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Average Value And Average Density

Use a double integral divided by area to find an average, and interpret the result with the correct units.

Intermediate 9 min read

What You'll Learn

  • Define average value for an integrable function on a planar region of positive area
  • Compute averages over rectangles, disks, and triangular regions
  • Relate the average value to the total double integral and area
  • Calculate average density from total mass and area
  • Track units and distinguish average quantities from totals
  • Identify common errors in interpreting or calculating an average

Why Average Value Uses Area

A double integral over a planar region adds up contributions across that region. Depending on the integrand, the integral might represent a signed volume beneath a surface or the total mass of a lamina. An average answers a different question: what single value, spread uniformly over the same area, would give the same total?

This question arises for any integrable function, not only one with a physical interpretation. The average value is the total accumulated by the integral divided by the area over which it accumulates. The area must be positive; otherwise, division by area is not defined.

Definition: Let \(R\) be a planar region with finite positive area \(A(R)\), and let \(f\) be integrable on \(R\). The average value of \(f\) over \(R\) is
$$ f_{\mathrm{avg}} = \frac{1}{A(R)}\iint_R f(x,y)\,dA. $$

The double integral is the total accumulated value, while \(f_{\mathrm{avg}}\) is that total per unit area. If \(f\) is a height above the region, the integral gives signed volume and the average gives average height. If \(f\) has another meaning, the average inherits its units.

Why the Formula Works

Imagine dividing \(R\) into many small pieces, each with area \(\Delta A_i\), and choosing a sample point in each piece. The sum of the function values multiplied by their piece areas approximates the total integral. Dividing that sum by the total area gives an area-weighted average of the sampled values.

$$ \frac{\sum_i f(x_i,y_i)\Delta A_i}{A(R)} = \sum_i f(x_i,y_i)\frac{\Delta A_i}{A(R)}. $$

Each weight \(\Delta A_i/A(R)\) is nonnegative, and the weights sum to \(1\), because the pieces have total area \(A(R)\). Thus, each sample contributes according to the fraction of the region it represents. As the pieces become finer, the sum in the numerator approaches the double integral, giving the average-value formula.

This reasoning also explains why the area is the denominator. The region is the set over which values are being averaged, so each part's weight is determined by its share of the total area—not by its perimeter or by the number of pieces in a particular subdivision.

Worked Example: Average Value on a Rectangle

Worked Example: Average Value on a Rectangle

Find the average value of \(f(x,y)=x+2y\) on \(R=[0,2]\times[0,1]\).

First, the area of the rectangle is \(A(R)=(2-0)(1-0)=2\). Using \(0\le x\le 2\) and \(0\le y\le 1\), compute the integral:

$$ \begin{aligned} \iint_R (x+2y)\,dA &=\int_0^2\int_0^1 (x+2y)\,dy\,dx\\ &=\int_0^2 \left[xy+y^2\right]_{y=0}^{y=1}\,dx\\ &=\int_0^2 (x+1)\,dx\\ &=\left[\frac{x^2}{2}+x\right]_0^2\\ &=(2+2)-0\\ &=4. \end{aligned} $$

Therefore, the average value is \(f_{\mathrm{avg}}=4/2=2\). Check the result by multiplying the average by the area: \(2\cdot2=4\), which recovers the integral over \(R\).

Average Value and Total Volume

When \(f(x,y)\ge 0\) on \(R\), the double integral \(\iint_R f\,dA\) is the volume between the surface \(z=f(x,y)\) and the planar region \(R\). The average value is not that volume. It is the constant height that would produce the same volume over the same base: a constant height \(c\) gives volume \(cA(R)\), so \(cA(R)=\iint_R f\,dA\) precisely when \(c=f_{\mathrm{avg}}\).

If \(f\) takes both positive and negative values, the integral represents signed accumulation: positive and negative contributions can cancel. The formula for average value still applies, but interpreting its integral as ordinary geometric volume would then be misleading.

Worked Example: Average Value on a Disk

Find the average value of \(f(x,y)=x^2+y^2\) on the disk \(R\) of radius \(2\) centered at the origin.

The area of the disk is \(A(R)=\pi(2)^2=4\pi\). In polar coordinates, \(x^2+y^2=r^2\), the area element is \(r\,dr\,d\theta\), and the disk is described by \(0\le r\le2\), \(0\le\theta\le2\pi\). Thus,

$$ \begin{aligned} \iint_R (x^2+y^2)\,dA &=\int_0^{2\pi}\int_0^2 r^2\cdot r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^4}{4}\right]_0^2\,d\theta\\ &=\int_0^{2\pi}4\,d\theta\\ &=4(2\pi)\\ &=8\pi. \end{aligned} $$

Divide by the disk's area: \(f_{\mathrm{avg}}=8\pi/(4\pi)=2\). As a check, a constant height of \(2\) over a disk of area \(4\pi\) has volume \(2(4\pi)=8\pi\), matching the integral.

A Result That Checks the Size of an Average

For a continuous function on a closed, bounded region of positive area, the function attains a minimum \(m\) and a maximum \(M\). At every point of the region, \(m\le f(x,y)\le M\). Integrating these inequalities over \(R\) preserves their order, since area is nonnegative:

$$ mA(R)\le \iint_R f(x,y)\,dA\le MA(R). $$

Because \(A(R)>0\), dividing through by it gives the following useful check.

Theorem: If \(f\) is continuous on a closed, bounded planar region \(R\) of positive area, then
$$ \min_R f \le f_{\mathrm{avg}} \le \max_R f. $$

The average cannot lie below every function value or above every function value. This bound is a helpful way to catch implausible answers, though it does not replace evaluating the integral.

Average Density of a Lamina

Suppose a lamina occupies a region \(R\) and has an integrable areal density \(\rho(x,y)\), measured as mass per unit area. Its total mass is the integral of density over the region. Dividing that mass by the area gives the average density—the uniform areal density that would give the same total mass over \(R\).

Definition: For a lamina occupying a region \(R\) of finite positive area, with integrable areal density \(\rho\), its average density is
$$ \rho_{\mathrm{avg}} = \frac{1}{A(R)}\iint_R \rho(x,y)\,dA = \frac{\text{total mass}}{\text{area}}. $$

Average density has the same units as \(\rho\), such as kilograms per square metre. Total mass has units of kilograms. They are related, but they are not interchangeable: multiplying average density by area gives total mass, while dividing total mass by area gives average density.

Worked Example: Average Density on a Rectangular Lamina

A rectangular lamina occupies \(0\le x\le2\) metres and \(0\le y\le1\) metre. Its density is \(\rho(x,y)=2+x\) kilograms per square metre, where \(x\) is measured in metres and the coefficient of \(x\) has units of kilograms per cubic metre. Find its total mass and average density.

The area is \(A(R)=(2\text{ m})(1\text{ m})=2\text{ m}^2\). Integrating the density gives the mass:

$$ \begin{aligned} m &=\int_0^2\int_0^1 (2+x)\,dy\,dx\\ &=\int_0^2 \left[(2+x)y\right]_{y=0}^{y=1}\,dx\\ &=\int_0^2 (2+x)\,dx\\ &=\left[2x+\frac{x^2}{2}\right]_0^2\\ &=(4+2)-0\\ &=6\text{ kg}. \end{aligned} $$

Therefore, \(\rho_{\mathrm{avg}}=6\text{ kg}/2\text{ m}^2=3\text{ kg/m}^2\). Check by multiplying: \((3\text{ kg/m}^2)(2\text{ m}^2)=6\text{ kg}\), the mass found by integration. The value \(6\) is the total mass, not the average density; the average density is \(3\text{ kg/m}^2\).

Worked Example: Average Value on a Triangle

Worked Example: Average Value on a Triangle

Find the average value of \(f(x,y)=x+y\) on the triangular region \(R\) bounded by \(x=0\), \(y=0\), and \(x+y=1\).

The region has bounds \(0\le x\le1\) and \(0\le y\le1-x\). Its area is \(\int_0^1(1-x)\,dx=[x-x^2/2]_0^1=1/2\). Now evaluate the integral:

$$ \begin{aligned} \iint_R(x+y)\,dA &=\int_0^1\int_0^{1-x}(x+y)\,dy\,dx\\ &=\int_0^1\left[xy+\frac{y^2}{2}\right]_{y=0}^{y=1-x}\,dx\\ &=\int_0^1\left(x(1-x)+\frac{(1-x)^2}{2}\right)\,dx\\ &=\int_0^1\left(x-x^2+\frac{1-2x+x^2}{2}\right)\,dx\\ &=\int_0^1\left(\frac12-\frac{x^2}{2}\right)\,dx\\ &=\left[\frac{x}{2}-\frac{x^3}{6}\right]_0^1\\ &=\frac12-\frac16\\ &=\frac13. \end{aligned} $$

Divide by the area: \(f_{\mathrm{avg}}=(1/3)/(1/2)=2/3\). Since \(0\le x+y\le1\) throughout the triangle, the result satisfies the bounds \(0\le2/3\le1\). Also, \((2/3)(1/2)=1/3\), recovering the integral.

Common Mistake: Reporting the Integral as the Average

In the rectangular example, the integral of \(x+2y\) was \(4\), and the area was \(2\). A common error is to report \(4\) as the average value. That confuses the total accumulated value with the value per unit area.

$$ \text{Incorrect: } f_{\mathrm{avg}}=4 \qquad \text{Correct: } f_{\mathrm{avg}}=\frac{4}{2}=2. $$

A useful correction is to check the units and reconstruct the total: average value times area must equal the integral. In this example, \(2\cdot2=4\); reporting \(4\) as the average would instead imply a total of \(4\cdot2=8\), contradicting the calculated integral.

Check Your Understanding

Use the definitions and checks in this tutorial to answer the following questions.

  1. What condition on the area of \(R\) is needed for the average-value formula to be defined?
  2. If \(\iint_R f\,dA=15\) and \(A(R)=5\), what is \(f_{\mathrm{avg}}\), and how can you check it?
  3. For a nonnegative height function, how does its average value relate to the volume under its graph?
  4. A lamina has total mass \(12\) kilograms and area \(4\) square metres. What is its average density, including units?
  5. Why must the average of a continuous function on a closed, bounded region lie between its minimum and maximum?

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