Proving $V_{k}(M)=\dfrac{1}{k} \int_{\partial M} \Phi$

Question

If $M$ is a piece-with-boundary of a k-manifold in $\mathbb{R}^{n}$ where $n\geq k$. I want to show that the k-volume $V_{k}(M)$ of $M$ is given by $$V_{k}(M)=\dfrac{1}{k} \int_{\partial M} \Phi$$

The first step from the solution manual of the textbook "Calculus: A Complete Course by Robert A. Adams, 8th Edition": \begin{aligned} \Phi &=\sum_{i=1}^{k}(-1)^{i-1} x_{i} d x_{1} \wedge \cdots \wedge \widehat{d x_{i}} \wedge \cdots \wedge d x_{k} \\ d \Phi &=\sum_{i=1}^{k}(-1)^{i-1} d x_{i} \wedge d x_{1} \wedge \cdots \wedge \widehat{d x_{i}} \wedge \cdots \wedge d x_{k} \\ &=\sum_{i=1}^{k} d x_{1} \wedge \cdots \wedge d x_{k}=k d x_{1} \wedge \cdots \wedge d x_{k} \end{aligned}

can someone explain to me where did the term $(-1)^{i-1}$ and $dx_{i}$ in the third line went? And where did the term $k$ come form? I just need to figure out this step.. I can go from there and do the proof.

Answer 1

The $(-1)^{i-1}$ is there precisely to move the $dx_i$ over $i-1$ slots so that it's sitting in the standard position. (Remember that $\omega\wedge\eta = (-1)^{k\ell}\eta\wedge\omega$ when $\omega$ is a $k$-form and $\eta$ is an $\ell$-form.) On the third line, they're writing $dx_1\wedge\dots\wedge dx_k$ for the wedge product of all the $dx_j$, including $dx_i$, in the correct order. The $k$ is there because you're adding up the same thing $k$ times! Right?