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&amp;tricky way to compute the area enclosed by such a curve?

Numerically, it is $\approx 3.09404630427286$.
Consider following parametrization of the first quadrant of the $(x,y)$ plane:

$$[1,\infty) \times [0,1] \ni (u,v) 
\quad\mapsto\quad (x,y) \in [0,\infty)^2 \quad\text{s.t.}\quad
\begin{cases}
r_1 &amp;= \sqrt{(x+1)^2+y^2} = u+v\\
r_2 &amp;= \sqrt{(x-1)^2+y^2} = u-v
\end{cases}$$
In this parametrization, the area element is given by
$$dx \wedge dy = -\omega\quad\text{ where }\quad \omega \stackrel{def}{=} \frac{u^2-v^2}{\sqrt{(u^2-1)(1-v^2)}} du \wedge dv$$
Let $D$ be the region in $(u,v)$ plane corresponds to the dipole in first quadrant. Its boundary $\partial D$ consists of 3 pieces


$C_1$ : a curve start at $(1,0)$, end at $(\phi,1)$ where $\phi = \frac{1+\sqrt{5}}{2}$ is the Golden ratio.
$$[1,\phi] \in t \quad\mapsto\quad (u,v) = \left(t,\sqrt{t(t-1)}\right) \in [1,\infty) \times [0,1]$$
$C_2$ : a line segment from $(\phi,1)$ to $(1,1)$.
$C_3$ : a line segment from $(1,1)$ to $(1,0)$.


Notice $\omega = \frac12 d\Omega$ where
$$\Omega \stackrel{def}{=} \frac{u (u^2 - 1) dv - v(1-v^2) du}{\sqrt{(u^2-1)(1-v^2)}}$$
The area $\mathcal{A}$ we seek equals to
$$\mathcal{A} = 4 \int_D \omega = 2\int_{\partial D}\Omega = 2\left(\int_{C_1} + \int_{C_2} + \int_{C_3}\right) \Omega$$

Introduce another parametrization for the region $[1,\infty) \times [0,1]$ in the $(u,v)$ plane:

$$[0,\infty) \times [0,\pi] \ni (\rho,\eta) \quad\mapsto\quad (u,v) = (\cosh\rho,\cos\eta) \in [1,\infty) \times [0,1]$$

One can use this to verify the line segments $C_2$ and $C_3$ contribute nothing to $\mathcal{A}$. As a result,

$$\mathcal{A} = 2\int_{C_1} \Omega 
= 2\int_{C_1} \left(u\sqrt{\frac{u^2-1}{1-v^2}} dv - v\sqrt{\frac{1-v^2}{u^2-1}} du\right)$$

Since $v^2 = u(u-1)$ on curve $C_1$, we can transform the $1^{st}$ piece of the integrand as
$$\begin{align}
u\sqrt{\frac{u^2-1}{1-v^2}} dv
&amp;= \sqrt{\frac{u(u+1)}{1-v^2}} vdv
= \sqrt{u(u+1)} d(-\sqrt{1-v^2})\\
&amp;= -d \sqrt{u(u+1)(1-v^2)} + \sqrt{1-v^2}d\sqrt{u(u+1)}\\
&amp;= -d\sqrt{u(u+1)(1-v^2)} + \sqrt{\color{red}{\frac{1-v^2}{u(u+1)}}}\left(\color{red}{u} + \frac12\right) du
\end{align}
$$
Notice the piece in red can be rewritten as
$\displaystyle\;\sqrt{\frac{u(1-v^2)}{u+1}} du = v\sqrt{\frac{1-v^2}{u^2-1}} du\;$
which is nothing but the $2^{nd}$ piece. This leads to

$$\begin{align}
\mathcal{A} 
&amp;= -2 \left[\sqrt{u(u+1)(1-v^2)}\right]_{(u,v) = (1,0)}^{(\phi,1)} +  \int_{C_1} \sqrt{\frac{1-v^2}{u(u+1)}} du\\
&amp;= \sqrt{8} + \int_1^\phi \sqrt{\frac{1+u-u^2}{u(u+1)}} du \tag{*1}
\end{align}
$$

As a double check, one can throw following command to wolfram alpha,  

Sqrt[8]+Integrate[Sqrt[(1+u-u^2)/(u*(u+1))],{u,1,GoldenRatio}]

to evaluate in $(*1)$ numerically. WA returns
$$\mathcal{A} \approx 
3.0940463058814386237217800770286020796565427678...$$
a number matching what has been stated on question.

This is as far as I can get. I hope someone can further simplify the integral in $(*1)$. Please note that WA do know how to compute the anti-derivative for the integral at $(*1)$. It is a page long expression in terms of elliptic integrals
and I won't reproduce it here.
HINT:

Almost sure area can be evaluated between limits $(u_1,u_2),(v_1,v_2)$ for holomorphic functions of complex variables in conformal maps.

The equipotentials and lines of force form an orthogonal net  of curvilinear rectangle curves (resemble Cassinian Ovals, but different.. one out of which is shown above), and force lines which are all concurrent hyperbolas passing through the pole of same polarity / charge and asymptotic to y-axis. For these $(u,v)$ parameter plots that satisfy Laplace equation all we need before evaluating integral of  area is to identify limits of u and v for any oval chosen. (shall upload equation , if it is found in sites e.g. 2d curves.com)

EDIT1:

The closest that I can get to the function is sketched by Mark McClure in another context for  $ f(z)= z^2 (z-1)^3 $:

Orth Traj to Hyperbolae

I think we should choose another complex variable part so that our hyperbolae pass through $ (\pm 1,0)$ instead of $ ( 0,0),(1,0)$ as given above. We are interested in a particular orthogonal trajectory doubly intersecting at origin. Calculation of area may be easier.