Find the centroid of the arc of the circle x²+y²=a² above

Find the centroid of the arc of the circle $x^2+y^2=a^2$ above the $x$-axis.

Step 1: Parametrize

$x=a\cos\theta,\ y=a\sin\theta,\ \theta\in[0,\pi]$; $ds=\sqrt{a^2\sin^2\theta+a^2\cos^2\theta}\,d\theta=a\,d\theta$.

Step 2: Total arc length

\[L=\int_0^\pi a\,d\theta=\pi a\]

Step 3: Moment about the $y$-axis

\[M_y=\int_0^\pi a\cos\theta\cdot a\,d\theta = a^2[\sin\theta]_0^\pi=0\]

(zero by the arc's symmetry about the $y$-axis.)

Step 4: Moment about the $x$-axis

\[M_x=\int_0^\pi a\sin\theta\cdot a\,d\theta = a^2[-\cos\theta]_0^\pi = a^2[1-(-1)]=2a^2\]

Step 5: Centroid

\[\bar x=\frac{M_y}L=0,\qquad \bar y=\frac{M_x}L=\frac{2a^2}{\pi a}=\frac{2a}\pi\]

\[\boxed{\text{Centroid: }\left(0,\frac{2a}\pi\right)\approx(0,0.637a)}\]

Step 6: Interpretation

The centroid sits at about $63.7\%$ of the radius above the diameter — higher than the geometric midpoint $\frac a2$, since the arc "spreads out" more (contributes more length per unit height) near the top of the semicircle than near its ends.