Why Reconstructing the Region Matters
A pair of polar limits can look plausible and still describe the wrong part of the plane. The angular limits determine which directions are included, while the radial limits determine how far each ray travels through the region. Checking both is essential: an incorrect lower radial limit can fill a hole, an incorrect upper limit can add points outside the region, and an incorrect angular interval can omit or include whole sectors.
A reliable check reverses the process of setting up the integral. Start with the proposed limits, follow rays from the origin, and identify the points where each ray enters and leaves the proposed region. Then compare those points with the boundaries in the original description. This tutorial is about verifying that geometric description, not evaluating an integral over it.
What Polar Bounds Describe
In polar coordinates, a point with radius \(r\geq 0\) and angle \(\theta\) has Cartesian coordinates \(x=r\cos\theta\) and \(y=r\sin\theta\). Fix an angle \(\theta\). As \(r\) increases from zero, the point moves outward along the ray in direction \(\theta\). Thus limits \(a(\theta)\leq r\leq b(\theta)\) specify the part of that ray between distances \(a(\theta)\) and \(b(\theta)\) from the origin.
Three checks follow directly from this definition:
- Angle coverage: Do the chosen angles point in every direction needed to reach the region, and only those directions?
- Radial feasibility: At each angle, are the lower and upper radii in the correct order and compatible with the original boundary conditions?
- Boundary identity: Does the inner radius correspond to the first boundary encountered and the outer radius to the last boundary encountered on that ray?
Testing a few angles is a useful diagnostic, not a complete proof by itself. A check at an interior angle may reveal an error quickly, but a correct setup also needs a reason that the bounds work throughout the entire angular interval. Endpoints deserve special attention because that is where a radial interval can collapse to a point or where one boundary ceases to be reachable.
A Ray-by-Ray Justification
The reconstruction method works because the polar coordinate map along a fixed direction is particularly simple: changing \(r\) changes only the distance from the origin, not the angle. For a fixed \(\theta\), substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) into the conditions defining the region. The points on that ray are then determined by the values of \(r\geq0\) satisfying those conditions.
For the justification, take any angle in the stated interval. By assumption, the points of \(D\) on its ray are exactly those with radius between \(a(\theta)\) and \(b(\theta)\). Taking all such angles collects exactly the points of \(D\) on those rays. The assumption that there are no points of \(D\) on other rays rules out omitted parts. Together, these statements show that the polar bounds neither add nor remove points.
In practice, substituting the polar formulas into the original inequalities is often the most direct way to verify the ray intersections. The resulting inequalities may impose both an angular restriction and a radial restriction. It is important to keep track of which condition does which: an inequality such as \(r\cos\theta\geq 1\) restricts both the allowed direction and the allowed radius.
Worked Example: A Disk Cut by a Vertical Line
Worked Example: A Disk Cut by a Vertical Line
Consider the part of the disk \(x^2+y^2\leq4\) lying to the right of the line \(x=1\). Check the proposed bounds \(-\pi/3\leq\theta\leq\pi/3\), \(\sec\theta\leq r\leq2\).
First translate each boundary condition. Since \(x^2+y^2=r^2\) and \(r\geq0\), the disk condition gives \(r\leq2\). The condition \(x\geq1\) becomes \(r\cos\theta\geq1\). For this to be possible with \(r\leq2\), we need \(\cos\theta>0\) and \(\sec\theta\leq2\). Therefore \(\cos\theta\geq1/2\), which in the range with positive cosine means \(-\pi/3\leq\theta\leq\pi/3\). On every such ray, the line is the inner boundary \(r=\sec\theta\), and the circle is the outer boundary \(r=2\).
Check an interior angle, \(\theta=0\). Then \(\sec(0)=1\), so the radial interval is \(1\leq r\leq2\). On the horizontal ray, these points satisfy \(1\leq x\leq2\), exactly the part of the ray inside the disk and to the right of the line.
At the endpoint \(\theta=\pi/3\), \(\sec(\pi/3)=2\), so the interval collapses to \(r=2\). The corresponding point is \((x,y)=(2\cos(\pi/3),2\sin(\pi/3))=(1,\sqrt{3})\), which lies on both boundaries. At \(\theta=-\pi/3\), the point is \((1,-\sqrt{3})\), also on both boundaries. The bounds reconstruct the intended region, with the line as the inner boundary and the circle as the outer boundary.
Worked Example: Between Two Circular Boundaries
Worked Example: Between Two Circular Boundaries
Let \(D\) be outside the unit disk and inside the circle \((x-1)^2+y^2\leq1\). Check the proposed bounds \( -\pi/3\leq\theta\leq\pi/3\), \(1\leq r\leq2\cos\theta\).
Outside the unit disk means \(x^2+y^2\geq1\), or \(r\geq1\). For the other circle, substitute \(x=r\cos\theta\) and \(y=r\sin\theta\):
Expanding and using \(\cos^2\theta+\sin^2\theta=1\) gives \(r^2-2r\cos\theta\leq0\). Since \(r\geq1\), we can divide by \(r>0\), obtaining \(r\leq2\cos\theta\). For both radial conditions to hold, the upper radius must be at least the lower radius: \(1\leq2\cos\theta\). This is equivalent to \(\cos\theta\geq1/2\), giving the proposed angular interval.
At \(\theta=0\), the radial interval is \(1\leq r\leq2\). The ray starts at the unit circle and ends at the circle centered at \((1,0)\). At \(\theta=\pi/3\), the outer radius is \(2\cos(\pi/3)=1\), so the interval collapses to the shared point \((1/2,\sqrt{3}/2)\). The lower endpoint \(-\pi/3\) similarly gives \((1/2,-\sqrt{3}/2)\). These checks confirm that the radial interval is nonempty exactly on the stated angles and that its two limits match the original circles.
Worked Example: A Change in the Outer Boundary
Worked Example: A Change in the Outer Boundary
Consider the part of the disk \(x^2+y^2\leq4\) in the first quadrant that also satisfies \(y\leq1\). Check these proposed bounds:
The first-quadrant condition gives \(0\leq\theta\leq\pi/2\). The disk gives \(r\leq2\), and the horizontal boundary condition \(y\leq1\) becomes \(r\sin\theta\leq1\). When \(0\leq\theta\leq\pi/6\), we have \(\sin\theta\leq1/2\), so \(r\leq2\) implies \(r\sin\theta\leq1\). The circle is the outer boundary there. When \(\pi/6\leq\theta\leq\pi/2\), the line condition requires \(r\leq\csc\theta\), and this is no larger than \(2\). The line is the outer boundary on that part.
At the switching angle \(\theta=\pi/6\), \(\csc(\pi/6)=2\), so the two formulas agree. At \(\theta=\pi/4\), the proposed outer radius is \(\sqrt{2}\), and the corresponding point has \(y=r\sin\theta=\sqrt{2}\cdot\sqrt{2}/2=1\), on the line. At \(\theta=\pi/2\), the outer radius is \(\csc(\pi/2)=1\), giving the point \((0,1)\). At \(\theta=0\), the radius reaches \(2\) along the positive \(x\)-axis, which remains below \(y=1\). Thus the bounds follow the circle where it is the limiting boundary and the line where that boundary takes over.
How a Sample Angle Exposes a Wrong Bound
Suppose the previous region is instead assigned the single outer bound \(0\leq r\leq2\) for every \(0\leq\theta\leq\pi/2\). This agrees with the disk boundary, but it ignores the condition \(y\leq1\). At \(\theta=\pi/4\), the proposed outer point has \(r=2\), so \(y=2\sin(\pi/4)=\sqrt{2}>1\). It is outside the original region. The angle test has found an extra part of the disk included by the proposed bounds.
The correction is not to change the angular range: the region still contains points in every first-quadrant direction. Instead, the outer radius must be reduced to \(\csc\theta\) after \(\theta=\pi/6\). A useful diagnostic is to pick an interior angle, substitute the proposed endpoint radius into the original inequalities, and see whether the resulting boundary point satisfies them all.
A Practical Verification Checklist
When checking a proposed description, work through the following steps. This keeps the geometric meaning of each limit visible rather than treating the bounds as symbols to accept at face value.
Substitute \(x=r\cos\theta\), \(y=r\sin\theta\), and \(x^2+y^2=r^2\) into the domain description.
Require that the radial conditions can hold for at least one \(r\geq0\). In particular, a lower radius cannot exceed an upper radius.
For a fixed angle, determine which boundary is met first and which is met last as \(r\) increases.
Check an interior direction, each angular endpoint, and any angle where the boundary formula changes.
Confirm that every direction meeting the region is included, and that each proposed radial interval excludes points violating any original condition.
A single interval \(a(\theta)\leq r\leq b(\theta)\) presumes that the intersection with each ray is one continuous radial segment. If a ray passes through two separated pieces of the region, one radial interval cannot represent that ray’s intersection exactly; the description must be split into multiple radial intervals or angular pieces. Likewise, an angle range that runs through a full revolution can represent points more than once, so use a range that covers the intended directions without unnecessary repetition.
Why This Check Is Useful
Reconstructing a region is a practical safeguard before any polar double integral is evaluated. It distinguishes errors in the geometry from errors in later algebra, and it makes the role of each limit explicit. The same method applies whether the boundaries are circles, lines, or other curves: impose the original conditions on a fixed ray, determine the allowed radii, and verify that the angular range captures exactly the needed directions.
Check Your Understanding
For each question, justify your answer by referring to the original boundary conditions.
- For the region inside \(x^2+y^2\leq9\) and to the right of \(x=2\), what feasibility condition determines the angular endpoints?
- Why is it not enough to check that the proposed outer radius lies on one boundary of the region?
- For the disk of radius \(2\) in the first quadrant with \(y\leq1\), at what angle do the outer-boundary formulas change?
- What does it mean geometrically if the proposed lower radial bound exceeds the upper bound at some angle?
- What adjustment may be necessary if a ray intersects the region in two separated radial intervals?