Find the force of water pressure on the inner lateral
Find the force of water pressure on the inner lateral side $x>0$ of a cylindrical vessel $x^2+y^2=a^2$, filled with water to depth $h$.
Explanation of Method
Parametrizing the half-cylinder by $\theta\in\left(-\frac\pi2,\frac\pi2\right)$ and $z\in[0,h]$, the pressure force per unit area at height $z$ is $\varrho g(h-z)$ acting along the outward radial normal; integrating its horizontal components over the half-cylinder gives the net force, which is then cross-checked via the projected-area shortcut.
- Set up the pressure and the surface parametrization.
- Integrate the $x$- and $y$-components of the force.
- Verify via the projected-area method.
Step 1 · Setup
With $z$ measured upward from the base, the pressure at height $z$ is $p(z)=\varrho g(h-z)$. Parametrize the half-cylinder ($x>0$) by $x=a\cos\theta,\,y=a\sin\theta,\,\theta\in\left(-\frac\pi2,\frac\pi2\right)$, $z\in[0,h]$, surface element $dS=a\,d\theta\,dz$, outward normal $\mathbf n=(\cos\theta,\sin\theta,0)$.
Step 2 · Force components
The pressure force on the wall acts along $\mathbf n$: $d\mathbf F=p(z)\,\mathbf n\,dS$.
$$ F_x = \rho ga\int_0^h(h-z)\,dz\cdot\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta = \rho ga\cdot\frac{h^2}2\cdot2 = \rho gah^2, $$
$$ F_y = \rho ga\int_0^h(h-z)\,dz\cdot\underbrace{\int_{-\pi/2}^{\pi/2}\sin\theta\,d\theta}_{=0} = 0, \qquad F_z=0. $$
Step 3 · Verification via projected area
The right half-cylinder projects onto a rectangle of width $2a$ and height $h$ in the $yz$-plane; the net horizontal force on a curved surface equals the pressure force on its projection:
$$ F_x = \int_0^h\rho g(h-z)\cdot2a\,dz = 2\rho ga\cdot\frac{h^2}2 = \rho gah^2.\ \checkmark $$
Additional Notes
- This derivation matches the source exactly. (The source's own working includes a brief self-correction of the sign of the normal direction mid-derivation, but arrives at the same final result shown here.)
Result
The force on the half-cylinder wall is directed horizontally, with magnitude: