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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?

Eric Auld
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  • This problem was also taken up here: http://math.stackexchange.com/questions/493104/evaluating-int-ab-frac12-r2-to-find-the-area-of-an-ellipse?rq=1 – colormegone Jan 20 '14 at 19:21

2 Answers2

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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 $t$.

Mark S.
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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!

Dmoreno
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