A thin metal dome, a soap film, a satellite dish: these are shells, surfaces with mass but almost no thickness. Their mass, balance point and resistance to spinning are surface integrals, just as the same quantities for a flat plate were double integrals and for a solid were triple integrals.
Key idea
Why it works
Compare shells with solids of the same shape. A uniform hemispherical shell balances at height $\frac{a}{2}$, while a solid half-ball balances lower, at $\frac{3a}{8}$, because the solid has more of its mass near the flat base. A thin spherical shell has $I_z = \frac{2}{3}Ma^2$, while a solid ball of the same mass and radius has $\frac{2}{5}Ma^2$: the shell's mass is all far from the axis, so it is harder to start spinning. That is why a hollow ball and a solid ball rolled down the same ramp arrive at different times, the solid one first.
Worked examples
Worked example: A uniform dome
- By symmetry $\bar{x} = \bar{y} = 0$. With constant $\delta$, $M = \delta \cdot 2\pi a^2$.
- $$\iint_S z\,\delta\,dS = \delta\int_0^{2\pi} \int_0^{\pi/2} (a\cos\varphi)(a^2\sin\varphi)\,d\varphi\,d\theta = \delta \cdot 2\pi a^3 \cdot \frac{1}{2} = \delta\pi a^3.$$
- $\bar{z} = \frac{\delta\pi a^3}{2\pi a^2\delta} = \frac{a}{2}$. (A solid hemisphere balances lower, at $\frac{3a}{8}$.)
Answer: $\left(0, 0, \frac{a}{2}\right)$
Worked example: A can with varying density
- $dS = d\theta\,dz$. $M = \int_0^{2\pi} \int_0^2 z\,dz\,d\theta = 2\pi \cdot 2 = 4\pi$.
- $\iint z \cdot z\,dS = 2\pi \cdot \frac{8}{3} = \frac{16\pi}{3}$, so $\bar{z} = \frac{16\pi/3}{4\pi} = \frac{4}{3}$, above the middle because the top is heavier.
Answer: $M = 4\pi$, $\bar{z} = \frac{4}{3}$
Worked example: Spinning a spherical shell
- $\delta = \frac{M}{4\pi a^2}$, and on the sphere $x^2 + y^2 = a^2\sin^2\varphi$.
- $$I_z = \delta\int_0^{2\pi} \int_0^\pi a^2\sin^2\varphi \cdot a^2\sin\varphi\,d\varphi\,d\theta = \delta \cdot a^4 \cdot 2\pi \cdot \frac{4}{3}.$$
- Substituting $\delta$: $I_z = \frac{2}{3}Ma^2$.
Answer: $\frac{2}{3}Ma^2$