What will be the complete area of curve $a^2x^2=y^3(2a-y)$

Question

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

Answer 1

With $y-2a=2a\sin t$ then $$A=\dfrac{2}{a}\int_{-\frac{\pi}{2}}^0(2a)^3(1+\sin t)\cos^2tdt=16a^2\int_{-\frac{\pi}{2}}^0\dfrac12+\dfrac12\cos2t+\sin t\cos^2t\,dt=\color{blue}{4a^2(\pi-2)}$$

Answer 2

Try to use $$\int (f(x))^a f^{(1)}(x) = \frac{f(x)^{a+1}}{(a+1)}$$ and $$\int \sqrt{a^2-x^2}=\frac{a^2}{2} \arcsin(x/2) + \frac{x}{2}\sqrt{a^2-x^2}$$