reversing the order of integration I have the double integral
$$ \int_{0}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} f(x,y) \, dx \, dy. $$
which I want to reverse the order of integration. 
Is it the double integral: 
$$\int_{-1}^{1}\int_{0}^{\sqrt{1-x^2}} f(x,y) \, dy \, dx$$
?
 A: Here your region is 
$$D=\{(x,y):0\le y \le 1,-\sqrt{1-y^2} \le x \le \sqrt{1-y^2}\}$$
then 
$$D=\{(x,y):0\le y \le 1,x^2+y^2\le 1\}$$, you see this region is upper part of unit circle.
If you want to reverse the order of integration.
You must find this $f(x,y)$ satisfying Fubini's condition: $f(x,y)$ is Lebesgue-integrable on the region. i.e.
$$\int_D |f(x,y)|dxdy < \infty$$
Suppose that the function $f(x,y)$ is integrable, then
your answer $$\int_{-1}^{1}\{\int_{0}^{\sqrt{1-x^2}} f(x,y)dy\}dx$$ is correct.
A: Here your region of integral is 
\begin{align}
D&=\{(x,y):0<y<1,-\sqrt y < x <\sqrt y\} \\ \implies D&=\{(x,y):0<x^2<y<1\} \\ \implies D&=\{(x,y):-1<x<1, 0<y<x^2 \}
\end{align}
So your double integral will be $$\int_{-1}^{1}\int_{0}^{x^2} f(x,y)\,dy\,dx$$
Answer for edited question:
Here your region of integral is 
\begin{align}
D&=\{(x,y):0<y<1,-\sqrt{1-y^2} < x <\sqrt {1-y^2}\} \\ \implies D&=\{(x,y):0<x^2<1-y^2<1\} \\ \implies D&=\{(x,y):-1<x<1, -\sqrt {1-x^2}<y<\sqrt {1-x^2} \}
\end{align}
So your double integral will be $$\int_{-1}^{1}\int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}} f(x,y)\,dy\,dx$$
