potential - Potenial difference from electric field and line integral

I am really confused about the relation of potential difference and the electric field.

The relation between potential and electric field is $$V_{ab} \equiv V_a - V_b = -\int_{b}^a{\vec{E}\cdot\vec{dr}} \, .$$

Now let's look at a solid sphere uniformly charged with $q$ and let's find the potential difference $V_{ab}$ while $b\to\infty$.

The electric field of a solid sphere is in the $\vec{r}$ direction hence if we are going from $b$ to $a$ we are going against the electric field (because $b>a$).

With that we get $$V_{ab} = -\int_{b}^a{\vec{E}\cdot d\vec{r}} = -\int_{b}^a E \, dr \, \cos(\pi) = \int_{b\to\infty}^a \frac{kq}{r^2} \, dr = -\frac{kq}{a} < 0 \, .$$

The result makes no sense because it means that $V(a) < V(b)$ which means that the electric field of the solid sphere is in the $-\vec{r}$ direction, which is wrong.

What I'm doing wrong? I saw many solutions of exercises which are using this relation and it seems that each one of them is just solving the integral without considering the dot product.