I want to calculate the Electric Field that is felt at the origin $O$ provoked by a hemisphere of radius $R$ with uniform charge density $\sigma$.
I used spherical coordinates: $\vec{r} = -R(\sin(\phi)\cos(\theta), \sin(\phi)\sin(\theta), \cos(\phi)) = -R\vec{e_r}$. The minus sign is because the $\vec{r}$ is pointing inwards from the surface to the origin. For a small area $dq = \sigma dA = \sigma R^2 \sin(\phi) d\phi d\theta$.
Intuitively I know that only the $e_z$ component will be relevant but want to get it from the calculations so I'll consider also the other dimensions for now.
Now, for a small area segment, a small part of the Field is felt by $$d\vec{E} = \cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^2} \sigma R^2 \sin(\phi) d\phi d\theta \vec{r} = \cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^3} \sigma R^2 \sin(\phi) d\phi d\theta \vec{e_r}$$
Throughout the upper hemisphere:
\begin{equation} \begin{aligned} \iint d\vec{E} &= - \cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^3} \sigma R^2 \int_0^{\pi/2} \int_0^{2\pi} \sin(\phi) d\phi d\theta \vec{e_r} \\ &= -\cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^3} \sigma R^2 \int_0^{\pi/2} \int_0^{2\pi} \sin(\phi) d\phi d\theta (\sin(\phi)\cos(\theta)\vec{e_x} + \sin(\phi)\sin(\theta)\vec{e_y} + \cos(\phi)\vec{e_z}) \\ &= -\cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^3} \sigma R^2 \int_0^{\pi/2} \int_0^{2\pi} \sin(\phi) d\theta d\phi (\sin(\phi)\cos(\theta)\vec{e_x} + \sin(\phi)\sin(\theta)\vec{e_y} + \cos(\phi)\vec{e_z}) \\ &= -\cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^3} \sigma R^2 \int_0^{\pi/2} (\int_0^{2\pi} \sin^2(\phi) \cos(\theta) d\theta \vec{e_x} + \sin^2(\phi)\sin(\theta)\vec{e_y} d \theta + \sin(\phi)\cos(\phi) d\theta \vec{e_z}))d\phi \\ &= -\cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^3} \sigma R^2 \int_0^{\pi/2} 0 \vec{e_x} + 0\vec{e_y} + 2 \pi \sin(\phi)\cos(\phi) d\theta \vec{e_z}d\phi \\ &= -\cfrac{1}{4\pi \epsilon_o} \cfrac{1}{R^3} \sigma 2 \pi R^2 \cfrac{\sin^2(\pi / 2) - 0}{2}\vec{e_z} \\ &= -\cfrac{1}{4\epsilon_o} \cfrac{1}{R} \sigma \vec{e_z} \end{aligned} \end{equation}
As you can see, the Field felt at the origin gives me $$\vec{E} = -\cfrac{1}{4\epsilon_o} \cfrac{1}{R} \sigma \vec{e_z}$$ But it should be $$\vec{E} = -\cfrac{1}{4\epsilon_o} \sigma \vec{e_z}$$ according to the solutions.
I have been over this for hours now and can't find the mistake. The only way I can see to remove the $R$ from the denominator is to integrate over $r$ also, but it doesn't make physical sense to me since it is the surface and not a volume that is producing the Electric Field.
Thank you in advance.