Explanation of incorrect argument for Ellipse Area?

Question

Why is the following argument incorrect?

I was thinking that if only either a or b were positive, then the line of reasoning 2ab=a2 + b2 could not be used to begin with, but I am unsure as to whether or not that answers the question.

Answer 1

This method for computing the ellipse's area can be made to work, but $r$ is the wrong Jacobian because the $\theta$ in the ellipse parameterization isn't the usual polar angle. Let's rename it $t$, and use $\theta$ for the true polar angle. Then$$\frac{b}{a}\tan t=\frac{y}{x}=\tan\theta\implies d\theta =\frac{\frac{b}{a}\sec^2tdt}{1+\frac{b^2}{a^2}\tan^2t}=\frac{abdt}{a^2\cos^2t+b^2\sin^2t}=\frac{abdt}{r_\max^2(t)},$$where a ray from the origin to the perimeter is $r\in[0,\,r_\max(t)]$. Since$$dxdy=rdrd\theta=\frac{abrdrdt}{r_\max^2(t)},$$the ellipse's area is$$\int_0^{2\pi}dt\int_0^{r_\max(t)}\frac{abrdr}{r_\max^2(t)}=\int_0^{2\pi}dt\frac{ab}{2}=\pi ab.$$