Base on the graph, make a conjecture about whether the circulation and flux of $\vec{F}=\langle-y, x\rangle$ on $C: \vec{r}(t)=\langle 3 \cos (t), 3 \sin (t)\rangle$ for $\frac{\pi}{6} \leq t \leq \frac{3 \pi}{4}$ is positive, negative, or zero.
Then
a. Compute the circulation and interpret the result.
Circulation $=$
b. Compute the flux and interpret the result.
\[
\text { Flux }=
\]

Step 1 ：The problem is asking for the circulation and flux of a vector field \(\vec{F}=\langle-y, x\rangle\) on a curve \(C: \vec{r}(t)=\langle 3 \cos (t), 3 \sin (t)\rangle\) for \(\frac{\pi}{6} \leq t \leq \frac{3 \pi}{4}\).

Step 2 ：Circulation is the line integral of the vector field along the curve, and flux is the line integral of the normal component of the vector field along the curve.

Step 3 ：To compute the circulation, we need to compute the line integral of \(\vec{F}\) along \(C\).

Step 4 ：To compute the flux, we need to compute the line integral of the normal component of \(\vec{F}\) along \(C\).

Step 5 ：Given \(t = t\), \(F = \begin{bmatrix}-3\sin(t) \\ 3\cos(t) \end{bmatrix}\), and \(r' = \begin{bmatrix}-3\sin(t) \\ 3\cos(t) \end{bmatrix}\), we find that the circulation is \(\frac{21\pi}{4}\) and the flux is \(\frac{7\pi}{4}\).

Step 6 ：The positive circulation indicates that the vector field is circulating in the same direction as the curve. The positive flux indicates that the vector field is flowing outwards from the curve.

Step 7 ：Final Answer: The circulation of the vector field along the curve is \(\boxed{\frac{21\pi}{4}}\) and the flux of the vector field along the curve is \(\boxed{\frac{7\pi}{4}}\).