{$r={\rm const}$}: \qquad && d\SS=\pm \,\rhat\, r^2\sin\theta\,d\theta\,d\phi \end{eqnarray*} where we must distinguish between the the cylindrical unit vector $\rhat=(x\,\ii+y\,\jj)/r $ (with the polar coordinate $r$) and the spherical unit vector $\rhat=(x\,\ii+y\,\jj+z\,\kk)/r $ (with the spherical coordinate $r$, sometimes written as $\rho$).

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

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