How does this Dirac delta constrain the region of integration? I have a function of two two-component vectors with this shape
$$F(\vec{p},\vec{q})=G(\vec{p},\vec{q})\times\delta\Big(a+b(|\vec{p}|+|\vec{q}|)-\vec{v}\cdot(\vec{p}+\vec{q})\Big),\tag{1}$$
with $a>0$ and $0<b<|\vec{v}|<1$. Besides, $G(\vec{p},\vec{q})$ is symmetric in both arguments. I want to compute the following integral
$$I=\int d^2\vec{p}\int d^2\vec{q}~F(\vec{p},\vec{q})\tag{2}.$$
The delta function constrains the integration region, but I can't see how it does completely. My attempt is the following:
First, I have the freedom to choose $\vec{v}=(v,0)$, so the argument of the delta function is
$$\chi(\vec{p},\vec{q})=a+b\left(\sqrt{p_x^2+p_y^2}+\sqrt{q_x^2+q_y^2}\right)-v(p_x+q_x)=0.\tag{3}$$
Now I can use
$$\delta\Big(g(x)\Big)=\sum_i\frac{\delta(x-x_i)}{|g'(x_i)|}~~~~,~~~~\text{with 
  }g(x_i)=0\tag{4}$$
to get simple deltas. For example, if I want to fix the value of $p_y$, I get
$$p_y^{\pm}=\pm\sqrt{\left[\frac{v}{b}(p_x+q_x)-\frac{a}{b}-|\vec{q}|\right]^2-p_x^2}\tag{5}.$$
However, this expression takes imaginary values in the integration region. How can I correctly compute $(2)$?
 A: I think I worked it out:
With $(5)$, if I want that $(3)$ is satisfied, I need
$$\left[\frac{v}{b}(p_x+q_x)-\frac{a}{b}-|\vec{q}|\right]\geq0.$$
With that, if I want $p_y$ to be real, I also need
$$\frac{v}{b}(p_x+q_x)-\frac{a}{b}-|\vec{q}|\geq|p_x|.$$
Now I define $s(\vec{q})=a+b|\vec{q}|-vq_x$, so the condition is
$$\text{If}~~p_x>0~:~~p_x\geq\frac{s(\vec{q})}{v-b}\tag{C1}$$
$$\text{If}~~p_x<0~:~~p_x\geq\frac{s(\vec{q})}{v+b}\tag{C2}$$
A priori the integral in $p_x$ is from $-\infty$ to $\infty$, but now this condition constrains the integration region: If $s(\vec{q})>0$, we see from (C$2$) that there won't be contribution of negative $p_x$ to the integral. Therefore, if $s(\vec{q})>0$, the region of integration for $p_x$ will be from $s(\vec{q})/(v-b)$ to $\infty$. On the other hand, if $s(\vec{q})<0$, the integration region for $p_x<0$ is from $s(\vec{q})/(v+b)$ to 0, and for positive $p_x$ from $0$ to $\infty$, so
$$\int\limits_{-\infty}^\infty dp_x\rightarrow\Theta\left[-s(\vec{q})\right]\hspace{-4mm}\int\limits_{\displaystyle{\frac{s(\vec{q})}{v+b}}}^\infty\hspace{-3mm}dp_x+\Theta\left[s(\vec{q})\right]\hspace{-4mm}\int\limits_{\displaystyle{\frac{s(\vec{q})}{v-b}}}^\infty\hspace{-3mm}dp_x$$
