Came across the classical result where seemingly plain problem of circle deformation leads to elliptic integrals. Wanted to derive the relation of the semi-axes of the deformed circle (ellipse, after that).
Say, we have a circle as described below: \begin{cases} x = r \sin \theta \\ y = r \cos \theta \end{cases}
which has perimeter of $L = 2 \pi r $.
After plastic deformation we shall have the ellipse as described below: \begin{cases} x = a \sin \theta \\ y = b \cos \theta \end{cases}
with the same perimeter $L$.
Defining $L$ of the deformed circle (now ellipse) will lead us to the following relation: $$L = \int_{0}^{2\pi} \sqrt{ \left(\frac{d(a \sin \theta)}{d \theta}\right)^2 + \left(\frac{d(b \cos \theta)}{d \theta}\right)^2} \,d \theta = 4 \int_{0}^{\frac{\pi}{2}} \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} \, d \theta = 4 a \int_{0}^{\frac{\pi}{2}} \sqrt{1-\left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta} \, d \theta % = 4 a E(k) = 2 \pi r. $$
After some manipulations, we may achieve the following relation where $b = f(a)$:
$$ \int_{0}^{\frac{\pi}{2}} \sqrt{1-\left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta} \, d \theta = \frac{\pi r}{2 a} \implies \int_{\sin 0}^{\sin \frac{\pi}{2}} \sqrt{1-\left(1 - \frac{b^2}{a^2}\right) t^2} \, d \arcsin t = \frac{\pi r}{2 a} \implies \int_{0}^{1} \sqrt{\frac{1-\left(1 - \frac{b^2}{a^2}\right) t^2}{1 - t^2}} \, d t = \frac{\pi r}{2 a} \implies \int_{0}^{1} \sqrt{\frac{1-\left(1 - \frac{f^2(a)}{a^2}\right) t^2}{1 - t^2}} \, d t = \frac{\pi r}{2 a} $$
Question: Is there any chance of arriving at the form of $b = f(a)$?
Update: Added Important Notice That the Solution Was Retrieved from Wolfram Alpha Differentiated $L$ with respect to both $a$ and $b$ and re-assembled the differentials: $$ \begin{cases} \frac{\partial L}{\partial a} = \frac{\partial}{\partial a}4 \int_{0}^{\frac{\pi}{2}} \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} \, d \theta \\ \frac{\partial L}{\partial b} = \frac{\partial}{\partial b}4 \int_{0}^{\frac{\pi}{2}} \sqrt{a^2 \cos^2 \theta + b^2 \sin^2 \theta} \, d \theta \end{cases} \implies \begin{cases} \frac{\partial L}{\partial a} = \frac{4}{a^2 - b^2}\left( a^2 E \left(\sqrt{1 - \frac{b^2}{a^2}}\right) - b^2 K \left(\sqrt{1 - \frac{b^2}{a^2}}\right) \right) \\ \frac{\partial L}{\partial b} = 4 \frac{b}{a} \frac{a^2}{a^2 - b^2}\left( K \left(\sqrt{1 - \frac{b^2}{a^2}}\right) - E \left(\sqrt{1 - \frac{b^2}{a^2}}\right) \right) \end{cases} \implies $$ $$ \frac{\partial b}{\partial a} = \frac{ a^2 E \left(\sqrt{1 - \frac{b^2}{a^2}}\right) - b^2 K \left(\sqrt{1 - \frac{b^2}{a^2}}\right) }{ ab \left( K \left(\sqrt{1 - \frac{b^2}{a^2}}\right) - E \left(\sqrt{1 - \frac{b^2}{a^2}}\right) \right) } \underset{b = a \sqrt{1 - k^2}}{\implies} \frac{\partial a \sqrt{1 - k^2}}{\partial a} = \frac{ \frac{1}{\sqrt{1 - k^2}} E \left(k\right) - \sqrt{1 - k^2} K \left(k\right) }{ K \left(k\right) - E \left(k\right) } \implies \frac{\partial k}{\partial a} = \frac{2}{a k} - \left( \frac{2 K \left(k\right) - E \left(k\right) }{K \left(k\right) - E \left(k\right)} \right) \frac{k}{a} \underset{\text{According to Wolfram Alpha}}{\implies} C - \ln(a) = \int_{1}^{\sqrt{1 - \frac{b^2}{a^2}}} \frac{\zeta (E(\zeta) - K(\zeta))}{-2 \zeta^2 K(\zeta) + 2 K(\zeta) + \zeta^2 E(\zeta) - 2 E(\zeta)} \, d\zeta $$
Follow-Up to the Previous Update: Now that we have $C$ independent on both $a$ and $b$, we may determine it for the future use using extreme case where $b \to 0$.
$$ \begin{cases} \frac{\pi r}{2 a} = \lim_{b \to 0} {\int_{0}^{\frac{\pi}{2}} \sqrt{1-\left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta} \, d \theta} \\ C - \ln(a) = \lim_{b \to 0} {\int_{1}^{\sqrt{1 - \frac{b^2}{a^2}}} \frac{\zeta (E(\zeta) - K(\zeta))}{-2 \zeta^2 K(\zeta) + 2 K(\zeta) + \zeta^2 E(\zeta) - 2 E(\zeta)} \, d\zeta} \end{cases} \implies \begin{cases} \frac{\pi r}{2 a} = \int_{0}^{\frac{\pi}{2}} \lim_{b \to 0} {\sqrt{1-\left(1 - \frac{b^2}{a^2}\right) \sin^2 \theta}} \, d \theta \\ C - \ln(a) = \int_{1}^{\lim_{b \to 0} {\sqrt{1 - \frac{b^2}{a^2}}}} \frac{\zeta (E(\zeta) - K(\zeta))}{-2 \zeta^2 K(\zeta) + 2 K(\zeta) + \zeta^2 E(\zeta) - 2 E(\zeta)} \, d\zeta \end{cases} \implies \begin{cases} \frac{\pi r}{2 a} = \int_{0}^{\frac{\pi}{2}} \sqrt{1 - \sin^2 \theta} \, d \theta \\ C - \ln(a) = \int_{1}^{1} \frac{\zeta (E(\zeta) - K(\zeta))}{-2 \zeta^2 K(\zeta) + 2 K(\zeta) + \zeta^2 E(\zeta) - 2 E(\zeta)} \, d\zeta \end{cases} \implies \begin{cases} a = \frac{\pi r}{2} \\ C = \ln(\frac{\pi r}{2}) \end{cases} \implies \frac{\pi r}{2 a} = \exp{\left(\int_{1}^{\sqrt{1 - \frac{b^2}{a^2}}} \frac{\zeta (E(\zeta) - K(\zeta))}{-2 \zeta^2 K(\zeta) + 2 K(\zeta) + \zeta^2 E(\zeta) - 2 E(\zeta)} \, d\zeta\right)} $$
On the other hand, once $b \to a$, we have: $$ \int_{0}^{1} \frac{\zeta (E(\zeta) - K(\zeta))}{-2 \zeta^2 K(\zeta) + 2 K(\zeta) + \zeta^2 E(\zeta) - 2 E(\zeta)} \, d\zeta = \ln\left(\frac{2}{\pi}\right) $$
For the time being, I have not delved that deep into elliptic inverse functions, but already it looks daunting to see that there is an independent variable which depends on function from itself within the integral.
Please may someone offer any piece of advice to make these things more comprehensible?
Thank you in advance!
