Suppose we have $U=g_1(X,Y)$, $V=g_2(X,Y)$ for two random variables, X,Y.
I want to prove this formula: $$f_{U,V}(u,v)=f_{X,Y}(h_1(u,v), h_2(u,v))|J|$$
Question 1 Is this right?
$$f_{U,V} = \frac{\partial}{\partial u \partial v} P(U < u, V < v)$$ by the FTC.
$$= \frac{\partial}{\partial u \partial v} \int_{\{(x,y) : \\ g_1(x,y) < u,\\ g_2(x,y) < v \}} f_{X,Y}(x,y)dxdy$$
$$= \frac{\partial}{\partial u \partial v} \int_{\{(x,y) : \\h_1(g_1(x,y),g_2(x,y)) < u, \\ h_2(g_1(x,y),g_2(x,y)) < v \}} f_{X,Y}(x,y)dxdy$$
$$= \frac{\partial}{\partial u \partial v} \int_{\{(x,y) : \\h_1(g_1(x,y),g_2(x,y)) < u, \\ h_2(g_1(x,y),g_2(x,y)) < v \}} f_{X,Y}(x,y) \left|\frac{\partial(x,y)}{\partial(u,v)} \right| dudv$$
$$= \frac{\partial}{\partial u \partial v} \int_{\{(x,y) : \\h_1(g_1(x,y),g_2(x,y)) < u, \\ h_2(g_1(x,y),g_2(x,y)) < v \}} f_{X,Y}(x,y) \left|J \right| dudv$$
$$= \frac{\partial}{\partial u \partial v} \int_{\{(u,v) : \\h_1(u,v) < u, \\ h_2(u,v) < v \}} f_{X,Y}(h_1(u,v),h_2(u,v)) \left|J \right| dudv$$
$$=f_{X,Y}(h_1(u,v), h_2(u,v))|J|$$
Question 2 I don't really understand what's going on in the last step. I think it's a certain application of the FTC, but I'm not quite making sense of how it works on the specified region of integration.