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Polar Double Integrals · Tutorial 554 of 999

Full Disks as Polar Integration Regions

Convert a full disk centered at the origin into polar bounds, include the area factor, and evaluate representative double integrals.

Intermediate 9 min read

What You'll Learn

  • Identify the polar bounds for a disk of radius \(R\) centered at the origin.
  • Explain why the angular interval must cover a full revolution.
  • Include the polar area factor when converting a double integral.
  • Evaluate integrals over full disks by integrating first with respect to radius.
  • Check bounds and calculations using area, antiderivatives, and units.

From a Disk to Polar Bounds

A disk centered at the origin has a simple description in Cartesian coordinates: its points lie no farther than a fixed distance \(R\) from the origin. In polar coordinates, that distance is the radial coordinate \(r\). The direction angle must make one complete revolution to include the entire disk.

This makes a full disk one of the most direct regions to describe with polar bounds. The radial bound is constant, and the angular interval has length \(2\pi\). To convert an integral, however, the bounds are only part of the calculation: the polar area factor \(r\) must also be included.

Definition: For \(R>0\), the full disk of radius \(R\) centered at the origin is $$ D_R=\{(x,y):x^2+y^2\le R^2\}. $$ In polar coordinates, the disk is described by \(0\le r\le R\) and \(0\le\theta\le 2\pi\).

Here \(r\) is the distance from the origin, so \(r\) cannot be negative. At each distance from \(0\) through \(R\), the angle ranges through a full turn. The choice \(0\le\theta\le2\pi\) includes every direction; the origin is represented by every angle when \(r=0\), but this duplication at a single point does not affect a double integral.

The Polar Integral over a Full Disk

In polar coordinates, \(x=r\cos\theta\) and \(y=r\sin\theta\). The area element is \(dA=r\,dr\,d\theta\). Consequently, if \(f\) is continuous on \(D_R\), its integral over the disk is

$$ \iint_{D_R} f(x,y)\,dA = \int_0^{2\pi}\int_0^R f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$

The \(r\) at the end of the integrand is the area factor; it is separate from any \(r\) that appears when \(x\) or \(y\) is substituted into \(f\). The inner integral is with respect to \(r\), from the center out to the edge. The outer integral is with respect to \(\theta\), over the complete range of directions.

Key takeaway: For a full disk centered at the origin, use \(0\le r\le R\), \(0\le\theta\le2\pi\), and multiply the transformed integrand by \(r\) for the area element.

Why These Bounds and the Area Factor Work

Every point of the disk other than the origin has a distance \(r\) between \(0\) and \(R\), and an angle that specifies its direction. Letting \(\theta\) range from \(0\) to \(2\pi\) sweeps through every direction once, apart from the repeated endpoint direction. Thus the two bounds describe the entire disk.

The area factor can be understood by considering a small polar patch. A small change \(dr\) gives it approximately radial thickness \(dr\). A small change \(d\theta\) gives it arc length approximately \(r\,d\theta\), because the arc is farther and longer at larger \(r\). Its area is therefore approximately \(r\,dr\,d\theta\). This is why integrating over polar bounds requires the factor \(r\), not just \(dr\,d\theta\).

Derivation: The polar coordinate map is \(x=r\cos\theta,\ y=r\sin\theta\). Its Jacobian determinant is $$ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta+r\sin^2\theta = r. $$ The change-of-variables formula therefore gives \(dA=|r|\,dr\,d\theta\). On the disk’s radial range \(r\ge0\), this is \(dA=r\,dr\,d\theta\), yielding the stated integral formula.

In practical work, the order of operations is: substitute \(r\cos\theta\) for \(x\) and \(r\sin\theta\) for \(y\), include the extra area factor \(r\), and then integrate over the disk’s polar bounds. The following examples illustrate these steps.

Worked Example: Find the Area of a Disk of Radius 3

Area is the integral of the constant function \(1\). For \(D_3\), the radial bound is \(0\le r\le3\), and the full angular interval is \(0\le\theta\le2\pi\). Include the area factor:

$$ \begin{aligned} \iint_{D_3}1\,dA &=\int_0^{2\pi}\int_0^3 r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^2}{2}\right]_0^3\,d\theta\\ &=\int_0^{2\pi}\frac{9}{2}\,d\theta\\ &=\frac{9}{2}[\theta]_0^{2\pi}\\ &=\frac{9}{2}(2\pi-0)\\ &=9\pi. \end{aligned} $$

As a check, the geometric area of a disk of radius \(3\) is \(\pi(3)^2=9\pi\), matching the integral.

Evaluating an Integral with Cartesian Variables

The same bounds apply when the original integrand is not written in polar coordinates. For example, \(2+x\) becomes \(2+r\cos\theta\). After substitution, remember that the area factor multiplies the entire transformed function.

Worked Example: Integrate \(2+x\) over a Disk of Radius 2

On \(D_2\), substitute \(x=r\cos\theta\) and include the area factor \(r\):

$$ \begin{aligned} \iint_{D_2}(2+x)\,dA &=\int_0^{2\pi}\int_0^2(2+r\cos\theta)r\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2(2r+r^2\cos\theta)\,dr\,d\theta. \end{aligned} $$

Integrate with respect to \(r\) first. The angle is constant during this inner integration:

$$ \begin{aligned} \int_0^2(2r+r^2\cos\theta)\,dr &=\left[r^2+\frac{r^3}{3}\cos\theta\right]_0^2\\ &=4+\frac{8}{3}\cos\theta. \end{aligned} $$

Now integrate over the full angular interval:

$$ \begin{aligned} \int_0^{2\pi}\left(4+\frac{8}{3}\cos\theta\right)\,d\theta &=\left[4\theta+\frac{8}{3}\sin\theta\right]_0^{2\pi}\\ &=8\pi+\frac{8}{3}(0-0)\\ &=8\pi. \end{aligned} $$

To check, the constant part \(2\) contributes \(2\) times the disk’s area, or \(2(4\pi)=8\pi\); the \(x\)-part integrates to zero over this centered disk. The result \(8\pi\) is consistent.

A Further Example with a Nonconstant Integrand

When the integrand contains both \(x\) and \(y\), convert each variable separately. This example keeps the full-disk bounds fixed while showing how products of the polar substitutions appear inside the integral.

Worked Example: Integrate \(xy+1\) over a Disk of Radius 1

Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\). The product \(xy\) becomes \(r^2\cos\theta\sin\theta\), and the area factor supplies one additional \(r\):

$$ \begin{aligned} \iint_{D_1}(xy+1)\,dA &=\int_0^{2\pi}\int_0^1 \left(r^2\cos\theta\sin\theta+1\right)r\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^1 \left(r^3\cos\theta\sin\theta+r\right)\,dr\,d\theta. \end{aligned} $$

Integrating in \(r\) gives

$$ \begin{aligned} \int_0^1\left(r^3\cos\theta\sin\theta+r\right)\,dr &=\left[\frac{r^4}{4}\cos\theta\sin\theta+\frac{r^2}{2}\right]_0^1\\ &=\frac14\cos\theta\sin\theta+\frac12. \end{aligned} $$

Integrate this expression over \(\theta\). Since \(\frac{d}{d\theta}(\sin^2\theta)=2\sin\theta\cos\theta\), the angular integral is

$$ \begin{aligned} \int_0^{2\pi}\left(\frac14\cos\theta\sin\theta+\frac12\right)\,d\theta &=\left[\frac18\sin^2\theta+\frac{\theta}{2}\right]_0^{2\pi}\\ &=\left(0+\pi\right)-\left(0+0\right)\\ &=\pi. \end{aligned} $$

For a check, the constant \(1\) contributes the area \(\pi\). The \(xy\) contribution is zero: replacing \(x\) by \(-x\) preserves the disk but changes the sign of \(xy\), so contributions cancel in pairs. Thus the computed result \(\pi\) is consistent.

Common Mistake: Omitting the Area Factor

A frequent error is to substitute polar expressions into the function and use the radial and angular bounds, but forget the factor \(r\). That factor is not optional: it accounts for how the area of a polar patch depends on its distance from the origin.

Worked Example: Correcting an Omitted Factor

Suppose the area of \(D_3\) is incorrectly set up as \(\int_0^{2\pi}\int_0^3 1\,dr\,d\theta\), with the factor \(r\) omitted. This gives

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

which is not the area of the disk. The corrected setup includes \(r\):

$$ \int_0^{2\pi}\int_0^3 r\,dr\,d\theta = \int_0^{2\pi}\frac{9}{2}\,d\theta = 9\pi. $$

This agrees with \(\pi(3)^2=9\pi\). The missing-factor calculation incorrectly treats equal changes in angle as having equal area at every radius; in fact, the arc length and corresponding area increase with \(r\).

A Reliable Setup Check

Before evaluating a full-disk integral, check that each part of the setup has the correct role. The bounds describe the region, the substitutions describe the function, and the area factor adjusts the area element.

  • Region: Is the disk centered at the origin with radius \(R\), so that \(0\le r\le R\)?
  • Full revolution: Does the angle range through \(2\pi\), such as \(0\le\theta\le2\pi\)?
  • Function: Has every \(x\) and \(y\) in the integrand been replaced by \(r\cos\theta\) and \(r\sin\theta\), respectively?
  • Area element: Has the whole transformed integrand been multiplied by \(r\)?
  • Evaluation: Have the radial limits been applied before integrating the resulting expression over the angular limits?

A quick check is to integrate \(1\) using the proposed bounds. If the result is not \(\pi R^2\), the setup or calculation needs attention. This test catches a missing area factor and many incorrect radial bounds.

Check Your Understanding

Use the full-disk bounds and the polar area factor in each question.

  1. Write the polar bounds for the disk \(x^2+y^2\le16\).
  2. Set up, but do not evaluate, the integral of \(x+y\) over \(D_R\) in polar coordinates.
  3. What factor must multiply \(dr\,d\theta\) when converting area to polar coordinates, and why?
  4. Set up and evaluate the area integral for a disk of radius \(4\).
  5. Explain why the angular interval must cover a full revolution when the region is a full disk.

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