I am trying to compute the stress tensor defined as $\vec{\Pi}=\eta(\nabla{\vec{u}}+\nabla{\vec{u}}^T)$ where $T$ indicates the transpose. The vector field $\vec{u}$ is defined as follows: $\vec{u}(\vec{r})=(\frac{a}{r})^3(\vec{\omega} \times \vec{r})$ with $a$ being a constant, $\eta$ being a constant, $\vec{r}$ a position vector and $\omega$ the angular speed constant in modulus.
Doing calculations I've obtained $\vec{\Pi}=-\frac{6\eta a^3}{r^4}\hat{r}(\vec{\omega} \times \vec{r})$ with $\hat{r}$ being a unit vector.
First is this result right?
Then I want to compute a surface integral: I wanna compute $\int_S\vec{\Pi} \cdot d\vec{S}$ over a sphere of radius $R>0$ with $d\vec{S}$ being $r^2sin(\theta)d\theta d\phi \hat{r}$.
$\hat{r}$ point in the same direction of the radius of the sphere over which I'm integrating.
To do this I thought to compute the dot product as $-\frac{6\eta a^3}{r^4}(\vec{\omega} \times \vec{r})\hat{r} \cdot \hat{r}r^2sin(\theta)d\theta d\phi=-\frac{6\eta a^3}{r^4}(\vec{\omega} \times \vec{r}) r^2sin(\theta)d\theta d\phi$ then $\vec{\omega} \times \vec{r}=\omega rsin(\theta)$ and the integral becomes $-\frac{6\eta a^3}{r}\omega\int_0^{2\pi}d\phi\int_0^{\pi}d\theta sin^2(\theta)$ but this integral should be equal to zero.
Where am I wrong?