Derivative of area of ellipse with respect to axis Suppose $A(a,b)$ defines the area of an ellipse with axes $a,b$. We know that $A(a,b)=\pi ab$, and so $\partial_bA(a,b) = \pi a$. 
But suppose I parameterize the ellipse in polar coordinates as $r(\theta) = \sqrt{a^2\cos^2 \theta + b^2\sin^2 \theta}$. Then the area is given by $$\int_0^{2\pi}\frac12 r^2 d\theta$$ and so $$\partial_bA(a,b)=\partial_b \int_0^{2\pi}\frac12 a^2\cos^2 \theta + b^2 \sin^2 \theta\,d\theta\\=\int_0^{2\pi}\partial_b \frac12( a^2\cos^2 \theta + b^2 \sin^2 \theta) \, d\theta=\pi b.$$
What am I doing wrong?
 A: Your polar coordinate parametrization is incorrect. See: https://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center
The issue is that the theta parameter for the cartesian parametrization $(a*\cos\theta,b*\sin\theta)$ is not the same as the angle $\theta$ in polar. The polar angle $\theta$ for $(a*\cos t,b*\sin t)$ (when $t\in(0,\pi/2)$) is $\tan^{-1}\left(\frac{b}{a}\tan t\right)$, which is not $t$.
A: Why don't you compute the area as follows:
$$A(a,b) = 4 I(a,b), \quad I(a,b) =  \int^a_0 b \sqrt{1-(x/a)^2}\,dx \, ?$$ 
$I(a,b)$ is the area of the 1st quad ellipse defined by $(x/a)^2+(y/b)^2-1=0$. Then, substituting $(x/a)^2 = \cos\alpha$ in the integral, it yields to $A(a,b) = \pi a b$.
Edit: as Mark indicated, the parametrization you used is not correct. In fact, the 1st quad ellipse can be parametrized as follows:
$$x = \rho \cos\theta, \quad y = \rho \sin\theta, \quad 0<\theta<\pi/2,$$ 
where the limits for $\rho$ are to be determined from the definition of the ellipse. In fact, it reads that:
$$0 < \rho < \left(\frac{\sin^2 \theta}{a^2} + \frac{\cos^2 \theta}{b^2} \right)^{-1/2} = \rho_E.$$
Therefore, the area of the ellipse can be obtained by:
$$A(a,b) = 4 \int_E 1 \, dE = 4 \int^{\pi/2}_0 \int^{\rho_E(\theta)}_0 |J| \, d\rho d\theta =  \pi a b,$$
where $J(\rho,\theta) = \rho$ is the jacobian matrix of the parametrization and $| \ |$ denotes determinant.
I must admit that Mathematica helped me a bit with this last calculation. 
I hope this is useful to you.
Cheers! 
