I would like to clarify the not quite correct statement that one hears every now and then $$ \tag{1}\label{1} \color{red}{\text{"The sphere has no boundary, and the integrand is a derivative so the integral is zero"}} $$
Consider for example $$ \int_{S^2}d^2x\sqrt{g} \nabla_\mu V^\mu $$ for some vector field $V^\mu$ on the tangent space of $S^2$. One could now try to use Stoke's theorem as follows $$ \int_{S^2}d^2x\sqrt{g} \nabla_\mu V^\mu = \int_{S^2}d^2x\;\partial_\mu(\sqrt{g}V^\mu) = \int_{\partial S^2}d\Sigma_\mu\sqrt{g}V^\mu $$ and conclude that the result is zero because there is no sphere boundary $\partial S^2$. But after taking the standard spherical coordinates where $\sqrt{g} = \sin\theta$, $d^2x = d\theta d\phi$ and setting for example $V^\theta = \frac{\theta}{\sin\theta}$ and $V^\phi = 0$ one sees that $\int_{S^2}d^2x\sqrt{g} \nabla_\mu V^\mu$ gives a non-zero result.
So what is the setting and the conditions so that the statement $\eqref{1}$ is correct? Or maybe better, what is the correct formulation of it? One might think that $\sqrt{g}$ is canceled by the corresponding term in the covariant derivative and one should consider regular derivatives instead of covariant ones. But take for example $\int_{S^2}d^2x\sqrt{g}\; \partial_\theta f(\theta)$ with $f(\theta) = \log\tan(\theta/2)$, $f'(\theta) = 1/\sin\theta$, which gives a non-zero result.