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A Complete Polar Lamina Calculation

Follow one annular-sector calculation from its polar bounds through area, mass, centroid, polar moment of inertia, and radius of gyration.

Intermediate 10 min read

What You'll Learn

  • Set polar bounds for an annular sector and identify its area element.
  • Derive area, mass, centroid, and polar moment integrals for radial density.
  • Calculate each requested quantity in a complete example with units.
  • Check centroid and radius-of-gyration results against their defining relations.
  • Recognize and correct the error of omitting the polar area factor.

From Polar Bounds to Lamina Properties

An annular sector is bounded by two circles centered at the origin and two rays from the origin. Its geometry is especially suited to polar coordinates: the radial bounds describe the inner and outer circles, while the angular bounds describe the rays. When the density depends only on distance from the origin, the same bounds also make the mass and inertia integrals direct.

In this tutorial, the density is a surface density, so mass is obtained by integrating density over area. We will use the same sector to calculate its area, mass, centroid, polar moment of inertia about the origin, and radius of gyration about the origin. Each result will be checked against its defining relation.

Definition: Let \(R\) be an annular sector described by \(a\le r\le b\) and \(\alpha\le\theta\le\beta\), where \(0\le a<b\), \(0<\beta-\alpha\le 2\pi\), and the angular interval describes the sector without overlap. Its radial surface density is a nonnegative function \(\rho(r)\). The mass is positive and finite for the centroid and radius of gyration to be defined.

The Polar Integrals for the Calculation

In polar coordinates, a point has Cartesian coordinates \(x=r\cos\theta\) and \(y=r\sin\theta\). The area element is \(dA=r\,dr\,d\theta\). The factor \(r\) is essential: a small polar region has radial thickness \(dr\) and approximate arc length \(r\,d\theta\).

Write \(\Delta\theta=\beta-\alpha\). Integrating the area element gives the area. Multiplying it by \(\rho(r)\) gives the mass element. Multiplying the mass element by \(x\) or \(y\) gives the first moments used to find the centroid. Finally, the squared distance from the origin is \(r^2\), so multiplying the mass element by \(r^2\) gives the polar moment of inertia.

Key formulas: For a radial density on this sector, the area \(A\), mass \(M\), centroid \((\bar{x},\bar{y})\), polar moment of inertia \(I_O\), and radius of gyration \(k_O\) about the origin are
$$ \begin{aligned} A&=\int_\alpha^\beta\int_a^b r\,dr\,d\theta =\frac{\Delta\theta}{2}(b^2-a^2),\\ M&=\int_\alpha^\beta\int_a^b \rho(r)r\,dr\,d\theta =\Delta\theta\int_a^b \rho(r)r\,dr,\\ \bar{x}&=\frac{1}{M} \left(\int_\alpha^\beta\cos\theta\,d\theta\right) \left(\int_a^b \rho(r)r^2\,dr\right),\\ \bar{y}&=\frac{1}{M} \left(\int_\alpha^\beta\sin\theta\,d\theta\right) \left(\int_a^b \rho(r)r^2\,dr\right),\\ I_O&=\int_\alpha^\beta\int_a^b r^2\rho(r)r\,dr\,d\theta =\Delta\theta\int_a^b \rho(r)r^3\,dr,\\ k_O&=\sqrt{\frac{I_O}{M}}. \end{aligned} $$

Why the Formulas Work

Here is the derivation for the centroid and polar moment, including the role of the radial-density assumption. By definition, the first moment about the \(y\)-axis is the integral of \(x\) times the mass element. Since \(x=r\cos\theta\), \(\rho=\rho(r)\), and \(dA=r\,dr\,d\theta\), the integrand separates into a factor depending on \(\theta\) and a factor depending on \(r\):

$$ \begin{aligned} \int_R x\rho\,dA &=\int_\alpha^\beta\int_a^b (r\cos\theta)\rho(r)r\,dr\,d\theta\\ &=\left(\int_\alpha^\beta\cos\theta\,d\theta\right) \left(\int_a^b\rho(r)r^2\,dr\right). \end{aligned} $$

The first moment about the \(x\)-axis follows in the same way, using \(y=r\sin\theta\). The centroid is defined by dividing each first moment by the total mass. For the polar moment, the squared distance to the origin is \(x^2+y^2=r^2\), so its mass-weighted integral is

$$ I_O=\int_R (x^2+y^2)\rho\,dA =\int_\alpha^\beta\int_a^b r^2\rho(r)r\,dr\,d\theta. $$

This derivation also explains why the integrals simplify into products: the sector has independent radial and angular bounds, and the density depends only on \(r\). If the density also depended on \(\theta\), that product separation would not generally apply.

Worked Example: A Quarter Annulus with Increasing Density

Worked Example: A Quarter Annulus with Increasing Density

Consider the sector \(1\le r\le3\), \(0\le\theta\le\pi/2\), measured in metres. Let \(\rho(r)=\kappa r\), where \(\kappa=1\ \mathrm{kg/m^3}\). Calculate its area, mass, centroid, polar moment of inertia about the origin, and radius of gyration about the origin.

Area. Here \(\Delta\theta=\pi/2\), \(a=1\ \mathrm m\), and \(b=3\ \mathrm m\). Thus

$$ A=\frac{\pi/2}{2}(3^2-1^2)\ \mathrm{m^2} =\frac{\pi}{4}(8)\ \mathrm{m^2} =2\pi\ \mathrm{m^2}. $$

Mass. Substitute \(\rho(r)=\kappa r\) into the mass integral. The radial integral is \(\int_1^3 r^2\,dr=(27-1)/3=26/3\), with the length units supplying \(\mathrm m^3\):

$$ \begin{aligned} M&=\frac{\pi}{2}\kappa\int_1^3 r^2\,dr\\ &=\frac{\pi}{2}(1\ \mathrm{kg/m^3}) \left(\frac{3^3-1^3}{3}\ \mathrm{m^3}\right)\\ &=\frac{13\pi}{3}\ \mathrm{kg}. \end{aligned} $$

Centroid. The radial factor in both first moments is \(\kappa\int_1^3r^3\,dr=\kappa(81-1)/4=20\ \mathrm{kg\,m}\). The angular factors are \(\int_0^{\pi/2}\cos\theta\,d\theta=1\) and \(\int_0^{\pi/2}\sin\theta\,d\theta=1\). Therefore each first moment is \(20\ \mathrm{kg\,m}\), and

$$ \bar{x}=\frac{20\ \mathrm{kg\,m}}{(13\pi/3)\ \mathrm{kg}} =\frac{60}{13\pi}\ \mathrm m, \qquad \bar{y}=\frac{20\ \mathrm{kg\,m}}{(13\pi/3)\ \mathrm{kg}} =\frac{60}{13\pi}\ \mathrm m. $$

Polar moment of inertia. The radial integral is \(\int_1^3r^4\,dr=(243-1)/5=242/5\), so

$$ \begin{aligned} I_O&=\frac{\pi}{2}\kappa\int_1^3r^4\,dr\\ &=\frac{\pi}{2}(1\ \mathrm{kg/m^3}) \left(\frac{3^5-1^5}{5}\ \mathrm{m^5}\right)\\ &=\frac{121\pi}{5}\ \mathrm{kg\,m^2}. \end{aligned} $$

Radius of gyration. Dividing the polar moment by mass gives the square of the radius of gyration:

$$ \begin{aligned} k_O&=\sqrt{\frac{I_O}{M}} =\sqrt{\frac{(121\pi/5)\ \mathrm{kg\,m^2}} {(13\pi/3)\ \mathrm{kg}}}\\ &=\sqrt{\frac{363}{65}}\ \mathrm m. \end{aligned} $$

Checks. The centroid reproduces the first moments: \(M\bar{x}=M\bar{y}=(13\pi/3)(60/(13\pi))=20\ \mathrm{kg\,m}\). The radius of gyration also satisfies \(Mk_O^2=(13\pi/3)(363/65)=121\pi/5\ \mathrm{kg\,m^2}=I_O\). Both checks recover the values calculated directly.

Worked Example: A Symmetric Annular Sector with Uniform Density

Worked Example: A Symmetric Annular Sector with Uniform Density

Let \(1\le r\le2\) metres and \(-\pi/4\le\theta\le\pi/4\), with constant surface density \(\sigma=2\ \mathrm{kg/m^2}\). Calculate all five quantities, using the origin as the reference point for inertia and radius of gyration.

The angular width is \(\Delta\theta=\pi/2\). The area is

$$ A=\frac{\pi/2}{2}(2^2-1^2) =\frac{3\pi}{4}\ \mathrm{m^2}. $$

Because the density is constant, \(M=\sigma A\):

$$ M=(2\ \mathrm{kg/m^2})\left(\frac{3\pi}{4}\ \mathrm{m^2}\right) =\frac{3\pi}{2}\ \mathrm{kg}. $$

For the first moments, the radial factor is \(\sigma\int_1^2r^2\,dr=2(7/3)=14/3\ \mathrm{kg\,m}\). The cosine angular factor is \(\sin(\pi/4)-\sin(-\pi/4)=\sqrt{2}\), while the sine angular factor is zero by symmetry. Hence

$$ \bar{x}=\frac{(14\sqrt{2}/3)\ \mathrm{kg\,m}}{(3\pi/2)\ \mathrm{kg}} =\frac{28\sqrt{2}}{9\pi}\ \mathrm m, \qquad \bar{y}=0\ \mathrm m. $$

The polar moment and radius of gyration are

$$ \begin{aligned} I_O&=\frac{\pi}{2}(2\ \mathrm{kg/m^2}) \left(\frac{2^4-1^4}{4}\ \mathrm{m^4}\right) =\frac{15\pi}{4}\ \mathrm{kg\,m^2},\\ k_O&=\sqrt{\frac{15\pi/4}{3\pi/2}}\ \mathrm m =\sqrt{\frac52}\ \mathrm m. \end{aligned} $$

Checks. The first moment in the \(y\)-direction is zero because \(\int_{-\pi/4}^{\pi/4}\sin\theta\,d\theta=0\), agreeing with \(\bar y=0\). Also, \(M\bar{x}=(3\pi/2)(28\sqrt2/(9\pi))=14\sqrt2/3\ \mathrm{kg\,m}\), the calculated \(x\)-first moment. Finally, \(Mk_O^2=(3\pi/2)(5/2)=15\pi/4\ \mathrm{kg\,m^2}=I_O\).

Worked Example: A Wider Sector with Radial Density

Worked Example: A Wider Sector with Radial Density

Take \(2\le r\le4\) metres and \(0\le\theta\le\pi/3\), with \(\rho(r)=\kappa r\) and \(\kappa=1\ \mathrm{kg/m^3}\). The angular width is \(\pi/3\). The area is

$$ A=\frac{\pi/3}{2}(4^2-2^2)=2\pi\ \mathrm{m^2}. $$

For mass, \(\int_2^4r^2\,dr=(64-8)/3=56/3\ \mathrm{m^3}\), giving

$$ M=\frac{\pi}{3}(1\ \mathrm{kg/m^3}) \left(\frac{56}{3}\ \mathrm{m^3}\right) =\frac{56\pi}{9}\ \mathrm{kg}. $$

The radial factor for each first moment is \(\kappa\int_2^4r^3\,dr=(256-16)/4\ \mathrm{kg\,m}=60\ \mathrm{kg\,m}\). The angular factors are \(\int_0^{\pi/3}\cos\theta\,d\theta=\sqrt3/2\) and \(\int_0^{\pi/3}\sin\theta\,d\theta=1/2\). Therefore

$$ \bar{x}=\frac{30\sqrt3}{56\pi/9} =\frac{135\sqrt3}{28\pi}\ \mathrm m, \qquad \bar{y}=\frac{30}{56\pi/9} =\frac{135}{28\pi}\ \mathrm m. $$

For inertia, \(\int_2^4r^4\,dr=(1024-32)/5=992/5\ \mathrm{m^5}\). Thus

$$ \begin{aligned} I_O&=\frac{\pi}{3}(1\ \mathrm{kg/m^3}) \left(\frac{992}{5}\ \mathrm{m^5}\right) =\frac{992\pi}{15}\ \mathrm{kg\,m^2},\\ k_O&=\sqrt{\frac{992\pi/15}{56\pi/9}}\ \mathrm m =\sqrt{\frac{372}{35}}\ \mathrm m. \end{aligned} $$

Checks. Multiplication by the mass gives \(M\bar{x}=30\sqrt3\ \mathrm{kg\,m}\) and \(M\bar{y}=30\ \mathrm{kg\,m}\), the two first moments used above. Also, \(Mk_O^2=(56\pi/9)(372/35)=992\pi/15\ \mathrm{kg\,m^2}\), as required.

A Common Error: Dropping the Polar Area Factor

A frequent mistake is to use \(dr\,d\theta\) as the polar area element instead of \(r\,dr\,d\theta\). For the quarter annulus in the first example, omitting \(r\) would give the incorrect area

$$ \int_0^{\pi/2}\int_1^3 1\,dr\,d\theta =\frac{\pi}{2}(3-1)=\pi\ \mathrm{m^2}, $$

rather than \(2\pi\ \mathrm{m^2}\). The correction is to include the arc-length factor \(r\):

$$ \int_0^{\pi/2}\int_1^3 r\,dr\,d\theta =\frac{\pi}{2}\left(\frac{3^2-1^2}{2}\right) =2\pi\ \mathrm{m^2}. $$

The same area factor must be included in mass and inertia. For example, the mass integrand is \(\rho(r)r\), and the polar-inertia integrand is \(r^2\rho(r)r\). Writing the mass element first as \(\rho\,dA=\rho(r)r\,dr\,d\theta\) helps keep the extra factor visible.

Putting the Results in Context

These quantities answer related but distinct questions. Area depends only on the sector's geometry. Mass depends on both geometry and density. The centroid describes the balance point, while the polar moment measures how mass is distributed with respect to distance from the origin. The radius of gyration expresses that same inertia as though the total mass were concentrated at one distance: \(I_O=Mk_O^2\).

For any such calculation, identify the bounds and density first, then use the correct area element and keep the reference point for inertia explicit. Check centroid coordinates by recovering the first moments, and check the radius of gyration by recovering the polar moment. These checks catch many errors in factors, bounds, and units.

Check Your Understanding

Use the definitions and checks above to answer the following questions.

  1. What factor must appear in the polar area element, and what geometric feature accounts for it?
  2. For a radial density, why can each centroid first-moment integral be separated into radial and angular factors?
  3. For a sector symmetric about the \(x\)-axis with radial density, what does symmetry imply about \(\bar{y}\)?
  4. How can you check a computed centroid using its first moments?
  5. What equation relating \(I_O\), \(M\), and \(k_O\) should hold for the radius of gyration about the origin?

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