A sphere of radius a is submerged in a liquid of density
A sphere of radius $a$ is submerged in a liquid of density $\varrho$ to depth $h$ (measured from the center), $h>a$. Find the force of liquid pressure on the lower and upper hemispheres of the sphere's surface.
Explanation of Method
With the center of the sphere at the origin and the free surface at $z=h$, only the vertical component of the pressure force survives by rotational symmetry; splitting the resulting surface integral over the lower and upper hemispheres (and checking their sum against Archimedes' principle) gives both forces.
- Set up the pressure and note that only $F_z$ survives.
- Compute $F_z$ over the whole sphere and check against Archimedes' principle.
- Compute $F_z$ on the lower hemisphere.
- Compute $F_z$ on the upper hemisphere, and verify the sum.
Step 1 · Setup
Pressure at $(x,y,z)$ on the sphere: $p=\varrho g(h-z)$. Outward normal $\mathbf n=\frac1a(x,y,z)$; force on a surface element acts along $-\mathbf n$ times $p$ (inward, from liquid onto sphere), i.e. $d\mathbf F=-p\,\mathbf n\,dS$. By rotational symmetry about the $z$-axis, $F_x=F_y=0$.
Step 2 · Total vertical force (Archimedes check)
In spherical coordinates ($dS=a^2\sin\phi\,d\phi\,d\theta$): $\iint_Sz\,dS=0$ and $\iint_Sz^2\,dS=\dfrac{4\pi a^4}3$, so:
$$ F_z = -\frac{\varrho g}a\left[h\cdot0-\frac{4\pi a^4}3\right] = \frac{4\pi\varrho ga^3}3 = \varrho gV,$$
matching Archimedes' principle for the whole sphere.
Step 3 · Lower hemisphere ($\phi\in[\pi/2,\pi]$)
$$ \iint_{\text{lower}}z\,dS = 2\pi a^3\!\int_{\pi/2}^{\pi}\!\sin\phi\cos\phi\,d\phi = -\pi a^3, \qquad \iint_{\text{lower}}z^2\,dS = 2\pi a^4\!\int_{\pi/2}^{\pi}\!\cos^2\phi\sin\phi\,d\phi = \frac{2\pi a^4}3. $$
$$ F_z^{\text{lower}} = -\frac{\varrho g}a\left[-\pi a^3h-\frac{2\pi a^4}3\right] = \pi\varrho ga^2\left(h+\frac{2a}3\right), $$
directed upward.
Step 4 · Upper hemisphere and verification
$$ \iint_{\text{upper}}z\,dS = \pi a^3, \qquad \iint_{\text{upper}}z^2\,dS = \frac{2\pi a^4}3, \qquad F_z^{\text{upper}} = -\pi\varrho ga^2\left(h-\frac{2a}3\right), $$
directed downward, with magnitude $\pi\varrho ga^2\left(h-\dfrac{2a}3\right)$. Sum of the two vertical contributions: $\pi\varrho ga^2\left(h+\frac{2a}3\right)-\pi\varrho ga^2\left(h-\frac{2a}3\right)=\dfrac{4\pi\varrho ga^3}3=\varrho gV$, matching Step 2.
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=1.7,\,h=3.2$): $F_{\text{lower}}\approx39.343$, $F_{\text{upper}}\approx18.764$, matching $\pi\varrho ga^2(h\pm2a/3)$ exactly.
Result
The force on the lower hemisphere (upward) and the upper hemisphere (downward) are: