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 $\approx 3.09404630427286$.

At the first step, I will introduce a proper curve linear coordinates for this problem. This will help to construct the integral for area. We can write the equation of these equi-potential curves as
$$\frac{1}{r_1}+\frac{1}{r_2}=C \tag{1}$$
where $C$ is some real constant and $r_1$ and $r_2$ are defined as
$$r_1=\sqrt{(x-a)^2+y^2} \\
r_2=\sqrt{(x+a)^2+y^2} \tag{2}$$
where $2a$ is the distance between the two like charges on the $x$-axis placed at $x=a$ and $x=-a$. Equation $(2)$ is introducing a new curve-linear coordinates $(r_1,r_2)$ which is called the two-center bipolar coordinates. The geometric interpretation is easy as it just describes the coordinates of a point in $xy$ plane via the distance of that point through two other points which are called the centers. You can take look at this link in WIKI or this post on MSE. However, they doesn't contain that much information.

Then we find $x$ and $y$ from equations in $(2)$ in terms of $r_1$ and $r_2$. For this purpose, subtract and add the equations in $(4)$ to get
$$\begin{align}
r_2^2-r_1^2&=4ax \\
r_2^2+r_1^2&=2(x^2+y^2+a^2) 
\end{align}
\tag{3}$$
after some simplifications, $x$ and $y$ in terms of $r_1$ an $r_2$ will be
$$
\begin{align}
x &= \frac{1}{4a} (r_{2}^{2}-r_{1}^{2})\\
y &= \pm \frac{1}{4a} \left( \sqrt{16 a^2 r_{2}^{2}-(r_{2}^{2}-r_{1}^{2}+4a^2)^2} \right)
\end{align} 
\tag{4}$$
For a given $(r_1,r_2)$ we will find two pairs $(x,y)$ and $(x,-y)$. It is evident from the above picture that why this happens.
The next step will be the construction of the integral for area. We set $C=2a=2$ and find the intersection of the $\infty$ shaped curve with the $x$-axis
$$y=0 \qquad \to \qquad \frac{1}{\sqrt{(x-1)^2}}+\frac{1}{\sqrt{(x+1)^2}}=2 \qquad \to \qquad x=-\phi,0,\phi \tag{5}$$ 
where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio number. So according to $(1)$, $(4)$, and $(5)$ the parametric equations of the curve in the second quadrant of $xy$ plane will be
$$
\begin{align}
x &= \frac{1}{4} \left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}\right)\\
y &= \frac{1}{4} \sqrt{16 r_{2}^{2}-\left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}+4\right)^2} 
\end{align}  
\qquad \qquad \phi-1 \le r_2 \le 1  
\tag{6}$$

Finally, the integral to be evaluated for the area will be
$$\begin{align}
\text{Area} &=4 \int_{-\phi}^{0} y dx \\
&=4 \int_{\phi-1}^{1}y \frac{dx}{dr_2}dr_2 \\
&= \int_{\phi-1}^{1} \sqrt{16 r_{2}^{2}-\left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}+4\right)^2} \left(\frac{r_2}{2}-\frac{r_2}{2(2r_2-1)^2}+\frac{r_{2}^{2}}{(2r_2-1)^3}\right) dr_2
\end{align}$$
The numerical value of area up to fifty digits is
$$\text{Area}=3.0940463058814386237217800770286020796565427678113$$
as stated in the question. However, finding a tricky way to evaluate the definite integral of the area in a closed form should be investigated. I did the computations in this MAPLE file which may be useful for anyone who reads this post.
