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Area, Mass, and Moments · Tutorial 585 of 999

Polar Integrals For Area And Mass

Use polar bounds and the area element \(dA=r\,dr\,d\theta\) to calculate the area and mass of planar regions.

Intermediate 9 min read

What You'll Learn

  • Explain why the polar area element is \(r\,dr\,d\theta\).
  • Set up area integrals using angular and radial bounds.
  • Express sector and annulus regions with polar bounds.
  • Convert a density function into polar coordinates.
  • Calculate mass when density varies with distance from the origin.
  • Detect the common error of omitting the factor \(r\).

From a Planar Region to an Integral

A double integral can calculate the area of a planar region or the mass of a thin lamina occupying that region. In Cartesian coordinates, these calculations use \(dA=dx\,dy\). In polar coordinates, the same small patch of area has a different expression: \(dA=r\,dr\,d\theta\).

That factor of \(r\) matters. At a fixed angle, a small change in angle sweeps out a longer arc when it is farther from the origin. Consequently, a polar integral must account for both a change in radius and a change in arc length.

The main task is to describe the region accurately: identify the angles it covers, then specify the inner and outer radius at each angle. Once the bounds are correct, area and mass follow from the same integral structure.

Key formulas: If a region \(D\) is described by \(a\leq\theta\leq b\) and \(r_{\mathrm{in}}(\theta)\leq r\leq r_{\mathrm{out}}(\theta)\), then its area is the integral of \(r\) over those bounds. Its mass is the integral of the density multiplied by \(r\) over the same bounds.

Why the Polar Area Element Contains \(r\)

Polar coordinates are related to Cartesian coordinates by \(x=r\cos\theta\) and \(y=r\sin\theta\). The local area scale for a change of coordinates is the absolute value of the Jacobian determinant. Here, the derivative matrix is

$$ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{pmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{pmatrix}. $$

Its determinant is

$$ r\cos^2\theta+r\sin^2\theta = r\bigl(\cos^2\theta+\sin^2\theta\bigr) = r. $$

Since \(r\geq 0\), the absolute value of this determinant is \(r\). The change-of-variables formula therefore gives \(dA=r\,dr\,d\theta\). The origin, where polar coordinates are singular, has area zero and does not affect the integral.

Area and mass in polar coordinates: If \(D\) is covered once by the bounds \(a\leq\theta\leq b\) and \(r_{\mathrm{in}}(\theta)\leq r\leq r_{\mathrm{out}}(\theta)\), then $$ \operatorname{Area}(D)=\int_a^b\int_{r_{\mathrm{in}}(\theta)}^{r_{\mathrm{out}}(\theta)} r\,dr\,d\theta. $$ If the lamina has surface density \(\rho(x,y)\), its mass is $$ m=\int_a^b\int_{r_{\mathrm{in}}(\theta)}^{r_{\mathrm{out}}(\theta)} \rho(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$

The density in the mass formula is evaluated at the Cartesian point represented by \((r,\theta)\). If the density is already given as a function of \(r\) and \(\theta\), use that expression directly. A density depending only on distance from the origin has the form \(\rho(r)\).

Choosing Bounds for Sectors and Annuli

A sector is described by an angular interval and a radial interval. For a sector of radius \(R\) between angles \(\alpha\) and \(\beta\), the bounds are \(\alpha\leq\theta\leq\beta\) and \(0\leq r\leq R\). An annular sector excludes an inner disk: its bounds are \(\alpha\leq\theta\leq\beta\) and \(r_1\leq r\leq r_2\), where \(0\leq r_1\leq r_2\).

More generally, the inner or outer radius may vary with angle. The same setup applies provided the radial bounds describe the region at each angle. If the region has different radial bounds over different angular intervals, split it into pieces and integrate each piece.

Worked Example: Area of an Annular Sector

Find the area of the annular sector with \(1\text{ m}\leq r\leq3\text{ m}\) and \(0\leq\theta\leq\frac{\pi}{3}\).

The bounds include every angle in the sector and every radius between the inner and outer circles. Integrating the polar area element gives

$$ \begin{aligned} A &=\int_0^{\pi/3}\int_1^3 r\,dr\,d\theta\\ &=\int_0^{\pi/3}\left[\frac{r^2}{2}\right]_1^3\,d\theta\\ &=\int_0^{\pi/3}\left(\frac{9}{2}-\frac{1}{2}\right)\,d\theta\\ &=\int_0^{\pi/3}4\,d\theta\\ &=4\left(\frac{\pi}{3}-0\right) =\frac{4\pi}{3}\text{ m}^2. \end{aligned} $$

As a check, this is the area of a sector of radius \(3\) minus the area of a sector of radius \(1\), both with angle \(\frac{\pi}{3}\):

$$ \frac{1}{2}\left(\frac{\pi}{3}\right)(3^2) - \frac{1}{2}\left(\frac{\pi}{3}\right)(1^2) = \frac{3\pi}{2}-\frac{\pi}{6} = \frac{4\pi}{3}\text{ m}^2. $$

Mass with Radial Density

The mass integral uses the same bounds as the area integral, but weights each small area element by its density. In particular, for a density depending only on radius, the integrand is \(\rho(r)r\), not just \(\rho(r)\). The additional \(r\) is still the area-element factor.

Worked Example: Mass of a Sector with Increasing Density

A quarter-disk lamina has radius \(2\text{ m}\), with \(0\leq\theta\leq\frac{\pi}{2}\). Its density is \(\rho(r)=2+r\), measured in \(\text{kg}/\text{m}^2\), where \(r\) is the numerical distance in meters. Find its mass.

The radial bounds are \(0\leq r\leq2\), and the angular bounds are \(0\leq\theta\leq\frac{\pi}{2}\). Include the area-element factor \(r\):

$$ \begin{aligned} m &=\int_0^{\pi/2}\int_0^2 (2+r)r\,dr\,d\theta\\ &=\int_0^{\pi/2}\int_0^2 (2r+r^2)\,dr\,d\theta\\ &=\int_0^{\pi/2} \left[r^2+\frac{r^3}{3}\right]_0^2\,d\theta\\ &=\int_0^{\pi/2}\left(4+\frac{8}{3}\right)\,d\theta\\ &=\int_0^{\pi/2}\frac{20}{3}\,d\theta\\ &=\frac{20}{3}\left(\frac{\pi}{2}-0\right) =\frac{10\pi}{3}\text{ kg}. \end{aligned} $$

To check the radial antiderivative, differentiate \(r^2+\frac{r^3}{3}\): its derivative is \(2r+r^2=(2+r)r\), the required integrand. The angular integration then multiplies the constant \(\frac{20}{3}\) by the angular width \(\frac{\pi}{2}\), as shown.

For a density given in Cartesian form, substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) before integrating. For example, \(\rho(x,y)=1+x\) becomes \(\rho(r,\theta)=1+r\cos\theta\). The bounds still describe the same region; only the density expression changes.

Worked Example: Mass of an Annulus with Density Proportional to Radius

An annulus has inner radius \(1\text{ m}\), outer radius \(2\text{ m}\), and covers a full revolution. Its density is \(\rho(r)=3r\), where the coefficient is \(3\text{ kg}/\text{m}^3\) and \(r\) is measured in meters, so the surface density is in \(\text{kg}/\text{m}^2\). Find its mass.

Use \(1\leq r\leq2\) and \(0\leq\theta\leq2\pi\). Multiplying the density by the area-element factor gives \(3r^2\):

$$ \begin{aligned} m &=\int_0^{2\pi}\int_1^2 3r\cdot r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[r^3\right]_1^2\,d\theta\\ &=\int_0^{2\pi}(8-1)\,d\theta\\ &=\int_0^{2\pi}7\,d\theta\\ &=7(2\pi-0) =14\pi\text{ kg}. \end{aligned} $$

The radial antiderivative checks because \(\frac{d}{dr}(r^3)=3r^2\), which is the radial integrand. The full angular interval contributes \(2\pi\), giving the stated mass.

A Common Error: Omitting the Jacobian Factor

A frequent mistake is to replace \(dx\,dy\) by \(dr\,d\theta\) and leave out the factor \(r\). That does not account for the change in area scale with distance from the origin.

Worked Error Check: Area of a Quarter-Disk

Consider a quarter-disk of radius \(3\text{ m}\), with \(0\leq\theta\leq\frac{\pi}{2}\). An incorrect setup that omits \(r\) gives

$$ \int_0^{\pi/2}\int_0^3 1\,dr\,d\theta = \frac{3\pi}{2}. $$

This is not the area. The correct area element includes \(r\):

$$ \begin{aligned} A &=\int_0^{\pi/2}\int_0^3 r\,dr\,d\theta\\ &=\int_0^{\pi/2}\left[\frac{r^2}{2}\right]_0^3\,d\theta\\ &=\int_0^{\pi/2}\frac{9}{2}\,d\theta =\frac{9\pi}{4}\text{ m}^2. \end{aligned} $$

As an independent check, a quarter of a disk of radius \(3\text{ m}\) has area \(\frac14\pi(3\text{ m})^2=\frac{9\pi}{4}\text{ m}^2\). The incorrect result differs because it treats equal radial and angular increments as equal-area contributions at every radius.

Setting Up a Reliable Integral

Before evaluating, make sure the bounds describe the region once and only once. An angular interval that goes around a full circle should have total width \(2\pi\); an interval wider than that may count some points more than once. Radial bounds should be nonnegative and should run from the region's inner edge to its outer edge at each angle.

1
Describe the angles.
Choose an interval that covers the region without unnecessary repetition.
2
Set the radial limits.
For each angle, identify the inner and outer radius. Split the region if these limits change by angular interval.
3
Choose the quantity to integrate.
Use \(1\) for area, or use the density for mass.
4
Include the area factor.
Multiply the chosen integrand by \(r\), since \(dA=r\,dr\,d\theta\).
5
Check the result.
Confirm that the units are appropriate and, when possible, compare with a geometric or other simple check.

Area has units of length squared. If density is mass per unit area, multiplying it by \(dA\) gives mass. A result with the wrong units often signals that a factor or unit conversion has been missed.

Polar bounds are especially useful when a region is naturally bounded by circles, circular arcs, or rays from the origin. The same description handles area and mass; only the integrand changes. Careful bounds and the Jacobian factor \(r\) turn the geometric description into the correct integral.

Check Your Understanding

For each question, identify the appropriate polar setup before carrying out any evaluation.

  1. Write the area integral for a sector with \(0\leq r\leq5\) and \(\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}\).
  2. Write the area integral for a full annulus with inner radius \(2\) and outer radius \(4\).
  3. For density \(\rho(r)=1+r\) on \(0\leq r\leq3\), \(0\leq\theta\leq\pi\), write the mass integral, including the area-element factor.
  4. What is the Jacobian determinant of the transformation \(x=r\cos\theta,\ y=r\sin\theta\), and what does it imply for \(dA\)?
  5. Explain why an integral of density alone with respect to \(dr\,d\theta\) does not generally give mass in polar coordinates.

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