Obtaining the area of a loop of the curve

Question

I have the following curve:

$x^4=a^2(x^2-y^2)$

Prove that the area of its loop is $\frac{2a^2}{3}$.

My approach

This curve has four loops. So the required area should be:

$4\int_{0}^{a}\frac{x}{a}\sqrt{a^2-x^2} dx$

But, After solving, the area turned out to be $\frac{4a^2}{3}$. What am I doing wrong here?

Thanks

Answer 1

Your derivation seems correct, maybe the original question is simply asking for the area of one loop that is precisely

$$2\int_{0}^{a}\frac{x}{a}\sqrt{a^2-x^2} dx=\frac{2a^2}{3}$$

Answer 2

There are only two loops, not four.

First, note that the LHS is always nonnegative, therefore, any solution $(x,y)$ satisfying the given relation must have $x^2 \ge y^2$. This precludes any points for which $y > |x|$ or $y < -|x|$.

Since the relation is also symmetric about the coordinate axes, we conclude that the behavior of the relation in the first quadrant, and specifically, in the wedge-shaped region $0 \le y \le x$, is sufficient to characterize the behavior in the entire plane.

From this, we simply express $y$ in terms of $x$: $$y = \sqrt{x^2 - \frac{x^4}{16}} = \frac{x}{4} \sqrt{16 - x^2}, \quad 0 \le y \le x.$$ Then integration of this function on $[0,4]$ gives, by symmetry, one-fourth of the total area enclosed.

Of course, you have an $a$ somewhere that is not reflected in the first equation you posted, so I have chosen to ignore it.

Answer 3

The solution you have is for a half of one loop ; there are two loops. One loop area can be obtained by doubling what you got.