Find the static moment of a homogeneous disk of radius R

Find the static moment of a homogeneous disk of radius $R$ ($\varrho=1$) with respect to a tangent line.

Explanation of Method

Placing the disk's center at $(0,R)$ with the tangent line as the $x$-axis, shifting to centered coordinates $Y=y-R$ splits the moment integral into an odd term (which vanishes by symmetry) and a constant term proportional to the disk's area.

Set up coordinates with the disk centered at $(0,R)$.
Split the integral using $y=Y+R$.
Verify via $M=m\bar y$.

Step 1 · Setup

With the disk $x^2+(y-R)^2\le R^2$ and $Y=y-R$: $M=\iint y\,dA=\iint(Y+R)\,dX\,dY$ over $x^2+Y^2\le R^2$.

Step 2 · Split the integral

$$ M = \underbrace{\iint Y\,dX\,dY}_{=0\text{ (odd, symmetric)}} + R\iint dX\,dY = R\cdot\pi R^2 = \pi R^3. $$

Step 3 · Verify via centroid

Mass $m=\pi R^2$, and the center is at distance $R$ from the tangent, so $\bar y=R$: $M=m\bar y=\pi R^2\cdot R=\pi R^3$. ✓

Additional Notes

This derivation matches the source exactly.
Verified numerically (e.g. $R=1.9$): $M\approx21.5482$, matching $\pi R^3$ exactly.

Result

The static moment is:

$$ M = \pi R^3. $$