# How to show the identity $\int_0^T \int_{\Gamma(t)}f(s,t)\;dsdt = \int_S f(\sigma)(1+(\mathbf w \cdot \mathbf n)^2)^{-\frac{1}{2}}\;d\sigma$?

Let $\Gamma(t)$ be a smooth closed connected oriented hypersurface for each $t \in [0,T]$. Define the set $$S = \bigcup_{t \in (0,T)}\Gamma(t) \times \{t\}.$$

On page 5 of the document, the authors say

the identity $$\int_0^T \int_{\Gamma(t)}f(s,t)\;dsdt = \int_S f(\sigma)(1+(\mathbf w \cdot \mathbf n)^2)^{-\frac{1}{2}}\;d\sigma$$ holds

where $\mathbf n$ is the unit normal on $\Gamma(t)$ and $\mathbf w$ is a velocity field that advects $\Gamma(t)$ (these definitions found in page 2).

QUESTION How does one prove this identity? In fact, I thought that $$\int_0^T \int_{\Gamma(t)}f(s,t)\;dsdt = \int_S f(\sigma)\;d\sigma$$ is supposed to be true.

The square rooted term reminds me of using a parametrisation, but not sure what or how. I wonder what it even means to write down a product integral as on the RHS...

I also posted this on MSE.

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You are right that, essentially, this is Fubini's theorem, except that you should be careful about the volume element. Probably, the easiest thing to do is to write the integrals locally in coordinates (say, defining $\Gamma(t)$ as a family of graphs $y=g(\mathbf{x},t)$; you know that an ugly factor will appear in the surface integral) and then see how this factor pops up on the right hand side. Alas, I do not quite understand the detail of your terminology/notation, but this seems doable. – Alex Degtyarev Feb 8 '14 at 18:27