Regions Between Two Polar Curves

Reading a Region Between Polar Curves

A region between two polar curves is described by an angular range and, for each angle, a radial interval. At a fixed angle, moving outward from the origin first reaches the inner boundary and then the outer boundary. Identifying which curve plays each role is essential: reversing them gives incorrect bounds or an empty radial interval.

Suppose the two curves are \(r=g(\theta)\) and \(r=f(\theta)\), and consider angles from \(\alpha\) to \(\beta\). If both radial values are nonnegative and \(g(\theta)\le f(\theta)\) throughout that interval, the region consists of the points whose angle lies in the interval and whose radius lies between the curves.

Definition: If \(0\le g(\theta)\le f(\theta)\) for every \(\theta\in[\alpha,\beta]\), the region between the curves on that angular interval is \[ \{(r,\theta):\alpha\le\theta\le\beta,\quad g(\theta)\le r\le f(\theta)\}. \] Here \(g\) is the radial inner boundary and \(f\) is the radial outer boundary.

“Inner” and “outer” refer to distance from the origin at the same angle. They do not mean lower and upper on the page. For example, the curve with the larger radial value is farther from the origin along that ray, even if its plotted point appears lower than another point.

From Radial Bounds to an Integral

In polar coordinates the area element is \(dA=r\,dr\,d\theta\). Therefore, for an integrable function \(H\) over the region above, its double integral is set up by integrating first along each radial interval and then over the angular interval.

\iint_R H(x,y)\,dA = \int_{\alpha}^{\beta}\int_{g(\theta)}^{f(\theta)} H(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

The factor \(r\) is part of the polar area element; it is not an additional boundary or part of the function \(H\). When finding area, set \(H=1\). The inner integral then measures the radial contribution at each angle.

\operatorname{Area}(R) = \int_{\alpha}^{\beta}\int_{g(\theta)}^{f(\theta)} r\,dr\,d\theta = \frac12\int_{\alpha}^{\beta}\bigl(f(\theta)^2-g(\theta)^2\bigr)\,d\theta.

Why the Area Formula Works

For each fixed angle \(\theta\), the radial coordinate ranges from \(g(\theta)\) to \(f(\theta)\). Integrating the area element \(r\,dr\,d\theta\) over that radial interval gives its contribution per angular increment \(d\theta\):

\int_{g(\theta)}^{f(\theta)}r\,dr = \left[\frac{r^2}{2}\right]_{g(\theta)}^{f(\theta)} = \frac12\bigl(f(\theta)^2-g(\theta)^2\bigr).

Integrating these contributions from \(\alpha\) to \(\beta\) gives the total area. This derivation also explains the subtraction: the radial slice contributes the area out to the outer curve, with the area out to the inner curve removed.

Determine Which Curve Is Outer

To compare two curves, subtract their radial values. On an interval where \(f(\theta)-g(\theta)\ge 0\), \(f\) is the outer curve and \(g\) is the inner curve. If the difference is negative, the roles are reversed. The curves meet wherever their radial values are equal, so solving \(f(\theta)=g(\theta)\) can identify angles where their ordering may change.

After finding any intersections in the angular range, check the sign of the difference on the interval of interest. A convenient test is to choose an angle strictly inside the interval and compare the two radial values there. The test is useful only when the curves do not cross elsewhere in that interval; the intersection equation and the sign comparison together establish the ordering.

Choose the angular interval.
Specify which rays are part of the region and identify the endpoints \(\alpha\) and \(\beta\).

Compare the radial values.
Find intersections if needed, then check which curve has the larger \(r\)-value over the interval.

Write the radial interval.
Place the smaller nonnegative radial value at the lower limit and the larger at the upper limit.

Include the polar area element.
For a double integral in polar coordinates, use \(r\,dr\,d\theta\).

Worked Examples

Worked Example: A Region with a Constant Outer Radius

Describe the region between \(r=4\) and \(r=2+\cos\theta\) for \(0\le\theta\le\frac{\pi}{2}\), and set up and evaluate its area.

On the angular interval, \(0\le\cos\theta\le1\). Thus \(2\le 2+\cos\theta\le3\), while the other curve has radius \(4\). Consequently, \(2+\cos\theta\) is the inner boundary and \(4\) is the outer boundary. The region is

0\le\theta\le\frac{\pi}{2}, \qquad 2+\cos\theta\le r\le4.

Using the polar area element, its area is

\begin{aligned} A &=\int_0^{\pi/2}\int_{2+\cos\theta}^{4}r\,dr\,d\theta\\ &=\frac12\int_0^{\pi/2}\left(16-(2+\cos\theta)^2\right)\,d\theta\\ &=\int_0^{\pi/2}\left(6-2\cos\theta-\frac12\cos^2\theta\right)\,d\theta\\ &=3\pi-2-\frac{\pi}{8}\\ &=\frac{23\pi}{8}-2. \end{aligned}

For the last step, \(\int_0^{\pi/2}\cos\theta\,d\theta=1\) and \(\int_0^{\pi/2}\cos^2\theta\,d\theta=\frac{\pi}{4}\). Checking the setup, the integrand before angular integration is \(\frac12(4^2-(2+\cos\theta)^2)\), exactly the difference of the squared outer and inner radii, divided by two.

Worked Example: Comparing Two Curves on a Quarter-Turn

For \(0\le\theta\le\frac{\pi}{2}\), determine the inner and outer curves \(r=2+\cos\theta\) and \(r=1+\sin\theta\), then set up and evaluate the area between them.

Compare the radii by subtraction:

(2+\cos\theta)-(1+\sin\theta)=1+\cos\theta-\sin\theta.

At \(\theta=0\), this difference is \(2\), so \(2+\cos\theta\) is the outer curve near that endpoint. The curves intersect when \(\cos\theta-\sin\theta=-1\). On this interval, that equality occurs at \(\theta=\frac{\pi}{2}\); for \(0\le\theta<\frac{\pi}{2}\), the difference is positive. Thus \(1+\sin\theta\) is inner and \(2+\cos\theta\) is outer throughout the interval, meeting at its endpoint.

The area is therefore set up as

\begin{aligned} A &=\int_0^{\pi/2}\int_{1+\sin\theta}^{2+\cos\theta}r\,dr\,d\theta\\ &=\frac12\int_0^{\pi/2}\left((2+\cos\theta)^2-(1+\sin\theta)^2\right)\,d\theta\\ &=\int_0^{\pi/2}\left(\frac32+2\cos\theta-\sin\theta +\frac12(\cos^2\theta-\sin^2\theta)\right)\,d\theta\\ &=\frac{3\pi}{4}+2-1+\frac12\int_0^{\pi/2}\cos(2\theta)\,d\theta\\ &=\frac{3\pi}{4}+1. \end{aligned}

The final trigonometric integral is zero because \(\int_0^{\pi/2}\cos(2\theta)\,d\theta=\frac12[\sin(2\theta)]_0^{\pi/2}=0\). The bounds check against the comparison: the lower limit is \(1+\sin\theta\), and the upper limit is \(2+\cos\theta\), with equality at \(\theta=\frac{\pi}{2}\).

Worked Example: A Nonconstant Integrand

Set up and evaluate \(\iint_R x\,dA\), where \(R\) is between \(r=2+\sin\theta\) and \(r=3\) for \(-\frac{\pi}{2}\le\theta\le0\).

On this interval, \(-1\le\sin\theta\le0\), so \(1\le2+\sin\theta\le2<3\). Thus \(2+\sin\theta\) is inner and \(3\) is outer. In polar coordinates, \(x=r\cos\theta\), and \(dA=r\,dr\,d\theta\). The integral is

\begin{aligned} \iint_R x\,dA &=\int_{-\pi/2}^{0}\int_{2+\sin\theta}^{3} (r\cos\theta)r\,dr\,d\theta\\ &=\int_{-\pi/2}^{0}\cos\theta \left[\frac{r^3}{3}\right]_{2+\sin\theta}^{3}\,d\theta\\ &=\frac13\int_{-\pi/2}^{0}\cos\theta \left(27-(2+\sin\theta)^3\right)\,d\theta. \end{aligned}

Let \(u=\sin\theta\), so \(du=\cos\theta\,d\theta\). The endpoints become \(u=-1\) and \(u=0\). Therefore,

\begin{aligned} \iint_R x\,dA &=\frac13\int_{-1}^{0}\left(27-(2+u)^3\right)\,du\\ &=\frac13\left[27u-\frac{(2+u)^4}{4}\right]_{-1}^{0}\\ &=\frac13\left(-4-\left(-27-\frac14\right)\right)\\ &=\frac13\cdot\frac{93}{4} =\frac{31}{4}. \end{aligned}

The radial antiderivative differentiates back to \(r^2\cos\theta\), the transformed integrand. The positive result is also consistent with the region: \(\cos\theta\ge0\), so \(x=r\cos\theta\ge0\) throughout it.

A Common Boundary Error

A frequent mistake is to put the curve with the smaller radius at the upper radial limit. For example, in the first worked example, reversing the limits would produce

\int_0^{\pi/2}\int_{4}^{2+\cos\theta}r\,dr\,d\theta = -\left(\frac{23\pi}{8}-2\right),

a negative value for an area. The sign exposes the reversed bounds, but the correct fix is not to take the absolute value and keep the mistaken limits. Write the radial interval in the geometric order, from the smaller radius to the larger radius: \(2+\cos\theta\le r\le4\).

Another check is to choose an angle in the interval and substitute it into both radial functions. If the proposed lower bound gives the larger \(r\)-value, the bounds have been reversed. Also verify that the radii are nonnegative for the chosen interval; an expression that is negative cannot serve as an ordinary radial distance in these bounds.

Why These Bounds Matter

The correct bounds describe exactly which points belong to the region. Choosing the wrong inner curve can exclude part of the intended region or include points beyond its boundary. Once the angular interval and radial ordering are correct, the integral follows systematically: substitute the polar form of the integrand, include the factor \(r\), and integrate over the specified radial and angular ranges.

If the curves exchange order within an angular range, a single pair of radial bounds cannot describe that entire range in the same order. The equality \(f(\theta)=g(\theta)\) identifies where an exchange could occur, and a sign comparison determines the ordering on either side. Each integral must use bounds that are correctly ordered on the angular interval it describes.

Check Your Understanding

For each question, identify the radial ordering before writing an integral.

On \(0\le\theta\le\frac{\pi}{2}\), which curve is outer: \(r=5\) or \(r=3+\sin\theta\)?
For a region with \(g(\theta)\le r\le f(\theta)\), write the polar double-integral setup for \(\iint_R H(x,y)\,dA\).
Why must the polar area element include a factor of \(r\)?
What equation identifies angles where two radial curves intersect?
Write the area integral for the region between \(r=1\) and \(r=2\cos\theta\) on \(0\le\theta\le\frac{\pi}{3}\), after checking which curve is outer.