How to integrate $\oint_{c}(\textbf{A}\cdot\textbf{r})d\textbf{r}$ where $(\textbf{A}$ is constance and $c$ is a circular path

Question

I think this should give some vector in the plane of the circle (which I will call the x-y plane for simplicity), with the direction of the vector dependent on the vector, $\textbf{A}$. More specifically, think the integral should be given by $$\oint_{c}(\textbf{A}\cdot\textbf{r})d\textbf{r}=\pi R^{2}\hat{\textbf{z}}\times\textbf{A}$$where $R$ is the radius of the circle, however my attempts to integrate have given zero. My method is as follows, express $$\textbf{r}=R\hat{\textbf{r}}$$ and $$d\textbf{r}=R\hat{\boldsymbol{\theta}}d\theta$$ where I have used the basis vectors, $$\hat{\textbf{r}}=(\cos\theta,\sin\theta,0)$$ and $$\hat{\boldsymbol{\theta}}=(-\sin\theta,\cos\theta,0)$$and then integrating $\theta$ over $2\pi$, $$R^{2}\int_{0}^{2\pi}\left(A_{x}\cos\theta+A_{y}\sin\theta\right)\begin{pmatrix}-\sin\theta \\ \cos\theta\\ 0 \end{pmatrix}d\theta$$ since trig functions evaluate to zero over a full period. Have I done something wrong here?