Area between $y^2=2ax$ and $x^2=2ay$ inside $x^2+y^2\le3a^2$ I need to find the area between $y^2=2ax$ and $x^2=2ay$ inside the circle $x^2+y^2\le3a^2$. I know it's an integral but I can't seem to find the right one.

The parabolas will intersect the circle at the points $(a,\sqrt{2}a)$ and $(\sqrt{2}a,a)$ giving the following region:

So you want to find
$$ \int_0^a\sqrt{2ax}-\frac{x^2}{2a}\,dx+\int_a^{\sqrt{2}a}\sqrt{3a^2-x^2}-\frac{x^2}{2a}\,dx $$
In polar coordinates you can use the symmetry of the region and find the area by evaluating
\begin{eqnarray}
A&=&2\int_{\pi/4}^{\arctan(\sqrt{2})}\frac{r^2}{2}\,d\theta+2\int_{\arctan(\sqrt{2})}^{\pi/2}\frac{r^2}{2}\,d\theta\\
&=&\int_{\pi/4}^{\arctan(\sqrt{2})}3a^2\,d\theta+\int_{\arctan(\sqrt{2})}^{\pi/2}(2a\cot\theta\csc\theta)^2d\theta
\end{eqnarray}
which can be finished by elementary means.
