I am trying to understand how to apply integration by substitution in this specific case. My domain is a triangle $K$ in a 3-dimensional space, its vertices are $v_i=(x_i,y_i,z_i)$ for $i=1,2,3$. The reference triangle $R$ is a right triangle with vertices $(0,0),(0,1)$ and $(1,0)$. Mapping $\phi:R\rightarrow K$ is $\phi(s_1,s_2)=s_1(v_2-v_1)+s_2(v_3-v_1)+v_1$. $(s_1,s_2)\in R$, $s_1,s_2\geq 0$ and $s_1+s_2\leq 1$ (this does not seem relevant to me, but it is here for the sake of being complete).
I found that, in this case $\int_K f(v)dv = 2|K|\int_R f(\phi(s))ds$, where $v=(x,y,z)$ and $s=(s_1,s_2)$. However I do not see how this is the case and it is not explained why it should be.
If somebody can explain I would really appreciate it.
Best, S. Karman
You do not want to do a change of variables here. You have a parametrized surface, and there are lots of references for setting up surface integrals. Here's the formula:
$$\int_K f(v)\,dv = \int_R f(\phi(s))\left\|\frac{\partial \phi}{\partial s_1}\times \frac{\partial \phi}{\partial s_2}\right\|ds.$$