How can I solve the integral equation $\pi a^2 (f(x))^2 = \int_{0}^{f(x)/a} \sqrt{(f(x))^2 -a^2t^2} \ \mathrm dt $ for the function $f(x)$, where $a$ is a constant?
I have tried turning this into a differential equation by differentiating both sides and applying Leibniz's rule, to obtain $2\pi a^2 f(x) f'(x) = \int_{0}^{f(x)/a} \frac{\partial}{\partial x} \sqrt{(f(x))^2 -a^2t^2} \ \mathrm dt = \int_{0}^{f(x)/a} \frac{f(x) f'(x)}{\sqrt{(f(x))^2 -a^2t^2}} \ \mathrm dt$, which cancels to give $2\pi a^2 = \int_{0}^{f(x)/a} \frac{1}{\sqrt{(f(x))^2 -a^2t^2}} \ \mathrm dt$. At this point, using Leibniz's rule again results in a division by zero from plugging the bounds of the integral into the integrand. I'm not really sure what to do next, or whether there is an easy shortcut out of this mess that I'm missing. The fact that the integrand is also a function of $x$ seems to be making this problem a lot less tractable than it otherwise would be.
Thank you for answering! I appreciate the advice.
Your integral on the RHS has a closed form, which is relatively easy to find after making the substitution $t\mapsto t |f(x)|/a$ (I'll assume $a>0$). The resulting equation you obtain is $$\pi a^2f(x)^2=\frac{f(x)^2}{a}\int_0^1\sqrt{1-t^2}dt=\frac{f(x)^2}{a}\pi/4$$ So, assuming you're not interested in the solution $f=0$ identically, the solution to your equation is $f(x)$ arbitrary but non-identically-zero and $a=2^{-2/3}.$