I am trying to proof that $V_{sphere}=\frac{4}{3}{\pi}R^3$.
I think that a sphere should be composed by infinite cylinders with an height that tends to zero and a ray that varies from $0\to R$ but I do not obtain the former equation for the Volume.
I came up with this equation but it's wrong.
$V=2\int_0^R{\pi R^2dr}$
On
Another way to prove it is to generalize the method of cylindrical shells to a method of spherical shells. The idea is the same, to take round slices and integrate over them, exploiting a symmetry. Then life is good, since the surface area of a sphere is $4 \pi R^2$, integrating $\int_0^R 4 \pi x^2 dx$ clearly gives the desired formula.
The surface area formula can also be found directly also using integration, by doing the double integral $$\int_0^{2\pi} \int_0^\pi r^2 \sin \theta\ d \theta \ d \phi$$
Which can easily be worked out to give the formula $S = 4 \pi R^2$, which we can then re-integrate. Alternatively, you can do this in one-fell-swoop by just integrating three times all at once.
$$\int_0^R \int_0^{2 \pi} \int_0^\pi r^2 \sin \theta \ d \theta \ d \phi \ dr$$
The surface of a slice is $\pi$ times its squared radius, and by the implicit formula for the sphere, the squared radius of the slice is the squared radius of the sphere minus the squared altitude.
So in terms of the altitude, the squared radius of the slice follows a downward parabolic curve, the area of which is easily found to be the two thirds of the enclosing rectangle.