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\,\yhat-y\,\xhat}{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 (1) of § {Other Coordinate Systems} and insert $dr=0$, so that $d\rr=r\,d\phi\,\phat$, and the integral becomes \begin{equation} \Lint \BB\cdot d\rr = \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 is given by \begin{equation} \EE = \frac{q}{4\pi\epsilon_0} \frac{\rhat}{r^2} = \frac{q}{4\pi\epsilon_0} \frac{x\,\xhat+y\,\yhat+z\,\zhat}{(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. 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.