$$\int^{\infty}_{0}dx\int^{\infty}_{0}dy \; \delta(\sqrt{y^2-x^2}-a)$$ Here, $$ a>0 $$ and using the Jacobian $$x=x , \quad y=\sqrt{z+x^2}$$ Then, $$\int^{\infty}_{0}dx\int^{\infty}_{0}dy \; \delta(\sqrt{y^2-x^2}-a)$$ $$=\int^{\infty}_{0}dx \int^{\infty}_{-x^2}\frac{dz}{2\sqrt{z+x^2}} \delta(\sqrt{z}-a)$$ $$=\int^{\infty}_{0}dx \left[\int^{\infty}_{0} \frac{dz}{2\sqrt{z+x^2}} \delta(\sqrt{z}-a) +\int^{x^2}_{0} \frac{dz}{2\sqrt{x^2-z}} \delta(i\sqrt{z}-a) \right] $$ Why did this problem come to me??
$$\int^{x^2}_{0} \frac{dz}{2\sqrt{x^2-z}} \delta(i\sqrt{z}-a) = ?? $$
Your initial problem is ill-defined. Since the radicand is $y^2-x^2$, this means that (for $x,y$ positive), $y$ should be larger than $x$. The integral you are after is in all likelihood $$ \int^{\infty}_{0}dx\int^{\infty}_{x}dy \; \delta\left(\sqrt{y^2-x^2}-a\right)\ . $$