Centre of Mass of a Hemispherical Shell

Result

The CM of a thin uniform hollow hemispherical shell of radius $R$ lies at:

$y_{cm} = \frac{R}{2}$

exactly halfway between the base and the top.

Derivation

Divide the shell into thin horizontal rings. A ring at angle $\theta$ from the vertical (polar angle from top):

• Radius of ring: $R\sin\theta$
• Width of ring along surface: $R \, d\theta$
• Area of ring: $2\pi R\sin\theta \cdot R \, d\theta = 2\pi R^2 \sin\theta \, d\theta$
• Mass: $dm = \sigma \cdot 2\pi R^2 \sin\theta \, d\theta$ where $\sigma = \frac{M}{2\pi R^2}$
• Height of ring above base: $y = R\cos\theta$ (measuring from base, $\theta = 0$ at top, $\theta = \pi/2$ at base... )

Let me use a cleaner setup: $\theta$ measured from the axis (top = 0, equator = $\pi/2$). Height of ring from base = $R\cos\theta$ (wait, let me define from base):

Taking $\theta$ as polar angle from the axis of symmetry (top of hemisphere):

• $\theta = 0$: top of hemisphere, height $= R$ above base
• $\theta = \pi/2$: rim, height $= 0$

Height above base: $y = R - R(1-\cos\theta)$... Let me use: height above base $= R\cos\theta$ where $\theta$ goes from $0$ at top to $\pi/2$ at the rim.

Actually, cleanest: measure $\theta$ from the flat base upward. At angle $\theta$ from the base plane, the ring has $y = R\sin\theta$ (height) and ring radius $= R\cos\theta$.

Ring area: $2\pi R\cos\theta \cdot R \, d\theta$

$dm = \sigma \cdot 2\pi R^2\cos\theta \, d\theta, \quad \sigma = \frac{M}{2\pi R^2}$

$y_{cm} = \frac{1}{M}\int_0^{\pi/2} R\sin\theta \cdot \frac{M}{2\pi R^2} \cdot 2\pi R^2\cos\theta \, d\theta$

$= R\int_0^{\pi/2} \sin\theta\cos\theta \, d\theta = R \cdot \frac{1}{2}\int_0^{\pi/2}\sin 2\theta \, d\theta$

$= \frac{R}{2}\left[-\frac{\cos 2\theta}{2}\right]_0^{\pi/2} = \frac{R}{2} \cdot \frac{1+1}{2} = \frac{R}{2}$

$\boxed{y_{cm} = \frac{R}{2}}$

Comparison with solid hemisphere

The hollow shell's CM is higher — its mass is concentrated at the curved surface (farther from the base). The solid hemisphere has mass distributed throughout the volume, including near the base, pulling the CM lower.