So here I got such integral. How to change the order of integration? I am not asking you for the answer, but what should be the first step?
$\displaystyle \int_{-1}^{1}dy \int_{{y^2 \over 2}}^{ \sqrt{3-y^2}} f(x,y)dx$
I assume this is actually
$\displaystyle \int_{-1}^{1}{\int_{\frac{y^2}{2}}^{\sqrt{3 - y^2}}{f(x, y)\,dx}\,dy}$.
It's important to realise that the terminals actually give you a boundary for $\displaystyle x$ and $\displaystyle y$.
So $\displaystyle \frac{y^2}{2}\leq x \leq \sqrt{3 - y^2}$
and $\displaystyle -1 \leq y \leq 1$.
You should be able to rearrange the inequalities so that you can change the order of integration.
Square both sides and you get $\displaystyle x^2 = 3 - y^2$, or $\displaystyle x^2 + y^2 = 3$, which is a circle centered at the origin with radius $\displaystyle \sqrt{3}$. Note that $\displaystyle x=\sqrt{(3-y^2)}$ is only part of this circle. The best way to draw it would be with pencil and compass.
OK. Is is something like this? http://img692.imageshack.us/img692/3816/kolejnosc.png