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I'm learning about differential forms and have seen that the wedge product of $k$ 1-forms $\omega_{1},\cdots,\omega_{k}$ acting on $k$ vectors $\mathbf{v}_{1},\ldots,\mathbf{v}_{k}$ is given by $$\left(\omega_{1}\wedge\cdots\wedge\omega_{k}\right)\left(\mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right)=\left|\begin{array}{ccc} \omega_{1}\left(\mathbf{v}_{1}\right) & \cdots & \omega_{1}\left(\mathbf{v}_{k}\right)\\ \vdots & & \vdots\\ \omega_{k}\left(\mathbf{v}_{1}\right) & \cdots & \omega_{k}\left(\mathbf{v}_{k}\right) \end{array}\right|.$$This provides me with a nice mental picture that the value of the $k$-form $\omega=\omega_{1}\wedge\cdots\wedge\omega_{k}$ acting on the $k$ vectors $\mathbf{v}_{1},\ldots,\mathbf{v}_{k}$ is given by the signed volume of the $k$-dimensional parallelotope spanned by the column vectors of the matrix$$\left[\begin{array}{ccc} \omega_{1}\left(\mathbf{v}_{1}\right) & \cdots & \omega_{1}\left(\mathbf{v}_{k}\right)\\ \vdots & & \vdots\\ \omega_{k}\left(\mathbf{v}_{1}\right) & \cdots & \omega_{k}\left(\mathbf{v}_{k}\right) \end{array}\right].$$

My question is, can that nice mental picture be extended to integrating differential forms? In other words, can I regard the integral of $$\intop_{M}\omega=\intop_{D}\omega\left(\frac{\partial\Phi}{\partial u^{1}},\ldots,\frac{\partial\Phi}{\partial u^{k}}\right)du^{1}\wedge\cdots\wedge du^{n}$$

as in some way the signed volume of the sum of all the little $k$-dimensional parallelotopes spanned by the column vectors of$$\left[\begin{array}{ccc} \omega_{1}\left(\frac{\partial\Phi}{\partial u^{1}}\right) & \cdots & \omega_{1}\left(\frac{\partial\Phi}{\partial u^{k}}\right)\\ \vdots & & \vdots\\ \omega_{k}\left(\frac{\partial\Phi}{\partial u^{1}}\right) & \cdots & \omega_{k}\left(\frac{\partial\Phi}{\partial u^{k}}\right) \end{array}\right].$$

Or have I got this wrong?

Late in the day edit

If my intuition is correct, could anyone provide a deeper explanation as to what it means for the integral to equal the sum of all the little $k$-dimensional parallelotopes? I'm having trouble visualising what that actually means.

Peter4075
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    Yes, you can view it intuitively in that way. – md2perpe Apr 03 '20 at 09:40
  • @md2perpe - So the determinant (aka the signed volume of each $k$-dimensional parallelotope) at each point of the region being integrated can be thought of a some sort of “density” function and the calculated integral is some sort of “mass” of the region? And that applies in any dimension? Is that reasonable? – Peter4075 Apr 03 '20 at 16:31
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    I would say that the determinant multiplied with $du^1 \cdots du^n$ is the signed volume. – md2perpe Apr 03 '20 at 18:32
  • @md2perpe - From the answer, in relation to triple integrals, given by BigbearZzz here: https://math.stackexchange.com/questions/1747317/what-does-a-triple-integral-represent - “You can think of the integrand as the ‘density’ of the region and the value of the integral as the ‘mass’ of the object.” Hence my thought that the integrand $\omega\left(\frac{\partial\Phi}{\partial u^{1}},\ldots,\frac{\partial\Phi}{\partial u^{k}}\right)$, which is a determinant (aka a signed volume), is a density function. Any chance of expanding your comments into an answer with maybe a reference I can look up? – Peter4075 Apr 04 '20 at 13:35

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