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Given that

$$g(\epsilon) = 2 \int \frac{d^3k}{(2\pi^3)} \delta[\epsilon - \epsilon_n(k)]$$

Find $\bigtriangledown \epsilon_n(k)$ and show it is a vector normal to $\epsilon = \epsilon_n(k)$.

Don't know how to evaluate the integral with the $\delta$ in it, and even if I could, how could I show it's normal to this particular type of surface?

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    You may find udeful $g(\epsilon)=\frac{2}{(2\pi)^3}\int{\delta(\epsilon-\epsilon_{n}(k))d^{3}k}=\frac{2}{(2\pi)^3}\int_{\epsilon_{n}(k)=\epsilon}\frac{1}{|\nabla\epsilon_{n}(k)|}{dS}$ – 2017-02-26
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    @KirylPesotski Thanks! Although that isn't obvious to me, how do you get there? – 2017-02-26
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    The integrand is zero unless $\epsilon_{n}(k)=\epsilon$. this defines a surface of codimension 1 in the momentum space, so really the integral is equal to the surface integral over the surface which is given by $\epsilon_{n}(k)=\epsilon$, and the factor of $|\nabla\epsilon_{n}(k)|^{-1}$ comes from the Coarea formula. – 2017-02-27

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