The area of the unit circle (with the equation $x^2+y^2=1$) can be calculated in several ways. One way would be to solve for $y$ in terms of $x$ and then take two times the integral of the resulting equation: \begin{align*}A&=2\int_{-1}^{1}\sqrt{1-x^2}\,dx\\&=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(u)\,du\\&=2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos(2u)+1}{2}\,du\\&=\pi.\end{align*} Another way would be to parameterize the circle, for example using $x(t)=\cos(t)$ and $y(t)=\sin(t)$ on the interval $t\in[0,2\pi]$. Then, use Green's theorem: \begin{align*}A&=\frac{1}{2}\iint_{R}\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}\,dy\,dx\\&=\frac{1}{2}\oint_{C}-y\,dx+x\,dy\\&=\frac{1}{2}\int_{0}^{2\pi}\sin^2(t)+\cos^2(t)\,dt\\&=\pi.\end{align*} I would like to know if it is possible to extend either of these techniques to find the areas bounded by arbitrary closed curves of the form $f(x,y)=c$, for some constant $c$, or if there is some other technique that would be more appropriate for calculating the general case. I am not necessarily looking for an elementary expression of the area, but rather some expression (most likely involving some kind of integral) that would express such an area.
For example, for the curve $x^6+x+y^6+y=1$, neither variable can be solved in terms of the other using standard functions, and I cannot think of a way to parameterize it like the unit circle.