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If I have an integral $\int_{A}{u(x_1,x_2,x_3)}$ where $A \subset \mathbb{R}^3$, and I have a function $X:B \subset \mathbb{R}^2 \to A$ with $X(t_1, t_2) = (x_1, x_2, x_3)$ then I can substitute this into the integral to get

$$\int_B{u(X(t_1,t_2))|J|}$$

where $J$ is the Jacobian matrix but I don't know how to work out what $J$ is. Please advise me. This I find confusing because I am going from $\mathbb{R}^3$ to $\mathbb{R}^2$.

(Suppose here that this is doable, eg., $A$ could be a graph on the domain $B$). Edit: $X(B) = A$ if this makes any difference. So X is a parametrisation.

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The first is ostensibly a volume integral, while the second is implicitly a surface ($X(B)\subset A$) integral... –  anon Apr 19 '12 at 22:12
    
@anon: Sure: but we can compute volumes (e.g., of solids of revolution) using one-variable integrals (areas). There is nothing wrong in principle: one would be computing using an accumulation function. –  Arturo Magidin Apr 19 '12 at 22:14
    
@Arturo: The function $u$ is the same inside both integrals, indicating that principle is unlikely to underlie this situation. | Edit: In my comment above I should instead say integral over a volume and integral over a surface respectively, because they aren't necessarily measuring volume or area. –  anon Apr 19 '12 at 22:17

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