Electric field in the center of hemisphere shell without double/triple integrals I'm trying to derive the electric field in the centre of a solid hemisphere of radius $ R $ where the charge is distributed uniformly. I have seen different methods involving double/triple integrals but my current knowledge is beyond that and anyway I want to do it the "hard way".
My approach is:

*

*Calculate the electric field produced by a semi-circle (think of it as a bent rod) of uniform charge. Caring only about the electric field component that doesn't cancel

*Knowing the electric field of a semicircle, calculate the electric field of the hemisphere shell, someway creating that shell from the adding lots of semicircles

*Knowing the electric field of a hemisphere shell, calculate the solid hemisphere integrating the shells from 0 to R

I don't have issues with the first step. The electric field in the axis that doesn't cancel, according to my calculations and other resources gives this result:
$$ E_y = \frac{2k\lambda}{R} = \frac{\lambda}{2\pi\epsilon_0R} = \frac{Q}{2\pi^2\epsilon_0R^2} $$
However, I have issues with how to approach the second step. Where now $ Q $ is the charge of the entire hemisphere shell, the electric field component produced by an entire semicircle in the centre that won't end up cancelled is:
$$ dE_y = \frac{dQ}{2\pi^2\epsilon_0R^2} \sin(\theta) $$
But I'm sure in how to express $ dQ $ as a function of the angle or even if the approach of adding infinitesimal thin semicircles while rotating them around the center works to create a shell.
 A: If I understand your plan properly, you are viewing the hemispherical shell as a collection of slices  between "lines of longitude" running from a point on the rim of the hemisphere to its antipode.  You are then replacing each thin slice by a uniformly charged rod.
The problem with this plan is that the thin "slices" you are envisioning do not have a uniform amount of charge per unit length. They are narrower near the "poles" of the hemisphere (the point on the rim where all the wedges meet), and thicker in the middle. This means that the effective charge debit per unit length will go to zero at the ends of the semicircles but be non-zero in the middle. (More specifically, the charge per length will be proportional to $\sin \theta$, where $\theta $ is the angle along the semicircle. )
Perhaps a better way of doing this, short  of doing a true triple integral, would be to view the hemisphere as the "northern" hemisphere of a sphere (rather than the "eastern" hemisphere as in your original approach), and imagine carving it up into a set of disks along the lines of latitude. Then use the result for the electric field along the axis of a uniformly charged disk.
A: It could work, your $dQ$ could be the charge on a small volume of length $\pi R$, width $Rd \theta$ and thickness $dr$, if you include a constant $k$ that is charge per unit volume then $$dQ = (k\times\pi R^2  dr) d\theta$$
Later when you integrate all the shells to get a volume, you can then use $$\frac{2\pi R^3}{3}k = Q $$ due to the total charge on the volume of a hemisphere being $Q$.
