Why Horizontal Bounds Need Care
A horizontal slice fixes a height \(y\) and runs across the base region from left to right. To describe that slice, we need two pieces of information: the range of heights that actually meet the region, and the leftmost and rightmost \(x\)-coordinates at each such height.
The y-range is not always obvious from the equations describing the boundary. Nor are the left and right boundaries necessarily given by the same curves at every height. A curve can turn, a line can meet it, or a corner can change which boundary is farthest left or right. Careful bounds come from examining the whole region, not from solving one boundary equation in isolation.
If each slice is one interval, its bounds can be written \(x_{\mathrm{left}}(y)\le x\le x_{\mathrm{right}}(y)\). If the region has a gap at a particular height, the slice may consist of more than one interval; do not silently replace such a slice by a single interval that includes points outside the region.
Find the Range of Heights First
To find the y-range, imagine projecting every point of the region straight onto the y-axis. The smallest and largest y-coordinates give the ends of the range when the region is closed and bounded. Candidate heights include boundary intersections, endpoints, vertices, and any highest or lowest points along a curved boundary.
Boundary intersections matter because they can mark where the outline of the region begins or ends, or where the curve defining a horizontal bound changes. However, intersections alone may not give the full y-range: a curved boundary can reach a maximum or minimum between intersections. Check the boundary as a whole.
Why the Projection Fills an Interval
For the usual connected, closed, bounded base regions in these problems, every height between the minimum and maximum occurs. Here is the reason. Choose a point of minimum height and a point of maximum height. Because the region is connected by paths, there is a continuous path inside it joining those points. The y-coordinate along the path varies continuously, so the intermediate value theorem guarantees that it takes every value between the two endpoint heights. No height outside the minimum and maximum can occur.
This argument justifies using one interval \([y_{\min},y_{\max}]\) for the projection of such a region. It does not say that each horizontal slice is a single interval. That must still be checked from the geometry.
Worked Example: Between a Parabola and a Line
Worked Example: Between a Parabola and a Line
Find the horizontal-slice bounds for the region between \(y=x^2\) and \(y=2x\).
First find the intersections:
Between these x-values, the line lies above the parabola: \(2x-x^2=x(2-x)\ge 0\) for \(0\le x\le2\). The region therefore has \(x^2\le y\le2x\) and \(0\le x\le2\). Its lowest height is \(0\), and its highest height is \(4\), attained where \(x=2\) and both curves meet. Thus the projection is \(0\le y\le4\).
At a fixed height \(y\), the inequality \(x^2\le y\) gives \(0\le x\le\sqrt y\), using the region’s restriction \(x\ge0\). The inequality \(y\le2x\) gives \(x\ge y/2\). Combining these gives:
Check the bounds at \(y=1\): they give \(1/2\le x\le1\). At the endpoints \(y=0\) and \(y=4\), both bounds agree: they give \(x=0\) and \(x=2\), respectively. Substitution confirms the boundary values: at \(x=y/2\), \(2x=y\); at \(x=\sqrt y\), \(x^2=y\).
Changing Left or Right Boundaries
A horizontal slice can have one boundary curve on part of its y-range and a different curve on the rest. To find where the boundary changes, compare the candidate x-values at the same height. The switch occurs where they are equal, which usually corresponds to an intersection or corner on the region’s boundary.
When a curve is given as \(y=f(x)\), solving for \(x\) may produce more than one branch. Choose the branch that lies on the actual boundary of the region. For example, \(y=x^2\) gives \(x=\pm\sqrt y\); the correct choice depends on which side of the parabola bounds the region.
Worked Example: A Right Boundary That Switches
Worked Example: A Right Boundary That Switches
Find the horizontal-slice bounds for the region enclosed by \(y=x^2\) and \(y=2-x\).
The intersections satisfy:
Thus \(x=-2\) or \(x=1\). On \([-2,1]\), the line is above the parabola because \(2-x-x^2=-(x+2)(x-1)\ge0\). The lowest point of the region has \(y=0\), at \(x=0\); the highest has \(y=4\), at the intersection \(x=-2\). Hence \(0\le y\le4\).
At a fixed height, the parabola condition \(x^2\le y\) requires \(-\sqrt y\le x\le\sqrt y\). The line condition \(y\le2-x\) requires \(x\le2-y\). The region also has \(-2\le x\le1\). Across \(0\le y\le4\), the left boundary is \(-\sqrt y\). The right boundary is the smaller of \(\sqrt y\), \(2-y\), and \(1\).
For \(0\le y\le1\), \(\sqrt y\le2-y\) and \(\sqrt y\le1\), so the right boundary is \(\sqrt y\). For \(1\le y\le4\), \(2-y\le\sqrt y\) and \(2-y\le1\), so the right boundary is \(2-y\). The switch is at \(y=1\): both candidates equal \(1\). Therefore:
Check the first interval at \(y=1/4\): the bounds are \(-1/2\) and \(1/2\), both on the parabola \(x^2=1/4\). Check the second at \(y=2\): the bounds are \(-\sqrt2\) and \(0\); the left endpoint satisfies \(x^2=2\), and the right endpoint satisfies \(2-x=2\). At the switch \(y=1\), the right bounds agree at \(x=1\). At \(y=4\), both bounds give \(x=-2\), the intersection point.
Worked Example: Slices of a Semicircle
Worked Example: Slices of a Semicircle
Let \(D\) be the upper half of the disk \(x^2+y^2\le9\). Find its y-range and horizontal bounds.
The upper half has \(y\ge0\). Since \(x^2+y^2\le9\), the largest possible height is \(y=3\), attained at \((0,3)\). Thus the projection is \(0\le y\le3\). For a fixed \(y\), rearrange the disk inequality:
These are the horizontal bounds for every \(y\) in the projection. At \(y=0\), they give \(-3\le x\le3\), and the endpoints satisfy \(x^2+0^2=9\). At \(y=3\), they both give \(x=0\), and \(0^2+3^2=9\). Substitution into the circle equation confirms both boundary branches.
A Reliable Procedure
When setting up horizontal bounds, use this order of checks:
Mark the relevant curves, intersections, endpoints, and corners. Decide which side of each boundary the region occupies.
Determine the least and greatest heights attained, checking curved extrema as well as intersections and endpoints.
At a fixed height, list the x-inequalities contributed by the region’s boundaries and combine them.
Compare candidate left or right bounds, split the y-range where the chosen boundary changes, and check endpoint heights by substitution.
Common Pitfall: Choosing a Branch Without Checking the Region
A frequent error is to solve \(y=x^2\) as \(x=\sqrt y\) and assume that this is always the left or right boundary. The equation actually has two branches, \(x=-\sqrt y\) and \(x=\sqrt y\). In the semicircle example, the horizontal slice has both branches as bounds. In the first worked example, the region lies only to the right of the y-axis, so the parabola contributes \(x=\sqrt y\), not \(-\sqrt y\).
Another error is to choose one right boundary for an entire y-range even after the curves cross. In the second worked example, using \(x=\sqrt y\) for all \(0\le y\le4\) would put the right edge outside the region for \(y>1\). The correction is to compare the candidates at a common height and split at the switch \(y=1\).
Why These Bounds Matter
For a solid described by cross-sections perpendicular to the y-axis, the horizontal bounds identify the chord of the base region at each height. The chord’s endpoints determine its length and other geometric dimensions. If the endpoints are wrong, the resulting slice description—and any volume calculation built from it—will describe the wrong solid. Getting the projection and boundary changes right is therefore a geometric step, not merely a matter of notation.
Check Your Understanding
For each question, identify the relevant projection or boundary choice before writing bounds.
- For the region between \(y=x^2\) and \(y=2x\), what is the y-projection, and what are the horizontal bounds?
- For the region between \(y=x^2\) and \(y=2-x\), at what height does the right boundary change?
- Why can solving a boundary equation for \(x\) require choosing between two branches?
- What information besides curve intersections may be needed to determine the minimum and maximum heights?
- For the upper half of \(x^2+y^2\le9\), what are the horizontal bounds at height \(y\)?
