Using an appropriate change of variables, calculate

Using an appropriate change of variables, calculate the area bounded by the curve: $$ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2 = \frac{xy}{c^2}. $$

Step 1 · Elliptical coordinates

Let $x=ar\cos\theta,\ y=br\sin\theta$; the Jacobian is $\dfrac{\partial(x,y)}{\partial(r,\theta)}=abr$. Then $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=r^2$ and $xy=abr^2\cos\theta\sin\theta$, so the curve becomes:

$$ (r^2)^2 = \frac{abr^2\cos\theta\sin\theta}{c^2} \quad\Longrightarrow\quad r^2=\frac{ab\sin2\theta}{2c^2} \quad(r\neq0). $$

Step 2 · Identify the two lobes

Since $r^2\geqslant0$ requires $\sin2\theta\geqslant0$, the curve is real for $\theta\in\left[0,\frac\pi2\right]$ (a lobe in the first quadrant) and $\theta\in\left[\pi,\frac{3\pi}2\right]$ (a symmetric lobe in the third quadrant) — a figure-eight shape through the origin.

Step 3 · Set up and evaluate the area integral

With area element $dx\,dy=abr\,dr\,d\theta$, the area of one lobe is $\dfrac{ab}2\displaystyle\int_0^{\pi/2}r^2(\theta)\,d\theta$; doubling for both lobes:

$$ A = ab\int_0^{\pi/2} r^2(\theta)\,d\theta = ab\int_0^{\pi/2}\frac{ab\sin2\theta}{2c^2}\,d\theta = \frac{a^2b^2}{2c^2}\int_0^{\pi/2}\sin2\theta\,d\theta. $$

$$ \int_0^{\pi/2}\sin2\theta\,d\theta = \left[-\frac{\cos2\theta}2\right]_0^{\pi/2} = \frac12(1+1) = 1, $$

so:

$$ A = \frac{a^2b^2}{2c^2}. $$

Additional Notes

• This derivation matches the source exactly. The curve is a lemniscate-like figure-eight with two symmetric lobes through the origin.
• Verified numerically: direct quadrature of $ab\int_0^{\pi/2}\frac{ab\sin2\theta}{2c^2}\,d\theta$ gives exactly $\dfrac{a^2b^2}{2c^2}$, matching to full precision.