Area enclosed by an equipotential curve for an electric dipole on the plane I am currently teaching Physics in an Italian junior high school. Today, while talking about the electric dipole generated by two equal charges in the plane, I was wondering about the following problem:

Assume that two equal charges are placed in $(-1,0)$ and $(1,0)$.
There is an equipotential curve through the origin, whose equation is
given by:
$$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 $$ and
whose shape is very lemniscate-like:



Is there a fast&tricky way to compute the area enclosed by such a curve?

Numerically, it is $\approx 3.09404630427286$.

Here is another method based on the curve-linear coordinates introduced by Achille Hui. He introduced the following change of variables
$$\begin{align}
\sqrt{(x+1)^2+y^2} &= u+v\\
\sqrt{(x-1)^2+y^2} &= u-v
\end{align} \tag{1}$$
Then solving for $x$ and $y$ we shall get
$$\begin{align}
x &= u v\\
y &= \pm \sqrt{-(u^2-1)(v^2-1)}
\end{align} \tag{2}$$
required that
$$-(u^2-1)(v^2-1) \ge 0 \tag{3}$$
It does not look familiar but in fact it is! Taking into account the equations $(2)$ and $(3)$, we can consider the following as a parameterization for the first quadrant of the $xy$ plane
$$\boxed{
\begin{array}{}
x=uv & & 1 \le u \lt \infty \\
y=\sqrt{-(u^2-1)(v^2-1)} &  & 0 \le v \le 1
\end{array}} \tag{4}$$
I tried to draw the coordinate curves of this curve-linear coordinates and I just noticed that it is exactly the same as the Elliptic Coordinates and nothing else! You can show this analytically by the change of variables
$$\begin{align}
u &= \cosh p \\
v &= \cos q
\end{align} \tag{5}$$
I leave the further details in this avenue to the reader.

Let us go back to the problem of calculating the area. The equation of the $\infty$ curve was
$$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 \tag{6}$$
so combining $(1)$ and $(6)$ leads to
$$v=\pm \sqrt{u^2-u} \tag{7}$$ 
and hence the parametric equation of the $\infty$ curve in the first quadrant by considering $(4)$ and $(7)$ will be
$$\boxed{
\begin{array}{}
x=u\sqrt{u^2-u} & & 1 \le u \lt \phi \\
y=\sqrt{-(u^2-1)(u^2-u-1)} 
\end{array}} 
\tag{8}$$
and finally the integral for the area is
$$\begin{align}
\text{Area} &=4 \int_{0}^{\phi} y dx \\
&=4 \int_{1}^{\phi}y \frac{dx}{du}du \\
&=2 \int_{1}^{\phi} (4u-3)\sqrt{-u(u+1)(u^2-u-1)}du \\
&\approx 3.09405
\end{align} \tag{9}$$
