Let $Q_0 = \{ x : x\in \mathbb{Q} , 0\leq x \leq 1 \}$ and let $f:R_Q \to \mathbb{R}$ with $R_Q = Q_0 \times Q_0$. f defined as:
$$ f(x,y)= \left\{ \begin{array}{lc} \sin(x) & \text{if} \; 0\leq y \leq x^2 \\ \\ \hspace{0.5cm} 0 & \text{in other case} \\ \end{array} \right. $$
Is this function integrable over $R_Q$?
My attempt:
For $0\leq y \leq x^2 $:
$$ f(x,y) = g(x) \ h(y) \ \text{ with } \ g:\mathbb{Q} \to \mathbb{R} , \ \ g(x)=\sin(x), \ h:\mathbb{Q} \to \mathbb{R} \ \ h(y) = 1 $$
Restriction of continuous functions is continuous then $g$ and $h$ are continuous functions, consequently $f$ is continuous because is a product of continuous functions. Therefore $f$ is integrable over $R_Q$.
Is the above correct? If so, how is the integral computed?
Your argument is not quite complete. You've shown that the function $\ f\ $ is continuous, and hence integrable, over the set $\ A=\left\{(x,y)\in\mathbb{Q}^2\right.\,\left|\,0\le x\le 1, 0\le y\le x^2\right\}\ $. However, $\ f\ $ is not continuous over $\ R_Q\ $, so this is not sufficient to establish its integrability over that set. It is not continuous at the point $\ \left(\frac{1}{2}, \frac{1}{4}\right)\in R_Q\ $, for instance, because $\ f\left(\frac{1}{2}, \frac{1}{4}\right)\ne0\ $, but $\ \lim_\limits{\epsilon\rightarrow0^+}f\left(\frac{1}{2}, \frac{1}{4}+\epsilon\right)=0\ $.
The easiest way to establish the integrability of $\ f\ $ over $\ R_Q\ $ is to note that $\ R_Q\ $ is countable, and therefore has measure $\ 0\ $. All real-valued functions are integrable over any set of measure $\ 0\ $, and their integrals over such a set are always $\ 0\ $.