<s>[INST] 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!
 [/INST] $$
\begin{align}
\int_{-1}^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x
&=2\int_0^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x\\
&=4\int_0^{\pi/4}\cos^2(\theta)\,\mathrm{d}\theta-\frac23\\
&=2\int_0^{\pi/4}(\cos(2\theta)+1)\,\mathrm{d}\theta-\frac23\\[3pt]
&=\frac\pi2+\frac13\\
\end{align}
$$
</s>