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.
 A: The entire figure scales proportionally to $a$, so we may set $a=1$ and multiply the area thus obtained by $a^2$ at the end.

The desired area when $a=1$ is twice the area of the blue region $B$ above plus the area of the circular sector $C$ bounded by the black lines $y=\sqrt2x$ and $x=\sqrt2y$. $B$ is bounded by $y=\frac x{\sqrt2}$ from above and $y=\frac{x^2}2$ from below, so
$$B=\int_0^{\sqrt2}\left(\frac1{\sqrt2}x-\frac12x^2\right)\,dx$$
$$=\left[\frac1{2\sqrt2}x^2-\frac16x^3\right]_0^{\sqrt2}=\frac{\sqrt2}6$$
The angle $\theta$ between the two black lines satisfies
$$\tan\theta=\frac{\sqrt2-1/\sqrt2}{1+\sqrt2\cdot1/\sqrt2}=\frac{\sqrt2}4$$
which implies $\cos\theta=\frac{2\sqrt2}3$ and (since the sector radius is $\sqrt3$)
$$C=\frac{r^2\theta}2=\frac32\cos^{-1}\frac{2\sqrt2}3$$
Finally, for a given $a$ the total area of the region in question is
$$(2B+C)a^2=\left(\frac{\sqrt2}3+\frac32\cos^{-1}\frac{2\sqrt2}3\right)a^2$$
$$=0.981159\dots×a^2$$
A: You could probably use this integral:
$$ \int_0^a 2 \pi r \,dr $$

A: 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.
