For example, to find the area of a hemisphere of radius $R$, I think of stacking discs with radii $r=Rcos(\theta)$ and side length $Rd\theta$, so the area of each disc is $dA=2\pi R^2cos(\theta)d\theta$. Doing the integral from $0$ to $\frac{\pi}{2}$ here gives me the correct answer for the area, but the same method doesn't seem to work to find the volume.
To find the volume, I similarly thought of each disk as having volume $dV = \pi(Rcos(\theta))^2Rd\theta$, but the integral gives the volume as $\frac{\pi^2R^3}{4}$. However, I get the correct answer if I use discs with 'straight' edges (blue part in the picture).
Assuming I haven't made a mistake somewhere, why is this the case? Shouldn't the discs with 'slanted' edges be a better approximation?
I'm not going to be too careful in this post, just trying to get the idea across.
The purple volume does not have volume equal to $\pi(R\cos\theta)^2\cdot R\delta\theta$. Its exact volume is hard to determine without doing another round of integration, which is a bit circular and is also not at all necessary. We need to determine an approximating function $f$ such that, in loose notation, the error terms: $$\tag{$\ast$}\sum_j|f(\theta_j)\cdot\delta\theta_j-V_j|$$Tend to zero as the partitions get finer and finer. There, $V_j$ would represent the true volume of the $j$th "slanted disk" slice.
In your working, I am not convinced that your chosen $f$ makes $(\ast)$ vanish as the partitions grow finer (indeed, the fact that your method didn't work is proof that your errors in $(\ast)$ do not disappear). I admit it's reasonable to think that it would, but without checking one cannot be sure. The base of the purple volume multiplied by the arc length is potentially non-negligibly different from the true volume.
As you know, choosing $f$ to compute the 'straight' disk area works. Why does that work? Because the errors will (essentially) be the size of the volumes of the triangle that you see, with $h$ and $\ell$, and this is roughly of size $(\delta\theta)^2$ and the corresponding error $(\ast)$ vanishes.
However, if you want, we can try a "better" approximation. Let's choose $f$ to compute the (approximate) volume of the frustums (the volume captured by the top and bottom plates and the straight slant edge $\ell$). Determining $h$ precisely should be possible but is a bit of a pain, we can cut some corners. I claim the following (recall a frustum's volume is given by $\frac{1}{3}\pi h(R^2+Rr+r^2)$)
Then we can pass to the Riemann integral, and one finds: $$V=\int_{-\pi/2}^{\pi/2}(\pi R^3\cdot\cos^3(\theta))\,\mathrm{d}\theta=\frac{4\pi R^3}{3}$$As expected. We used a "better" approximating shape (the frustum) but notice that we end up with the same integral one arrives at if we just use the "straight" disks. That's actually because the errors between the frustum and cylinder volumes vanishes, so the different choice of shape becomes irrelevant.
Again, your method did not work because you did not carefully check (to be fair, I didn't here, but I know that I could) that your approximation made $(\ast)$ vanish. If $(\ast)$ doesn't vanish, then the Riemann integral will not give the right value.