Stokes' Theorem Example I am reading Wade's Introduction to Analysis.  One of the exercises is to show that 
$$
\int_{\partial M}\sum_{k=1}^n \, dx_1dx_2\cdots \hat{dx_i}\cdots dx_n
$$
is equal to the volume of $M$ if $n$ is odd and $0$ if $n$ is even.
Let's take $n=3$.  Then the integral is
$$
\int_{\partial M}\,dydz+dxdz+dxdy
$$
By Stokes' Theorem we can take the differential of 
$$
\omega=dydz+dxdz+dxdy
$$
and integrate it over all of $M$.  My question is why $d\omega\neq 0$.  Shouldn't I be taking partial derivatives of $1$, which would all be $0$?
 A: I believe there is a small typo in this problem, because indeed
$$
\int_{\partial M}\sum_{i=1}^n \color{red}{x_i} \,dx_1\wedge dx_2\wedge \cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n = \mathrm{Vol} (M).
$$ 
The computation is pretty straightforward
$$
d(x_i \,dx_1\wedge dx_2\wedge \cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n)
\\
= d x_i \wedge dx_1\wedge dx_2\wedge \cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n
\\
= (-1)^{i-1} dx_1\wedge dx_2\wedge \cdots \wedge dx_i\wedge \cdots \wedge dx_n.
$$
When $n$ is even, everything cancels out for the sum of $(-1)^{i-1}$ is just zero. When $n$ is odd, there is $1$ term survived luckily to produce that volume.
Like in your example:
$$
\int_{\partial M}\,x\,dy\wedge dz+y\,dx\wedge dz+z\,dx\wedge dy
\\
= \int_{\partial M}\,x\,dy\wedge dz - y\,dz\wedge dx+z\,dx\wedge dy
\\
= \int_{  M}\,d(x\,dy\wedge dz) - d(y\,dz\wedge dx)+ d(z\,dx\wedge dy)
\\
=  \int_{  M}dx\wedge dy\wedge dz = \mathrm{Vol}(M).
$$
It doesn't matter which term you choose, just one term will survive.
