Reading a Region One Ray at a Time
In Cartesian coordinates, a region is often described by vertical or horizontal slices. Polar bounds describe the same region using rays from the origin: choose a direction, then record which distances along that ray lie inside the region. The angular interval tells us which directions occur; the radial interval tells us how far the region extends in each direction.
This viewpoint is especially useful when the boundaries or the shape of a region are naturally organized around the origin. It also makes a key issue visible: if the region does not contain the origin, a ray may first enter the region at a positive distance rather than at radius zero.
What the Bounds Mean
Fix an angle \(\theta\). The points with that angle lie on one ray from the origin. As \(r\) increases from zero, \(x=r\cos\theta\) and \(y=r\sin\theta\) move along that ray. To find the radial bounds, substitute these expressions into the region’s boundary equations or inequalities and determine which values of \(r\) satisfy the description.
The angular bounds are the directions for which the ray meets the region. They can often be read from a sketch by identifying the extreme rays, or found from boundary points using \(\tan\theta=y/x\) when \(x\ne 0\), together with the correct quadrant.
The origin needs particular care. When it lies in the region, the radial interval may begin at zero. When it lies outside, the region may be reached only after a positive distance along a ray. The origin itself has no unique angle: \(r=0\) gives \((x,y)=(0,0)\) for every \(\theta\). This does not prevent a region containing the origin from having a polar description; it means only that the origin is represented redundantly.
Why Ray Bounds Describe the Region
The logic behind polar bounds follows directly from the coordinate equations. A point \((x,y)\ne(0,0)\) has a distance \(r=\sqrt{x^2+y^2}\) from the origin and a direction \(\theta\). Thus, to decide whether it lies in a region, we can first select its direction and then test its distance along that ray.
Fix \(\theta\) in the proposed angular range.
Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) into its inequalities and boundary equations.
Solve the resulting conditions for \(r\geq0\). Their solution is the radial interval on that ray.
If this procedure gives exactly the proposed radial interval for every angle in the proposed angular range, and gives no points outside that range, then the bounds describe the region. This is an equivalence: the original conditions hold exactly for the points generated by the bounds.
Worked Example: A Triangle Containing the Origin
Worked Example: A Triangle Containing the Origin
Describe the region \(D\) given by \(x\geq0\), \(y\geq0\), and \(x+y\leq2\) using polar bounds.
The first two inequalities place the region in the first quadrant, so the directions range from the positive \(x\)-axis to the positive \(y\)-axis:
For a fixed angle in this range, substitute \(x=r\cos\theta\) and \(y=r\sin\theta\). The line condition becomes
On \(0\leq\theta\leq\pi/2\), both sine and cosine are nonnegative, and their sum is positive. Dividing by that sum gives \(r\leq2/(\cos\theta+\sin\theta)\). The region begins at the origin along every ray, so the lower radial bound is zero:
Check the endpoints of the angular range. At \(\theta=0\), the ray lies on the positive \(x\)-axis and the upper radius is \(2\), reaching \((2,0)\). At \(\theta=\pi/2\), the upper radius is also \(2\), reaching \((0,2)\). At \(\theta=\pi/4\), it is \(2/(\sqrt2)=\sqrt2\), which gives the boundary point \((1,1)\). The origin is included because \(r=0\) is allowed.
Worked Example: A Rectangle Whose Radial Bound Changes
Worked Example: A Rectangle Whose Radial Bound Changes
Describe the rectangle \(D=\{(x,y):0\leq x\leq2,\ 0\leq y\leq1\}\) in polar bounds.
The rectangle lies in the first quadrant, so \(0\leq\theta\leq\pi/2\). For a ray in this range, the right edge \(x=2\) gives
and the top edge \(y=1\) gives
The ray must satisfy both conditions, so its upper radial bound is the smaller of these two values. Find where they are equal:
The change occurs at \(\theta=\arctan(1/2)\). Below this angle, the ray reaches the right edge first: at \(\theta=0\), the two candidate bounds are \(2\) and an unbounded top-edge constraint, respectively. Above the change, it reaches the top edge first: as \(\theta\) approaches \(\pi/2\), the top edge limits the ray. Therefore the bounds are
At the transition angle, \(\tan\theta=1/2\), so a direction vector is proportional to \((2,1)\). Its point on the rectangle’s top-right corner is \((2,1)\), at distance \(\sqrt{2^2+1^2}=\sqrt5\). The formulas agree there: \(\cos\theta=2/\sqrt5\) and \(\sin\theta=1/\sqrt5\), so \(2/\cos\theta=\sqrt5\) and \(1/\sin\theta=\sqrt5\). This confirms both the split angle and the matching radial limits.
Worked Example: A Region That Does Not Reach the Origin
Worked Example: A Region That Does Not Reach the Origin
Describe the region \(D\) defined by \(x\geq1\), \(y\geq0\), and \(x+y\leq3\).
The region is in the first quadrant. Its furthest upper-left vertex is \((1,2)\), whose direction has \(\tan\theta=2/1=2\). Thus the relevant angles run from the positive \(x\)-axis to the ray through \((1,2)\):
The condition \(x\geq1\) becomes \(r\cos\theta\geq1\). Since \(\cos\theta>0\) on this angular interval, it gives the lower radial limit
The other slanted boundary gives \(r(\cos\theta+\sin\theta)\leq3\), or \(r\leq3/(\cos\theta+\sin\theta)\). Hence
Check that the two limits meet at the final angle. There, \(\cos\theta=1/\sqrt5\) and \(\sin\theta=2/\sqrt5\), so the lower limit is \(\sqrt5\) and the upper limit is \(3/(3/\sqrt5)=\sqrt5\). This is the vertex \((1,2)\). At \(\theta=0\), the interval is \(1\leq r\leq3\), which gives the portion of the \(x\)-axis from \((1,0)\) to \((3,0)\). The positive lower limit correctly records that the region does not include the origin.
A Common Mistake: Forgetting That Both Boundaries Apply
In the rectangle example, a tempting but incorrect description is to use \(0\leq r\leq2/\cos\theta\) for every angle from \(0\) to \(\pi/2\). That condition enforces \(x\leq2\), but it does not enforce \(y\leq1\). For example, at \(\theta=\pi/4\), this incorrect upper limit is \(2\sqrt2\). Choosing \(r=2\) gives \((x,y)=(\sqrt2,\sqrt2)\), which is outside the rectangle because \(\sqrt2>1\).
The correction is to impose both boundary conditions and take the more restrictive radial bound. At \(\pi/4\), the two candidates are \(2/\cos(\pi/4)=2\sqrt2\) and \(1/\sin(\pi/4)=\sqrt2\), so the actual upper limit is \(\sqrt2\), reaching \((1,1)\). A proposed bound should always be checked against every defining condition, not just one visible edge.
Choosing and Checking the Angular Range
A correct radial formula on the wrong angular range does not describe the intended region. Use the sketch or the region’s inequalities to identify which rays can meet it. Then verify the extreme directions against boundary points. When a boundary point has coordinates \((a,b)\) with \(a>0\), its direction is \(\arctan(b/a)\); the quadrant must still be checked. For points on the vertical axis, the direction is \(\pi/2\) in the upper half-plane.
Finally, check whether each ray meets the region in one continuous radial interval. If it enters, leaves, and later re-enters, one pair of radial bounds is not enough for that angle: the radial set must be described as multiple intervals, or the region must be split into pieces. Similarly, if the active boundary changes as the angle varies, split the angular range at the direction where the formulas meet. These checks prevent gaps, overlaps, and unintended points.
- Do the angles cover exactly the directions that meet the region?
- Does every radial interval begin and end on the correct boundary?
- Does the lower limit reflect whether the origin is included?
- Do the formulas agree at any angle where the active boundary changes?
- Does a test point satisfy both the bounds and the original region conditions?
Check Your Understanding
For each question, use the ray-by-ray viewpoint and check your proposed bounds against the region’s conditions.
- For \(x\geq0\), \(y\geq0\), and \(x+2y\leq4\), what are the angular and radial bounds?
- For \(0\leq x\leq1\), \(0\leq y\leq2\), which boundary limits a ray close to the positive \(x\)-axis, and which limits one close to the positive \(y\)-axis?
- Why is a positive lower radial limit needed for the region \(x\geq2\), \(y\geq0\), \(x+y\leq4\)?
- For the rectangle \(0\leq x\leq2\), \(0\leq y\leq1\), at what angle does the active upper radial boundary change?
- What should be done if a ray meets a region in two separate radial intervals?