Analysis of electric field lines

Equation for a dipole field line is $r=R {sin}^{2} \theta$ (this is an equation in polar form) because as we already know for tiny dipole :

$\frac { dr }{ rd\theta } =\frac { E\quad component\quad along\quad r }{ E\quad component\quad along\quad theta(polar\quad angle) } =2cot\theta$

Now taking $d\theta$ to other side and integrating we get :

$ln(r)=2ln(sin\theta )+d$ or $r=c {sin}^ {2}\theta$ .

At $\frac{\pi}{2},r= R$

so $r= R{sin}^{2}\theta$

where $x$ is the polar angle and $R$ is the radius at equator, or at $\theta= 90$ degree,

(note -- the above equation is same as it is for magnetic field line ) it is valid for both as magnetic field line is simply a dipole field line, no difference

Now at $x= \frac{\pi}{2}, r =1$ as it is said that it passes through $(0,1)$ hence $R=1$

Now in polar co-ordinates area is given as integral of :

$\int _{ 0 }^{ \theta }{ \frac { 1 }{ 2 } } { r }^{ 2 }d\theta$

integrate from $0$ to $\frac{\pi}{2}$ and multiply it by $2$ to get area of the closed loop as :

$\frac { 3\pi }{ 16 }$

thus answer $19$

Here's the cartesian approach (please refer to deepanshu's diagram.

$\frac{dy}{dx}=\frac{{E}_{y}}{{E}_{x}}=tan(\alpha+\theta)$

$tan\theta = \frac{y}{x}$

Also it is a well known fact that :

$tan\alpha=tan\frac{\theta}{2}$ (i)

using all this we will be forming our differential equation and it becomes :

$\frac{dy}{dx}=\frac{3xy}{2{x}^{2}-{y}^{2}}$

Since it is a homogenous differential equation of degree 2 we will put $y=vx$ to get :

$x\frac{dv}{dx}=\frac{v+{v}^{3}}{2-{v}^{2}}$

It is now become an variable seperable form hence we have :

$\frac{2-{v}^{2}}{v(1+{v}^{2})}dv=\frac{dx}{x}$

Integrating both sides we get :

$2ln(v)-\frac{3}{2}ln(1+{v}^{2})=ln(x)+C$

Simplifying we get that as :

$\frac{{v}^{4}}{{(1+{v}^{2})}^{3}}=C{x}^{2}$

Putting back $v=\frac{y}{x}$ we get :

${y}^{4}=C{({y}^{2}+{x}^{2})}^{3}$

We have got the function but now how to find the area is the main problem .

What we do is that we will switch to polar coordinates and my seemingly complicated function becomes :

$R=C{sin}^{2}\theta$

From initial conditions $C$ is coming out to be $1$ .

Now the thing left is integration in polar co-ordinates :

$Area=A=\int _{ 0 }^{ \pi }{ \frac { { r }^{ 2 } }{ 2 } d\theta }$

Putting the value of $R$ as function of $\theta$ and integrating it we get :

$A=\frac{3\pi }{16}$