If $M$ is connected and $\int_M \omega = 0,$ then $\omega = 0$? Let $M$ be a connected, orientable, and compact $(r+1)$-dimensional manifold and $\omega$ a $(r+1)$-differential form on $M$. If $\int_M \omega = 0$, then $\omega =0$?
I want to say that the answer is yes. The intuitive reason behind it is that if $\omega\neq 0$, since $M$ is connected then "it must be $\omega>0$ or $\omega<0"$, so we have $\int_M \omega \neq 0$. I've been trying to formalize this intuition by stating what means $\omega>0$ first. Since we can write $\omega(x) = a(u) du_1\wedge \cdots \wedge du_{r+1},x\in M,$ for $x = \varphi(u)$ and some chart $\varphi$, then $\omega>0$ would mean that $a(u)>0$ for every $x\in M$. But i'm struggling to use the fact that $M$ is connected to ensure that $a(u)>0$ should happen.
How can I proper formalize this intuition?
 A: Let's go back to basics, where $M = [0,1]$.
I'm sure you can construct a non-zero function $f: M \to \Bbb R$ such that $\int_M f(x) \ \mathrm dx = 0$.
A: This is not true. For instance, take $M$ to be a manifold (without boundary) and $\omega$ an exact form, so that $\omega=d\eta$. Then by Stokes' Theorem
$$
\int_{M}d\eta=\int_{\partial M}\eta=0.
$$
but being exact does not imply being zero. E.g. we can find $\eta$ so that $d\eta\ne 0$. For a simple example, take any nonconstant function $f$ on $S^1$, then $df\in \Omega^1(S^1)$ is exact but not zero.
A: Let $f:[0,2\pi]\to\mathbb{R}$, note that $M=[0,2\pi]$ connected, orientable and compact, and define $f(x)=\sin(x)$ a $0$-form, then
$$\int_M f(x) \,dx=0$$
A: The correct statement that corresponds to the intuition that you had is as follows: if $\omega$ is a nowhere-vanishing top degree form on an orientable closed manifold $M$, then the integral of $\omega$ over $M$ is nonzero.
To put it differently, if $\int_M \omega =0$, then $\omega$ vanishes at some point of $M$.
