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Polar Double Integrals · Tutorial 574 of 999

Worked Evaluations on Annular Sectors

Learn to set up and evaluate annular-sector integrals accurately, including how the polar area factor affects the calculation.

Intermediate 10 min read

What You'll Learn

  • Identify the radial and angular bounds of a concentric annular sector.
  • Convert a Cartesian area integral into polar form.
  • Explain why the polar area element includes a factor of the radius.
  • Evaluate integrals with radial and angular dependence in either order.
  • Detect and correct the common error of omitting the area factor.

Introduction

An annular sector is the part of a ring between two concentric circles and two radial boundary lines. Integrals over these regions are especially useful for practicing a central feature of polar double integrals: the integrand and the area element must both be expressed in polar coordinates.

The circular boundaries give constant radial bounds, while the rays give constant angular bounds. Once these bounds are identified, the main calculation is to integrate the polar form of the function with the correct area factor. The order of integration can be chosen to make that calculation simpler.

Describing an Annular Sector

Let \(a\) and \(b\) be constants with \(0\leq a<b\), and let \(\alpha<\beta\) be angular bounds. The corresponding annular sector consists of points whose polar coordinates satisfy \(a\leq r\leq b\) and \(\alpha\leq\theta\leq\beta\). Here \(a\) is the inner radius, \(b\) is the outer radius, and the two angles specify the radial sides of the sector.

Definition: An annular sector with radial bounds \(a\leq r\leq b\) and angular bounds \(\alpha\leq\theta\leq\beta\) is described in polar coordinates by \(a\leq r\leq b\), \(\alpha\leq\theta\leq\beta\), where \(0\leq a<b\) and \(\beta-\alpha\) is the angular width of the sector.

The angles must describe the intended sector without tracing it more than once. For example, an angular interval of width less than \(2\pi\) describes a sector that does not wrap around the origin multiple times. The boundary curves themselves do not affect the value of an ordinary double integral, so it is immaterial whether the inequalities on the boundary are written as strict or non-strict.

In Cartesian coordinates, polar coordinates are given by \(x=r\cos\theta\) and \(y=r\sin\theta\). The area element is not simply \(dr\,d\theta\): it is \(dA=r\,dr\,d\theta\). That factor is essential in every polar double integral.

Polar integral over an annular sector: If \(f\) is continuous on the sector, then $$ \iint_D f(x,y)\,dA = \int_{\alpha}^{\beta}\int_a^b f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$ The same integral can be evaluated in the reverse order: $$ \iint_D f(x,y)\,dA = \int_a^b\int_{\alpha}^{\beta} f(r\cos\theta,r\sin\theta)\,r\,d\theta\,dr. $$

Why the Area Element Contains \(r\)

The factor \(r\) comes from how a small polar-coordinate rectangle maps to the plane. A small change \(dr\) in radius gives a radial length of approximately \(dr\). A small change \(d\theta\) at radius \(r\) gives an arc length of approximately \(r\,d\theta\). The corresponding small area is therefore approximately \(dr\cdot r\,d\theta\).

This factor can also be derived from the Jacobian of the coordinate transformation. For \(x=r\cos\theta\) and \(y=r\sin\theta\), the determinant of the matrix of partial derivatives is

$$ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta+r\sin^2\theta = r. $$

Since \(r\geq 0\), the absolute value of this determinant is \(r\), so the change-of-variables formula gives \(dA=r\,dr\,d\theta\). Thus, even if the function being integrated has no visible factor of \(r\), the area element contributes one.

A Reliable Setup

For an annular sector bounded by concentric circles and rays, the limits in \(r\) and \(\theta\) are constant. Before integrating, separate the task into three decisions:

1
Identify the radii.
The inner circle supplies the lower radial bound \(a\); the outer circle supplies the upper radial bound \(b\).
2
Identify the rays.
Choose angular limits \(\alpha\) and \(\beta\) that cover the intended sector once.
3
Transform the integrand and area.
Replace \(x,y\) by \(r\cos\theta,r\sin\theta\), then include the factor \(r\) from \(dA\).

After setting up the integral, choose the order that makes the inner integration easiest. For continuous integrands on a bounded annular sector, both orders give the same value. The limits do not change when the order is reversed because the region is described by independent constant bounds for \(r\) and \(\theta\).

Worked Examples

Worked Example: A Product of Radial and Angular Factors

Evaluate \(\iint_D r^2(1+\sin\theta)\,dA\) over the sector \(1\leq r\leq3\), \(0\leq\theta\leq\frac{\pi}{2}\).

The radial bounds are \(1\) and \(3\), and the angular bounds are \(0\) and \(\frac{\pi}{2}\). Multiplying the integrand by the polar area factor gives

$$ I=\int_0^{\pi/2}\int_1^3 r^2(1+\sin\theta)\,r\,dr\,d\theta =\int_0^{\pi/2}\int_1^3 r^3(1+\sin\theta)\,dr\,d\theta. $$

Integrate first with respect to \(r\). Since \(1+\sin\theta\) is constant during that integration,

$$ \int_1^3 r^3(1+\sin\theta)\,dr =(1+\sin\theta)\left[\frac{r^4}{4}\right]_1^3 =(1+\sin\theta)\frac{81-1}{4} =20(1+\sin\theta). $$

Now integrate with respect to \(\theta\):

$$ I=20\int_0^{\pi/2}(1+\sin\theta)\,d\theta =20\left[\theta-\cos\theta\right]_0^{\pi/2} =20\left(\frac{\pi}{2}+1\right). $$

As a check, integrating in the other order gives the angular integral \(\int_0^{\pi/2}(1+\sin\theta)\,d\theta=\frac{\pi}{2}+1\), followed by the radial integral \(\int_1^3r^3\,dr=20\). Their product is again \(20(\frac{\pi}{2}+1)\).

Worked Example: Choosing the Order for a Mixed Integrand

Evaluate \(\iint_D r\theta(1+r)\,dA\) on \(1\leq r\leq2\), \(0\leq\theta\leq\frac{\pi}{3}\).

The integrand depends on both variables. Including the area factor first gives

$$ I=\int_0^{\pi/3}\int_1^2 r\theta(1+r)\,r\,dr\,d\theta =\int_0^{\pi/3}\int_1^2\theta(r^2+r^3)\,dr\,d\theta. $$

Integrate with respect to \(r\):

$$ \int_1^2\theta(r^2+r^3)\,dr =\theta\left[\frac{r^3}{3}+\frac{r^4}{4}\right]_1^2 =\theta\left(\frac{8}{3}+4-\frac{1}{3}-\frac{1}{4}\right) =\frac{73}{12}\theta. $$

Then integrate with respect to \(\theta\):

$$ I=\frac{73}{12}\int_0^{\pi/3}\theta\,d\theta =\frac{73}{12}\left[\frac{\theta^2}{2}\right]_0^{\pi/3} =\frac{73}{12}\cdot\frac{\pi^2}{18} =\frac{73\pi^2}{216}. $$

For a check, reverse the order. The angular integral is \(\int_0^{\pi/3}\theta\,d\theta=\frac{\pi^2}{18}\), and the radial integral is \(\int_1^2(r^2+r^3)\,dr=\frac{73}{12}\), as computed above. Their product is \(\frac{73\pi^2}{216}\), matching the result.

Worked Example: An Angular Interval Away from Zero

Evaluate \(\iint_D (r\sin\theta+2\cos\theta)\,dA\) on \(2\leq r\leq4\), \(\frac{\pi}{6}\leq\theta\leq\frac{\pi}{2}\).

The angular interval starts at \(\frac{\pi}{6}\), not at zero. The polar setup, including the area factor, is

$$ I=\int_{\pi/6}^{\pi/2}\int_2^4 (r\sin\theta+2\cos\theta)r\,dr\,d\theta =\int_{\pi/6}^{\pi/2}\int_2^4 (r^2\sin\theta+2r\cos\theta)\,dr\,d\theta. $$

Integrating with respect to \(r\), the two terms give

$$ \int_2^4r^2\,dr=\left[\frac{r^3}{3}\right]_2^4 =\frac{64-8}{3}=\frac{56}{3}, \qquad \int_2^4 2r\,dr=[r^2]_2^4=16-4=12. $$

Therefore,

$$ I=\frac{56}{3}\int_{\pi/6}^{\pi/2}\sin\theta\,d\theta +12\int_{\pi/6}^{\pi/2}\cos\theta\,d\theta. $$

Evaluate each angular integral using its antiderivative:

$$ \int_{\pi/6}^{\pi/2}\sin\theta\,d\theta =[-\cos\theta]_{\pi/6}^{\pi/2} =\frac{\sqrt{3}}{2}, \qquad \int_{\pi/6}^{\pi/2}\cos\theta\,d\theta =[\sin\theta]_{\pi/6}^{\pi/2} =\frac{1}{2}. $$

Thus,

$$ I=\frac{56}{3}\cdot\frac{\sqrt{3}}{2} +12\cdot\frac{1}{2} =\frac{28\sqrt{3}}{3}+6. $$

Checking in the reverse order, the angular integral at a fixed \(r\) is \(r^2\frac{\sqrt{3}}{2}+r\). Its radial integral is \(\frac{\sqrt{3}}{2}\cdot\frac{56}{3}+6=\frac{28\sqrt{3}}{3}+6\), which confirms the answer.

A Common Error: Omitting the Area Factor

A frequent mistake is to substitute polar expressions into the function but leave the area element as \(dr\,d\theta\). In the first example, that error would produce

$$ \int_0^{\pi/2}\int_1^3r^2(1+\sin\theta)\,dr\,d\theta = \frac{26}{3}\left(\frac{\pi}{2}+1\right), $$

because \(\int_1^3r^2\,dr=\frac{27-1}{3}=\frac{26}{3}\). This is not the desired integral over area: the calculation has omitted the Jacobian factor \(r\). The correct integrand for integration with respect to \(dr\,d\theta\) is \(r^2(1+\sin\theta)\cdot r=r^3(1+\sin\theta)\), giving \(20(\frac{\pi}{2}+1)\). A useful habit is to write \(dA=r\,dr\,d\theta\) explicitly before simplifying the integrand.

Why the Setup Matters

The bounds and the area factor represent different parts of the geometry. The angular and radial bounds describe which points belong to the annular sector. The factor \(r\) accounts for how area changes with radius. Neither can replace the other.

When the integrand is a sum of terms with different radial and angular behavior, it can help to integrate first with respect to the variable that makes the inner antiderivative simplest. For an annular sector the limits remain constant in either order, so the choice is guided by the integrand rather than by changing boundary descriptions. Keeping the region, transformed integrand, and area element distinct makes the setup easier to check and the evaluation more reliable.

Check Your Understanding

For each question, write the polar integral carefully before evaluating it.

  1. For \(2\leq r\leq5\) and \(0\leq\theta\leq\frac{\pi}{4}\), what are the inner and outer radii and the angular width?
  2. Set up, but do not evaluate, the integral of \(r\cos\theta\) over \(1\leq r\leq3\), \(\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2}\).
  3. Why does a polar double integral include a factor of \(r\), even when the function being integrated does not contain \(r\)?
  4. For a continuous integrand on a bounded annular sector, what permits the order of integration to be reversed?
  5. Identify the missing factor in the incorrect setup \(\int_{\alpha}^{\beta}\int_a^b f(r\cos\theta,r\sin\theta)\,dr\,d\theta\).

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