Double Integral 255
Easy Applications

Find the centroid of a homogeneous semicircle of radius a

Find the centroid of a homogeneous semicircle of radius $a$.

Explanation of Method

Placing the diameter on the $x$-axis makes the region symmetric about the $y$-axis, so only the static moment about the diameter needs to be computed — done cleanly in polar coordinates.

  1. Use symmetry to get $\bar x=0$.
  2. Compute the area $A$.
  3. Compute the static moment $M_x$ in polar coordinates.
  4. Divide to get $\bar y$.

Step 1 · Setup and symmetry

With $D:x^2+y^2\le a^2,\,y\ge0$, the region is symmetric about the $y$-axis, so $\bar x=0$.

Step 2 · Area

$$ A = \frac{\pi a^2}2. $$

Step 3 · Static moment $M_x$

In polar coordinates $x=r\cos\theta,\,y=r\sin\theta$, $dA=r\,dr\,d\theta$, with $0\le r\le a$, $0\le\theta\le\pi$:

$$ M_x = \int_0^\pi\int_0^ar\sin\theta\cdot r\,dr\,d\theta = \int_0^\pi\sin\theta\,d\theta\cdot\int_0^ar^2\,dr = 2\cdot\frac{a^3}3 = \frac{2a^3}3. $$

Step 4 · Centroid coordinate $\bar y$

$$ \bar y = \frac{M_x}A = \frac{2a^3/3}{\pi a^2/2} = \frac{4a}{3\pi}. $$


Additional Notes

Result

The centroid of the homogeneous semicircle lies on the axis of symmetry at:

$$ \bar x = 0, \qquad \bar y = \frac{4a}{3\pi}. $$