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

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?

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

pasmith said:

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

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

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