Problem: Let $\omega\in\Omega^r(M^n)$ suppose that $\int_\sum \omega = 0$ for every oriented smooth manifold $\sum \subseteq M^n$ that is diffeomorphic to $S^r$. Show that $d\omega = 0$.
Proof: Assume $d\omega \neq 0$. Then there exists $v_1, \ldots, v_{r+1}\in T_pM$ such that $d\omega_p(v_1, \ldots, v_{r+1}) \neq 0$.
$D^{r+1}\subseteq \mathbb{R}^{r+1}$ a smooth submanifold of $\mathbb{R}^{n}$ with boundary $S^r$. Let $(h,U)$ be a chart around $p$ such that $D^{r+1}$ (with some radius) is mapped to $N = h^{-1}(D^{r+1})$ around $p$. Then $N$ is a smooth submanifold of $M^n$ with boundary equal to $\partial N = h^{-1}(S^r)$ (diffeomorphic to $S^r$).
By definition of the integral and Stokes' theorem:
$\int_{\partial N} \omega = \int_N d\omega = \int_{D^{r+1}}(h^{-1})^* (d\omega)$.
Now let $\alpha = (h^{-1})^*(d\omega)$. Then $\alpha = f(x)dx_1\wedge\ldots\wedge dx_{r+1}$(topform in $\mathbb{R}^{r+1}$). Since $f(x) \neq 0$, it has to be different from zero on a small domain. Assume that $f(x) > 0$. Then $\int_{D^{r+1}}\alpha = \int_{D^{r+1}}f(x)d\mu_{r+1} > 0$.
Contradiction.
-- I feel my idea is correct, but I'm not fully sure this is a full good proof. Could this have been done easier? I'm grateful for any feedback.