Initially assume that the area $a$ and circumference $c$ of a radius $R>0$ circle are given axiomatically as:
$a=πR^2$ and $c=2πR$
Now in order to calculate the surface area of the top hemisphere (without its base) of a radius $R$ sphere, we approximate it with the following sum:
$$A=\sum_{i=1}^n2\pi r_iR\Delta\theta _i$$
Where $r_i$ is the radius of the disk on the plane $z=Rcos{\theta}_i$ and ${\theta}_i$ is the complementary to the polar angle of a point that is both on that plane and the hemisphere surface.
$A'$ as it can easily be proven, when our partition length $\Delta\theta_i$ tends to zero for every $i$, converges to the integral:
$$\int_{0}^{\frac{π}{2}}2πR^2cos\theta d\theta=2πR^2$$
The surface area of the entire sphere is naturally twice that, hence the familiar formula:
$$A=4πR^2$$
Now it would be tempting to approximate the volume of the hemisphere by partitioning it as indicated by the following sum:
$$V'=\sum_{i=1}^nπ{r_i}^2R\Delta{\theta}_i$$
We get the area of the disk at a given point and we multiply that by the arc length that is directly in-between that disk and the one directly after it, with respect to our partitioning. Basically, a direct continuation of the process we followed for the derivation of the surface area of the sphere, instead there we had the circumference of the disk, whereas now we substitute that for its area.
But if we get the Riemann integral that corresponds to the sum $V'$, we get an obviously false result for the volume of the hemisphere:
$$V=πR^3\frac{π^2}{4}$$
When we know that actually: $$V=πR^3\frac{2}{3}$$
(Again, the last two values are associated with the hemisphere, ie half the sphere)
So, it is evident that there is something wrong with the assumption that we can generalize the first process, to derive the volume of the hemisphere. My question in a few words is:
Where, exactly, does that assumption fail?
