What is a curve? What is a surface?

We find it productive to ask students to answer these questions. There are of course many possible answers, but the one we like best is that a curve is simply a 1-dimensional set of points, while a surface is a 2-dimensional set of points. In particular, we view parameterizations as a subsidiary concept, which is not part of the definition.

With the important exception of particle motion, most physical curves and surfaces are specified by their shape, not by a parameterization. Many curves and surfaces, especially those with high symmetry, can be more easily described by an equation rather than an explicit parameterization. For instance, the unit circle can be described by $x^2+y^2=1$, or better yet $r=1$. For many problems, this is enough, and there is no need to parameterize $x$ and $y$ in terms of an angle.

Yet the standard techniques for line and surface integrals assume that a parameterization is given. This leads the student to an algebraic formalism for computing such integrals, rather than using geometric reasoning. The heart of our efforts to reverse this tendency is to use vector differentials and to “use what you know”, in whatever form you know it.

Perhaps the key problem-solving strategy we teach our students is to use what they know, rather than trying to apply a set strategy to all problems of a given type. In particular, we discourage students from explicitly parameterizing curves unless they have to.

To illustrate this approach, consider the problem of trying to find the length of a quarter of a circle. What do you know? In polar coordinates, a circle is given by $r=\hbox{constant}$, so that $dr=0$. Inserting this fact into the expression (6) of Chapter 3 for arclength in polar coordinates, 1) one immediately obtains \begin{equation} ds^2 = r^2 \, d\phi^2 \end{equation} and finally \begin{equation} \int_C \, ds = \int_0^{\frac{\pi}{2}} r\,d\phi = \frac{\pi r}{2} \label{answer} \end{equation}

But what if you didn't remember (6) of Chapter 3? The calculation is not much harder in rectangular coordinates: You know that $x=r\cos\phi$ and $y=r\sin\phi$, with $r=\hbox{constant}$, so that $dx=-r\sin\phi\,d\phi$ and $dy=r\cos\phi\,d\phi$. Inserting this into (4) of Chapter 3 again leads to ($\ref{answer}$).

But what if you didn't even remember how to parameterize a circle, or, equivalently, how to use polar coordinates? Well, you still know that $x^2+y^2=r^2=\hbox{constant}$, so that $2x\,dx+2y\,dy=0$. Solving for $dy$ and inserting this into (4) of Chapter 3 yields \begin{equation} ds^2 = \left(1+\frac{x^2}{y^2}\right) dx^2 = \frac{r^2}{r^2-x^2} \> dx^2 \end{equation} This leads to the (improper!)\ integral \begin{equation} \int_0^r {{dx}\over\sqrt{1-{{x^2}\over{r^2}}}} \end{equation} which is, of course, most easily computed via a trig substitution — or numerically — yielding the same answer.

The point is that using what you know will always yield correct answers — eventually.

The “use what you know” philosophy is especially powerful in the context of vector line integrals. Suppose you want to find the work done by the force $\FF=x\,\ii+y\,\jj$ when moving along a given curve $C$, using the formula \begin{equation} W = \Lint \FF\cdot d\rr \end{equation} Curves can be specified in several different ways; let us consider some examples, all of which start at $(1,0)$ and end at $(0,1)$.

Consider first the parametric curve $\rr=(1-t^2)\,\ii+t\,\jj$. It is straightforward to compute \begin{equation} d\rr = (-2t\,\ii + \jj) \, dt \end{equation} Since in this case the given vector field $\FF$ is just $\rr$ itself, we get \begin{equation} \Lint \FF\cdot d\rr = \int_0^1 \left( (1-t^2)(-2t)+t \right) dt = \int_0^1 (2t^3-t) dt = 0 \end{equation}

In physical applications, one is rarely given an explicit parameterization of the curve, but rather some other description. For instance, the curve just discussed might have been defined by the equation $x=1-y^2$. Finding the differential of both sides of this expression yields $dx=-2y\,dy$, and substituting into (2) of Chapter 3 leads to \begin{equation} d\rr = (-2y\,\ii + \jj) \,dy \end{equation} The computation is exactly the same as before, using $y$ instead of $t$.

Alternatively, one can solve for $y$, obtaining $y=\sqrt{1-x}$, then compute $dy$ in terms of $dx$, then substitute into (2) of Chapter 3, obtaining 1) \begin{equation} d\rr = dx\,\ii - \frac{dx}{2\sqrt{1-x}}\,\jj \end{equation} so that \begin{equation} \int\FF\cdot d\rr = \int_1^0 \left(x-\frac12\right)\,dx = 0 \end{equation}

It is empowering for the student to learn that all of these approaches will work, some more quickly than others. This skill can be nicely emphasized by an appropriate group activity, during which different groups report on different approaches.

In a “use what you know” strategy, students sometimes don't know when to stop. For example, they may correctly substitute for $dy$ in terms of $dx$ in $d\rr$, but forget to substitute for $y$ in terms of $x$ in $\FF$. The rule of thumb which students should learn for line integrals is that they shouldn't start integrating until they have the integral in terms of a single parameter, including correctly determining the limits in terms of that parameter. Lines are one-dimensional!

This strategy works well for straight lines: Simply determine the changes in each coordinate, and use ratios to relate the corresponding differentials.

One final point: Vector line integrals of the form $\int\FF\cdot d\rr$ are directed integrals (unlike integrals with respect to arclength, of the form $\int f\,ds$); the sign of the answer depends on which way you traverse the curve. Students will obtain the correct sign automatically if they integrate from the beginning point to the final point, without putting in any artificial signs. As in the last example, this may result in an integral which goes from a larger value of the integration variable to a smaller one.

The approach we have set up for line integrals can easily be extended to surface integrals. The key ingredient is to note that, for a given surface $S$, the vector surface element $d\SS$ is just the (appropriately ordered) cross product of the $d\rr$'s computed for (any!) two non-collinear families of curves lying in the surface. That is \begin{equation} d\SS = \nn\,\dS = d\rr_1 \times d\rr_2 \end{equation} where $\dS$ is the (scalar) surface element of $S$, and $\nn$ is the unit normal vector to $S$ (with given orientation). While any families of curves will work, in practice it is important to choose ones which make it as easy as possible to express the limits of integration.

Furthermore, from this point of view the scalar surface element $\dS$ is just the magnitude of the vector surface element $d\SS$, that is \begin{equation} \dS = |d\SS| \end{equation} Formulas for surface area can be easily obtained using this approach. 1)

We first illustrate this approach on a problem typical of those in calculus textbooks. Consider finding the flux of the vector field $\FF=z\,\kk$ up through the part of the plane $x+y+z=1$ lying in the first octant. We begin with the infinitesimal vector displacement in rectangular coordinates in 3 dimensions, namely \begin{equation} d\rr = dx\,\ii + dy\,\jj + dz\,\kk \end{equation} A natural choice of curves in this surface is given by setting $y$ or $x$ constant, so that $dy=0$ or $dx=0$, respectively. We thus obtain \begin{eqnarray} d\rr_1 &=& dx\,\ii + dz\,\kk = (\ii-\kk)\,dx \\ d\rr_2 &=& dy\,\jj + dz\,\kk = (\jj-\kk)\,dy \end{eqnarray} where we have used what we know (the equation of the plane) to determine each expression in terms of a single parameter. The surface element is thus \begin{equation} d\SS = d\rr_1\times d\rr_2 = (\ii+\jj+\kk)\,dx\,dy \end{equation} and the flux becomes 2) \begin{equation} \Sint \FF\cdot d\SS = \Sint z \,dx\,dy = \int_0^1 \int_0^{1-y} (1-x-y) \,dx\,dy =\frac16 \end{equation} The limits were chosen by visualizing the projection of the surface into the $xy$-plane, which is a triangle bounded by the $x$-axis, the $y$-axis, and the line whose equation is $x+y=1$. Note that this latter equation is obtained from the equation of the surface by using what we know — namely that $z=0$.

Just as for line integrals, there is a rule of thumb which tells you when to stop using what you know to compute surface integrals: Don't start integrating until the integral is expressed in terms of two parameters, and the limits in terms of those parameters have been determined. Surfaces are two-dimensional!

Two of the most fundamental examples in electromagnetism are the magnetic field around a wire and the electric field of a point charge. We consider each in turn.

The magnetic field along an infinitely long straight wire along the $z$ axis, carrying uniform current $I$, is given by \begin{equation} \BB = \frac{\mu_0I}{2\pi} \frac{\phat}{r} = \frac{\mu_0I}{2\pi} \frac{x\,\jj-y\,\ii}{x^2+y^2} \end{equation} where $\mu_0$ and $I$ are constant. Note that the first expression clearly indicates both the direction of $\BB$ and its $\frac{1}{r}$ fall-off behavior, while the second expression does neither. Given the magnetic field, Ampère's Law allows one to determine the current flowing through (not around) any loop $C$, namely \begin{equation} \mu_0 I = \Lint \BB\cdot d\rr \end{equation} which is just Stokes' Theorem. We verify this for the case of a circle around the $z$-axis.

What do we know? Our circle lies in the $xy$-plane; it is enough to use polar coordinates. We therefore start with the formula (5) of Chapter 3 and insert $dr=0$, so that $d\rr=r\,d\phi\,\phat$, and the integral becomes \begin{equation} \Lint \BB\cdot d\rr % = \Lint \frac{\mu_0I}{2\pi} \frac{\phat}{r} \cdot r\,d\phi\,\phat = \int_0^{2\pi} \frac{\mu_0I}{2\pi} \,d\phi = \mu_0 I \end{equation} as expected; the only current flowing through this loop is that flowing along the $z$-axis.

This was too easy! Yes, of course, one can solve this problem using rectangular coordinates, but more work is required, involving both algebra and trigonometry; the geometry is lost.

The electric field of a point charge $q$ at the origin, which is given by \begin{equation} \EE = \frac{q}{4\pi\epsilon_0} \frac{\rhat}{r^2} = \frac{q}{4\pi\epsilon_0} \frac{x\,\ii+y\,\jj+z\,\kk}{(x^2+y^2+z^2)^{3/2}} \end{equation} where $\rhat$ is now the unit vector in the radial direction in spherical coordinates. Note that the first expression clearly indicates both the spherical symmetry of $\EE$ and its $\frac{1}{r^2}$ fall-off behavior, while the second expression does neither. % Given the electric field, Gauss' Law allows one to determine the total charge inside any closed surface, namely \begin{equation} \frac{q}{\epsilon_0} = \Sint \EE\cdot d\SS \end{equation} which is of course just the Divergence Theorem.

It is easy to determine $d\SS$ on the sphere by inspection; we nevertheless go through the details of the differential approach for this case. We use “physicists' conventions” for spherical coordinates, so that $\theta$ is the angle from the North Pole, and $\phi$ the angle in the $xy$-plane; see §7.2 in Chapter 7. We use the obvious families of curves, namely the lines of latitude and longitude. Starting either from the general formula for $d\rr$ in spherical coordinates, namely \begin{equation} d\rr = dr\,\rhat + r\,d\theta\,\that + r\sin\theta\,d\phi\,\phat \end{equation} or directly using the geometry behind that formula, one quickly arrives at \begin{eqnarray} d\rr_1 &=& r\,d\theta\,\that \\ d\rr_2 &=& r\sin\theta\,d\phi\,\phat \\ d\SS &=& d\rr_1 \times d\rr_2 = r^2\sin\theta\,d\theta\,d\phi\,\rhat \end{eqnarray} so that \begin{equation} \Sint \EE\cdot d\SS = \int_0^{2\pi} \int_0^\pi \frac{q}{4\pi\epsilon_0} \frac{\rhat}{r^2} \cdot r^2 \sin\theta\,d\theta\,d\phi \,\rhat = \frac{q}{\epsilon_0} \end{equation} as expected.

Finally, we note that this approach can be extended to volume integrals using the triple product. The volume element becomes \begin{equation} dV = (d\rr_1\times d\rr_2)\cdot d\rr_3 \end{equation} for the $d\rr$'s computed for (any!) 3 non-coplanar families of curves. 1) Using the natural families of curves in spherical coordinates, we could take $d\rr_1$ and $d\rr_2$ as above, and add \begin{equation} d\rr_3 = dr \, \rhat \end{equation} For any orthogonal coordinate system, this is of course equivalent to visualizing a small coordinate box, and multiplying the lengths of the 3 sides.

The reader may have the feeling that two quite different languages are being spoken here. The tilted plane was treated in essentially the traditional manner found in calculus textbooks, using rectangular coordinates. While the “use what you know” strategy may be somewhat unfamiliar, the basic idea should not be. On the other hand, the examples in the previous section will be quite unfamiliar to most mathematicians, due to their use of adapted basis vectors such as $\rhat$. Mathematicians should realize that mastering these examples helps students learn to look for symmetry and to argue geometrically. At the same time, mathematicians may not be satisfied by an approach which seems to be so intimately dependent on the particular problem being studied. After all, not all problems have symmetry!

We argue, however, that the approach being presented here is much more flexible than may appear at first sight. We demonstrate this flexibility by integrating over a paraboloid, the classic example found in calculus textbooks.

We compute the flux of the axially symmetric vector field \begin{equation} \FF = r\,\rhat = x\,\ii + y\,\jj \end{equation} “outwards” through the part of the paraboloid $z=r^2$ lying below the plane $z=4$. The first thing we need is the formula for $d\rr$ in cylindrical coordinates, which is a straightforward generalization of (5) of Chapter 3 in polar coordinates, namely \begin{equation} d\rr = dr\,\rhat + r\,d\phi\,\phat + dz\,\zhat \label{dr3} \end{equation}

Next, we need two families of curves on the paraboloid; the natural choices are those with $r=\hbox{constant}$ or $\phi=\hbox{constant}$, respectively. If $r$ is constant, so is $z$, and we have simply \begin{equation} d\rr_1 = r\,d\phi\,\phat \end{equation} If $\phi$ is constant, there will be no $\phat$ term in $d\rr_2$, but we must still use what we know to compute $dz=2r\,dr$, thus obtaining \begin{equation} d\rr_2 = dr\,\rhat + 2r\,dr\,\zhat \end{equation} Taking the cross product leads to \begin{equation} d\SS = d\rr_1 \times d\rr_2 = (2r^2\,\rhat - r\zhat) \,dr\,d\phi \end{equation} and at this point we must check that we have chosen the correct orientation. (We have, since the coefficient of $\zhat$ is negative.) The rest is easy: Compute the dot product, determine the limits, and do the integral. This results in \begin{equation} \Sint \FF \cdot d\SS % = \Sint r\,\rhat \cdot (2r^2\,\rhat - r\zhat) \,dr\,d\phi = \int_0^{2\pi} \int_0^2 2r^3 \,dr\,d\phi = 16\pi \end{equation}

It is of course also possible to use rectangular coordinates to determine $d\SS$, choosing curves on the paraboloid $z=x^2+y^2$ with $x=\hbox{constant}$ or $y=\hbox{constant}$. The resulting integral cries out for polar coordinates — which turns it into the same integral as the above.

There is however a significant difference between these two approaches: Drawing the curves on the paraboloid with $r$ or $\phi$ held constant is easy; drawing the curves with $x$ or $y$ held constant is not. Try it! Could you do this for students without preparation? 1) The rectangular coordinate approach relies for its geometric intuition on the generic figure found in most textbooks which maps the rectangular “parameter space” into the surface, whereas the cylindrical coordinate approach starts by choosing the geometrically obvious curves in the given surface.

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.

The traditional approach to curves and surfaces involves parameterization, which we have deliberately saved for last. Recall from (2) of Chapter 3 that for a parametric curve we have \begin{eqnarray*} d\rr = {d\rr\over du} \,du \end{eqnarray*} On a parametric surface described by \begin{equation} \rr=\rr(u,v) \end{equation} it is natural to construct $d\SS$ using curves of constant parameter. But along a curve with $v=\hbox{constant}$, we have \begin{equation} d\rr_1 = \Partial{\rr}{u} \,du \end{equation} and similarly along a curve with $u=\hbox{constant}$ we have \begin{equation} d\rr_2 = \Partial{\rr}{v} \,dv \end{equation} Thus, \begin{equation} d\SS = d\rr_1\times d\rr_2 = \left(\Partial{\rr}{u} \times \Partial{\rr}{v}\right) \> du\,dv \end{equation} This formula can be found in most calculus texts, and may help students to relate our approach to the more traditional one.