First Moments About Coordinate Axes

Why First Moments Keep Track of Position

A lamina has mass distributed across a two-dimensional region. Its total mass tells us how much material there is, but not how that mass is positioned relative to an axis. A first moment combines these two pieces of information: each small amount of mass is weighted by its signed distance from the axis.

The word signed matters. A point above the \(x\)-axis has positive \(y\)-coordinate, while a point below it has negative \(y\)-coordinate. Consequently, mass on opposite sides of an axis can contribute with opposite signs. The first moment is not generally the mass multiplied by an ordinary, nonnegative distance.

Definition of the First Moments

Let \(R\) be a bounded region in the plane occupied by a lamina. Suppose its area density is an integrable function \(\rho(x,y)\geq 0\). The mass of a small area element \(dA\) is \(dm=\rho(x,y)\,dA\), and the total mass is \(m=\iint_R \rho(x,y)\,dA\).

Definition: The first moment of a lamina about the \(x\)-axis is the integral of the signed \(y\)-coordinate against mass. The first moment about the \(y\)-axis is the integral of the signed \(x\)-coordinate against mass:

M_x=\iint_R y\rho(x,y)\,dA =\iint_R y\,dm, \qquad M_y=\iint_R x\rho(x,y)\,dA =\iint_R x\,dm.

The coordinate multiplying the mass identifies the perpendicular signed distance to the axis: \(y\) for the \(x\)-axis and \(x\) for the \(y\)-axis. If coordinates are measured in metres and density in kilograms per square metre, then each first moment has units of kilogram-metres.

Because \(\rho\) is nonnegative, the sign of each contribution is determined by the relevant coordinate. If the lamina lies entirely above the \(x\)-axis and has positive mass, then \(M_x\) is positive; if it lies entirely below that axis and has positive mass, then \(M_x\) is negative. If the density is zero everywhere, the mass and both moments are zero. When mass occurs on both sides of an axis, positive and negative contributions may partially or completely cancel.

Deriving the Formulas from Thin Strips

The double-integral definition can be understood by dividing the lamina into thin strips. Each strip has a small mass, and its contribution to a first moment is that mass multiplied by the appropriate signed coordinate. Adding the contributions and taking the limit gives the integral.

For a region described vertically by \(a\leq x\leq b\) and \(g(x)\leq y\leq f(x)\), a vertical strip at \(x\) has thickness \(dx\). Within the strip, a horizontal piece at height \(y\) has area \(dy\,dx\), so its mass is \(\rho(x,y)\,dy\,dx\). Its contribution to \(M_x\) is weighted by \(y\); its contribution to \(M_y\) is weighted by \(x\). Integrating through the strip and then along the region gives:

M_x=\int_a^b\left(\int_{g(x)}^{f(x)}y\rho(x,y)\,dy\right)dx, \qquad M_y=\int_a^b x\left(\int_{g(x)}^{f(x)}\rho(x,y)\,dy\right)dx.

For a horizontally described region, \(c\leq y\leq d\) and \(L(y)\leq x\leq R(y)\), a horizontal strip has thickness \(dy\). The same mass-weighting argument yields:

M_x=\int_c^d y\left(\int_{L(y)}^{R(y)}\rho(x,y)\,dx\right)dy, \qquad M_y=\int_c^d\left(\int_{L(y)}^{R(y)}x\rho(x,y)\,dx\right)dy.

These formulas require that the region and density permit the indicated integrals; in particular, the moments must be integrable. They express the same double integrals as the definition, with the order of integration matched to the strip direction.

Partition the lamina.
Divide it into small area pieces. A piece near \((x_i,y_i)\) has mass approximately \(\rho(x_i,y_i)\Delta A_i\).

Weight each piece.
Its approximate contribution to \(M_x\) is \(y_i\rho(x_i,y_i)\Delta A_i\); its contribution to \(M_y\) is \(x_i\rho(x_i,y_i)\Delta A_i\).

Take the limit.
As the pieces become arbitrarily small, the sums of these contributions converge to \(\iint_R y\rho\,dA\) and \(\iint_R x\rho\,dA\), respectively. Evaluating those integrals by strips gives the formulas above.

This derivation also explains why the coordinate belongs inside the integral. The position of the mass varies from one part of the lamina to another, so each small mass element must be weighted by its own coordinate before the contributions are added.

Worked Examples

Worked Example: A Uniform Rectangular Lamina

A rectangular lamina occupies \(0\leq x\leq4\), \(1\leq y\leq3\), with constant density \(\rho=2\text{ kg/m}^2\). Find both first moments. Treat the coordinates as measured in metres.

For vertical strips, the moment about the \(x\)-axis is

\begin{aligned} M_x &=\int_0^4\int_1^3 2y\,dy\,dx\\ &=\int_0^4 [y^2]_1^3\,dx\\ &=\int_0^4(9-1)\,dx\\ &=\int_0^4 8\,dx\\ &=32\text{ kg}\cdot\text{m}. \end{aligned}

The moment about the \(y\)-axis is

\begin{aligned} M_y &=\int_0^4\int_1^3 2x\,dy\,dx\\ &=\int_0^4 2x(3-1)\,dx\\ &=\int_0^4 4x\,dx\\ &=[2x^2]_0^4\\ &=32\text{ kg}\cdot\text{m}. \end{aligned}

As a check, use horizontal strips for \(M_x\). Each strip has width \(4\) and density \(2\), so its mass per unit height is \(8\text{ kg/m}\). Thus \(M_x=\int_1^3 8y\,dy=4[y^2]_1^3=4(9-1)=32\text{ kg}\cdot\text{m}\), agreeing with the vertical-strip calculation. Both results are positive, as expected because all the mass has positive \(x\)- and \(y\)-coordinates.

Worked Example: A Uniform Lamina Below the \(x\)-Axis

Consider the triangular region with vertices \((0,0)\), \((3,0)\), and \((3,-2)\), and constant density \(3\text{ kg/m}^2\). Use vertical strips to find \(M_x\) and \(M_y\).

For \(0\leq x\leq3\), the strip runs from \(y=-2x/3\) up to \(y=0\). Therefore

\begin{aligned} M_x &=\int_0^3\int_{-2x/3}^{0}3y\,dy\,dx\\ &=\int_0^3 3\left[\frac{y^2}{2}\right]_{-2x/3}^{0}dx\\ &=\int_0^3 3\left(0-\frac{1}{2}\cdot\frac{4x^2}{9}\right)dx\\ &=\int_0^3-\frac{2x^2}{3}\,dx\\ &=-\frac{2}{3}\left[\frac{x^3}{3}\right]_0^3\\ &=-\frac{2}{3}\cdot9\\ &=-6\text{ kg}\cdot\text{m}. \end{aligned}

For the moment about the \(y\)-axis, the strip height is \(0-(-2x/3)=2x/3\):

\begin{aligned} M_y &=\int_0^3\int_{-2x/3}^{0}3x\,dy\,dx\\ &=\int_0^3 3x\left(\frac{2x}{3}\right)dx\\ &=\int_0^3 2x^2\,dx\\ &=2\left[\frac{x^3}{3}\right]_0^3\\ &=18\text{ kg}\cdot\text{m}. \end{aligned}

To check the signs and values using horizontal strips, \(y\) runs from \(-2\) to \(0\), and \(x\) runs from \(-3y/2\) to \(3\). The width is \(3+3y/2\). Hence

\begin{aligned} M_x &=\int_{-2}^0 3y\left(3+\frac{3y}{2}\right)dy\\ &=\left[\frac{9y^2}{2}+\frac{3y^3}{2}\right]_{-2}^{0} =-(18-12)=-6\text{ kg}\cdot\text{m}, \\[4pt] M_y &=\int_{-2}^0 3\left(\int_{-3y/2}^3x\,dx\right)dy\\ &=\int_{-2}^0 3\left(\frac{9}{2}-\frac{9y^2}{8}\right)dy\\ &=\left[\frac{27y}{2}-\frac{9y^3}{8}\right]_{-2}^{0} =18\text{ kg}\cdot\text{m}. \end{aligned}

The triangle has positive mass, and all its mass lies below the \(x\)-axis, so a negative \(M_x\) is appropriate. Its mass also has positive \(x\)-coordinates, so \(M_y\) is positive.

Worked Example: Variable Density on Both Sides of an Axis

Let \(R\) be the rectangle \(1\leq x\leq3\), \(-1\leq y\leq1\), with density \(\rho(x,y)=2+y\text{ kg/m}^2\). This density is nonnegative throughout the region because \(2+y\geq1\). Find the first moments.

For \(M_x\), integrate across each horizontal strip first:

\begin{aligned} M_x &=\int_{-1}^1\int_1^3 y(2+y)\,dx\,dy\\ &=\int_{-1}^1 2y(2+y)\,dy\\ &=\int_{-1}^1(4y+2y^2)\,dy\\ &=\left[2y^2+\frac{2y^3}{3}\right]_{-1}^{1}\\ &=\left(2+\frac{2}{3}\right)-\left(2-\frac{2}{3}\right)\\ &=\frac{4}{3}\text{ kg}\cdot\text{m}. \end{aligned}

The term \(4y\) integrates to zero because the contributions at positive and negative heights cancel. The \(2y^2\) term remains positive. For \(M_y\), use the inner integral \(\int_1^3x\,dx=(9-1)/2=4\):

\begin{aligned} M_y &=\int_{-1}^1\int_1^3x(2+y)\,dx\,dy\\ &=\int_{-1}^1 4(2+y)\,dy\\ &=4[2y+y^2/2]_{-1}^{1}\\ &=4\left(\frac{5}{2}-\left(-\frac{3}{2}\right)\right)\\ &=16\text{ kg}\cdot\text{m}. \end{aligned}

As a check, the density integrated over the rectangle gives mass \(\int_{-1}^1 2(2+y)\,dy=8\text{ kg}\), which is positive. The computed \(M_x\) is also positive: the density is greater at positive \(y\) than at negative \(y\), so the positive-side contribution outweighs the negative-side contribution. The positive \(M_y\) is consistent with the rectangle lying at positive \(x\).

Reading the Sign and Size of a Moment

A first moment records a signed balance of mass with respect to an axis. For \(M_x\), mass above the \(x\)-axis contributes positively and mass below it contributes negatively. For \(M_y\), mass to the right of the \(y\)-axis contributes positively and mass to the left contributes negatively. Contributions exactly on the relevant axis are zero because their signed distance is zero.

The magnitude of a contribution increases with both the amount of mass and its distance from the axis. For example, a small mass far from the axis can contribute as much as a larger mass closer to it. This is why total mass alone cannot determine a first moment.

If contributions from the two sides cancel exactly, the corresponding first moment is zero even when the lamina has positive mass. A zero moment therefore does not by itself mean there is no mass. The density-weighted positions on opposite sides may simply balance in the signed integral.

A Common Sign Error

A frequent mistake is to replace the signed coordinate by its absolute value. In the triangular example, using \(|y|\) would produce a positive value for the moment about the \(x\)-axis:

\int_0^3\int_{-2x/3}^{0}3|y|\,dy\,dx=6\text{ kg}\cdot\text{m}.

That calculation measures mass weighted by ordinary distance below the axis, not the first moment as defined here. The correct first moment uses \(y\), not \(|y|\), and is \(-6\text{ kg}\cdot\text{m}\). A reliable check is to identify which side of the axis contains the mass before integrating: if the lamina lies entirely below the \(x\)-axis and has positive mass, then \(M_x\) must be negative.

Why First Moments Matter

First moments summarize how mass is distributed relative to chosen axes. They are useful in applications involving balance and in calculations that locate a lamina’s center of mass. Their value depends on both the geometry of the region and the density within it: changing either can change the magnitude or even the sign of a moment.

The essential procedure is to identify the signed coordinate for the chosen axis, multiply it by the mass density, and integrate over the lamina. When using strips, the same rule applies strip by strip: compute the strip’s mass distribution and weight it by the appropriate coordinate. Keeping the sign throughout the calculation preserves the physical meaning of the result.

Check Your Understanding

Answer each question using the signed-coordinate definition of first moment.

Which coordinate weights mass when computing the first moment about the \(x\)-axis, and which coordinate weights mass about the \(y\)-axis?
A lamina lies entirely below the \(x\)-axis and has positive mass. What is the sign of its first moment about that axis, and why?
Why can a lamina have positive mass but a zero first moment about the \(y\)-axis?
For a region described by horizontal strips, which inner integral gives the strip’s contribution to \(M_y\)?
What quantity would be computed by replacing \(y\) with \(|y|\) in the formula for \(M_x\), and why is it not the signed first moment?