The ideas from plates carry over to solid objects: integrate density to get mass, weight by position to find the balance point, and weight by squared distance from an axis to get the moment of inertia, which measures how hard the object is to spin.
Key idea
Why it works
Two shortcuts save work. Symmetry: if the solid and its density are symmetric across a plane, the center of mass lies in that plane, so for a solid of revolution about the $z$-axis only $\bar{z}$ needs computing. Coordinates: moments about the $z$-axis contain $x^2 + y^2 = r^2$, which is simplest in cylindrical coordinates, while balls and cones centered at the origin are simplest in spherical ones. For instance, a solid ball of radius $a$ and mass $m$ has $I_z = \frac{2}{5}ma^2$, a standard result found in spherical coordinates, where $x^2 + y^2$ becomes the square of the distance to the axis.
Worked examples
Worked example: Mass of a cube
- $m = \int_0^1\int_0^1\int_0^1 (x + y + z)\,dz\,dy\,dx$.
- Each of the three terms integrates to $\frac{1}{2}$ over the cube, so $m = \frac{3}{2}$.
Answer: $\frac{3}{2}$
Worked example: Balance point of a half-ball
- Mass (density 1): the volume $\frac{2}{3}\pi a^3$.
- Moment: $\iiint z\,dV = \frac{\pi a^4}{4}$ (from the spherical-coordinates lesson).
- $\bar{z} = \frac{\pi a^4/4}{2\pi a^3/3} = \frac{3a}{8}$: below the middle height $\frac{a}{2}$, since the half-ball is wider at the bottom.
Answer: $\bar{z} = \frac{3a}{8}$
Worked example: Spinning a cylinder
- Cylindrical coordinates: $I_z = \int_0^{2\pi}\int_0^R\int_0^H r^2 \cdot r\,dz\,dr\,d\theta$.
- $= 2\pi \cdot H \cdot \frac{R^4}{4} = \frac{\pi R^4H}{2}$.
- With mass $M = \pi R^2H$, that is $\frac{1}{2}MR^2$, the formula physics books quote.
Answer: $\frac{\pi R^4H}{2}$
Common mistake