I need to numerically evaluate an integral of the following form:
$$ I(\omega) = \int_0^\infty\int_0^\infty\int_0^\infty f(\omega_1,\omega_2,\omega_3,\omega)\delta(\omega_1+\omega_2-\omega_3-\omega)\delta(\omega_1^2+\omega_2^2+\omega_3^2-\omega^2)d\omega_1d\omega_2d\omega_3 $$
I am assuming this integral converges. The function $f$ and variables $\omega,\omega_1,\omega_2,\omega_3,$ are real and non-negative.
So far I have determined that the delta functions describe the intersection of the plane $\omega_1+\omega_2-\omega_3=\omega$ and the sphere $\omega_1^2+\omega_2^2 + \omega_3^2 =\omega^2$ in $\mathbb{R}^3_+$, where $\mathbb{R_+}=[0,\infty)$. I believe this intersection can be parameterized by two curves given by $$ \vec{r}(\omega_3) = [r_1,r_2,r_3]= \left[\frac{1}{2}(\omega_3+\omega)(1+q),\frac{1}{2}(\omega_3+\omega)(1-q),\omega_3\right] $$ with $$ q = \pm\sqrt{\frac{2(\omega^2-\omega_3^2)}{(\omega+\omega_3)^2}-1}. $$ and $\omega_3\in[0,\omega/3]$. Due to the symmetry of $f$, integration over the first of these contours will produce the same value as integration over the second. Let's choose $q$ to be positive. My limited knowledge of calculus has me thinking that $$ I(\omega) = 2\int_0^{\omega/3} f(r_1,r_2,r_3)|r'(\omega_3)|d\omega_3. $$ where $|r'(\omega_3)| = \sqrt{r_1'^2+r_2'^2+1}$. This is the expression I would like to integrate numerically.
I am very unsure of the correct way to integrate these delta functions, however, and I can't seem to find a reference for this. Given that my parameterization of the contours is correct, is my expression for $I(\omega)$ valid? I can also envision that the $|r'(\omega_3)|$ might be dropped.