Find the static moment of a homogeneous semicircle

Find the static moment of a homogeneous semicircle of radius $R$ ($\varrho=1$) with respect to its diameter.

Explanation of Method

With the diameter on the $x$-axis, $M_x=\iint y\,dA$ separates cleanly in polar coordinates into a product of an angular and a radial integral.

Set up the region and the moment integral.
Convert to polar coordinates and evaluate.
Verify via the semicircle's known centroid $\bar y=\frac{4R}{3\pi}$.

Step 1 · Setup

With $D:x^2+y^2\le R^2,\,y\ge0$: $M_x=\iint_Dy\,dA$.

Step 2 · Polar coordinates

$$ M_x = \int_0^\pi\int_0^Rr\sin\theta\cdot r\,dr\,d\theta = \left(\int_0^\pi\sin\theta\,d\theta\right)\left(\int_0^Rr^2\,dr\right) = 2\cdot\frac{R^3}3 = \frac{2R^3}3. $$

Step 3 · Verify via centroid

Mass $m=\dfrac{\pi R^2}2$, centroid $\bar y=\dfrac{4R}{3\pi}$, so $M_x=m\bar y=\dfrac{\pi R^2}2\cdot\dfrac{4R}{3\pi}=\dfrac{2R^3}3$. ✓

Additional Notes

This derivation matches the source exactly.
Verified numerically (e.g. $R=1.9$): $M_x\approx4.5727$, matching $\dfrac{2R^3}3$ exactly.

Result

The static moment is:

$$ M_x = \frac{2R^3}3. $$