Find the area bounded by a curve by changing variables

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?

• It's a quartic curve not a circle. May 17 '16 at 12:18

You were almost there. You can perform directly $$(x,y) \mapsto (a r \cos (\theta), b r \sin (\theta))$$ with $(r,\theta) \in \left[0, \sqrt{\sin (\theta ) \cos (\theta ) \frac{a b }{c^2}} \right] \times \left\{\left[0, \frac{1}{2}\pi \right] \cup \left[\pi, \frac{3}{2}\pi \right] \right\}$.
As we assume $r\in\mathbb{R}_+$, it yields $\cos (\theta)\sin (\theta) \geq 0$, which implies $\theta \in \left\{\left[0, \frac{1}{2}\pi \right] \cup \left[\pi, \frac{3}{2}\pi \right] \right\}.$
• Should be $\frac{a^2 b^2}{2c^2}$ instead. May 17 '16 at 13:28