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On $\mathbb{R}^3$, consider a compactly supported $2$-form $$\omega = f_1 \, dx_2 \wedge dx_3 + f_2 \, dx_3 \wedge dx_1 + f_3 \, dx_1 \wedge dx_2.$$

Choose for $S$ the parametrization $h: \mathbb{R}^2 \to S$ defined by $$h(x_1, x_2) = (x_1, x_2, G(x_1, x_2)).$$

The I start to lost at why we are doing this:

Compute $$h^* dx_1 \wedge dx_2 = dx_1 \wedge dx_2$$ $$h^* dx_2 \wedge dx_3 = -\frac{\partial G}{\partial x_1} dx_1 \wedge dx_2$$ $$h^* dx_3 \wedge dx_1 = -\frac{\partial G}{\partial x_2} dx_1 \wedge dx_2$$

So I get totally lost here

And we emerge with the formula $$\int_s \omega = \int_{\mathbb{R}^2}(n_1 f_1 + n_2 f_2 + n_3 f_3) dx_1 dx_2,$$ where $$\vec{n} = (n_1, n_2, n_3) = (-\frac{\partial G}{\partial x_1},-\frac{\partial G}{\partial x_2}, 1).$$

Can someone give me some explanation/intuition here? Thank you.

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It might help if you use different notation for the variables in 2D. For instance $u_1, u_2$ for 2D and $x_1, x_2, x_3$ for 3D. Otherwise, the formulas look a bit strange. – Andrey Sokolov Aug 8 '13 at 3:54
It quoted from Guillemin and Pollack's Differential Topology. But I agree with you, @AndreySokolov. – WishingFish Aug 8 '13 at 4:32
up vote 3 down vote accepted

They're pulling back oriented areas in 3d space to areas in the 2d coordinate plane. This allows you to convert the very complicated 3d surface integral into a simpler 2d integral on a flat domain.

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