Find area of curve $(x^2+y^2)^2=2a^2xy$ How can I find the area of this curve? 
$(x^2+y^2)^2=2a^2xy$
Should I first  assume that $x=a\cos t$, or $y=\sin t$? 
And then use the formula with $x$ and $y$ derivative?  

Let's convert to polar coordinates.
We get $\displaystyle (r^2)^2 = 2(a^2)(r\cos(\theta))(r\sin(\theta))$.
That boils to $r^4 = a^2r^2\sin(2\theta)$.
We then get $r^2 = a^2\sin(2\theta)$.
So that means $r = \pm a\sqrt{\sin(2\theta)}$.

Graph analysis.
Notice that it actually doesn't matter which root we take, because $-r$ corresponds to a reflection about the origin and $\sin(2(\theta+π))=\sin(x)$, since $π$ is the period.
What does that mean? That means that taking the negative root traces out the same region in reverse.

Now to the integration.
We have two intervals of $\sin(2\theta)$, where it is positive and the root is defined. Because these two portions of sine are identical, so is the graph. We will therefore compute the area enclosed by one petal and mutliply by $2$.
We have that $\sin(2\theta)$ has roots at $\theta=0$ and $\displaystyle \theta = \frac{π}{2}$.
Then, $A_{\text{petal}} = \displaystyle \frac{1}{2}\int_0^{\frac{π}{2}} r^2\,d\theta= \frac{1}{2}\int_0^{\frac{π}{2}} (a\sqrt{\sin(2\theta}))^2\,d\theta= \frac{a^2}{2}\int_0^{\frac{π}{2}} \sin(2\theta)\,d\theta=\frac{a^2}{2}$.
Therefore, two petals enclose an area of $\boxed{a^2}$.
