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Area, Mass, and Moments · Tutorial 598 of 999

A Complete Cartesian Lamina Calculation

Learn to set up and evaluate Cartesian integrals for the geometric and mass properties of a lamina with nonconstant density.

Intermediate 9 min read

What You'll Learn

  • Express a bounded planar region using Cartesian iterated-integral bounds.
  • Calculate area and mass from a region and its density function.
  • Find average density and the mass-weighted centroid.
  • Compute moments of inertia about the coordinate axes and their sum about the origin.
  • Check results using moment identities and the geometry of the region.

From a Cartesian Region to Lamina Properties

A lamina is a thin, flat object occupying a planar region. Its density may vary from point to point, so different parts of the same region can contribute different amounts of mass. Cartesian iterated integrals provide a systematic way to combine the region’s geometry with its density.

In a complete calculation, the bounds describe the region, the area integral measures its size, and the density-weighted integrals determine its mass, centroid, and moments of inertia. Keeping those roles distinct helps prevent common errors.

Definitions and Cartesian Bounds

Suppose a bounded region \(R\) is described by vertical slices, with \(a\le x\le b\) and \(g_1(x)\le y\le g_2(x)\). If the boundaries and integrands are sufficiently well behaved for the iterated integrals to exist, then an integral over \(R\) can be written by integrating \(y\) first and then \(x\).

$$ \iint_R f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx. $$

Here \(dA\) denotes an area element. The density function \(\rho(x,y)\) gives mass per unit area at \((x,y)\). For a physical lamina, density is nonnegative. The formulas below assume the region has finite area and the relevant integrals exist; calculations of average density and centroid also require positive area and positive total mass, respectively.

Definition: The area \(A\), mass \(M\), average density \(\rho_{\mathrm{avg}}\), and centroid \((\bar{x},\bar{y})\) of a lamina with density \(\rho\) are given by $$ A=\iint_R 1\,dA,\qquad M=\iint_R \rho(x,y)\,dA,\qquad \rho_{\mathrm{avg}}=\frac{M}{A}, $$ $$ \bar{x}=\frac{1}{M}\iint_R x\rho(x,y)\,dA,\qquad \bar{y}=\frac{1}{M}\iint_R y\rho(x,y)\,dA. $$

Define the first moments about the coordinate axes by \(M_x=\iint_R y\rho\,dA\) and \(M_y=\iint_R x\rho\,dA\). The distance from a point \((x,y)\) to the \(x\)-axis is \(|y|\), while its distance to the \(y\)-axis is \(|x|\). Squaring those distances gives the corresponding moments of inertia.

Definition: The moments of inertia about the \(x\)-axis and \(y\)-axis are $$ I_x=\iint_R y^2\rho(x,y)\,dA,\qquad I_y=\iint_R x^2\rho(x,y)\,dA. $$ The polar moment of inertia about the origin is \(I_O=\iint_R (x^2+y^2)\rho(x,y)\,dA=I_x+I_y\).

Why the Density-Weighted Formulas Work

The formulas for the mass properties express the idea of a weighted average. Divide the region into small pieces. A piece near \((x_i,y_i)\) with area \(\Delta A_i\) has approximate mass \(\rho(x_i,y_i)\Delta A_i\). Its contribution to the total mass is that amount; its contribution to the horizontal coordinate’s mass-weighted sum is approximately \(x_i\rho(x_i,y_i)\Delta A_i\).

Adding these contributions over a partition gives finite sums. As the pieces are refined, the sums approach the corresponding double integrals. Thus the total mass approaches \(\iint_R\rho\,dA\), and the horizontal mass-weighted sum approaches \(\iint_R x\rho\,dA\). Dividing the latter by the former gives the horizontal coordinate of the centroid. The same argument using \(y_i\) gives its vertical coordinate. Replacing the coordinate by the square of the distance to an axis gives the moment-of-inertia formulas.

$$ \bar{x} = \lim\frac{\sum_i x_i\,\rho(x_i,y_i)\Delta A_i} {\sum_i \rho(x_i,y_i)\Delta A_i} = \frac{\iint_R x\rho(x,y)\,dA}{\iint_R\rho(x,y)\,dA}. $$

This also explains why the centroid generally differs from the geometric center: regions with greater density contribute more mass and therefore have greater influence on the weighted average.

Worked Example: Area of a Triangular Region

Worked Example: Area of a Triangular Region

Let \(R\) be bounded by \(x=0\), \(y=0\), and \(y=1-\frac{x}{2}\). Find its area using Cartesian bounds.

The top boundary meets the \(x\)-axis where \(1-\frac{x}{2}=0\), so \(x=2\). The region is therefore described by \(0\le x\le2\) and \(0\le y\le1-\frac{x}{2}\). Its area is

$$ \begin{aligned} A &=\int_0^2\int_0^{1-x/2}1\,dy\,dx\\ &=\int_0^2\left[y\right]_0^{1-x/2}\,dx\\ &=\int_0^2\left(1-\frac{x}{2}\right)\,dx\\ &=\left[x-\frac{x^2}{4}\right]_0^2\\ &=2-1=1. \end{aligned} $$

The result agrees with the triangle’s base-times-height calculation: the intercepts give base \(2\) and height \(1\), so the area is \(\frac12(2)(1)=1\).

Worked Example: Mass and Average Density

Worked Example: Mass and Average Density

On the triangular region \(0\le x\le1\), \(0\le y\le x\), let the density be \(\rho(x,y)=2+x\). Find the area, mass, and average density.

First integrate \(1\) over each vertical slice. Then integrate the density over the same slices to find mass:

$$ \begin{aligned} A &=\int_0^1\int_0^x1\,dy\,dx =\int_0^1x\,dx =\left[\frac{x^2}{2}\right]_0^1 =\frac12,\\[4pt] M &=\int_0^1\int_0^x(2+x)\,dy\,dx =\int_0^1(2+x)x\,dx\\ &=\int_0^1(2x+x^2)\,dx =\left[x^2+\frac{x^3}{3}\right]_0^1 =1+\frac13=\frac43. \end{aligned} $$

The average density is mass divided by area:

$$ \rho_{\mathrm{avg}} =\frac{M}{A} =\frac{4/3}{1/2} =\frac83. $$

Since \(2\le\rho(x,y)\le3\) throughout the region, the average density must lie between \(2\) and \(3\). The computed value \(\frac83\) passes this check.

Worked Example: A Complete Cartesian Lamina Calculation

Worked Example: A Complete Cartesian Lamina Calculation

Consider the rectangle \(R=\{(x,y):0\le x\le2,\ 0\le y\le1\}\) with density \(\rho(x,y)=1+x+y\). Find its area, mass, average density, centroid, and moments of inertia about both coordinate axes and the origin.

Area. Integrating \(1\) over the rectangle gives

$$ A=\int_0^2\int_0^1 1\,dy\,dx =\int_0^2 1\,dx=2. $$

Mass and average density. Integrate the density first with respect to \(y\), then with respect to \(x\):

$$ \begin{aligned} M &=\int_0^2\int_0^1(1+x+y)\,dy\,dx\\ &=\int_0^2\left[(1+x)y+\frac{y^2}{2}\right]_0^1\,dx\\ &=\int_0^2\left(x+\frac32\right)\,dx\\ &=\left[\frac{x^2}{2}+\frac{3x}{2}\right]_0^2 =2+3=5. \end{aligned} $$

Consequently,

$$ \rho_{\mathrm{avg}}=\frac{M}{A}=\frac52. $$

First moments and centroid. The first moment \(M_y\) gives the horizontal centroid coordinate, while \(M_x\) gives the vertical coordinate:

$$ \begin{aligned} M_y &=\int_0^2\int_0^1x(1+x+y)\,dy\,dx\\ &=\int_0^2 x\left(x+\frac32\right)\,dx\\ &=\left[\frac{x^3}{3}+\frac{3x^2}{4}\right]_0^2 =\frac83+3=\frac{17}{3},\\[4pt] M_x &=\int_0^2\int_0^1y(1+x+y)\,dy\,dx\\ &=\int_0^2\left(\frac{1+x}{2}+\frac13\right)\,dx\\ &=\int_0^2\left(\frac{x}{2}+\frac56\right)\,dx\\ &=\left[\frac{x^2}{4}+\frac{5x}{6}\right]_0^2 =1+\frac53=\frac83. \end{aligned} $$

Dividing each first moment by the mass gives

$$ \bar{x}=\frac{M_y}{M}=\frac{17/3}{5}=\frac{17}{15}, \qquad \bar{y}=\frac{M_x}{M}=\frac{8/3}{5}=\frac{8}{15}. $$

Moments of inertia. For \(I_x\), weight each mass element by \(y^2\); for \(I_y\), weight it by \(x^2\):

$$ \begin{aligned} I_x &=\int_0^2\int_0^1 y^2(1+x+y)\,dy\,dx\\ &=\int_0^2\left(\frac{1+x}{3}+\frac14\right)\,dx\\ &=\int_0^2\left(\frac{x}{3}+\frac{7}{12}\right)\,dx\\ &=\left[\frac{x^2}{6}+\frac{7x}{12}\right]_0^2 =\frac23+\frac76=\frac{11}{6},\\[4pt] I_y &=\int_0^2\int_0^1 x^2(1+x+y)\,dy\,dx\\ &=\int_0^2 x^2\left(x+\frac32\right)\,dx\\ &=\left[\frac{x^4}{4}+\frac{x^3}{2}\right]_0^2 =4+4=8. \end{aligned} $$

The polar moment of inertia about the origin is

$$ I_O=I_x+I_y=\frac{11}{6}+8=\frac{11}{6}+\frac{48}{6}=\frac{59}{6}. $$

Checks. The centroid reproduces the first moments: \(M\bar{x}=5\cdot\frac{17}{15}=\frac{17}{3}=M_y\), and \(M\bar{y}=5\cdot\frac{8}{15}=\frac83=M_x\). Also, \(I_x+I_y=\frac{11}{6}+8=\frac{59}{6}\), as required by the definition of \(I_O\).

A Common Error: Interchanging the Inertia Formulas

The distance to the \(x\)-axis is measured in the \(y\)-direction, so \(I_x\) uses \(y^2\), not \(x^2\). The distance to the \(y\)-axis is measured in the \(x\)-direction, so \(I_y\) uses \(x^2\).

For the rectangle and density in the complete example, an incorrect interchange would claim \(I_x=\iint_R x^2\rho\,dA=8\) and \(I_y=\iint_R y^2\rho\,dA=\frac{11}{6}\). The correction is to match each axis with its perpendicular coordinate: \(I_x=\iint_R y^2\rho\,dA=\frac{11}{6}\), and \(I_y=\iint_R x^2\rho\,dA=8\). These values are unequal here, so the interchange changes the answer.

A Reliable Calculation Order

For a bounded region described by vertical slices, a consistent order of work makes it easier to catch mistakes:

1
Write the region bounds.
State the \(x\)-interval and the lower and upper \(y\)-boundaries before setting up any integrals.
2
Find the area and mass.
Integrate \(1\) for area and \(\rho\) for mass, using the same region bounds.
3
Compute first moments.
Use \(x\rho\) for \(M_y\) and \(y\rho\) for \(M_x\), then divide by mass to obtain the centroid.
4
Compute moments of inertia.
Use \(y^2\rho\) for \(I_x\) and \(x^2\rho\) for \(I_y\); add them for \(I_O\).
5
Check the results.
Confirm the centroid reproduces the first moments and that \(I_O=I_x+I_y\). Use the region and density to assess whether the results are plausible.

Why the Checks Matter

Each quantity answers a different question. Area depends only on the region, while mass depends on both the region and density. Average density compares those two quantities. The centroid records the balance point of the mass distribution, and each moment of inertia measures how that mass is distributed relative to a specified axis.

Checks do not replace the integrals, but they can reveal setup errors. For example, a centroid coordinate for a region lying between \(x=0\) and \(x=2\) should lie within those bounds when density is nonnegative and total mass is positive. Likewise, the moments of inertia are nonnegative because their integrands are nonnegative. A negative result indicates an error in the bounds, integrand, or arithmetic.

Check Your Understanding

Use the definitions and calculation order above to answer these questions.

  1. For a region \(0\le x\le b\), \(g_1(x)\le y\le g_2(x)\), write the iterated integral for its area.
  2. Why is average density \(M/A\), rather than \(A/M\)?
  3. Which first moment determines \(\bar{x}\), and why?
  4. For a density \(\rho(x,y)\), write the integrands for \(I_x\) and \(I_y\).
  5. What identity checks the polar moment of inertia about the origin?

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