Choosing Bounds For Mass And Moment Integrals

From a Lamina Description to Integral Bounds

A mass or moment integral is only as useful as its description of the lamina. The integrand records the density and the quantity being measured; the bounds must describe every point of the region exactly once, without including points outside it. Choosing bounds is therefore a geometric task before it is a calculation.

For regions bounded by curves, a practical first step is to imagine moving a vertical or horizontal line across the region. If each vertical slice is one interval, Cartesian bounds with \(y\) between a lower and an upper curve are often direct. If the region is more naturally described by distance from the origin and angle, polar coordinates can simplify the bounds. A change in boundary or density may require splitting the integral.

Definition: A lamina occupies a planar region \(R\) and has surface density \(\rho(x,y)\), measured as mass per unit area. Its mass is \(m=\iint_R\rho(x,y)\,dA\). Its first moments about the coordinate axes are \(M_x=\iint_R y\rho(x,y)\,dA\) and \(M_y=\iint_R x\rho(x,y)\,dA\). Its moments of inertia about those axes are \(I_x=\iint_R y^2\rho(x,y)\,dA\) and \(I_y=\iint_R x^2\rho(x,y)\,dA\). The polar moment about the origin is \(I_O=\iint_R(x^2+y^2)\rho(x,y)\,dA\).

The coordinate in a first moment is signed: points with negative \(x\) contribute negatively to \(M_y\), and points with negative \(y\) contribute negatively to \(M_x\). Inertia instead uses squared distances from the relevant axis. Keep these different integrands distinct when setting up the same bounds.

Reading Cartesian Bounds from a Region

An \(x\)-simple description specifies an interval of \(x\)-values and, for each such \(x\), a vertical interval in \(y\). If the lower and upper boundaries are \(y=g_1(x)\) and \(y=g_2(x)\), the description is \(a\le x\le b\), \(g_1(x)\le y\le g_2(x)\). The corresponding integral of a function \(f\) is \(\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx\).

A \(y\)-simple description instead uses horizontal slices: \(c\le y\le d\), \(h_1(y)\le x\le h_2(y)\), giving \(\int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy\). These descriptions are alternatives when both apply. Prefer the one whose slice endpoints are simplest and whose intervals do not need unnecessary splitting.

Rule: The bounds must cover the region \(R\), not the support of the integrand alone. For each slice, identify its actual endpoints from the boundary curves, and split the integral whenever an endpoint formula or the density changes.

The reason iterated bounds work is that the inner integral adds contributions along one slice, while the outer integral adds the slices across the region. When a region has separate pieces or a density has separate formulas, add the integrals over those pieces. A boundary shared by adjacent pieces has area zero, so including it in both pieces does not change the integral.

Worked Example: A Rectangle with a Density Change

Let \(R=[0,2]\times[0,1]\). Suppose \(\rho=1\) for \(0\le x\le1\) and \(\rho=2\) for \(1<x\le2\). The density changes at \(x=1\), so divide the rectangle into two vertical strips. Both have \(0\le y\le1\).

The mass is the sum of the two strip integrals:

m=\int_0^1\int_0^1 1\,dy\,dx+\int_1^2\int_0^1 2\,dy\,dx =1+2=3.

For the first moments, use \(y\rho\) for \(M_x\) and \(x\rho\) for \(M_y\):

M_x=\left(\int_0^1 1\,dx+\int_1^2 2\,dx\right)\int_0^1y\,dy =(1+2)\frac12=\frac32,

M_y=\int_0^1x\,dx+2\int_1^2x\,dx =\frac12+2\left(\frac{4-1}{2}\right)=\frac72.

For the coordinate-axis inertias, integrate squared distances from the respective axes:

I_x=(1+2)\int_0^1 y^2\,dy=3\cdot\frac13=1, \qquad I_y=\int_0^1x^2\,dx+2\int_1^2x^2\,dx =\frac13+\frac{14}{3}=5.

As a check, the mass of the first strip is \(1\) and its \(x\)-moment is \(1\cdot\frac12=\frac12\); the second strip has mass \(2\) and \(x\)-moment \(2\cdot\frac32=3\). Their sum is \(M_y=\frac72\), agreeing with the integral. The split is essential: using density \(1\) throughout would omit the extra mass on the right strip.

When the Slice Direction Changes

Curved boundaries can make one direction awkward. To choose between vertical and horizontal slices, sketch or reason from the boundary equations, then ask whether a typical slice intersects the region in one interval. If it intersects in two separated intervals, either describe both intervals or change the slice direction if that produces a simpler description.

For example, the region between \(y=x^2\) and \(y=2x\) lies where \(x^2\le2x\), or \(x(x-2)\le0\). Thus \(0\le x\le2\), and vertical slices give \(x^2\le y\le2x\). To reverse the order, solve the boundary equations for \(x\): the region has \(0\le y\le4\), with \(\frac y2\le x\le\sqrt y\). Both descriptions follow from the same inequalities; checking the intersections \(x=0,2\) and \(y=0,4\) helps confirm the limits.

Worked Example: Mass and Moments Between Two Curves

Use the region \(R\) between \(y=x^2\) and \(y=2x\), with constant density \(\rho=1\). Their intersections satisfy \(x^2=2x\), so \(x=0\) or \(x=2\). On \(0\le x\le2\), \(x^2\le2x\), giving vertical bounds \(0\le x\le2\), \(x^2\le y\le2x\).

The mass is the area:

m=\int_0^2\int_{x^2}^{2x}1\,dy\,dx =\int_0^2(2x-x^2)\,dx =\left[x^2-\frac{x^3}{3}\right]_0^2 =\frac43.

For the first moments, integrate the signed coordinate in each case:

M_x=\int_0^2\int_{x^2}^{2x}y\,dy\,dx =\frac12\int_0^2(4x^2-x^4)\,dx =\frac12\left(\frac{32}{3}-\frac{32}{5}\right) =\frac{32}{15},

M_y=\int_0^2\int_{x^2}^{2x}x\,dy\,dx =\int_0^2x(2x-x^2)\,dx =\frac{16}{3}-4 =\frac43.

The inertias use squared coordinates:

I_x=\int_0^2\int_{x^2}^{2x}y^2\,dy\,dx =\frac13\int_0^2(8x^3-x^6)\,dx =\frac13\left(32-\frac{128}{7}\right) =\frac{32}{7},

I_y=\int_0^2\int_{x^2}^{2x}x^2\,dy\,dx =\int_0^2x^2(2x-x^2)\,dx =8-\frac{32}{5} =\frac85.

Check the mass by reversing the order. Here \(0\le y\le4\) and \(y/2\le x\le\sqrt y\), so

m=\int_0^4\left(\sqrt y-\frac y2\right)\,dy =\frac{16}{3}-4 =\frac43.

The reversed calculation gives the same mass, confirming that the two descriptions cover the same region.

Polar Bounds for Radial Geometry

Use polar coordinates when the region is naturally described by distance \(r\) from the origin and angle \(\theta\). In polar coordinates, \(x=r\cos\theta\), \(y=r\sin\theta\), and area is \(dA=r\,dr\,d\theta\). The extra factor \(r\) is required in every polar area integral.

A sector of an annulus often has bounds \(\alpha\le\theta\le\beta\) and \(a\le r\le b\). More generally, a radial slice may start or end at a curve, so its bounds can be \(\alpha\le\theta\le\beta\), \(r_1(\theta)\le r\le r_2(\theta)\). The polar integrands for mass and moments follow by substituting the coordinates into the definitions, as well as including the area factor.

Polar-coordinate formulas: In a polar description, \(m=\iint_R \rho(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta\). The first-moment integrands become \(M_x=r\sin\theta\,\rho\,r\) and \(M_y=r\cos\theta\,\rho\,r\). The inertia integrands become \(I_x=r^2\sin^2\theta\,\rho\,r\), \(I_y=r^2\cos^2\theta\,\rho\,r\), and \(I_O=r^2\rho\,r\).

Worked Example: A Quarter-Annulus with Angularly Changing Density

Consider \(1\le r\le2\), \(0\le\theta\le\pi/2\). Let \(\rho=1\) on \(0\le\theta\le\pi/4\) and \(\rho=3\) on \(\pi/4<\theta\le\pi/2\). Split at \(\theta=\pi/4\), where the density changes. The common radial factor for mass is \(\int_1^2r\,dr=(4-1)/2=3/2\). Therefore

m=\frac32\left(\int_0^{\pi/4}1\,d\theta+ 3\int_{\pi/4}^{\pi/2}1\,d\theta\right) =\frac32\left(\frac\pi4+\frac{3\pi}{4}\right) =\frac{3\pi}{2}.

For the first moments, the radial integral is \(\int_1^2r^2\,dr=(8-1)/3=7/3\). Thus

M_x=\frac73\left(\int_0^{\pi/4}\sin\theta\,d\theta+ 3\int_{\pi/4}^{\pi/2}\sin\theta\,d\theta\right) =\frac73(1+\sqrt2),

M_y=\frac73\left(\int_0^{\pi/4}\cos\theta\,d\theta+ 3\int_{\pi/4}^{\pi/2}\cos\theta\,d\theta\right) =\frac73(3-\sqrt2).

For \(I_x\) and \(I_y\), the radial factor is \(\int_1^2r^3\,dr=(16-1)/4=15/4\). The angular integrals give

I_x=\frac{15}{4}\left(\int_0^{\pi/4}\sin^2\theta\,d\theta+ 3\int_{\pi/4}^{\pi/2}\sin^2\theta\,d\theta\right) =\frac{15(\pi+1)}{8},

I_y=\frac{15}{4}\left(\int_0^{\pi/4}\cos^2\theta\,d\theta+ 3\int_{\pi/4}^{\pi/2}\cos^2\theta\,d\theta\right) =\frac{15(\pi-1)}{8}.

As a check, \(I_O=I_x+I_y\) by adding their integrands. The results give \(I_O=15\pi/4\). Directly, the polar integral is \(\int_1^2r^3\,dr\) times the density-weighted angular width, \((15/4)(\pi/4+3\pi/4)=15\pi/4\), as required.

A Reliable Bound-Choosing Procedure

Identify the region.
Record each boundary and find relevant intersections. Determine which curve lies above, below, inside, or outside the others.

Choose a coordinate description.
Try vertical or horizontal slices for Cartesian bounds; use radial and angular descriptions when the geometry is centered or sector-like.

Mark every change.
Split at changes in boundary formulas, density formulas, or disconnected slice intervals.

Insert the correct integrand.
Use \(\rho\) for mass, signed \(y\rho\) or \(x\rho\) for first moments, and squared distances times \(\rho\) for inertias. In polar coordinates include the area factor \(r\).

Check the description.
Test boundary points, verify endpoint order, and compare with another slice direction or an identity when practical.

A Common Sign Error

Distance from the \(y\)-axis is \(|x|\), but the first-moment integrand about the \(y\)-axis is \(x\rho\), not \(|x|\rho\). The inertia integrand about that axis is \(x^2\rho\). These quantities answer different questions: the first moment records signed location, while inertia weights mass by squared distance.

For a concrete check, take the rectangle \(-2\le x\le-1\), \(0\le y\le1\), with \(\rho=1\). The incorrect calculation replaces \(x\) in \(M_y\) with \(|x|\), giving \(\int_{-2}^{-1}|x|\,dx=3/2\). The correct first moment is

M_y=\int_{-2}^{-1}\int_0^1 x\,dy\,dx =\int_{-2}^{-1}x\,dx =\left[\frac{x^2}{2}\right]_{-2}^{-1} =-\frac32.

The negative sign correctly records that the lamina lies to the left of the \(y\)-axis. In contrast, its inertia is \(I_y=\int_{-2}^{-1}x^2\,dx=(8-1)/3=7/3\), which is positive because squared distance cannot be negative.

Why Careful Bounds Matter

Once the bounds correctly describe the lamina and the integrand matches the physical quantity, the integral combines local contributions across the entire region. A wrong endpoint can omit or add mass; an unsplit density can assign the wrong weight; and a missing polar factor changes area itself. Sketching the region, identifying slice endpoints, and checking one independently computable quantity are efficient safeguards.

Check Your Understanding

For each question, focus first on what points the bounds include and what quantity the integrand measures.

For the region \(0\le x\le3\), \(x\le y\le2x\), write bounds for the mass integral with density \(\rho(x,y)\).
What changes in the integral setup if the density in Question 1 has one formula for \(x<1\) and another for \(x\ge1\)?
For a quarter-annulus \(2\le r\le4\), \(0\le\theta\le\pi/2\), write the polar mass integral for density \(\rho\).
State the integrands for \(M_y\) and \(I_y\) in Cartesian coordinates, and explain how their meanings differ.
Why can a single vertical-slice description fail for a region whose vertical slices have two separated intervals?