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 $1$.
Let $f(x)=1\_{(0,1)}(x)=F\_{X}'(x)$. Note that
$$F\_{Y}(y)=P(X\geq e^{-y/2})=1-P(X\leq e^{-y/2})=1-F\_{X}(e^{-y/2}).$$
Then the density is
\begin{align}
F\_{Y}'(y)&=(1-F\_{X}(e^{-y/2}))'\\
&=-F\_{X}'(e^{-y/2})\frac{\mathrm{d} }{\mathrm{d} y}(e^{-y/2})\\
&=-1\_{(0,1)}(e^{-y/2})\cdot \left ( -\frac{1}{2}e^{-y/2} \right )\\
&=1\_{(0,\infty)}(y)\frac{1}{2}e^{-y/2}.
\end{align}