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Problem:

Let's consider the collection of $C^1$ functions, where $k=1,2,\ldots,(n-1)$: $$g_k:\mathbb{R}^k\rightarrow \mathbb{R},$$ where: $$ g_k=g_k(x_1,x_2,\ldots,x_k)$$

Then a new map $f$ is defined as follows: $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ such that: $$f(x_{1},x_{2},\ldots,x_n)=(x_1, g_1(x_1)+2x_2, g_2(x_1,x_2) + 3x_3, \ldots, g_{n-1}(x_1,x_2,\ldots,x_{n-1})+nx_n)$$

How can find the volume of $f((0,1)^n)$ where $(0,1)^n$ is an open unit cube $(0,1)^n$?

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Where does it come from? –  draks ... Apr 26 '12 at 15:21
    
@draks: I am solving this problems for practice –  Boyan Klo Apr 26 '12 at 15:42
    
Can you use the change of variables formula? The derivative of $f$ is triangular, so the Jacobian might be easy to compute. In fact, the Jacobian seems to be $n!$ (do you have a typo. in your expression for $f$, if not, I'm not sure what the general term is?). –  copper.hat Apr 26 '12 at 15:55
    
The change of variables formula gives $\int_{\phi(U)} g(y) dy = \int_U g(\phi(x)) J_{\phi} (x) dx$. Use $\phi = f$, $g = 1$. –  copper.hat Apr 26 '12 at 16:06
    
Looks good to me. But you should check to make sure you are comfortable with the application. –  copper.hat Apr 26 '12 at 16:21
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1 Answer 1

up vote 1 down vote accepted

By change of variables formula, if $\Omega \subset \mathbb{R^{n}}$ is open and $G: \Omega \mapsto \mathbb{R^{n}}$ is a $C^{1}$ diffeomorphism, we have \begin{equation} \int_{G(\Omega)} f(x) dx = \int_{\Omega)}f \circ G(x)| det D_xG| dx \end{equation} Then $$ L^{n}(f((0,1)^{n})) = \int_{f((0,1)^{n})} 1 dx = \int_{(0,1)^{n}} |det D_x f| dx $$

But, $$ D_xf = \left[ \begin{array}{cccccc} 1 & 0 & 0 & \cdots & 0 & 0 \\ D_1 g_1(x_1) & 2 & 0 & \cdots & 0 & 0\\ D_1 g_2(x_1,x_2) & D_2 g_2(x_1,x_2)& 3 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & 0 & 0\\ D_1 g_{n-1}(x*) & D_2 g_{n-1}(x*) & D_3 g_{n-1}(x*)& \cdots & D_{n-1}g_{n-1}(x*)& n \\ \end{array} \right] $$ Where $x* = (x_1, x_2, \cdots , x_{n-1})$

Hence $$ L^{n}(f((0,1)^{n}))= n!L^{n}(0,1)^{n}.$$

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