[Physics] Dipole Moment of a Ring

Question

Consider a ring with radius $R$ and charge density $\lambda=\lambda_0\cos\phi$, where $\phi$ is the angular coordinate in the cylindrical coordinate. If I want to find the dipole moment of this charge distribution, then I put it into
$$\vec{p}=\int{\mathrm d^3r~\rho(\vec{r})\vec{r}},$$ where I tried
$$\rho(\vec{r})=\lambda_0 \cos\phi \times \delta(s-R) \delta(z)$$
and
$$\vec{r}=s\hat{s}+z\hat{z}$$
So, the dipole then becomes:
$$\vec{p}=\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}[\lambda_0 \cos\phi \times \delta(s-R) \delta(z)](s\hat{s}+z\hat{z})s~\mathrm ds~\mathrm d\phi ~\mathrm dz$$
The delta function kills the $z$ component and leave the $s$ component, so:
$$\vec{p}=\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}\lambda_0 \cos\phi \times \delta(s-R) \delta(z)s^2~\mathrm ds~\mathrm d\phi ~\mathrm dz\hat{s}=\lambda_0R^2\int_{0}^{2\pi}\cos\phi~\mathrm d\phi~\hat{s}=\vec{0}$$
The answer is
$$\vec{p}=\frac{1}{2}\lambda_0R^2\hat{x}$$
What's wrong with my solution?
Actually this is problem 4.1 from Zangwill's Modern Electrodynamics. I read it's solution, but don't get why it works under Cartesian coordinates but not under cylindrical coordinate.

Answer 1

The previous answer is just a repeat of whats in the solution manual and the p provided in this is incorrect. I am taking a class for grins and was assigned this problem and beat myself up trying to get the stated answer. I finally did it in both spherical and cylindrical coordinates and got the same answer in both. I emailed prof and he confirmed the answer in the book is incorrect. The actual answer is $\pi \lambda_0R^2\hat{x}$

note that the $\phi$ integration is $\int_0^{2\pi}cos^2 \phi = \pi$

the $\theta$ integral is $\int_0^{\pi} sin(\theta)\delta(\theta - \frac{\pi}{2})d\theta = sin(\frac{\pi}{2}) = 1$

the $r$ integral is $\int_0^{\infty} r^2\delta(r-R) = R^2$

so the answer is $\pi\lambda_0R^2\hat{x}$