Given a defined region, write down the integral and then evaluate 
Let the region be defined by $x \geq 0$, $x^2 + y^2 \leq 2$, and $x^2 + y^2 \geq 1$. Write down the integral over the region in each of the two possible orders of $f(x,y) = x^2$. Evaluate both integrals.

I was wondering before I try to evaluate the integral if I set up the bounds correctly for the two integrals
\begin{align}
\int_{\sqrt{1-y^2}}^{\sqrt{2-y^2}}dx \int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2dy
\end{align}
The other order would be
\begin{align}
 \int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}dy \int_{\sqrt{1-y^2}}^{\sqrt{2-y^2}}x^2dx
\end{align}
edit: Updated with picture of the bounds


None of your integrals has correct bounds.  For the first one, since the integral w.r.t. $y$ is inside, $x$ should not depend on $y$ so the bounds of the outer integrals should have no $y$ at all.  Similarly, for the second one, you have the integral w.r.t. $x$ inside; therefore the outer integral should have no $x$ appearing in the bounds.  I recommend that you really draw the picture to see what the proper bounds in each case are.
The correct bounds for the first integral would be
$$\int_0^{1}dx\ \left(\int_{-\sqrt{2-x^2}}^{-\sqrt{1-x^2}}x^2 dy+\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2 dy\right) +\int_1^{\sqrt2}dx\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}x^2dy.$$
Similarly for the second integral, you should have
$$\int_{-\sqrt{2}}^{-1}dy\ \int_{0}^{\sqrt{2-y^2}}x^2dx+\int_{-1}^{1}dy\ \int_{\sqrt{1-y^2}}^{\sqrt{2-y^2}}x^2dx+\int_1^{\sqrt{2}}dy\ \int_0^{\sqrt{2-y^2}}x^2dx.$$
However, it is much easier to evaluate this integral in polar coordinates.  If $I$ is the required integral, then by symmetry, $$I=\frac12\iint_{A} x^2 da$$ where $A$ is the annulus $\big\{(x,y)\ :\ 1\le x^2+y^2\le 2\big\}$, and $da$ is the area element.  By symmetry again, $$I=\frac12\iint_{A}y^2 da.$$ So $$I=\frac14\iint_A(x^2+y^2)da=\frac14\iint_A r^2da=\frac14\int_{0}^{2\pi} \ \int_1^{\sqrt{2}} r^2\cdot r dr\ d\theta.$$ Hence $I=\frac{\pi}{2}\left(\frac{(\sqrt{2})^4-1^4}{4}\right)=\frac{3\pi}{8}$.
