Rate of change of a volume integral over a volume form

Question

Using Cartan's formula, it can be shown that the rate of change of the volume integral over a p-form is given by

$$ \frac{d}{dt}\int_{V(t)} \alpha^p = \int_{V(t)}i_Xd \alpha^p + di_X\alpha^p $$

Here $d$ is the exterior derivative and i_X is the inner product with a vector field X. However, if $\alpha$ is a volume form and $V^n$ a compact region in a manifold $M^n$ we have

$$ \frac{d}{dt}\int_{V(t)} vol^n = \int_{V(t)} d i_X vol^n $$

Why does in case of a volume form the first term vanish?

Answer 1

vol has top degree so d of it is 0