Derivation of Area for an Ellipse & Implicit Integration I was contemplating deriving the area for an ellipse utilising integration. I'm aware we can make the usual relation a function in terms of $y$, ($y=\frac{b}{a}\sqrt{a^2-x^2}$) and thus the area for an ellipse can be found by evaluating the integral of, $\int_a^b 4\frac{b}{a}\sqrt{a^2-x^2}dx$.
Though that would be similar to finding the derivative of an ellipse by making it a function and differentiating explicitly.
The Question: Is there such a thing as "implicit integration" (similar to implicit differentiation)? i.e. you can integrate the relation as a relation, rather than having to split it up into a function and consider the various cases. As a result, can you then derive the area for an ellipse?
Thanks
 A: You may want to consider parametrics. For example we can express an ellipse as $$x=a\cos\theta$$ $$y=b\sin\theta$$ $\forall x\in (0,2\pi)$ and your average integral (pardon the pun) looks like $$\int_{a}^{b} f(x)\text{d}x$$ and $y=f(x)$ so we're really looking at $\int_{a}^{b} y\text{d}x$ from our parametrization we know that this integral can be rewritten as $\left |\int_{0}^{2\pi}b\sin\theta (-a\sin\theta)\text{d}\theta\right |$ by direct substitution. Which will get us the area. However without parameterizations such as this we can only use multiple integrals, which may be a bit advanced. So technically we still had to split the function up in order to evaluate it parametrically, as for multiple integrals you will end up with essentially the same process/result.
A: I am not sure whether I understand your question completely, but I guess that you are wondering if we can do something like this: $$\int x^2/a^2 dx+\int y^2/b^2 dx=\int1 dx .$$ Let's try this, $$x^3/a^2+xy^2/b^2=x$$ or $$x(x^2/a^2+y^2/b^2-1)=0,$$ which is indeed true. But did we compute the area of an ellipse? No. Why? I'll let you think about it.     
