Integrating over a y-simple region $D$? Let $D=\{(x,y)\space|\space1\leq x^2+y^2 \leq 2 \text{ and }y\geq0\}$.
Evaluate $\int\int_D(1+xy) dA$.
So I stated that $D$ is a y-simple region because for all $(x,y)\in D$, $\sqrt{1-x^2} \leq y \leq \sqrt{2-x^2}$.
My book states that for a y-simple region $D$, where $\phi_1(x)\leq y \leq \phi_2(x)$, the integral
$\int\int_Df(x,y) dA=\int_a^b\int_{\phi_1(x)}^{\phi_2(x)}f(x,y)dydx$.
I tried doing this with $$\int_{-\sqrt{2}}^{\sqrt{2}}\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}(1+xy)dydx$$ but I am not getting the correct result.
Could someone explain what I am doing wrong?

The mistake I made was assuming that $\phi_1(x)=\sqrt{1-x^2}$ over the entire interval of $-\sqrt{2} \leq  x \leq \sqrt{2}$.
Graphing the region helps, it will make you realize that $\phi_1(x)=0$ over the intervals of 
$[-\sqrt{2},-1)$ and $(1,\sqrt{2}]$.
You end up with the evaluation of
$$\int_{-\sqrt{2}}^{-1}\int_0^{\sqrt{2-x^2}}f(x,y)dydx+\int_{-1}^{1}\int_{\sqrt{-x^2}}^{\sqrt{2-x^2}}f(x,y)dydx+\int_{1}^{\sqrt{2}}\int_0^{\sqrt{2-x^2}}f(x,y)dydx=\frac{\pi}{2}$$
