Area of $\left( \frac{x^2}{9}+\frac{y^2}{25} \right)^2 \le x^2 + y^2$

Question

I've used the modified polar coordinates: $x = 3r \cos \theta$, $y =5r \sin \theta$, which got me to

$$r^2 \le 9 \cos^2 \theta + 25 \sin^2 \theta$$

What now?

Answer 1

You are on a right path. The following limits are ones you are looking for:

$$r|_0^{\sqrt{9+16\sin^2(\theta)}},\theta|_0^{2\pi}$$