The problem defines:
velocity in the local base $\{e_r , e_\theta , e_z\}$, the gradient and acceleration in the local curvilignear base $e_{i} \otimes e_{j}$ avec $i, j \in\{r, \theta, z\}$
In a problem's solution, this result is used to calculate the stress resultant
$\sigma \vec{n}=\left[\begin{array}{l} 0 \\ \mu w/H \\ -p_{atm} \end{array}\right]$
$R=\int_{0}^{2 \pi} $ $\int_{\theta}^{R_{a}}(\sigma \vec{n}) r d r d \theta =$ $\int_{0}^{2 \pi} \int_{0}^{0}(\mu \omega /H) \vec{e}_{\theta}(\theta) rdr d \theta$
$+\int_{0}^{2 \pi} \int_{0}^{R_{d}} p_{atm} \cdot \vec{e}_{z } rd r d \theta=$
$=\int_{0}^{2 \pi}(\vec{e}_\theta (\theta) d \theta \int_{0}^{R_d} \frac{\mu \omega }{H} r d r-\left(2 \pi p_{\text {atm}} \int_{0}^{R_{d}} rdr\right)\vec{e}_{z}$
then it says $\int_{0}^{2 \pi}(\vec{e}_\theta (\theta) d \theta = 0$ which I've been stuck for one hour looking where it came from ?
My first answer is an appeal to symmetry. The integral in question is, in some sense, a sum of all the radially-pointing unit vectors pointing out from the same "anchor point". Since they all have equal magnitudes, all possible directions are included in this "sum", and all angles are given equal weight, these vectors "cancel out" one another.
An explicit demonstration uses the fact that
\begin{equation} \vec{\mathbf{e}}_{\theta}(\theta) = -\sin\theta\widehat{\mathbf{i}} + \cos\theta\widehat{\mathbf{j}}, \end{equation} where $\widehat{\mathbf{i}}$ and $\widehat{\mathbf{j}}$ are the unit vectors (anchored at the same point) pointing along the $x$- and $y$-axes, respectively.
\begin{equation} \begin{split} \int_{0}^{2\pi}\vec{\mathbf{e}}_{\theta}(\theta)\textrm{d}\theta &=~ \int_{0}^{2\pi}(-\sin\theta\widehat{\mathbf{i}} + \cos\theta\widehat{\mathbf{j}})\textrm{d}\theta\\ &=~ -\widehat{\mathbf{i}}\int_{0}^{2\pi}\sin\theta\textrm{d}\theta + \widehat{\mathbf{j}}\int_{0}^{2\pi}\cos\theta\textrm{d}\theta\\ &=~ -\widehat{\mathbf{i}}\times 0 + \widehat{\mathbf{j}}\times 0\\ &=~ \mathbf{0}, \end{split} \end{equation} the zero vector.