I have the following equation
$$(x+y)^{4} = ax^{2}y$$
I need to find the area limited by the equation above. I know I have to transform x and y in polar coordinates:
$$\begin{align*} &x = r\cos^2\theta\\ &y = r\sin^2\theta \end{align*}$$
I also know the double integral formula. But what should I do after is unknown to me... Thanks!
The main thing is that you need to compute the Jacobian $J$ of your transformation and use it in the area integral. For $x=r \cos^2{\theta}$, $y=r \sin^2{\theta}$ you get
$$J(r,\theta) = 2 r \cos{\theta} \sin{\theta} $$
The curve you are given is $r = a \cos^4{\theta} \sin^2{\theta}$. The area integral is then
$$\begin{align}A &= \int_0^{\pi} d \theta \: \int_0^{a \cos^4{\theta} \sin^2{\theta}} dr \: J(r,\theta) \\ &= a^2 \int_0^{\pi/2} d \theta \: \cos^9{\theta} \sin^5{\theta} \\ &= a^2 \int_0^1 du \: u^9 (1-u^2)^2 \end{align}$$
And I'll leave it for you to fill in the gaps and finish this off.