Why Horizontal Bounds Need Care
A horizontal slice through a region is a short horizontal segment. At a fixed height \(y\), it begins at the region’s left boundary and ends at its right boundary. To describe the entire region with horizontal slices, we need the range of y-values the region reaches and the correct left and right boundaries at each height.
The boundaries may not stay the same throughout that range. A curve can meet another curve, a boundary can end, or the region can reach a corner. At such a height, the description of the horizontal slice may need to change. Identifying those heights is the main task of building horizontal-slice bounds.
What the Bounds Describe
Suppose a region can be described at each height \(y\) by one horizontal segment. Write \(x_L(y)\) for the x-coordinate of its left endpoint and \(x_R(y)\) for the x-coordinate of its right endpoint. The region then has the description
The values \(y_{\min}\) and \(y_{\max}\) are the lowest and highest y-coordinates in the region. They are not necessarily found by taking the x-coordinates of curve intersections. A horizontal slice is indexed by y, so its outer bounds must be y-values.
If the left or right boundary changes at an intermediate height, describe the region in separate y-intervals. At the change height, the adjacent boundary descriptions should meet at the same point or agree along the relevant edge. For a complicated shape, a horizontal line might cut the region into more than one segment; such a region cannot be represented by a single left-to-right interval at every y without splitting the region into components.
How to Locate the Important Heights
Begin by sketching or carefully examining the boundary curves and the enclosed region. Then identify the y-coordinates at which the horizontal-slice description could change. These include:
- the y-values of the region’s highest and lowest points;
- the y-values where two candidate boundary curves intersect;
- the y-values where a boundary segment starts or ends; and
- the y-values where the boundary turns in a way that changes which points are the leftmost or rightmost points of a slice.
Use these values to divide the full vertical range into intervals. Within each interval, choose a test y-value and compare the x-coordinates of the candidate boundaries. The smaller x-coordinate is the left boundary; the larger is the right boundary. A test value helps identify the order, but the intersections and endpoints establish where that order can change.
Determine the lowest and highest y-values reached by the enclosed region.
Find the y-values of relevant curve intersections, endpoints, and boundary turns.
At a y-value inside each interval, compare the candidate x-coordinates to identify left and right.
Substitute the interval endpoints into the boundary descriptions and confirm they match the region.
Worked Example: A Right Boundary That Changes
Worked Example: A Right Boundary That Changes
Consider the region enclosed by the y-axis, the parabola \(y=x^2\) on its right-hand branch, and the line \(y=2-x\). Find its horizontal-slice bounds.
First locate the relevant boundary intersections. The parabola meets the y-axis at \(x=0\), so \(y=0^2=0\), giving \((0,0)\). The line meets the y-axis at \(x=0\), so \(y=2-0=2\), giving \((0,2)\).
To find where the parabola and line meet, set their y-values equal:
Move all terms to one side and factor:
Thus \(x=-2\) or \(x=1\). The stated region uses the right-hand branch and is enclosed with the y-axis, so the relevant intersection is \(x=1\). Its y-coordinate is \(y=1^2=1\). The region therefore extends from \(y=0\) to \(y=2\), and the boundary can change at \(y=1\).
For \(0<y<1\), the parabola’s right-hand branch is to the right of the y-axis, while the line lies farther right. For example, at \(y=\frac14\), the parabola has \(x=\sqrt{\frac14}=\frac12\), and the line has \(x=2-\frac14=\frac74\). The slice starts at \(x=0\) and ends at the parabola.
For \(1<y<2\), the line lies between the y-axis and the parabola’s right-hand branch. At \(y=\frac32\), the line has \(x=2-\frac32=\frac12\), while the parabola has \(x=\sqrt{\frac32}\), which is greater than \(\frac12\). The enclosed slice ends at the line. Therefore the bounds are
Check the joins: at \(y=1\), the parabola gives \(x=\sqrt{1}=1\), and the line gives \(x=2-1=1\). At the lower endpoint, the parabola gives \(x=\sqrt0=0\), meeting the y-axis at \((0,0)\). At the upper endpoint, the line gives \(x=2-2=0\), meeting the y-axis at \((0,2)\). The endpoint checks agree with the described enclosure.
Worked Example: A Left Boundary That Changes
Worked Example: A Left Boundary That Changes
Consider the region with \(0\le y\le 2\), right boundary \(x=2\), and left edge formed by the lower of the two lines \(x=y-1\) and \(x=1-y\). Find its horizontal-slice bounds and the height where the left boundary changes.
Set the two candidate left-boundary coordinates equal:
The lines meet at \(y=1\). Substitution gives \(x=1-1=0\), so their intersection is \((0,1)\). This is the possible change height.
Test a height below the intersection, \(y=\frac12\). The two x-coordinates are
Since \(-\frac12<\frac12\), the line \(x=y-1\) is the left boundary below \(y=1\). Test a height above the intersection, \(y=\frac32\):
Now \(1-y\) is the left boundary. The right boundary stays \(x=2\) throughout. Thus the horizontal-slice bounds are
Check the change height: at \(y=1\), both left-boundary expressions give \(x=0\). At \(y=0\), the first gives \(x=-1\); at \(y=2\), the second gives \(x=-1\). The left edge and its joins are consistent with the stated region.
Worked Example: One Pair of Boundaries on the Whole Range
Worked Example: One Pair of Boundaries on the Whole Range
Consider the region between \(x=y^2\) and \(x=2-y\). Find its y-range and determine whether the left and right boundaries change within that range.
The boundary curves meet when their x-coordinates are equal:
The intersection heights are \(y=-2\) and \(y=1\). Between these values, compare the candidate x-coordinates by subtracting:
For \(-2<y<1\), both factors \(1-y\) and \(y+2\) are positive. Therefore \((2-y)-y^2>0\), so \(y^2\) is the smaller x-coordinate and \(2-y\) the larger one throughout the interval. For instance, at \(y=0\), the two values are \(0\) and \(2\), confirming that order.
The bounds are therefore
Check both endpoints: at \(y=-2\), the expressions give \((-2)^2=4\) and \(2-(-2)=4\); at \(y=1\), they give \(1^2=1\) and \(2-1=1\). The boundaries meet at both ends, as required. The factorization also confirms that their order does not reverse between those endpoints.
Why These Bounds Produce the Area
Once the horizontal bounds are correct, they also identify the width of each slice: at height \(y\), its width is \(x_R(y)-x_L(y)\). If the boundary pair stays the same from \(y=a\) to \(y=b\), thin horizontal strips have approximate area \([x_R(y)-x_L(y)]\Delta y\). Adding these strip areas and taking the limit gives the area integral
This reasoning assumes the boundaries describe a bounded region and the width is integrable over the stated interval. If the left or right boundary changes at an intermediate height, split the integral there and use the appropriate pair of bounds on each part. The bounds are therefore not just labels: they determine which width is accumulated at every height.
A Common Mistake: Keeping One Boundary After It Stops Being Left
In the second example, a tempting but incorrect description is to use \(x=y-1\) as the left boundary for all \(0\le y\le2\). At \(y=\frac32\), this gives
But the other candidate boundary gives
and \(-\frac12<\frac12\). Thus \(x=y-1\) is to the right, not the left, at that height. The correction is to split at the intersection height \(y=1\): use \(x=y-1\) below it and \(x=1-y\) above it. A test point in each interval catches this error before the bounds are used.
Check Your Understanding
For each question, focus on identifying the vertical range and the correct left and right boundaries.
- In the region from the first worked example, why must the horizontal-slice description be split at \(y=1\)?
- For the two candidate left boundaries \(x=y-1\) and \(x=1-y\), at what height do they intersect, and which is farther left when \(y=0\)?
- For boundaries \(x=y^2\) and \(x=2-y\), what factorization shows their order on \(-2<y<1\)?
- When checking a proposed split, what should be compared at a test y-value inside each interval?
- Why are x-coordinates of intersections not, by themselves, the outer bounds for horizontal slices?
