You are almost right but the final answer is $$ \int_0^{\sqrt{2}} \int_0^{\sqrt{ \frac{2-y^2}{2} } } f(x, y) dx dy. $$ instead of $$ \int_{\sqrt{2}}^{0} \int_0^{\sqrt{ \frac{2-y^2}{2} } } f(x, y) dx dy. $$

Note that if $f \equiv 1$ then your answer would be negative whereas the given integral is positive.