Find area bounded by $(\frac{x}{3}+\frac{y}{6})^3-9xy=0$ $C = \{(x,y) \mid (\frac{x}{3}+\frac{y}{6})^3-9xy=0\}$.
I want to find the area bounded by $C$ in first quadrant.
Can you tell me how to solve exercises like this?
I have no idea what integral I need to solve, because my curve is implicitly defined.
 A: See if you can parametrize the curve as $(x(t),y(t))$, and then integrate $\int_a^bx(t)y'(t)\,dt$ where $t=a$ and $t=b$ are the times when the curve hits the origin. Hyperbolic coordinates might help with the parametrization:
$$x=ve^{u}\quad y=ve^{-u}$$
$$
\begin{align}
\left(\frac{ve^u}{3}+\frac{ve^{-u}}{6}\right)^3-9v^2&=0\\
v\left(\frac{e^u}{3}+\frac{e^{-u}}{6}\right)^3-9&=0\\
v&=\frac{9}{\left(\frac{e^u}{3}+\frac{e^{-u}}{6}\right)^3}\\
\end{align}
$$
So $$x(t)=\frac{9e^t}{\left(\frac{e^t}{3}+\frac{e^{-t}}{6}\right)^3}\quad y(t)=\frac{9e^{-t}}{\left(\frac{e^t}{3}+\frac{e^{-t}}{6}\right)^3}$$ with $t$ ranging from $-\infty$ to $\infty$ parametrizes the curve. This parametrization runs the loop clockwise, so actually once you compute the integral, you'll need to negate the result to get the unsigned area.
$\int_{-\infty}^{\infty}x(t)y'(t)\,dt$ doesn't appear to be a friendly integral at first glance, but at least you now have an integral to work with. 
The integral is theoretically doable. Substituting $u=e^t$ will leave you with a rational function in $u$, and it appears that its denominator will be already factored into a quadratic raised to a power. So partial fraction decomposition will work.
(I was inclined to use hyperbolic coordinates because they naturally leave you in the first quadrant. But polar coordinates will probably work equally well, and you could use Weierstrass substitution to again get a rational function to integrate.)

I confirmed that this integral is quite doable without computer assistance. I get that the area is (exactly) $7873.2$.
A: Let's use (slightly modified) polar coordinates. Set $x = 3r\cos\theta$ and $y = 6r\sin\theta$. 
The Jacobian of this transformation is $\left|\begin{matrix}\tfrac{\partial x}{\partial r} & \tfrac{\partial y}{\partial r} \\ \tfrac{\partial x}{\partial \theta} & \tfrac{\partial y}{\partial \theta}\end{matrix}\right| = \left|\begin{matrix}3\cos\theta & 6\sin\theta \\ -3r\sin\theta & 6r\cos\theta\end{matrix}\right| = 18r$. 
The curve $C$ in $(r,\theta)$ coordinates is given by: 
$\left(\dfrac{3r\cos\theta}{3}+\dfrac{6r\sin\theta}{6}\right)^3-9(3r\cos\theta)(6r\sin\theta) = 0$
$r^3(\cos\theta+\sin\theta)^3 - 162r^2\cos\theta\sin\theta = 0$
$r = \dfrac{162\cos\theta\sin\theta}{(\cos\theta+\sin\theta)^3} =: R(\theta)$
Thus, the area of the region bounded by $C$ is: 
$\displaystyle\iint\limits_{C}\,dx\,dy$ $=\displaystyle\int_{0}^{\pi/2}\int_{0}^{R(\theta)}18r\,dr\,d\theta$ $=\displaystyle\int_{0}^{\pi/2}9R(\theta)^2\,d\theta$ $=\displaystyle\int_{0}^{\pi/2}\dfrac{9 \cdot 162^2\cos^2\theta\sin^2\theta}{(\cos\theta+\sin\theta)^6}\,d\theta$ $=\displaystyle9 \cdot 162^2\int_{0}^{\pi/2}\dfrac{\tan^2\theta\sec^2\theta}{(1+\tan\theta)^6}\,d\theta$ $= 9 \cdot 162^2\displaystyle\int_{0}^{\infty}\dfrac{u^2}{(1+u)^6}\,du$ $= 9 \cdot 162^2\displaystyle\int_{1}^{\infty}\dfrac{(v-1)^2}{v^6}\,dv$ $= 9 \cdot 162^2\displaystyle\int_{1}^{\infty}\left(\dfrac{1}{v^4}-\dfrac{2}{v^5}+\dfrac{1}{v^6}\right)\,dv$ $= 9 \cdot 162^2 \left[-\dfrac{1}{3u^3}+\dfrac{1}{2u^4}-\dfrac{2}{5u^5}\right]_{1}^{\infty}$ $= 7873.2$
