Calculate a tensor as the sum of gradients and compute a surface integral

[Salmone]

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?

 

[pasmith]

Your integral is $$f(r)r^2 \int_0^{\pi} \sin \theta \int_0^{2\pi} (\vec \omega \times \vec r)\,d\phi\,d\theta.$$ Without loss of generality, you can assume $\hat\omega = \omega \hat z$. Then $$\vec \omega \times \vec r = \omega(x \hat y - y \hat x).$$ Integrating either $x = r \cos \phi \sin \theta$ or $y = r \sin \phi \sin \theta$ over a full period of $\phi$ gives zero.

 

[Salmone]

Your integral is $$f(r)r^2 \int_0^{\pi} \sin \theta \int_0^{2\pi} (\vec \omega \times \vec r)\,d\phi\,d\theta.$$ Without loss of generality, you can assume $\hat\omega = \omega \hat z$. Then $$\vec \omega \times \vec r = \omega(x \hat y - y \hat x).$$ Integrating either $x = r \cos \phi \sin \theta$ or $y = r \sin \phi \sin \theta$ over a full period of $\phi$ gives zero.

Sorry I really don't understand your answer. However, if I didn't nothing wrong, why my integral is different from zero? What's wrong with it?

 

[pasmith]

Your error is in stating that $\vec \omega \times \vec r = r \omega \sin \theta$. That deals with the magnitude of the cross-product, but not its direction; that must also depend on $\phi$.

EDIT: Even the magnitude is only correct when $\sin \theta \geq 0$; outside of this range you must multiply by -1 to get the magnitude.