electrostatics - Calculation of electric potential in Geiger-Müller tube

We are given a cylindrical rod with linear charge density $λ$, and a coaxial cylinder with linear charge density $-λ$, as shown in the picture.

We are to show that the potential difference between $r_a$ and $r_b$ is $$ΔV=2 k_e λ \ln\big( \frac{r_a}{r_b} \big).$$

We can easily calculate from Gauss' Law that the magnitude of the electric field at a distance $r$ from the axis of symmetry is $E=\frac{2 k_e λ}{r}.$ From the definition of potential difference we have $$ΔV=- \int_{\vec{r_a}}^{\vec{r_b}} \vec{E} \cdot d\vec{s}.$$ Now, the field points radially outward, and $d\vec{s}=-dr \hat{r}$, so $\vec{E} \cdot d\vec{s}=-Edr$. Therefore we should get $$ΔV=- \int_{\vec{r_a}}^{\vec{r_b}} \vec{E} \cdot d\vec{s}= \int_{r_a}^{r_b} \frac{2 k_e λ}{r} dr=2k_eλ\ln \big( \frac{r_b}{r_a}\big).$$ Why do I get the sign wrong?