Why Mass and Centroid Integrals Go Wrong
Mass and centroid calculations can look like routine integration, but a correct antiderivative does not guarantee a correct answer. The integrand, the coordinate being measured, and the bounds must all match the physical object. A small mismatch can produce a plausible-looking number that describes a different region or quantity.
This tutorial focuses on common errors in lamina and solid calculations: choosing strip bounds, including density, labelling first moments, and handling cutouts. The central check is to ask what each small element contributes before integrating.
Set Up the Quantities Before Integrating
For a lamina in the \(xy\)-plane with density \(\rho(x,y)\), a small area element \(dA\) has mass \(dm=\rho(x,y)\,dA\). Its total mass is the integral of these contributions over the region \(R\). In the plane, the first moment about the \(x\)-axis weights each mass element by its \(y\)-coordinate; the first moment about the \(y\)-axis weights it by its \(x\)-coordinate.
Notice the order in the centroid formula: \(\bar{x}\) comes from \(M_y\), because \(M_y\) weights mass by \(x\); \(\bar{y}\) comes from \(M_x\), because \(M_x\) weights mass by \(y\). These are first moments, not moments of inertia.
For a solid occupying \(E\) in three-dimensional space, \(dm=\rho(x,y,z)\,dV\). To avoid confusing coordinate first moments with moments about axes, name the solid first moments by the coordinate planes: \(M_{yz}\) weights by \(x\), \(M_{xz}\) by \(y\), and \(M_{xy}\) by \(z\).
A moment about an axis is a different quantity if it means a moment of inertia: for example, the mass moment of inertia about the \(x\)-axis uses the squared distance \(y^2+z^2\), not just \(y\). Keep that distinction explicit whenever naming a moment.
Worked Example: Choosing Consistent Strips for a Lamina
Worked Example: Choosing Consistent Strips for a Lamina
Let \(R\) be the triangular region \(0\le y\le x\le2\), with density \(\rho(x,y)=1+x\). Find its mass and centroid. Since the density depends only on \(x\), it is constant along a vertical strip, where \(0\le y\le x\).
Using \(dA=dy\,dx\), the mass is
The first moment about the \(y\)-axis weights by \(x\), and the first moment about the \(x\)-axis weights by \(y\):
Therefore,
Check: the point lies in the triangle because \(0\le5/7\le10/7\le2\). The bounds can also be written using horizontal strips as \(0\le y\le2,\ y\le x\le2\). For example, this gives the same mass:
The reversed horizontal bounds \(0\le x\le y\) would describe the other side of the line \(y=x\), not the given triangle.
Why the Moment Formulas Give the Centroid
The centroid is the mass-weighted average of position. For a lamina, each element \(dm\) contributes \(x\,dm\) to the total weighted \(x\)-coordinate and \(y\,dm\) to the total weighted \(y\)-coordinate. Adding these contributions over the region gives the first moments. Dividing by total mass turns each weighted total into an average coordinate.
The same reasoning in three dimensions gives \(\bar{x}=M_{yz}/m\), \(\bar{y}=M_{xz}/m\), and \(\bar{z}=M_{xy}/m\). The plane labels identify which coordinate is being weighted: the \(yz\)-plane has \(x=0\), so the first moment about that plane weights by \(x\).
This argument also explains why a cutout is subtracted consistently. If a full region is the disjoint union of the retained part and the cutout, then the integral over the full region equals the sum of the integrals over those two parts. Rearranging shows that the retained mass and each retained first moment equal the corresponding full-object quantity minus the cutout quantity.
Worked Example: Subtracting a Lamina Cutout
Worked Example: Subtracting a Lamina Cutout
A rectangular lamina occupies \(0\le x\le4,\ 0\le y\le3\), except for a rectangular hole \(1\le x\le2,\ 1\le y\le2\). Its uniform areal density is \(2\). Find the mass and centroid of the remaining lamina by subtraction.
For the full rectangle, the mass is \(2(4)(3)=24\). Its centroid is at \((2,3/2)\), so its first moments are \(M_y=24(2)=48\) and \(M_x=24(3/2)=36\). The hole has area \(1\), mass \(2\), and centroid \((3/2,3/2)\). Its first moments are \(M_y=2(3/2)=3\) and \(M_x=2(3/2)=3\).
Subtract mass and each first moment separately:
Thus
Check: the original rectangle is symmetric about \(y=3/2\), and the removed hole is also centered on that horizontal line, so the remaining centroid must have \(\bar{y}=3/2\), as calculated. Also \(0<45/22<4\) and \(0<3/2<3\), so the coordinates lie within the outer rectangle. The hole shifts the \(x\)-coordinate left because its center is left of the full rectangle’s center.
Density and Bounds: Concrete Error Checks
A common density error is to integrate area or volume alone. In the triangular example, omitting \(\rho=1+x\) gives \(\int_0^2\int_0^x1\,dy\,dx=2\), which is the area, not the mass \(14/3\). The correction is to use \(dm=\rho\,dA\) for every element. The same rule applies to solids with \(dm=\rho\,dV\).
Bounds are another frequent source of error. Sketch the region, identify its outer limits, and then determine the inner limits from the boundary curves. For the triangle above, vertical strips have \(0\le x\le2\) and \(0\le y\le x\). Horizontal strips instead have \(0\le y\le2\) and \(y\le x\le2\). The inner limits must describe the actual slice at each fixed outer coordinate; swapping them changes the region or reverses the integral.
Worked Example: A Solid with a Box-Shaped Cutout
Worked Example: A Solid with a Box-Shaped Cutout
A solid of uniform density \(3\) occupies the box \(0\le x\le4,\ 0\le y\le2,\ 0\le z\le2\), except for the cutout \(1\le x\le2,\ 0\le y\le1,\ 0\le z\le2\). Find the mass and centroid. Use first moments about coordinate planes, not moments about axes.
The full box has volume \(4\cdot2\cdot2=16\), hence mass \(3\cdot16=48\). Its center is \((2,1,1)\), so its coordinate-plane first moments are \(M_{yz}=48(2)=96\), \(M_{xz}=48(1)=48\), and \(M_{xy}=48(1)=48\).
The cutout has volume \(1\cdot1\cdot2=2\), hence mass \(6\). Its center is \((3/2,1/2,1)\), so its first moments are \(M_{yz}=6(3/2)=9\), \(M_{xz}=6(1/2)=3\), and \(M_{xy}=6(1)=6\). Subtract:
Therefore,
Check: the cutout lies left of and below the full box’s center, so removing it shifts the \(x\)- and \(y\)-coordinates of the centroid upward, to \(29/14>2\) and \(15/14>1\). The cutout is centered at \(z=1\), as is the full box, so the centroid retains \(\bar{z}=1\). Each coordinate lies within the outer box. These checks support the subtraction and the coordinate-plane labels.
A Reliable Setup and Final Check
Before evaluating an integral, write down the region, density, and quantity being calculated. This makes it easier to catch mistakes before they are buried in algebra.
Write bounds that match its geometry, and distinguish the retained region from any cutout.
Use \(dm=\rho\,dA\) for a lamina or \(dm=\rho\,dV\) for a solid.
In a lamina, weight \(M_x\) by \(y\) and \(M_y\) by \(x\). In a solid, use plane labels \(M_{yz},M_{xz},M_{xy}\) for weights \(x,y,z\), respectively.
For a cutout, subtract its mass and each corresponding first moment from the full-object values.
Verify positive mass, plausible centroid coordinates, symmetry where applicable, and agreement with the geometry.
These checks are useful because mass and centroid formulas encode physical information as well as algebra. A centroid outside a simple filled rectangle, a nonzero centroid coordinate contradicted by symmetry, or an answer unchanged after removing an off-center hole can signal a setup error.
Check Your Understanding
For each question, identify which part of the setup or interpretation should be checked.
- For the region \(0\le y\le x\le3\), write valid bounds for vertical strips and for horizontal strips.
- A lamina has density \(\rho(x,y)\). What differential element should appear in its mass integral, and what does omitting \(\rho\) calculate instead?
- Which planar first moment determines \(\bar{x}\), and which coordinate does that moment weight?
- For a solid, which coordinate weights \(M_{xz}\)? Why is this not the same notation as a moment about the \(x\)-axis?
- When a hole is removed from a uniform lamina, which quantities must be subtracted to find the remaining centroid?
