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Polar Double Integrals · Tutorial 572 of 999

Common Errors in Polar Double Integrals

Check the area factor, coordinate convention, radial order, and angular interval before trusting a polar double integral.

Intermediate 9 min read

What You'll Learn

  • Explain why the polar area element includes the factor r.
  • Keep the angle convention consistent with the coordinate formulas and bounds.
  • Check that radial bounds run from the inner curve to the outer curve.
  • Choose an angular interval that covers the intended region exactly once.
  • Diagnose common mistakes by comparing the integral with the geometry.

Introduction

Polar coordinates can make a double integral much easier to set up, especially when the region has circular boundaries or radial symmetry. But a plausible-looking polar integral can still be wrong because of a small setup error.

Four errors deserve particular attention: leaving out the factor \(r\) in the area element, mixing angle conventions, reversing the radial bounds, and integrating over an angular interval that does not describe the intended region. Each can change the value, sometimes without producing an obviously unreasonable answer.

Reliable check: Before evaluating a polar double integral, verify the area element, the coordinate convention, the radial order, and the angular coverage. These checks concern the region and the coordinate transformation, not the difficulty of the antiderivative.

Why the Factor \(r\) Is Required

Use the standard polar convention throughout this tutorial: \(r\ge 0\), \(\theta\) is measured counterclockwise from the positive \(x\)-axis, and

$$ x=r\cos\theta,\qquad y=r\sin\theta. $$

The factor \(r\) in the area element is not optional and is not part of the integrand. It accounts for how polar coordinates scale area: a small change in angle sweeps out a wider arc when it is farther from the origin.

Derivation: The transformation from \((r,\theta)\) to \((x,y)\) has derivative matrix
$$ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial\theta}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\theta} \end{pmatrix} = \begin{pmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{pmatrix}. $$

Its determinant is

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

For a change of variables in a double integral, area is scaled by the absolute value of this determinant. Since \(r\ge 0\), the scale factor is \(r\), so \(dA=r\,dr\,d\theta\). At \(r=0\) the transformation is singular, but that single point has zero area and does not affect the integral.

Formula: If a region is described once by \(\alpha\le\theta\le\beta\) and \(a(\theta)\le r\le b(\theta)\), with \(0\le a(\theta)\le b(\theta)\), then
$$ \iint_R f(x,y)\,dA = \int_\alpha^\beta\int_{a(\theta)}^{b(\theta)} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$

The formula requires bounds that describe the intended region, with the radial coordinate running from its inner boundary to its outer boundary. If the radial boundary curves exchange order as \(\theta\) changes, the angular range must be split where needed.

Worked Example: Omitting the Area Factor

Worked Example: Omitting the Area Factor

Find the area of the disk \(x^2+y^2\le 9\) using polar coordinates. The disk is described by \(0\le\theta\le2\pi\) and \(0\le r\le3\). For area, the integrand is \(1\).

The incorrect setup omits the area factor and uses \(dA=dr\,d\theta\):

$$ \int_0^{2\pi}\int_0^3 1\,dr\,d\theta = \int_0^{2\pi}3\,d\theta =6\pi. $$

This is not the disk's area. The correct setup includes \(r\):

$$ \int_0^{2\pi}\int_0^3 r\,dr\,d\theta = \int_0^{2\pi}\left[\frac{r^2}{2}\right]_0^3d\theta = \int_0^{2\pi}\frac{9}{2}\,d\theta = \frac{9}{2}(2\pi) =9\pi. $$

As a check, the disk has radius \(3\), so its area is \(\pi(3)^2=9\pi\). The corrected integral agrees; the setup without \(r\) does not.

Keep the Angle Convention Consistent

With the convention used here, \(x=r\cos\theta\) and \(y=r\sin\theta\). The formulas for the integrand and the angular bounds must use the same convention. If instead an angle is measured from the positive \(y\)-axis, the coordinate formulas change and the bounds must be expressed in that new angle.

A common mistake is to keep the bounds from one convention but use a coordinate formula from another. The result may even have the right sign, so checking only whether the answer is positive or negative is not always enough.

Worked Example: Mixing Coordinate Formulas

Integrate \(x\) over the sector \(0\le r\le1\), \(0\le\theta\le\pi/3\), where \(\theta\) is measured from the positive \(x\)-axis. The correct conversion is \(x=r\cos\theta\), and the area element is \(r\,dr\,d\theta\). Thus

$$ \begin{aligned} \iint_R x\,dA &=\int_0^{\pi/3}\int_0^1(r\cos\theta)r\,dr\,d\theta\\ &=\int_0^{\pi/3}\cos\theta\left[\frac{r^3}{3}\right]_0^1d\theta\\ &=\frac13\int_0^{\pi/3}\cos\theta\,d\theta\\ &=\frac13[\sin\theta]_0^{\pi/3} =\frac{\sqrt3}{6}. \end{aligned} $$

If one mistakenly uses \(x=r\sin\theta\) while retaining these same bounds, the calculation instead gives

$$ \int_0^{\pi/3}\int_0^1(r\sin\theta)r\,dr\,d\theta = \frac13[-\cos\theta]_0^{\pi/3} = \frac13\left(1-\frac12\right) =\frac16. $$

This is the wrong coordinate conversion for the stated convention. For a check, every point in this sector has \(x\ge0\), and the value must match the integral of \(r\cos\theta\), not \(r\sin\theta\). The correct value is \(\sqrt3/6\).

Order Radial Bounds from Inner to Outer

For a fixed angle, \(r\) measures distance outward from the origin. The lower radial bound must therefore be the nearer boundary, and the upper radial bound the farther boundary. Reversing them reverses the sign of the radial integral; it does not describe the same region with a different order.

Worked Example: Reversing Radial Bounds

Find the area of the quarter-annulus \(1\le r\le2\), \(0\le\theta\le\pi/2\). The inner radius is \(1\), the outer radius is \(2\), and the integrand for area is \(1\). The correct integral is

$$ \begin{aligned} \text{Area} &=\int_0^{\pi/2}\int_1^2 r\,dr\,d\theta\\ &=\int_0^{\pi/2}\left[\frac{r^2}{2}\right]_1^2d\theta\\ &=\int_0^{\pi/2}\left(2-\frac12\right)d\theta\\ &=\frac32\cdot\frac{\pi}{2} =\frac{3\pi}{4}. \end{aligned} $$

If the radial bounds are mistakenly written from \(2\) down to \(1\), then

$$ \int_0^{\pi/2}\int_2^1 r\,dr\,d\theta = \int_0^{\pi/2}\left(\frac12-2\right)d\theta =-\frac{3\pi}{4}. $$

A negative value cannot be the area of this region. The sign flags the reversed order. The corrected value \(3\pi/4\) also agrees with one quarter of the annulus area \(\pi(2^2-1^2)=3\pi\).

When the boundaries are curves depending on \(\theta\), compare them at each angle rather than assuming their order from their names or appearance in an equation. If their order changes, divide the angular interval at the crossing and use the correct inner and outer bounds on each piece.

Choose an Angular Interval That Covers the Region

Polar angles are periodic: angles differing by \(2\pi\) point in the same direction. This makes it possible to write equivalent descriptions of a region, but it also makes incorrect intervals easy to overlook. The interval must cover the intended directions, and the region should be counted exactly once, apart from shared boundary rays, which have zero area.

Worked Example: Using the Wrong Angular Interval

Integrate \(x\) over the left-hand sector of radius \(2\), from \(\theta=3\pi/4\) to \(\theta=5\pi/4\). These angles lie in the left half-plane, where \(x\le0\). Using \(x=r\cos\theta\),

$$ \begin{aligned} \iint_R x\,dA &=\int_{3\pi/4}^{5\pi/4}\int_0^2(r\cos\theta)r\,dr\,d\theta\\ &=\left[\frac{r^3}{3}\right]_0^2 \int_{3\pi/4}^{5\pi/4}\cos\theta\,d\theta\\ &=\frac83[\sin\theta]_{3\pi/4}^{5\pi/4}\\ &=\frac83\left(-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\right) =-\frac{8\sqrt2}{3}. \end{aligned} $$

A mistaken interval \(-\pi/4\le\theta\le\pi/4\) describes a right-hand sector, not this left-hand one. Using that interval would give

$$ \frac83[\sin\theta]_{-\pi/4}^{\pi/4} = \frac83\left(\frac{\sqrt2}{2}-\left(-\frac{\sqrt2}{2}\right)\right) =\frac{8\sqrt2}{3}. $$

The sign check catches the mistake: \(x\) is nonpositive throughout the intended left-hand sector, so its integral cannot be positive. The corrected negative value is consistent with the region.

A Practical Error Check

Before evaluating, reconstruct the meaning of each part of the integral. The following checks are particularly useful when an answer seems numerically reasonable but the setup may be wrong.

1
Check the area element.
For standard polar coordinates with \(r\ge0\), the element is \(r\,dr\,d\theta\). Include that factor once, outside the converted integrand.
2
Check the coordinate conversion.
With angles measured from the positive \(x\)-axis, replace \(x\) by \(r\cos\theta\) and \(y\) by \(r\sin\theta\), and make sure the bounds use that same angle.
3
Check radial order.
At each included angle, verify that the lower radius is no larger than the upper radius and that they trace the inner and outer boundaries.
4
Check angular coverage.
Confirm that the interval covers the region's directions once, without replacing the intended sector by a different one or counting the region multiple times.

These checks can expose errors before any integration is done. A negative area, an answer with a sign inconsistent with the integrand, or a result that fails a simple geometric comparison is a reason to revisit the setup rather than merely repeat the arithmetic.

Check Your Understanding

For each question, focus first on the region and the coordinate setup.

  1. Why does \(dA\) equal \(r\,dr\,d\theta\) in standard polar coordinates rather than \(dr\,d\theta\)?
  2. For an angle measured counterclockwise from the positive \(x\)-axis, what are the polar formulas for \(x\) and \(y\)?
  3. What does reversing the radial bounds do to an iterated integral, and why does this not describe the same area?
  4. What geometric check can help identify an incorrect angular interval when integrating \(x\)?
  5. If two radial boundary curves exchange order over an angular range, what must be done before writing one polar integral over that range?

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