I'm wondering about something which should be trivial. Suppose we look at the $S^2$ in $\mathbb{R}^3$, with the volume form $\omega = \sin \phi \,d\phi \,d \theta$.
We know that $\int_{S^2}\omega = 4\pi$. But on the other hand $\omega$ seems to be a closed form, so the integral should be zero as $S^2$ is the boundary of $D^3$, the $3$ dimensional ball with radius $1$.
Where do I make a mistake in my reasoning?
First, the issue is not related to extending forms from the boundary to the whole manifold.
Proposition: Suppose $M^n$ is a manifold with boundary $\partial M$. Then any smooth $k$-form on $\partial M$ has a smooth extension to all of $M$.
Proof: By the collar neighborhood theorem, there is a neighborhood $U$ of $\partial M$ for which $U\cong \partial M\times [0,1)$ with the diffeomorphism taking $\partial M$ to $\partial M \times \{0\}$.
Let $\phi:\mathbb{R}\rightarrow \mathbb{R}$ be a $C^\infty$ function with $\phi(0) = 1$ and $\phi(x) = 0$ for $x\in [1/2,1]$. For example, one can use $$\phi(x) = \begin{cases} e^{- \frac{1}{1-4x^2} + 1} & x\in (-1/2,1/2)\\ 0 & x\notin (-1/2,1/2)\end{cases}.$$ Then we can extend $\omega$ to $U \cong \partial M \times [0,1)$ by the formula $\omega_{m,x} = \phi(x)\omega_m$, where I am thinking of $m\in \partial M$.
Then we simply define $\omega$ on $M\setminus U$ by declaring it to be identically zero. Because $\omega_{u,x}$ is identically $0$ for any $x \geq 1/2$, this is a smooth extension.$\square$
$ \ $
However, if the form $\omega$ is closed, there is no reason why the extension should be closed. Using the $\phi$ above, if $\omega$ is closed, then on $U$ we have $d(\omega_{m,x}) = d\phi(x)\omega_m = (d\phi) \wedge \omega_m$. On $(-1/2, 1/2)$, this is non-zero, unless $\omega_m = 0$.
Further, Stokes's theorem actually guarantees that some closed forms don't have extensions to closed forms. In your example, $\omega$ can be extended to all of $D^3$, but if $d\omega = 0$ on all of $D^3$ then Stokes's theorem gives you a contradiction as you noted.