chalte chalte find the area

is it so tough

The curve will be symmetric about the y-axis (since the given expression is quadratic in x). Also, since a²x² ≥ 0, we get
y³(2a-y) ≥ 0 => y(2a-y) ≥ 0.
Therefore, 0 ≤ y ≤ 2a. Solving for x we get (the positive root)
x=\dfrac{1}{a}y^{3/2}\sqrt{2a-y}%A=2\int_0^{2a} \dfrac{1}{a}\,y\sqrt{2a-y}\ \mathrm dy
Hence, the required area
A=2\int_0^{2a} \dfrac{1}{a}\,y^{3/2}\sqrt{2a-y}\ \mathrm dy
The result: A=\pi a^2

thanks sirmffor the solution ,but i thought changing x for y would make the prob look more elegant