Why Moments of Inertia Depend on Distance
A surface lamina is a thin sheet of matter distributed over a surface. Its mass depends on both the amount of material at each point and the area of the sheet. For rotation about an axis, however, mass alone is not enough to describe how strongly the lamina resists changes in its rotation: mass farther from the axis contributes more.
The distance matters quadratically. A small piece of mass twice as far from an axis contributes four times as much to the moment of inertia. To calculate the total effect, we weight each piece of surface area by its surface density and its squared perpendicular distance from the axis, then integrate over the surface.
From Distance to an Integral
The perpendicular distance from a point \((x,y,z)\) to the \(x\)-axis is \(\sqrt{y^2+z^2}\), because the closest point on that axis is \((x,0,0)\). Its squared distance is therefore \(y^2+z^2\). The corresponding squared distances to the \(y\)- and \(z\)-axes are \(x^2+z^2\) and \(x^2+y^2\).
A small patch of area \(dS\) near \((x,y,z)\) has mass \(dm=\rho\,dS\). Its contribution to the moment about the \(x\)-axis is its squared distance from that axis times its mass: \(\bigl(y^2+z^2\bigr)\rho\,dS\). Adding these contributions over the whole surface gives the defining integral. The same reasoning gives the formulas for the other axes.
To evaluate one of these integrals, we need the appropriate surface area element. For a parametrized surface \(\mathbf r(u,v)\), it is \(dS=\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\). For a graph \(z=g(x,y)\), it is \(dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy\). These elements account for the actual area of the surface, rather than merely the area of a parameter domain or its projection.
For example, the moment about the \(z\)-axis of a graph over a domain \(D\) is $$ I_z=\iint_D (x^2+y^2)\rho(x,y,g(x,y)) \sqrt{1+g_x^2+g_y^2}\,dx\,dy. $$ This formula applies when the graph and density are defined over the domain and the integral exists. The other axis moments use their corresponding squared-distance expressions instead.
Worked Example: A Uniform Rectangular Lamina
Worked Example: A Uniform Rectangular Lamina
Let \(S\) be the rectangle in the \(xy\)-plane given by \(0\le x\le 2\) metres and \(0\le y\le 1\) metre. Its density is constant, \(\rho=3\) kilograms per square metre. Find its moments about all three coordinate axes.
On this surface \(z=0\), and \(dS=dx\,dy\). Thus the three squared-distance factors reduce to \(y^2\), \(x^2\), and \(x^2+y^2\), respectively. About the \(x\)-axis, $$ I_x=3\int_0^2\int_0^1 y^2\,dy\,dx =3\int_0^2\left[\frac{y^3}{3}\right]_0^1dx =3\int_0^2\frac13\,dx =2\ \text{kg}\,\text{m}^2. $$ The antiderivative \(y^3/3\) is verified by differentiating it to obtain \(y^2\).
About the \(y\)-axis, $$ I_y=3\int_0^1\int_0^2 x^2\,dx\,dy =3\int_0^1\left[\frac{x^3}{3}\right]_0^2dy =3\int_0^1\frac83\,dy =8\ \text{kg}\,\text{m}^2. $$ Here, differentiating \(x^3/3\) gives \(x^2\), and the evaluated inner integral is \(2^3/3-0^3/3=8/3\).
For the \(z\)-axis, add the two contributions: $$ I_z=3\int_0^2\int_0^1(x^2+y^2)\,dy\,dx =3\left(\int_0^2x^2\,dx+\int_0^2\int_0^1y^2\,dy\,dx\right) =3\left(\frac83+\frac23\right) =10\ \text{kg}\,\text{m}^2. $$ The check is that the integrand is exactly the sum of the two squared distances used for \(I_x\) and \(I_y\), so \(I_z=I_x+I_y=2+8=10\), as calculated directly.
Worked Example: A Uniform Upper Hemisphere
Worked Example: A Uniform Upper Hemisphere
Consider the upper hemisphere \(x^2+y^2+z^2=R^2\), \(z\ge 0\), with constant density \(\sigma\). Find its moment about the \(z\)-axis. Use spherical coordinates with \(0\le\theta\le\pi/2\) measured from the positive \(z\)-axis and \(0\le\phi\le2\pi\) around that axis.
On the sphere, \(x^2+y^2=R^2\sin^2\theta\), and the surface area element is \(dS=R^2\sin\theta\,d\theta\,d\phi\). Therefore $$ I_z=\int_0^{2\pi}\int_0^{\pi/2} R^2\sin^2\theta\,\sigma R^2\sin\theta\,d\theta\,d\phi =\sigma R^4\left(\int_0^{2\pi}d\phi\right) \left(\int_0^{\pi/2}\sin^3\theta\,d\theta\right). $$ The angular integral in \(\phi\) is \(2\pi\). An antiderivative of \(\sin^3\theta\) is \(-\cos\theta+\cos^3\theta/3\), since its derivative is \(\sin\theta-\cos^2\theta\sin\theta=\sin^3\theta\). Hence $$ \int_0^{\pi/2}\sin^3\theta\,d\theta =\left[-\cos\theta+\frac{\cos^3\theta}{3}\right]_0^{\pi/2} =0-\left(-1+\frac13\right) =\frac23. $$ It follows that $$ I_z=\frac{4\pi}{3}\sigma R^4. $$ As a check, the surface area of this hemisphere is \(2\pi R^2\), so its mass is \(2\pi\sigma R^2\). The ratio \(I_z/\text{mass}=2R^2/3\) has units of squared distance, as it must. The moment itself has units \((\text{mass}/\text{area})\times(\text{length})^4=\text{mass}\times\text{length}^2\).
Worked Example: A Graph with Varying Density
Worked Example: A Graph with Varying Density
Let the surface be the graph \(z=x\) above the square \(0\le x\le L,\ 0\le y\le L\), where \(L\) is a length. Let its density be \(\rho(x,y)=\rho_0(1+x/L)\), where \(\rho_0\) is a constant surface density. Find its moments about the \(z\)- and \(y\)-axes.
Since \(g_x=1\) and \(g_y=0\), the area element is \(dS=\sqrt{2}\,dx\,dy\). For the \(z\)-axis the squared distance is \(x^2+y^2\). Separating the two terms gives $$ I_z=\sqrt{2}\rho_0 \left[ L\int_0^L x^2\left(1+\frac{x}{L}\right)dx +\left(\int_0^L\left(1+\frac{x}{L}\right)dx\right) \left(\int_0^L y^2\,dy\right) \right]. $$ The first integral is $$ \int_0^L x^2\left(1+\frac{x}{L}\right)dx =\left[\frac{x^3}{3}+\frac{x^4}{4L}\right]_0^L =\frac{L^3}{3}+\frac{L^3}{4} =\frac{7L^3}{12}. $$ The derivative of the displayed antiderivative is \(x^2+x^3/L\), the original integrand. The other two integrals are $$ \int_0^L\left(1+\frac{x}{L}\right)dx =\left[x+\frac{x^2}{2L}\right]_0^L =\frac{3L}{2},\qquad \int_0^L y^2\,dy=\left[\frac{y^3}{3}\right]_0^L=\frac{L^3}{3}. $$ Substituting these values, $$ I_z=\sqrt{2}\rho_0\left(\frac{7L^4}{12}+\frac{L^4}{2}\right) =\frac{13\sqrt{2}}{12}\rho_0L^4. $$
For the \(y\)-axis, \(x^2+z^2=2x^2\) on this graph. Therefore $$ I_y=2\sqrt{2}\rho_0L \int_0^L x^2\left(1+\frac{x}{L}\right)dx =2\sqrt{2}\rho_0L\left(\frac{7L^3}{12}\right) =\frac{7\sqrt{2}}{6}\rho_0L^4. $$ This result uses the same verified \(x\)-integral as above. Both moments have units \(\rho_0L^4\), or mass times length squared, which checks their dimensions.
Common Pitfall: Using Distance Rather Than Squared Distance
A frequent error is to use the perpendicular distance itself in the integrand. For example, a point’s distance from the \(z\)-axis is \(\sqrt{x^2+y^2}\), but the moment of inertia uses its square. The incorrect expression $$ I_z^{\text{wrong}}=\iint_S\sqrt{x^2+y^2}\,\rho\,dS $$ weights each mass element by distance, not squared distance. The correct expression is $$ I_z=\iint_S(x^2+y^2)\rho\,dS. $$ The dimensions expose the mistake: the incorrect integral has units of mass times length, whereas a moment of inertia must have units of mass times length squared.
Another error is to use only the two coordinates that appear in the surface parametrization without checking which axis is being considered. The reliable procedure is to identify the point’s perpendicular distance to the specified axis in three-dimensional space, square that distance, and only then insert the surface density and area element.
What the Moments Tell Us
Each moment measures how the mass of the lamina is distributed relative to a particular axis. The same lamina can have different moments about different axes because the squared distances differ. Moments of inertia are used in rotational mechanics, where they quantify the relation between applied torque and angular acceleration. In geometric calculations, they also provide a useful test of whether the correct axis, density, and surface area element have been used.
For a surface lamina, the key calculation is always: choose the axis, write its squared perpendicular distance, multiply by \(\rho\,dS\), and integrate over the entire surface. This separates the geometric effect of distance from the mass distribution and applies whether the surface is flat or curved.
Check Your Understanding
For each question, identify the axis and the relevant squared-distance factor before evaluating any integral.
- What is the squared perpendicular distance from \((x,y,z)\) to the \(y\)-axis?
- Write the formula for \(I_z\) of a surface lamina with density \(\rho\).
- For a graph \(z=g(x,y)\), what surface area factor multiplies \(dx\,dy\)?
- Why does a moment of inertia have units of mass times length squared?
- On the graph \(z=x\), what squared-distance factor should be used for a moment about the \(y\)-axis?