Setting Up Lamina Integrals With Vertical Slices

From a Lamina to Vertical Strips

A lamina is a thin, flat object whose mass is spread over a two-dimensional region. To find its mass and first moments, we can divide the region into narrow vertical strips, find the mass and moments of each strip, and add them across the region. This approach is useful when the region has a simple description in terms of its left and right edges and its top and bottom curves.

Let the region extend from \(x=a\) to \(x=b\), with top curve \(y=f(x)\) and bottom curve \(y=g(x)\). At each \(x\), the vertical part of the region runs from \(g(x)\) to \(f(x)\). The functions must satisfy \(g(x)\leq f(x)\) throughout the interval. We will allow the mass density to vary from point to point.

Suppose the lamina has areal density \(\rho(x,y)\), measured as mass per unit area. For example, if lengths are measured in meters and mass in kilograms, then \(\rho\) has units of \(\mathrm{kg/m^2}\). We assume the density is integrable over the region, and ordinarily nonnegative because it represents physical mass density.

Mass and First-Moment Integrals

A thin vertical strip at \(x\), with width \(dx\), has height \(f(x)-g(x)\) and area approximately \((f(x)-g(x))\,dx\). If density varies with \(y\) as well as \(x\), it need not be the same throughout that strip. Its mass per unit width is therefore obtained by integrating the density from the bottom to the top of the strip.

The coordinate in each moment has a specific role. The moment about the \(y\)-axis uses the horizontal distance \(x\) from that axis; the moment about the \(x\)-axis uses the vertical distance \(y\). These definitions are weighted sums of mass: each small amount of mass is multiplied by its distance from the relevant axis.

Why the Slice Formulas Work

For a fixed \(x\), the mass in a thin vertical strip is approximately its width \(dx\) times the density accumulated through its height. Its mass is thus approximately \(\left(\int_{g(x)}^{f(x)}\rho(x,y)\,dy\right)dx\). The strip lies at horizontal position \(x\), so its contribution to \(M_y\) is \(x\) times that mass. To obtain its contribution to \(M_x\), each point at height \(y\) must instead be weighted by \(y\), giving \(\left(\int_{g(x)}^{f(x)}y\rho(x,y)\,dy\right)dx\). Adding the contributions of all strips and taking the limit as their widths shrink gives the three iterated integrals in the theorem.

When Density Is Constant

If the density has the same value \(\sigma\) everywhere, it can be taken outside the integrals. The mass of a strip is \(\sigma(f(x)-g(x))\,dx\). For its first moment about the \(x\)-axis, integrate \(y\) across the strip: \(\int_{g(x)}^{f(x)}y\,dy=(f(x)^2-g(x)^2)/2\).

The expressions can also be checked against the general theorem: for example, the inner integral for \(M_x\) is \(\int_g^f y\sigma\,dy=\frac{\sigma}{2}(f^2-g^2)\). The order of the curve values matters: the top boundary supplies the upper limit, and the bottom boundary supplies the lower limit.

Worked Examples

Common Pitfall: Using the Wrong Coordinate

A frequent error is to use \(y\) in the moment about the \(y\)-axis, or \(x\) in the moment about the \(x\)-axis. For example, in the first example, the incorrect calculation \(M_y=3\int_0^2 y(4-2x)\,dx\) cannot represent the moment about the \(y\)-axis: the factor \(y\) varies along the strip, and the strip’s horizontal distance from the \(y\)-axis is \(x\). The correct calculation is \(M_y=3\int_0^2x(4-2x)\,dx=8\ \mathrm{kg\,m}\). For \(M_x\), the vertical distance is \(y\), so its contribution must be integrated across the strip’s height.

Checking a Vertical-Slice Model

Before evaluating, check that the bounds actually describe the lamina: \(a\leq x\leq b\), and \(g(x)\leq f(x)\) throughout that interval. If the curves cross inside the interval, a single top-minus-bottom description may fail; split the \(x\)-interval wherever needed so each piece has a consistent lower and upper boundary.

Units offer another useful check. Density times area has units of mass, while multiplying mass by a distance gives a first moment with units of mass times length. The integrand for \(M_y\) must include the horizontal coordinate \(x\); the integrand for \(M_x\) must include the vertical coordinate \(y\). For nonnegative density, mass must be nonnegative. Moments can be negative if part or all of the lamina lies on the negative side of the relevant axis, because the signed coordinate measures distance in that direction.

Vertical slices are especially convenient when each \(x\) gives one continuous interval from a bottom curve to a top curve. Once those boundaries are identified, the same setup handles constant density and density that varies throughout the lamina.