TITLE: Arnold Trivium Problem 39
QUESTION [12 upvotes]: We find in Arnold's Trivium the following problem, numbered 39. (The double integral should have a circle through it, but the command /oiint does not work here.)

Calculate the Gauss integral
$$\int \int \frac{(d\vec A, d\vec B, \vec A-\vec B)}{|\vec A-\vec
B|^3},$$
where $\vec A$ runs along the curve $x=\cos \alpha$, $y=\sin \alpha$,
  $z=0$, and $\vec B$ along the curve $x=2\cos^2 \beta$, $y=(1/2)\sin
 \beta$, $z=\sin 2\beta$.

First, what exactly is being asked here? We seem to have an inner product in the integral, except there are three arguments. I am not sure what to make of this -- the notation is unfamiliar to me. And I suppose $\alpha$ and $\beta$ both independently run from $0$ to $2\pi$?
Second, how does one solve this?
Update: David H. reports that the three-place function is probably the vector triple product. "More likely it represents the vector triple product. This is the integral one must evaluate to compute the magnetic force between two current-carrying wires." This interpretation makes sense, because these problems are directed physics students.

REPLY [5 votes]: Here is 'physical' derivation for Gauss's linking number formula. In contrast with David H's remark above, I will talk in terms of magnetic circulation rather than the magnetic force between two current-carrying wires.
Suppose I have two closed curves, with paths $C_1,C_2$ respectively. We can generate a magnetic field by running a uniform current $I$ through $C_1$, which by the Biot-Savart law is $$\mathbf{B}_{1}(\mathbf{r})=\frac{\mu_0 I}{4\pi}\oint_{C_1}\frac{d\mathbf{l}_1\times (\mathbf{r}-\mathbf{r}_1)}{|\mathbf{r}-\mathbf{r}_1|^3}.$$ The resulting circulation of this magnetic field relative to $C_2$ is $$\oint_{C_2}\mathbf{B}_1(\mathbf{r})\cdot d\mathbf{l}=\frac{\mu_0 I}{4\pi}\oint_{C_2}\oint_{C_1}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3}\cdot (d\mathbf{l}_2\times d\mathbf{l}_1)$$ where the cyclic invariance of the scalar triple product has been used to rearrange the integrand.
But Ampere's law informs us that such a circulation integral equal $\mu_0 I_{enc}$ where $I_{enc}$ is the current enclosed by $C_2$; since the current is uniform , this amounts to $I_{enc}=N_{12} I$ where $N_{12}$ is the linking number of $C_1$ relative to $C_2$. Solving for this linking number then finally gives Gauss's formula: $$\boxed{N_{12}=\frac{1}{4\pi}\oint_{C_2}\oint_{C_1}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3}\cdot (d\mathbf{l}_2\times d\mathbf{l}_1)}$$
Consequently your integral indeed amounts to finding the linking number, a task which the other answers have elaborated in sufficient detail.