There is a right handed orthonormal basis ${a_1,a_2,a_3}$ of $R^3$ (meaning $a_3=a_1\times a_2$) and a number $\omega \geq 0$. The rotation around the axis $a_3$ with the constant angular velocity $\omega$ is then described by $R_t,t\in R$ of the linear maps $R_t:R^3\to R^3$ with
$R_t(a_1)=a_1cos\omega t + a_2sin\omega t,\\ R_t(a_2)=-a_1sin\omega t+a_2cos\omega t,\\R_t(a_3)=a_3.$
Let the vector field $K$ on $R^3$ is defined by
$K(x):=(\frac{\partial}{\partial t}R_t(x))|_{t=0}$
1.1.:) Compute $K(x)$ and $curl~ K(x)$ explicitly.
1.2.:) Let $S$ be a green-region ($A\subseteq R^2$, $A=B_1\cup ...\cup B_m$, and $B^{\circ}_i\cap B^{\circ}_j=\emptyset$) in the x/y-plane. Show that:
$\oint_{\partial S}K=2\omega a_{33}A(S)$.
The boundary of $S$ is run through in a way that $S$ lies on the left, and $a_{33}$ is the z-component of the vector $a_3$.
Sorry for reposting but the other post seems to be off-radar and it's actually a very similar problem to one I have seen in a past exam our teacher gave us to exercise with. Anyway, here is the original one: https://math.stackexchange.com/questions/1318432/righthanded-onb .
Unlike the original poster I had a different approach in 1.1..
I was thinking of $x$ as a linear comibination between $R_t(a_1)$, $R_t(a_2)$ and $R_t(a_3)$ in such a way: $\alpha R_t(a_1)+\beta R_t(a_2) + \gamma R_t(a_3)$.
Using the given formula I got $K(x)=\alpha \omega a_2 -\beta \omega a_1$, which should be correct if I didn't make any arithmetic mistake.
Anyway, the problem with 1.1. lies within the curl. I thought that since $K(x)$ doesn't have any $x,y,z$ the curl is zero, so I got $curl ~K(x)=0$. But it ticked me off so I asked my prof and he said that it is definitely a non-zero vector. Or are the derivates taken after a different variable?
And I'm basically lost on 1.2. just like the author of the original post.