Find the area of the region bounded above by the curve... Find the area of the region bounded above by the curve $x^2+y^2=2$ and below by the curve $y=x^2$.
 (Large Version)
Here's my attempt:   
Since $y =x^2$, then $x^2+x^2=2$, which simplifies to give $x=1$ or $x=-1$. Since $y^2=2-x^2$, then $y=\sqrt( 2-x^2)$. Finally, I tried to integrate $\sqrt (2-x^2)-x^2$ and I got area $\pi/2 $ as my answer.         Please correct me if I'm wrong, thanks!
 A: $$
\begin{align}
\int_{-1}^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x
&=2\int_0^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x\\
&=4\int_0^{\pi/4}\cos^2(\theta)\,\mathrm{d}\theta-\frac23\\
&=2\int_0^{\pi/4}(\cos(2\theta)+1)\,\mathrm{d}\theta-\frac23\\[3pt]
&=\frac\pi2+\frac13\\
\end{align}
$$
A: This is the calculation of the area between $\;y=x^2\;$ below and $\;x^2+y^2=2\;$ above:
Since both $\;y=\sqrt{2-x^2}\;,\;\;y=x^2\;$ are even functions, and since also
$\;y=\sqrt{2-x^2}=x^2\iff x=\pm1\;$ , the area is given by
$$A=2\int_0^1\left(\sqrt{2-x^2}-x^2\right)dx=2\sqrt2\int_0^1\sqrt{1-\left(\frac x{\sqrt2}\right)^2}\,dx-\left.\frac23x^3\right|_0^1$$
Substitute in the last integral $\sin u=\frac x{\sqrt2}\implies \sqrt2\,\cos u\,du=dx\;$ , and we get
$$A=4\int_0^{\pi/4}\cos^2u\,du-\frac23=\left.4\left(\frac{u+\sin u\cos u}2\right)\right|_0^{\pi/4}-\frac23=2\left(\frac\pi4+\frac12\right)-\frac23=\frac\pi2+\frac13$$
Now, if you want the area inside the disk $\;x^2+y^2\le2\;$ and below the parabola $\;y=x^2\;$, we have to substract the above from the disk's area:
$$2\pi-\left(\frac\pi2+\frac13\right)=\frac{3\pi}2-\frac13$$
