Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A disk of radius 0.10 m is oriented with its normal unit vector $\hat{n}$ at 30$^{\circ}$ to a uniform electric field $\vec{E}$ of magnitude 2000 N/C. What is the electric flux through the disk?

I know that electric flux is given by the following equation:

$$ \Phi_E = \oint_{S} \vec{E}\cdot \hat{n} \: dA $$

My question is that if you're integrating over a region $S$ and it's with respect to $dA$, how is this not a double integral? Why is

$$ \Phi_E = \int_{0}^{0.10}\int_{0}^{2\pi}r \: \vec{E}\cdot \hat{n} \: d\theta \: dr $$

not right?

share|cite|improve this question
up vote 15 down vote accepted

It is, in fact, a double integral! The first notation used

$$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$

is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} \vec{A}$ represents an infinitesimal patch of a surface, which is a real, physical object, whereas $r \ \mathrm{d}r \ \mathrm{d}\theta$ and $\mathrm{d}x \ \mathrm{d}y$ are ways to express this surface element in a particular coordinate system, which is arbitrary.

share|cite|improve this answer

It is just a more compact notation. It is implied by the integration element $dA$ that you are integrating over the surface.

share|cite|improve this answer

The general formula is indeed a double integral, so the most technically correct way to write it is

$$\Phi_E = \iint_S \vec{E}\cdot\mathrm{d}^2\vec{A}$$

But when formulas start to involve four, five, or more integrals, it gets tedious to write them all out all the time, so there's a notational convention in which a multiple integration can be designated by a single integral sign. The number of integrations you have to do is then shown by the number of differentials in the expression. For example, this is a triple integral:

$$\int \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\,f(x,y,z)$$

share|cite|improve this answer

All the other answers that say the "single" integral is simply a shorthand notation are right, but it is well to remember that one can indeed construe the integral as a single integral as a Lebesgue integral (if you do nothing else, look up Lebesgue's very cute little half paragraph summary (on the Wiki page) of his idea in a letter to Paul Montel).

If you've not met this before, the idea is to define a function that assigns a measure (of "how big") to certain well behaved subsets of the two (or any dimensional region). Then one constructs the sum of the measures, weighted by $f$, of all the subsets which are preimages of intervals $(f,\,f+\Delta f)$ (i.e. subsets wherein the value of $f$ lies between certain values) and passes to certain bounds on these sums as $\Delta f$ is allowed to take any positive value to define the integral.

One of the upshots of this approach is that the domain of integration (in this case a two dimensional plane) does not have to be thought of as slices of lower dimensional spaces pasted together, thus leading to iterated, multiple integrals which one must tackle with the Riemann integral. One just thinks of a lone, non-iterated sum of measures of subsets weighted by the integrated object (in this case $f$). So in the Lebesgue sense $d\,A$ is simply the measure function defined for two-dimensional subsets of the plane.

However, Lebesgue integration over multidimensional manifolds can also be construed as an iterated, in this case, double integral, just like the Riemann integral is and Fubini's theorem is, given certain conditions on $f$, essentially that the iterated Lebesgue integral equals the one defined thinking of subsets of the plane as atomic entities: a corollary of which is that the order of the integration is not important (because the iterated integrals in any order are all equal to the single, multidimensional one).

Of course the two concepts of integral, Lebsegue and Riemann, co-incide given certain conditions on the integrated function. The Lebesgue is more general: Lebesgue integrals are always defined and equal to the Riemann integral when the latter exists, but the converse is not true (there are functions that have a Lebesgue integral, but no Riemann one). Lebesgue integration is extremely important for probability theory, where it is not always convenient to break down an "event" (i.e. measurable subset of the event space) into unions of "lower dimensional" events.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.