Polar integration confusion

Question

The integral $$ \int_{\phi = 0}^{2\pi}\int_{\theta = 0}^{\pi} (\vec{a}\cdot\hat{r})(\vec{b}\cdot\hat{r})\sin\theta\,d\theta\,d\phi $$ $\vec{a}$ and $\vec{b}$ being constant vectors, equals $\frac{4\pi}{3}\vec{a}\cdot\vec{b}$ as proven here, but if we write $\vec{a} = a_r\hat{r} + a_\theta\hat{\theta} + a_\phi\hat{\phi} $ and $\vec{b}$ similarily, we can simply calculate it to be $4\pi a_rb_r$. Am I doing anything wrong or $\vec{a}\cdot\vec{b} = 3 a_rb_r$?

Answer 1

Take the vector $(1,0)$ in 2D polar for simplicity's sake. At any point along the positive $x$ axis, this vector field is $+\hat{r}$. But on the negative $x$ axis, this would be $-\hat{r}$. Another example would be on the positive $y$ axis, where it would be $-\hat{\phi}$. Putting this information together, we have that

$$(1,0) = \hat{x} = \cos\phi \hat{r} -\sin\phi \hat{\phi}$$

which is no longer constant. You would have to integrate whatever function that may be.