How can I set the range of this integral? [integration by substitution for $\int\int_D \sin(x^2-y^2) dA$ ] Question) $D = \{(x,y) \in \mathbb{R}^2 \vert 2y \leq x \leq \sqrt{y^2+9} ,0 \leq y\leq \sqrt3  \}$. Find $\int\int_D \sin(x^2-y^2) dA$
I put $x=r\cosh\theta$ and $y=r\sinh\theta$. Then, only left is finding $\int\int_D r\sin(r^2) dr d\theta$.
From $x^2-y^2 \leq 9$, I got $r\leq 3$. Plus I could get $\tanh\theta \leq \frac{1}{2}$ from $2y \leq x$. But the real problem is I can't find the range of  $(r,\theta)$ from $0\leq y\leq \sqrt
3$. In fact the answer sheet claimed $0 \leq \theta \leq  \tanh^{-1}(\frac{1}{2})$ and $0\leq r\leq  3$. I can't understand how could he(or she) derive that. Why is the range of $(r,\theta)$ should like that?
 A: Since $x:=r\cosh(\theta)$ and $y:=r\sinh(\theta)$ and $r>0$ and $\theta\in[0,2\pi[$, then $$2y\leqslant x\leqslant \sqrt{y^{2}+3^{2}}\implies 2r\sinh(\theta)\leqslant r\cosh(\theta)\leqslant \sqrt{(r\sinh(\theta))^{2}+3^{2}},$$
and  $$0\leqslant y\leqslant \sqrt{3}, \implies 0\leqslant r\sinh(\theta)\leqslant \sqrt{3}$$

*

*Then, $$0\leqslant \underbrace{\frac{2r\sinh(\theta)}{r\cosh(\theta)}\leqslant \frac{r\cosh(\theta)}{r\cosh(\theta)}}_{\cosh(\theta)>0; \theta\in [0,2\pi[}\implies \boxed{0\leqslant\tanh(\theta)\leqslant \frac{1}{2}}$$

*Then,
$$0\leqslant r\sinh(\theta)\leqslant 2r\sinh(\theta)\leqslant r\cosh(\theta)\leqslant\sqrt{(r\sinh(\theta))^{2}+3^{2}},$$
so in particular $0\leqslant r\sinh(\theta)\leqslant r\cosh(\theta)$ and then $0\leqslant r^{2}\sinh^{2}(\theta)\leqslant r^{2}\cosh^{2}(\theta)$, so $0\leqslant r^{2}\cosh^{2}(\theta)-r^{2}\sinh^{2}(\theta)$. So,
$$0\leqslant r^{2}\cosh^{2}(\theta)\leqslant r^{2}\sinh^{2}(\theta)+3^{2},$$
and finally
$$0\leqslant r^{2}\cosh^{2}(\theta)-r^{2}\sinh^{2}(\theta)\leqslant 3^{2} \implies \boxed{0\leqslant r\leqslant 3}$$
