Determine the water pressure force on a vertical

Determine the water pressure force on a vertical semicircular plate of radius $a$ whose diameter $2a$ lies at the water surface (the semicircle hangs downward into the water).

Step 1: Width as a function of depth

\[w(y)=2\sqrt{a^2-y^2},\qquad0\le y\le a.\]

Step 2: Set up and evaluate the force integral

\[F=\int_0^a\rho gy\cdot2\sqrt{a^2-y^2}\,dy=2\rho g\int_0^ay\sqrt{a^2-y^2}\,dy.\]

With $u=a^2-y^2,\ du=-2y\,dy$:

\[F=-\rho g\int_{a^2}^0\sqrt u\,du=\rho g\left[\frac{2u^{3/2}}{3}\right]_0^{a^2}=\frac{2\rho ga^3}{3}.\]

\[\boxed{F=\frac{2\rho ga^3}{3}}\]

Step 3: Cross-check via the centroid

A semicircle of radius $a$ has area $A=\pi a^2/2$ and centroid at depth $4a/(3\pi)$ from the diameter. Then $F=\rho g\cdot\dfrac{4a}{3\pi}\cdot\dfrac{\pi a^2}{2}=\dfrac{2\rho ga^3}{3}$, matching Step 2.

Step 4: Center of pressure

Using $y=a\sin\theta$ in the numerator integral $\int_0^ay^2\sqrt{a^2-y^2}\,dy=\dfrac{\pi a^4}{16}$ together with the denominator $a^3/3$ from Step 2:

\[y_{\text{cp}}=\frac{\pi a^4/16}{a^3/3}=\frac{3\pi a}{16}\approx0.589a.\]

\[\boxed{y_{\text{cp}}=\frac{3\pi a}{16}\approx0.589a}\]

This is deeper than the centroid at $4a/(3\pi)\approx0.424a$, as expected: the center of pressure always lies at or below the centroid for a surface where pressure increases with depth.