Question about area calculation

Question

I would like to calculate the area of the ellipse $5x^2 + 11y^2 = 1$ using double integrals.

Now, I know the method of calculating areas using double integrals. We will double integrate $f(x,y) = 1$ and use polar coordinates but at first, we have to "turn" the ellipse into a circle, making the switch with the Jacobian determinant.

Question: What is the correct substitution; substitute for example $x = \sqrt{5}u$ or substitute $u = \sqrt{5}x$? ($u$ will be one of our new two coordinates) is there any difference and why?

Answer 1

Set

$$u=\frac{x}{\sqrt 5} \quad u=\frac{y}{\sqrt {11}} $$

$$5x^2+11y^2=1\implies u^2+v^2=1$$

Note that since we are dealing with an affine transformation from the original equation

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

we'll find that the area is $\pi ab$ and since

$$a=\frac1{\sqrt{5}} \quad b=\frac1{\sqrt{11}}$$

we have that $$A=\frac{\pi}{\sqrt{55}}$$

Answer 2

I would say:

$x = \frac {1}{\sqrt 5} rcos \theta\\ y = \frac 1{\sqrt {11}} r\sin\theta\\ dy\ dx = \frac {1}{\sqrt {55}} r \ dr\ d\theta$

$\frac {1}{\sqrt {55}} \int_0^{2\pi}\int_0^1 r \ dr\ d\theta$