Introduction
A planar region can be examined by moving a horizontal line through it. At each height, the line cuts out a segment running from the region’s left boundary to its right boundary. Describing the full region in terms of these segments gives a Type II region.
This viewpoint organizes a region by its range of \(y\)-values and by the \(x\)-coordinates present at each fixed height. The key question is whether every horizontal slice is one interval, with a well-defined left endpoint and right endpoint.
Definition of a Type II Region
The numbers \(c\) and \(d\) specify the lowest and highest heights in the region. At any height \(y\) between them, \(g_1(y)\) and \(g_2(y)\) give the two horizontal endpoints. The boundary functions may be curves, straight lines, or constants.
For a fixed value of \(y\), it is useful to name the corresponding slice: $$ D_y=\{x:(x,y)\in D\}. $$ For a Type II region, this slice is the interval \(D_y=[g_1(y),g_2(y)]\). At the lowest or highest height, the interval may reduce to one point if its endpoints coincide.
Why the Slice Description Works
The defining conditions specify membership one height at a time. A point \((x,y)\) belongs to \(D\) precisely when its height is between \(c\) and \(d\), and its horizontal coordinate lies between the two boundary values at that height. In other words, the definition collects exactly the horizontal intervals \(D_y\) over the full range of heights.
This description has a necessary condition: for every allowed \(y\), the slice must be a single interval. If a horizontal line cuts the region into two separated pieces, one interval from a left boundary to a right boundary would incorrectly include the gap between them. Such a region is not described by one Type II interval over that height range without dividing the region into simpler pieces.
Worked Example: A Rectangle
Worked Example: A Rectangle
Describe the rectangle with corners \((-2,1)\), \((3,1)\), \((-2,4)\), and \((3,4)\) as a Type II region, and identify some horizontal slices.
The rectangle extends from \(y=1\) to \(y=4\). At each height in this range, its left edge has \(x=-2\) and its right edge has \(x=3\). Thus the boundary functions are constant: $$ c=1,\qquad d=4,\qquad g_1(y)=-2,\qquad g_2(y)=3. $$ The region is $$ D=\{(x,y):1\le y\le4,\;-2\le x\le3\}. $$
At \(y=1\), the slice is \([-2,3]\). At \(y=\frac52\), it is also \([-2,3]\), since $$ g_1\!\left(\frac52\right)=-2,\qquad g_2\!\left(\frac52\right)=3. $$ At \(y=4\), the slice remains \([-2,3]\).
Check: The boundary values satisfy \(-2\le3\), and the full range \(1\le y\le4\) matches the rectangle’s bottom and top edges. The description therefore includes precisely its horizontal segments, including those on its boundary.
Worked Example: A Triangle
Worked Example: A Triangle
Consider the triangular region whose horizontal boundaries are \(x=\frac y2\) on the left and \(x=2-\frac y2\) on the right, for \(0\le y\le2\). Write its Type II description and check its slices.
The candidate description is $$ D=\left\{(x,y):0\le y\le2,\;\frac y2\le x\le2-\frac y2\right\}. $$ The left boundary is \(g_1(y)=\frac y2\); the right boundary is \(g_2(y)=2-\frac y2\).
Check that these functions are ordered throughout the height interval: $$ g_2(y)-g_1(y) =\left(2-\frac y2\right)-\frac y2 =2-y. $$ Since \(0\le y\le2\), we have \(2-y\ge0\), so \(g_1(y)\le g_2(y)\) everywhere in the stated range.
At the bottom, \(y=0\), the slice endpoints are $$ g_1(0)=\frac02=0,\qquad g_2(0)=2-\frac02=2. $$ The bottom slice is \([0,2]\). At the intermediate height \(y=1\), $$ g_1(1)=\frac12,\qquad g_2(1)=2-\frac12=\frac32, $$ so the slice is \([\frac12,\frac32]\). At the top, \(y=2\), $$ g_1(2)=\frac22=1,\qquad g_2(2)=2-\frac22=1. $$ The slice has collapsed to the single point \((1,2)\).
Check: The two boundaries meet where $$ \frac y2=2-\frac y2 \quad\Longrightarrow\quad y=2,\qquad x=1. $$ This agrees with the top endpoint found from the slice functions. The boundary difference \(2-y\) is nonnegative throughout \(0\le y\le2\), so the left and right functions do not switch order inside the region.
Worked Example: A Disk
Worked Example: A Disk
Describe the unit disk \(x^2+y^2\le1\) using horizontal slices.
First, the disk’s vertical extent is \(-1\le y\le1\). At a fixed height \(y\), its points satisfy \(x^2\le1-y^2\). On this height range \(1-y^2\ge0\), so taking square roots gives $$ -\sqrt{1-y^2}\le x\le\sqrt{1-y^2}. $$ Consequently, $$ D=\left\{(x,y):-1\le y\le1,\;-\sqrt{1-y^2}\le x\le\sqrt{1-y^2}\right\}. $$ Here \(g_1(y)=-\sqrt{1-y^2}\) is the left boundary and \(g_2(y)=\sqrt{1-y^2}\) is the right boundary.
At \(y=0\), $$ g_1(0)=-\sqrt{1-0^2}=-1,\qquad g_2(0)=\sqrt{1-0^2}=1, $$ so the middle slice is \([-1,1]\). At \(y=\frac35\), $$ 1-y^2=1-\left(\frac35\right)^2 =1-\frac9{25} =\frac{16}{25}, $$ and therefore $$ g_1\!\left(\frac35\right)=-\sqrt{\frac{16}{25}}=-\frac45,\qquad g_2\!\left(\frac35\right)=\sqrt{\frac{16}{25}}=\frac45. $$ That slice is \([-\frac45,\frac45]\). By symmetry, the slice at \(y=-\frac35\) has the same endpoints. At either extreme, \(y=\pm1\), both boundary functions equal \(0\), so the slice is a single point.
Check: The endpoints at \(y=\frac35\) lie on the circle because $$ \left(\frac45\right)^2+\left(\frac35\right)^2 =\frac{16}{25}+\frac9{25} =1. $$ The left and right boundaries have equal magnitude and opposite signs, as expected from the disk’s symmetry about the \(y\)-axis. Also, \(g_1(y)\le g_2(y)\) because the square root is nonnegative throughout \(-1\le y\le1\).
Checking Whether the Description Is Valid
When describing a candidate Type II region, check the height range and the ordering of the boundaries. The lower and upper limits must include all heights in the region, and the left boundary must stay at or to the left of the right boundary at every height.
Find the smallest and largest \(y\)-values in the region; these become \(c\) and \(d\).
At a fixed height \(y\), determine the first and last \(x\)-coordinates in the region.
Write the smaller endpoint as \(g_1(y)\) and the larger endpoint as \(g_2(y)\).
Verify \(g_1(y)\le g_2(y)\) across the entire interval \(c\le y\le d\), including its endpoints.
The last check matters even if a sketch looks correct. If two proposed boundary functions cross inside the stated height range, their left-to-right order changes there. The proposed functions then do not describe one Type II region on the whole interval in the stated way.
Common Mistake: Using Constant Bounds for Curved Slices
A frequent error is to take the largest horizontal width of a region and use it at every height. For the unit disk, this would incorrectly describe every slice as \([-1,1]\) for \(-1\le y\le1\). That gives the set $$ \{(x,y):-1\le y\le1,\;-1\le x\le1\}, $$ which is the whole square around the disk, not the disk itself.
For example, at \(y=\frac35\), the incorrect bounds include \(x=1\). But that point is outside the disk because $$ 1^2+\left(\frac35\right)^2 =1+\frac9{25} =\frac{34}{25}>1. $$ The correct slice at this height is \([-\frac45,\frac45]\), so \(x=1\) is excluded. The correction is to let the left and right endpoints depend on \(y\), rather than assuming they are constant.
When One Type II Description Is Not Enough
Some regions have a horizontal slice that breaks into separate intervals. A single pair of functions \(g_1(y)\) and \(g_2(y)\) would span from the leftmost piece to the rightmost piece and include the gap between them. That would add points that are not in the region.
Other regions have boundary functions whose identities change with height. In those cases, a description may need to be divided into height ranges so each piece has consistent left and right boundaries. The essential test remains the same: on every height range described, each horizontal slice must be exactly the interval between its stated endpoints.
Why Horizontal Slices Matter
Horizontal slices provide a direct geometric way to understand a region in terms of \(y\). The outer limits describe how far the region extends vertically, while the boundary functions record how its horizontal width and position change from one height to another. This makes the Type II description useful for curved regions as well as rectangles and polygons.
It also gives a practical test for a proposed description: choose a height, compute both boundary values, and compare the resulting interval with the actual cross-section of the region. Testing the bottom, top, and one or more interior heights often reveals a reversed boundary, an incorrect constant bound, or a missing height range.
Check Your Understanding
For each question, focus on what a horizontal slice looks like and what information the Type II description must provide.
- In a Type II description, what do the limits \(c\) and \(d\) represent?
- For a fixed \(y\), which boundary function gives the left endpoint of the slice, and which gives the right endpoint?
- For the triangle in the worked example, what are the horizontal slice endpoints at \(y=\frac12\)?
- Why is the condition \(g_1(y)\le g_2(y)\) required throughout the full height interval?
- What goes wrong if a horizontal slice consists of two separated intervals but is described using only one left boundary and one right boundary?