First Moments About The Coordinate Axes

Why First Moments Measure Weighted Distance

A lamina occupies a planar region and has a density that may vary from point to point. Its mass is not generally concentrated at one location, so a useful way to describe how that mass is distributed is to weight each small amount of mass by its distance from an axis.

For a point $(x,y)$, the perpendicular distance to the $y$-axis is $|x|$, while the perpendicular distance to the $x$-axis is $|y|$. In a first moment, we use the signed coordinates $x$ and $y$, rather than their absolute values. Thus mass on opposite sides of an axis contributes with opposite signs.

Definition: If a lamina occupies a region $R$ and has areal density $\rho(x,y)$, its first moment about the $y$-axis is $$ M_y=\iint_R x\rho(x,y)\,dA, $$ and its first moment about the $x$-axis is $$ M_x=\iint_R y\rho(x,y)\,dA. $$

For a physical lamina, the density is nonnegative. The first moments are finite when the indicated integrals exist and are finite; in particular, this holds for a bounded region with a bounded, integrable density. The mass element is $dm=\rho(x,y)\,dA$, so the same definitions can be read as $M_y=\iint_R x\,dm$ and $M_x=\iint_R y\,dm$.

Which Coordinate Belongs in Each Moment?

The name of the axis tells us which moment we are taking, but the integrand contains the coordinate measuring perpendicular distance to that axis. The $y$-axis is vertical, so its perpendicular coordinate is $x$; the $x$-axis is horizontal, so its perpendicular coordinate is $y$.

Moment	Coordinate factor	Reason
$M_y$, about the $y$-axis	$x$	$x$ is the signed horizontal distance from that axis.
$M_x$, about the $x$-axis	$y$	$y$ is the signed vertical distance from that axis.

These are signed moments. For example, if part of a lamina lies at $x<0$, that part contributes negatively to $M_y$. A moment can therefore be zero even when the lamina has positive mass: contributions on opposite sides of the relevant axis may cancel.

Why the Double-Integral Formulas Work

Imagine dividing $R$ into many small pieces. For a piece near $(x_i,y_i)$ with area $\Delta A_i$, its mass is approximately $\rho(x_i,y_i)\Delta A_i$. Its contribution to the moment about the $y$-axis is approximately its mass multiplied by its signed distance $x_i$. Adding these contributions gives a sum that approaches the double integral as the pieces become arbitrarily small.

\sum_i x_i\rho(x_i,y_i)\Delta A_i \longrightarrow \iint_R x\rho(x,y)\,dA=M_y.

For the $x$-axis, the signed distance is $y_i$, so the corresponding sums approach $\iint_R y\rho(x,y)\,dA$. This explains both the density factor and the choice of coordinate in each formula: density converts area into mass, and the coordinate weights that mass by its signed distance from the chosen axis.

If the region is described by $a\le x\le b$ and $g_1(x)\le y\le g_2(x)$, for example, the moments can be written as iterated integrals:

M_y=\int_a^b\int_{g_1(x)}^{g_2(x)}x\rho(x,y)\,dy\,dx, \qquad M_x=\int_a^b\int_{g_1(x)}^{g_2(x)}y\rho(x,y)\,dy\,dx.

The limits must describe the region being integrated over, and the density must be evaluated at the points in that region. When polar coordinates are appropriate, use $x=r\cos\theta$, $y=r\sin\theta$, and $dA=r\,dr\,d\theta$; the extra factor $r$ is the area element.

Worked Example: A Uniform Rectangular Lamina

Let $R=[0,2]\times[0,1]$, and let the constant density be $\rho(x,y)=3$ mass units per square unit. Find both first moments.

For the moment about the $y$-axis, use $x\rho=3x$. Integrating first with respect to $y$ gives

\begin{aligned} M_y &=\int_0^2\int_0^1 3x\,dy\,dx\\ &=\int_0^2 3x(1-0)\,dx\\ &=3\left[\frac{x^2}{2}\right]_0^2\\ &=3\left(\frac{4}{2}-0\right)=6. \end{aligned}

For the moment about the $x$-axis, use $y\rho=3y$:

\begin{aligned} M_x &=\int_0^2\int_0^1 3y\,dy\,dx\\ &=\int_0^2 3\left[\frac{y^2}{2}\right]_0^1\,dx\\ &=\int_0^2 \frac{3}{2}\,dx\\ &=\frac{3}{2}(2-0)=3. \end{aligned}

As a check, the mass is $3\cdot(2\cdot1)=6$ mass units. The rectangle extends farther horizontally than vertically, so the larger moment is about the $y$-axis: the computed values are $M_y=6$ and $M_x=3$. If lengths are measured in meters and density in kilograms per square meter, each moment has units of kilogram-meters.

Moments on a Region Bounded by a Curve

For a nonrectangular region, the bounds are part of the setup. A convenient order of integration often makes the inner integral straightforward, but the formulas remain the same: multiply density by $x$ for $M_y$, or by $y$ for $M_x$, and integrate over the entire region.

Worked Example: A Triangular Lamina

Let $R$ be the triangle $0\le x\le1,\ 0\le y\le x$, with constant density $\rho(x,y)=2$. Find $M_y$ and $M_x$.

For $M_y$, the integrand is $x\rho=2x$. The vertical bounds have length $x$, so

\begin{aligned} M_y &=\int_0^1\int_0^x 2x\,dy\,dx\\ &=\int_0^1 2x(x-0)\,dx\\ &=2\int_0^1x^2\,dx\\ &=2\left[\frac{x^3}{3}\right]_0^1 =2\left(\frac13-0\right)=\frac23. \end{aligned}

For $M_x$, the integrand is $y\rho=2y$:

\begin{aligned} M_x &=\int_0^1\int_0^x 2y\,dy\,dx\\ &=\int_0^1\left[y^2\right]_0^x\,dx\\ &=\int_0^1x^2\,dx\\ &=\left[\frac{x^3}{3}\right]_0^1 =\frac13. \end{aligned}

Check the region and density by calculating its mass: the triangle has area $\int_0^1 x\,dx=1/2$, so its mass is $2(1/2)=1$. Both moments are positive because the region lies in the first quadrant. The results, $M_y=2/3$ and $M_x=1/3$, reflect its greater horizontal than vertical spread.

Using Polar Coordinates

For a region naturally described by $r$ and $\theta$, substitute the Cartesian coordinates in the moment factor and include the polar area element. Thus the integrands become $r\cos\theta\,\rho\cdot r$ for $M_y$ and $r\sin\theta\,\rho\cdot r$ for $M_x$. The final factor of $r$ comes from $dA$, not from the definition of a first moment.

Worked Example: A Quarter-Disk with Variable Density

Consider the quarter-disk $0\le r\le2,\ 0\le\theta\le\pi/2$, with density $\rho(r,\theta)=1+r\cos\theta$. This density is nonnegative on the region. Compute both moments.

Since $x=r\cos\theta$ and $y=r\sin\theta$, the moment integrals, including the polar area factor, are

\begin{aligned} M_y &=\int_0^{\pi/2}\int_0^2 (r\cos\theta)(1+r\cos\theta)r\,dr\,d\theta\\ &=\int_0^{\pi/2}\int_0^2 \left(r^2\cos\theta+r^3\cos^2\theta\right)\,dr\,d\theta. \end{aligned}

Separate the two terms and evaluate each factor:

\begin{aligned} M_y &=\left(\int_0^2r^2\,dr\right) \left(\int_0^{\pi/2}\cos\theta\,d\theta\right) +\left(\int_0^2r^3\,dr\right) \left(\int_0^{\pi/2}\cos^2\theta\,d\theta\right)\\ &=\frac83(1)+4\left(\frac{\pi}{4}\right)\\ &=\frac83+\pi. \end{aligned}

For the moment about the $x$-axis, expand the integrand $y\rho\,dA$:

\begin{aligned} M_x &=\int_0^{\pi/2}\int_0^2 (r\sin\theta)(1+r\cos\theta)r\,dr\,d\theta\\ &=\left(\int_0^2r^2\,dr\right) \left(\int_0^{\pi/2}\sin\theta\,d\theta\right) +\left(\int_0^2r^3\,dr\right) \left(\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta\right)\\ &=\frac83(1)+4\left(\frac12\right)\\ &=\frac{14}{3}. \end{aligned}

For a check, the mass is $\int_0^{\pi/2}\int_0^2(r+r^2\cos\theta)\,dr\,d\theta=\pi+8/3$, which is positive and finite. Both moments are positive because $x$, $y$, and the density are nonnegative throughout the quarter-disk. The computed values are $M_y=8/3+\pi$ and $M_x=14/3$.

Common Mistakes and Their Corrections

A frequent error is to match the moment subscript to the coordinate in the integrand. That reverses the definitions. For the rectangular lamina in the first example, an incorrect setup would be

\text{Wrong: }\quad M_y=\iint_R y\rho\,dA=3, \qquad M_x=\iint_R x\rho\,dA=6.

The integrals themselves are evaluated correctly, but their labels are reversed. The correction is to use the perpendicular coordinate: $M_y=\iint_R x\rho\,dA=6$, and $M_x=\iint_R y\rho\,dA=3$.

Another error is to replace $x$ or $y$ by its absolute value without a reason. First moments are signed, so such a replacement changes the definition whenever the region crosses an axis. Absolute values may describe unsigned distance, but they do not compute the first moments defined here. A third common error in polar coordinates is omitting the factor $r$ in $dA=r\,dr\,d\theta$; doing so gives the wrong integral even if the coordinate and density substitutions are correct.

Interpreting the Results

A first moment combines mass and signed distance. Moving a given amount of mass farther from an axis increases the magnitude of its contribution; moving it to the opposite side changes the sign. Consequently, a moment records both how much mass is present and how it is distributed relative to the axis.

The units provide a useful check. Density has units of mass per area, and $dA$ has units of area, so $\rho\,dA$ has units of mass. Multiplying by $x$ or $y$, each a length, gives moment units of mass times length. These first moments are distinct from moments of inertia, which weight mass by the square of distance.

Key takeaway: To compute $M_y$, integrate $x\rho$; to compute $M_x$, integrate $y\rho$. Keep the coordinates signed, describe the region correctly, and include the appropriate area element for the coordinate system.

Check Your Understanding

Use the definitions and reasoning above to answer these questions.

For a lamina occupying $[0,1]\times[0,3]$ with constant density $2$, which first moment is larger? Compute both to check.
Write iterated-integral formulas for $M_y$ and $M_x$ when $0\le x\le2$, $0\le y\le x^2$, and density is $\rho(x,y)$.
Why can a lamina with positive mass have $M_y=0$?
In polar coordinates, what factor must accompany $dr\,d\theta$ in the area element, and where does it come from?
If the density is in kilograms per square meter and the coordinates are in meters, what are the units of $M_x$ and $M_y$?

Moment	Coordinate factor	Reason
\(M_y\), about the \(y\)-axis	\(x\)	\(x\) is the signed horizontal distance from that axis.
\(M_x\), about the \(x\)-axis	\(y\)	\(y\) is the signed vertical distance from that axis.