<s>[INST] What will be the complete area of curve $a^2x^2=y^3(2a-y)$ I want to calculate the area of  $a^2x^2=y^3(2a-y)$ using double integral.
I decided to guess the curve and wrote the equation is:-
$$\pm x=\frac{y\sqrt{y(2a-y)}}{a}$$
which means the curve is symmetric about $y$ axis and $x=0$ at $y=0,2a$.
Thus, area should be:-
$$Area= 2\times\int_0^{2a}\int_0^{\frac{y\sqrt{2ay-y^2}}{a}}dx.dy$$
$$=\frac{2}{a}\times\int_0^{2a}y\sqrt{4a^2-(y-2a)^2}.dy$$
Now, how do I integrate it? Please help
 [/INST] Try to use $$\int (f(x))^a f^{(1)}(x) = \frac{f(x)^{a+1}}{(a+1)}$$ 
and 
$$\int \sqrt{a^2-x^2}=\frac{a^2}{2} \arcsin(x/2) + \frac{x}{2}\sqrt{a^2-x^2}$$
</s>