Splitting Regions at Changes in Radial Bounds

Why a Polar Region May Need More Than One Integral

In polar coordinates, a region is often described by an angular interval and a radial interval. A single iterated integral is convenient when the same curve supplies the inner radial boundary throughout the angular range and the same curve supplies the outer boundary throughout.

But two candidate boundaries can cross as the angle changes. On one side of the crossing, one curve may be the outer boundary; on the other side, the other curve may be outer. The same issue can occur with inner boundaries. The region is still one region, but its description requires separate angular intervals.

The key is to find the angles where the boundary curves meet, then test which boundary applies on each resulting interval. Each piece can be integrated separately, and the pieces can be added because together they cover the original region without overlap in area.

Definition: A switching angle is an angle at which two candidate radial boundaries have equal radius and may exchange their roles as the inner or outer boundary of a region.

Polar Bounds and Switching Points

A polar description of a region uses an angular interval \(a\leq\theta\leq b\) and radial bounds \(r_{\mathrm{in}}(\theta)\leq r\leq r_{\mathrm{out}}(\theta)\). The radial bounds must be ordered: \(r_{\mathrm{in}}(\theta)\leq r_{\mathrm{out}}(\theta)\) throughout the interval. If a boundary changes at an angle inside the interval, the integral should be split there.

The polar area element is \(dA=r\,dr\,d\theta\). Thus, for an integrable function \(f\), the integral over a region with these bounds is

\iint_R f(x,y)\,dA = \int_a^b\int_{r_{\mathrm{in}}(\theta)}^{r_{\mathrm{out}}(\theta)} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

To locate a possible switch, set the candidate radial expressions equal and solve for \(\theta\). A solution is relevant only if it lies in the angular range of the region and the curves are both boundaries there. Then test one angle on each side of the solution to determine which expression is larger. For an outer boundary, use the smaller of two radial limits when the region must lie inside both; for an inner boundary, use the larger when the region must lie outside both.

Identify candidate boundaries.
Write the radial expressions that could supply the inner or outer limit.

Find their intersections.
Equate the radial expressions and retain the intersection angles within the region’s angular range.

Determine the ordering.
Compare the expressions on each interval between consecutive switching angles.

Integrate each piece.
Use the appropriate radial limits on each interval, then add the integrals.

Why Splitting and Adding Works

Suppose a region \(R\) is divided by the ray \(\theta=c\) into two pieces \(R_1\) and \(R_2\), with no overlap except possibly along their common boundary. A boundary curve or ray has area zero, so it contributes nothing to a double integral. Additivity of the double integral therefore gives

\iint_R f\,dA=\iint_{R_1}f\,dA+\iint_{R_2}f\,dA.

In polar coordinates, the same fact appears as splitting the angular integral at \(c\). To see this directly, suppose the radial bounds change there. On \(a\leq\theta\leq c\), let them be \(L_1(\theta)\) and \(U_1(\theta)\); on \(c\leq\theta\leq b\), let them be \(L_2(\theta)\) and \(U_2(\theta)\). Integrating first in \(r\) describes each angular slice, and then integrating over the two angular intervals accounts for every slice of the region:

\int_a^c\int_{L_1(\theta)}^{U_1(\theta)} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta + \int_c^b\int_{L_2(\theta)}^{U_2(\theta)} f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

The change in formula at \(c\) does not create a gap: both descriptions meet at the switching angle. Nor does it count a positive-area piece twice, because the two angular intervals are disjoint apart from their shared endpoint.

Worked Example: An Outer Boundary Switches

Worked Example: The Intersection of Two Circular Regions

Find the area in the first quadrant that lies inside both circles \(r=2\sin\theta\) and \(r=2\cos\theta\). In the first quadrant, being inside both means \(0\leq r\leq\min(2\sin\theta,2\cos\theta)\), for \(0\leq\theta\leq\pi/2\).

The switching angle satisfies

2\sin\theta=2\cos\theta \quad\Longrightarrow\quad \tan\theta=1 \quad\Longrightarrow\quad \theta=\frac{\pi}{4}.

For \(0\leq\theta\leq\pi/4\), \(\sin\theta\leq\cos\theta\), so the outer radius is \(2\sin\theta\). For \(\pi/4\leq\theta\leq\pi/2\), \(\cos\theta\leq\sin\theta\), so it is \(2\cos\theta\). The area is

\begin{aligned} A &=\int_0^{\pi/4}\int_0^{2\sin\theta}r\,dr\,d\theta +\int_{\pi/4}^{\pi/2}\int_0^{2\cos\theta}r\,dr\,d\theta\\ &=\int_0^{\pi/4}2\sin^2\theta\,d\theta +\int_{\pi/4}^{\pi/2}2\cos^2\theta\,d\theta\\ &=\left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{\pi/4} +\left[\theta+\frac{\sin(2\theta)}{2}\right]_{\pi/4}^{\pi/2}\\ &=\left(\frac{\pi}{4}-\frac12\right) +\left(\frac{\pi}{4}-\frac12\right) =\frac{\pi}{2}-1. \end{aligned}

Check: differentiating \(\theta-\sin(2\theta)/2\) gives \(1-\cos(2\theta)=2\sin^2\theta\); differentiating \(\theta+\sin(2\theta)/2\) gives \(1+\cos(2\theta)=2\cos^2\theta\). The bounds also agree at \(\theta=\pi/4\), where both radii equal \(\sqrt{2}\). Thus the two pieces meet at the switching ray.

When the Inner Boundary Switches

A switch need not occur at the outer radius. If a region lies outside two radial boundaries, both constraints must be satisfied, so the inner radius is the larger of the two. The curve supplying that larger value can change with angle. The same procedure applies: solve for equality, compare on either side, and split the integral.

Worked Example: A Changing Inner Radius

Consider the part of the first-quadrant disk \(0\leq r\leq2\) that lies outside both circles \(r=2\sin\theta\) and \(r=2\cos\theta\). Its inner radius is the larger of these two values. Find its area.

The curves intersect at \(\theta=\pi/4\). On \(0\leq\theta\leq\pi/4\), \(\cos\theta\geq\sin\theta\), so the inner radius is \(2\cos\theta\). On \(\pi/4\leq\theta\leq\pi/2\), the inner radius is \(2\sin\theta\). The outer radius remains \(2\). Therefore,

\begin{aligned} A &=\int_0^{\pi/4}\int_{2\cos\theta}^{2}r\,dr\,d\theta +\int_{\pi/4}^{\pi/2}\int_{2\sin\theta}^{2}r\,dr\,d\theta\\ &=\int_0^{\pi/4}2\sin^2\theta\,d\theta +\int_{\pi/4}^{\pi/2}2\cos^2\theta\,d\theta\\ &=\left[\theta-\frac{\sin(2\theta)}{2}\right]_0^{\pi/4} +\left[\theta+\frac{\sin(2\theta)}{2}\right]_{\pi/4}^{\pi/2}\\ &=\left(\frac{\pi}{4}-\frac12\right) +\left(\frac{\pi}{4}-\frac12\right) =\frac{\pi}{2}-1. \end{aligned}

Check: the radial intervals are valid because both candidate inner radii are at most \(2\) throughout the first quadrant. At the switch they both equal \(\sqrt{2}\); the outer radius is \(2\) on both sides. The differentiated antiderivatives give the displayed angular integrands, confirming the area evaluation.

Combining Pieces for a Nonconstant Integrand

The bounds are determined by the region, not by the particular integrand. Once the region has been split correctly, substitute the polar form of the integrand and include the factor \(r\) from the area element. This is important even when the integrand itself already contains a factor of \(r\).

Worked Example: Integrating \(x+y\) over the Intersection

Over the intersection of the two circular regions in the first worked example, evaluate \(\iint_R(x+y)\,dA\). In polar coordinates, \(x+y=r(\cos\theta+\sin\theta)\). The area element contributes another factor of \(r\), so the integrand in the iterated integral is \(r^2(\cos\theta+\sin\theta)\).

\begin{aligned} I &=\int_0^{\pi/4}\int_0^{2\sin\theta} r^2(\cos\theta+\sin\theta)\,dr\,d\theta\\ &\qquad+\int_{\pi/4}^{\pi/2}\int_0^{2\cos\theta} r^2(\cos\theta+\sin\theta)\,dr\,d\theta. \end{aligned}

The radial antiderivative is \(r^3/3\). The first angular piece is

\begin{aligned} I_1 &=\frac{8}{3}\int_0^{\pi/4}(\cos\theta+\sin\theta)\sin^3\theta\,d\theta\\ &=\frac{8}{3}\left( \int_0^{\pi/4}\sin^3\theta\cos\theta\,d\theta+ \int_0^{\pi/4}\sin^4\theta\,d\theta\right)\\ &=\frac{8}{3}\left( \frac{1}{16}+\frac{3\pi}{32}-\frac14\right) =\frac{\pi}{4}-\frac12. \end{aligned}

Here, \(\int\sin^3\theta\cos\theta\,d\theta=\sin^4\theta/4\), and an antiderivative of \(\sin^4\theta\) is \(3\theta/8-\sin(2\theta)/4+\sin(4\theta)/32\). The second angular piece has the same value: under the substitution \(u=\pi/2-\theta\), its integrand and limits become those of the first piece. Hence

I=I_1+I_2 =\left(\frac{\pi}{4}-\frac12\right) +\left(\frac{\pi}{4}-\frac12\right) =\frac{\pi}{2}-1.

Check: the derivative of \(r^3/3\) is \(r^2\), and differentiating the stated antiderivative of \(\sin^4\theta\) gives \(\sin^4\theta\). The substitution \(u=\pi/2-\theta\) maps the second piece’s interval \([\pi/4,\pi/2]\) to \([0,\pi/4]\), confirming that the two contributions are equal.

A Common Mistake: Using One Boundary on Both Sides

A frequent error is to solve for the intersection correctly but then use the same candidate curve as the outer boundary over the entire angular range. For the intersection in the first example, the incorrect choice would be to use \(r=2\sin\theta\) all the way from \(0\) to \(\pi/2\). It fails on \(\pi/4<\theta\leq\pi/2\), where \(2\sin\theta>2\cos\theta\); that radial interval extends outside the second circle.

\text{Incorrect area} =\int_0^{\pi/2}\int_0^{2\sin\theta}r\,dr\,d\theta =\int_0^{\pi/2}2\sin^2\theta\,d\theta =\frac{\pi}{2}.

The correct area is \(\pi/2-1\), so the incorrect setup overcounts by \(1\). The correction is to use the smaller candidate radius on each interval: \(2\sin\theta\) before \(\pi/4\), and \(2\cos\theta\) after it. For inner bounds, the corresponding mistake is to choose the smaller radius when the region must lie outside both curves; in that situation the inner boundary is the larger radius.

Practical Checks Before Integrating

A short check of the bounds can prevent a long calculation from describing the wrong region. For each interval, verify that the proposed inner radius does not exceed the outer radius. At a switch, substitute the switching angle into both expressions and confirm that they agree. Finally, test one angle on either side to ensure that the selected minimum or maximum is correct.

For multiple switches, repeat the same process at every relevant intersection angle and order the resulting intervals from left to right. Use the appropriate pair of radial bounds on each interval, then add all the iterated integrals. The method is unchanged whether the integrand is \(1\), a polynomial in \(x\) and \(y\), or another integrable function.

Key takeaway: A changing radial boundary is handled by partitioning the angular range at its switching angles. On each piece, use the correct inner and outer curves, and add the resulting integrals.

Check Your Understanding

Use the ordering of candidate radii, not just their intersection points, to answer each question.

If two curves give candidate outer radii and the region lies inside both, which radius should be used on each angular interval?
For \(r=3\sin\theta\) and \(r=3\cos\theta\) in the first quadrant, at what angle do the curves intersect?
Suppose the region lies outside both of two curves. Should its inner radius be the larger or the smaller of their radial values?
Why may the integrals over two angular pieces be added to obtain the integral over the full region?
What two checks should be made at a proposed switching angle and on the intervals around it?