Question: Just wondering why we have equations defined in the Cartesian coordinate for circles and ellipses: wouldn't graphing those shapes contradict the fundamental property of a function (i.e.: to each element in the domain corresponds a single element in the codomain)?

I know I can *algebraically* restate the equations as a function of y, but that would leave me with a square root whose domain I don't know, so the problem remains.

Answer: 
you are asked to find the probability distribution of the random variable $Y$ that is related to the random variable $X$ by the relation $Y=-2\ln X$, being $X$ uniform in the interval $[0,1]$. You can solve it considering a change of variable applied to the cumulative function $F(x)$:
$$F(x)=\int\_0^x f(x)dx = \int\_\infty^y f(y)\Big|\frac{dx}{dy}\Big|^{-1}dy$$
so that
$$f(y) = f(x)\Big|\frac{dx}{dy}\Big|\_y$$
and in your situation $x=\exp(-y/2)$, so $\Big|\frac{dx}{dy}\Big|\_y = \frac{1}{2}\exp(-y/2)$, so $$f(y)= \frac{1}{2}\exp(-y/2)$$

If you try to integrate $f(y)$ between $\infty$ and $0$ you can verify that it gives you $1$.
It was such a long time ago I did this, but this seems to work too. At least in my computer simulations. **Maybe it is just a coincidence** it works for this particular function, if that is the case, please show me why.

$Y = -2 \ln(X)$ means that if a sample of X is x, then the corresponding sample of Y is $y = -2\ln(x)$

$y = \ln(x^{-2})$ and then $e^y = x^{-2}$ and $e^{-y/2} = x$. We see that since x is between 0 and 1, y will be between 0 and infinity. Now we just need to normalize, i.e. find $k$ so that $\int\_0^\infty ke^{-y/2} dy = 1$ and that is an easy exercise in calculus.