Find the area of the inner loop of $r=2\tan\left(\frac{\theta}{2}\right)$

Question

I can graph this polar equation but using the polar curve integral formula, I can't seem to be able to find the answer. Also an additional question, what type of curve does this fall under? I can't seem to find much information on this.

Answer 1

The area of a region in polar coordinates defined by the equation $r=f(θ)$ with $α≤θ≤β$ is given by the integral $$\mathcal A=\frac 12∫^β_α[f(θ)]^2\,dθ$$

Now we have $$\mathcal A'=\frac 12∫^{\pi/2}_{0}\left[2 \tan \left(\frac{\theta}2\right)\right]^2\,dθ=\left[4\left(\tan\left(\dfrac{{\theta}}{2}\right)-\dfrac{{\theta}}{2}\right)\right]^{\pi/2}_{0}=4 − π$$ The figure has symmetry in relation to the $y$-axis then the area $\mathcal A=\color{magenta}{2}\mathcal A'=8-2\pi.$

Answer 2

Since we have the classic substitution of $r = 2\tan{\frac{\theta}{2}}$, we can express $x, y$ in terms of $r$ and integrate.

If $y = r\sin{\theta} \\ x= r\cos{\theta}$, and we have $\sin{\theta} = \frac{2\tan{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}} = \frac{r}{1+\frac{r^2}{4}} = \frac{4r}{4 + r^2} \\ \cos{\theta} = \frac{1-\tan^2{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}} = \frac{4 - r^2}{4 + r^2}$

then, $y(r) = \frac{4r^2}{4 + r^2} \\ x(r) = \frac{4r - r^3}{4 + r^2}$

It is easier to integrate w.r.t $y$, with one complete integral and, since $y = r$ at the intercepts, the bounds remain unchanged. That is,

$$2\int_{0}^{2}{x(y)dy} = 2\int_{0}^{2}{x(r)\frac{dy}{dr}dr} = 2\int_{0}^{2}{x(r)\frac{dy}{dr}dr} = 8 - 2\pi$$