How is the centroid of a semicircle and a quarter circle derived to be 4r/3π from the diameter?

In one line

Divide the area into thin triangular sectors from the centre, integrate their first moments, and divide by the area to get \( 4r/3\pi \).

Explained simply

Picture a semicircular pizza cut into very many thin slices from the centre. Each slice is almost a triangle whose own centroid is two-thirds of the way out along the slice. Averaging all these slice centroids, weighted by area, gives the centroid of the whole shape.

Technically, we use polar elements: area \( dA = \tfrac12 r^2\,d\theta \) with its centroid at distance \( \tfrac{2r}{3} \) from the centre.

How it works

1. Place the diameter on the x-axis with the centre at the origin; by symmetry \( \bar{x} = 0 \).
2. Elemental sector at angle \( \theta \), angle \( d\theta \): \( dA = \tfrac12 r^2 d\theta \).
3. Its centroid height: \( y = \tfrac{2r}{3}\sin\theta \).
4. \( \int y\,dA = \int_0^{\pi} \tfrac{2r}{3}\sin\theta \cdot \tfrac{r^2}{2} d\theta = \tfrac{r^3}{3}[-\cos\theta]_0^{\pi} = \tfrac{2r^3}{3} \).
5. \( A = \tfrac{\pi r^2}{2} \), so \( \bar{y} = \dfrac{2r^3/3}{\pi r^2/2} = \dfrac{4r}{3\pi} \).
6. Quarter circle: integrate from 0 to \( \pi/2 \): \( \int y\,dA = r^3/3 \), \( A = \pi r^2/4 \), so \( \bar{y} = 4r/3\pi \); by symmetry about the 45° line, \( \bar{x} = 4r/3\pi \) as well.

Key formula

\( \bar{y} = \dfrac{4r}{3\pi} \) from the diameter (semicircle) and from each straight edge (quarter circle), r = radius (mm).

Quick example

For r = 60 mm, \( 4r/3\pi = 240/9.425 = 25.46 \) mm. So a semicircular plate of radius 60 mm has its centroid 25.46 mm from the diameter on the axis of symmetry, and a quarter circle of radius 60 mm has it at (25.46, 25.46) mm from its corner.

Common mistakes

• Placing the element's centroid at r/2 or r instead of 2r/3 from the centre.
• Using limits 0 to \( 2\pi \) for the semicircle.
• Thinking the quarter circle's centroid is \( 4r/3\pi \) from the curved edge.

How MU asks it

Asked as a 4 to 5-mark derivation: 'Find the centroid of a semicircular lamina (or quarter circle) from first principles'. Draw the sector element, state dA and its centroid, integrate step by step, and state the final result with the reference edge.