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Calculate the area bounded by the following formula:

$$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2} \right)^2 = \frac{xy}{c^2}$$ where $a,b,c>0.$

I have used changing variable of $x=au$ and $y=vb$ to eliminate the $a^2$ and $b^2$, and found out the jacobian is $ab$, while it seems like I need to change to polar coordinate apart from this, so I came up with the formula:

$$\iint_D ab r\, dr\, d\theta$$

where $D$ is the region of the circle. The problem is that I cannot figure out the region of the circle. Am I doing something wrong?

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    $\begingroup$ It's a quartic curve not a circle. $\endgroup$ May 17 '16 at 12:18
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You were almost there. You can perform directly $$(x,y) \mapsto (a r \cos (\theta), b r \sin (\theta))$$ with $(r,\theta) \in \left[0, \sqrt{\sin (\theta ) \cos (\theta ) \frac{a b }{c^2}} \right] \times \left\{\left[0, \frac{1}{2}\pi \right] \cup \left[\pi, \frac{3}{2}\pi \right] \right\}$.

Those limits come from the solution of \begin{align} \left(\frac{x^2}{a^2}+\frac{y^2}{b^2} \right)^2 &= \frac{xy}{c^2}\\ \left(\frac{(a r \cos (\theta))^2}{a^2}+\frac{(b r \sin (\theta))^2}{b^2} \right)^2 &= \frac{a r \cos (\theta) b r \sin (\theta)}{c^2}\\ r^4 &= r^2\frac{ab}{c^2} \cos (\theta)\sin (\theta)\\ r^2 &= \frac{ab}{c^2} \cos (\theta)\sin (\theta). \end{align}

As we assume $r\in\mathbb{R}_+$, it yields $\cos (\theta)\sin (\theta) \geq 0$, which implies $\theta \in \left\{\left[0, \frac{1}{2}\pi \right] \cup \left[\pi, \frac{3}{2}\pi \right] \right\}.$

Straightforward integration yields \begin{align} A &= \int_D \,\mathrm{d}A \\ &= \int_0^{\frac{\pi}{2}} \int_0^{\sqrt{\sin (\theta ) \cos (\theta ) \frac{a b }{c^2}}} abr \,\mathrm{d}r\, \mathrm{d}\theta + \int_\pi^{\frac{3\pi}{2}} \int_0^{\sqrt{\sin (\theta ) \cos (\theta ) \frac{a b }{c^2}}} abr \,\mathrm{d}r\, \mathrm{d}\theta\\ &= \frac{a^2 b^2}{2 c^2} \end{align}

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    $\begingroup$ Should be $\frac{a^2 b^2}{2c^2}$ instead. $\endgroup$ May 17 '16 at 13:28

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