Using polar coordinates to find the area of an ellipse Considering an ellipse with the $x$ radius equal to $a$ and the $y$ radius equal to b$:$

I figured that some kind of parameterization might be: 

$x=a\cos\theta$

$y=b\sin\theta$

and then polar $r^2$ is just $x^2 + y^2$

But then I tried to come up with some unit of infinitesimal area using triangles $\left(\dfrac{d\theta r^2}{2}\right)$ which does not give the correct answer. I read somewhere that my polar coordinates are wrong and that they are actually 

$x=ar\cos\theta$

$y=br\sin\theta$

But this does not make sense to me as an engineer because that seems like it would have the dimension of area equal to the dimension of a distance. The integral also takes $r$ from $0$ to $1$ which I thought was eliminated because the equation for $r$ should be in terms of $\theta$ and the constants $a$ and $b.$ 

I would like some explanation of what I am doing wrong that would make some "physical" sense (or why physical intuition might fail for this problem)
Notice, since the ellipse: $x=a\cos\theta$ &amp; $y=b\sin\theta$ is equally divided into four symmetrical regions hence,  the area of ellipse in Cartesian coordinates is given as $$=4\int_0^ay\ dx$$
Now changing in Polar-coordinates by setting $y=b\sin\theta$ &amp; $x=a\cos\theta$ or $dx=-a\sin\theta\ d\theta$, one should get area of ellipse $$=4\int_{\pi/2}^0(b\sin\theta)(-a\sin\theta\ d\theta)$$
$$=4ab\int_0^{\pi/2}\sin^2\theta\ d\theta$$
$$=4ab\int_0^{\pi/2}\frac{1-\cos2\theta}{2}\ d\theta$$
$$=4ab\left(\frac{1}{2}\int_0^{\pi/2}\ d\theta-\frac 12\int_0^{\pi/2}\cos2\theta\ d\theta\right)$$

$$=2ab\int_0^{\pi/2}\ d\theta$$$$=2ab\frac{\pi}{2}=\color{red}{\pi ab}$$
I am quite surprised that no one pointed out the following fact: the parametrisation in question does not contain the usual plane polar angle; what is present there is the eccentric anomaly. That is why you can't use the normal techniques for finding areas in plane polar coordinates, and you have to resort to (different forms of) Green's theorem in a plane.