I am trying to derivate the Archimedes principle for a sphere with direct hydrostatic pressure calculation. I started with the asumptions that: $F_b=F_2-F_1$ and $F_2=2F_1$, where $F_b$ is a buoyancy force, $F_2$ is force of hydrostatic pressure on a lower half of the sphere and $F_1$ is a force of pressure on a upper half of the sphere. Therefore it should be $F_b=F_1$ and because of a spherical symmetry: $F_1 = 2F_{1/2}$, where $F_{1/2}$ is hydrostatic force on a quarter of the sphere in the direction of gravitational field. So the integral for $F_{1/2}$ should be: $$F_{1/2}=\int_{\pi/2}^0 \int_0^{\pi}\rho g r^3 sin^2(\phi) sin(\theta)d\phi d\theta$$ Here is a differential of a force in the direction of gravitational field: $dF=\rho gr \sin^2(\phi)dS$. With the preceding calculation I get that $$F_b=-\frac{\rho g r^3 \pi}{2}$$ I can not see my mistake and I already checked every part of the derivation multiple times.
1 Answer
You need to integrate the pressure at each position to recover Archimedes' law, using the fact that at a given height the pressure is $p(z)=\rho g z$ where $\rho$ is the water's density.
Here'a full derivation. At the end, I have formulas to compare my and yours formalism and see if your assumptions about $F_1$, $F_2$ etc are right!
Complete derivation
First of all, we have the weight of the sphere (with density $\rho_s$) which gives: $$F_w=-mg=-\rho_s {4\over 3}\pi R^3$$
pointed towards the bottom (minus sign). This is the first force acting on the sphere, we will need it later.
On the other hand, the force due to hydrostatic pressure is $$p(z)=\rho g(R-z)$$ where $z$ is the height. We set $z=0$ at the equatorial plane, which will be useful later on when we go to spherical coordinates, so that $p(z=R)=0$ at the north pole of the sphere and the pressure grows towards the bottom of the sphere, reaching $p(z=-R)=\rho g 2R$ at south pole (towards the top). We can always set the $z$ where we want as this simply yields an extra constant external pressure $p_0$ which balances on the top and bottom, so we neglect it.
The pressure acting on the sphere is directed radially (it's actually directed isotropically, but only the radial part acts on the sphere).
This means that the force due to water acting on an infinitesimal surface of the sphere $dA$ is given by
$$dF=p(z)dA$$
and we can write $dA=R^2 \sin(\theta) d\theta d\phi$ using spherical coordinates. $\theta$ is the angle with the vertical, as usual convention, with $\theta=0$ being the north pole and $\theta=\pi$ the south pole.
Because we are only interested in the $z$ direction we only consider the part of the force directed along the z direction (the other components disappear by symmetry as they all are on the same $z$-level).
So we only need $$dF_z=dF \cos(\theta)$$ where, as before, $\theta$ is the angle with respect to the vertical director. At $\theta=0$ (north pole) the force pushes towards the bottom.
Now we integrate on the sphere only along $\theta$ as the force does not depend on $\phi$. The integral on $\phi$ gives us a $2\pi$ factor.
$$F_z=\int_0^{2\pi}\int_0^\pi dF_z =\int_0^{2\pi}\int_0^\pi dF cos(\theta) = 2\pi \int_0^\pi p(z)dA(\theta)\cos(\theta)$$ where $dA$ has already been integrated in $\phi$.
Putting all together and using $z=R\cos(\theta)$ and $dA(\theta)=R^2 \sin(\theta) d\theta$ we get
$$F_z= 2\pi \int_0^\pi \rho g (z-R) R^2 \sin(\theta) \cos(\theta)d\theta =$$
$$ = 2\pi \int_0^\pi \rho g (R \cos(\theta)-R) R^2 \sin(\theta) \cos(\theta)d\theta$$
which we then rewrite as
$$F_z = 2\pi \int_0^\pi \rho g R(\cos(\theta)-1) R^2 \sin(\theta) \cos(\theta) d\theta$$ and so
$$F_z = 2\pi R^3 \rho g \int_0^\pi (\cos(\theta)-1)\sin(\theta)\cos(\theta) d\theta$$
The integral is not that hard, here is the solution so we get $$\int_0^\pi (\cos(\theta)-1)\sin(\theta)\cos(\theta) d\theta={2\over 3}$$ and this means
$$F_z= 2\pi R^3 \rho g \int_0^\pi (\cos(\theta)-1)\sin(\theta)\cos(\theta) d\theta = {4\over 3}\pi R^3 \rho g$$
Which means that the total force acting on the sphere, is $$F_b = F_w + F_z = -\rho_s {4\over 3}\pi R^3 + {4\over 3}\pi R^3 \rho g = (\rho-\rho_s){4\over 3}\pi R^3$$
i.e. Archimedes law: the sphere weight minus the equivalent weight in water.
Comparison with you computation
Using your approach, you can check your results by computing $F_1$, $F_2$ and $F_{1/2}$ using the integral for $F_z$ at different values of the angle $\theta$ i.e.
$$F_z(\alpha, \beta)= 2\pi R^3 \rho g \int_\alpha ^\beta (\cos(\theta)-1)\sin(\theta)\cos(\theta) d\theta$$
with $$F_1=F_z(0, \pi/2)=-{1\over3}\pi R^3 \rho g $$, $$F_2=F_z(\pi/2, \pi)={5\over 6}\pi R^3 \rho g $$ (so it is not true that $F_1=2F_2$ !!!).
You could also compute $F_{1/2}=Fz(0, \pi/4)$ (if I got it right).
