A geometry problem by Atul Shivam

We note that the curve $a^2y^2 = x^2(a^2-x^2)$ has two loops symmetrical to both the x- and y-axes for $x \in [-a,a]$ as follows.

The area bounded by the curve $A$ is therefore $4$ times that of the curve in the first quadrant.

$\begin{aligned} A & = 4 \int_0^a y \space dx \\ & = 4 \int_0^a x\sqrt{1-\frac{x^2}{a^2}} \space dx \quad \quad \small \color{#3D99F6}{\text{Let } \frac{x}{a} = \sin \theta \quad \Rightarrow dx = a \cos \theta \space d \theta } \\ & = 4a^2 \int_0^\frac{\pi}{2} \sin \theta \cos^2 \theta \space d \theta \quad \quad \small \color{#3D99F6}{\text{we note that } d \cos \theta = - \sin \theta \space d \theta} \\ & = 4a^2 \int_0^1 \cos^2 \theta \space d \cos \theta \\ & = 4a^2 \left[\frac{\cos^3 \theta}{3} \right]_0^1 \\ & = \frac{4}{3}a^2 \end{aligned}$

$\Rightarrow p+q+r = 4+3+2 = \boxed{9}$

$\color{#D61F06}{\text{Note: It should be } \frac{p}{q}a^r \text{ instead of } \frac{p}{q}x^r \text{ in the problem.}}$

guys sorry for your inconvenience,I have edited the problem