What Do Vertical-Slice Bounds Describe?
A vertical slice through a base region is taken at a fixed \(x\)-value. Its position is determined by \(x\), while its extent within the region is measured in the \(y\)-direction. Before using vertical slices to describe cross-sections of a solid, we must identify every \(x\)-value for which a slice meets the base.
Geometrically, this means projecting the base region onto the \(x\)-axis. Imagine dropping a vertical line through each point of the region and recording where it meets the \(x\)-axis. The resulting set of \(x\)-values is the set of valid slice positions. It may be a single interval, or it may consist of several separate intervals.
It is useful to keep two different questions separate. The slice bounds describe which \(x\)-values occur. The slice’s vertical extent describes which \(y\)-values are in the region at a particular \(x\). The curves that give the top and bottom of a slice help answer the second question; their intersections, together with any other boundary points, can help determine the first.
A General Principle
To justify the projection statement, take any \(x\in[a,b]\). Both \(u(x)\) and \(\ell(x)\) are defined, and the point \((x,\ell(x))\) belongs to the region. Thus every \(x\) in \([a,b]\) occurs in the projection. Conversely, by the definition of this region, every point in it has an \(x\)-coordinate between \(a\) and \(b\). No other \(x\)-values occur, so the projection is exactly \([a,b]\).
This simple argument also points to a limitation: when a region is not described by one pair of graphs over one interval, we cannot assume its projection is one interval. We must find the set of \(x\)-values directly, taking account of each part of the region.
Finding Candidate Endpoints
For a region bounded by curves, intersections often provide the endpoints of the \(x\)-intervals. At an intersection, two boundary descriptions meet. To find its \(x\)-coordinate, set their \(y\)-expressions equal and solve for \(x\). If a boundary is a vertical line \(x=c\), its \(x\)-coordinate is already specified.
Intersection calculations produce candidates, not an automatic answer. Check that the points belong to the boundary of the region being considered. An intersection may lie outside the relevant portion of a curve, or it may be an intersection of internal boundaries rather than an outer edge of the region. For a region described by inequalities, a reliable check is to ask whether a vertical slice can meet the region for \(x\)-values just inside and just outside each candidate endpoint.
Record its boundary curves and any inequalities or stated components.
Solve relevant boundary intersections and include any vertical boundary values.
Determine which \(x\)-values actually have at least one point in the region.
Use one interval for a continuous horizontal extent, or a union of intervals when there are gaps.
Worked Examples
Worked Example: A Parabola and a Horizontal Line
Find the \(x\)-interval for vertical slices of the region bounded by \(y=x^2\) and \(y=2\).
First locate their intersections by equating their \(y\)-values:
Between these values, \(x^2\le 2\). Therefore, at each \(x\) in this interval, the parabola is below the horizontal line, and the region includes the points with \(x^2\le y\le 2\). If \(x<-\sqrt{2}\) or \(x>\sqrt{2}\), then \(x^2>2\), so the parabola is above the line and there is no point between them with the specified ordering.
Thus the projection, and hence the range of vertical-slice positions, is \([-\sqrt{2},\sqrt{2}]\). At either endpoint, \(x^2=2\), so the two boundaries meet and the vertical slice has zero height. At \(x=0\), the slice runs from \(y=0\) to \(y=2\), confirming that this interior \(x\)-value is included.
Check: At the endpoints, the boundary equation gives \(y=2\); at \(x=0\), the inequalities give \(0\le y\le2\). These checks agree with the stated interval and slice extent.
Worked Example: Two Curves That Meet Twice
Find the \(x\)-interval for the bounded region between \(y=x^2\) and \(y=x\).
At an intersection, \(x^2=x\). Factor the equation:
To determine which curve is above the other between these endpoints, compare \(x\) and \(x^2\). For \(0\le x\le1\), both \(x\ge0\) and \(1-x\ge0\), so \(x(1-x)\ge0\), which is equivalent to \(x\ge x^2\). The region between the curves therefore has vertical slices for every \(x\in[0,1]\), with \(x^2\le y\le x\).
The \(x\)-interval is \([0,1]\). Outside this interval, the curves still exist, but that does not make those \(x\)-values part of this bounded region. For example, at \(x=-1\), the values are \(x^2=1\) and \(x=-1\); this lies outside the interval between the two intersection points.
Check: At \(x=0\), both curves give \(y=0\), and at \(x=1\), both give \(y=1\). At the interior value \(x=\tfrac12\), they give \(y=\tfrac14\) and \(y=\tfrac12\), so the slice lies between distinct boundary values.
Worked Example: A Region with Two Separate Projection Intervals
Let \(R\) be the union of the two disks \((x+2)^2+y^2\le1\) and \((x-2)^2+y^2\le1\). Find the \(x\)-intervals for vertical slices of \(R\).
For the left disk, a vertical slice can meet the disk exactly when there is some \(y\) satisfying \((x+2)^2+y^2\le1\). Since \(y^2\ge0\), this is possible only if \((x+2)^2\le1\). It is also sufficient: when that inequality holds, \(y=0\) satisfies the disk inequality. Solving gives
For the right disk, the same existence check gives \((x-2)^2\le1\), so
The projection of the union is therefore \([-3,-1]\cup[1,3]\), not the single interval \([-3,3]\). For example, at \(x=0\), neither disk contains a point: the left disk would require \(4+y^2\le1\), and the right disk would also require \(4+y^2\le1\). Hence the gap \((-1,1)\) contains no valid slice positions.
As an additional check, at \(x=-2\) the left disk permits \(-1\le y\le1\); at \(x=2\) the right disk permits the same range. At each listed endpoint, the corresponding disk’s slice reduces to \(y=0\). These checks confirm both intervals and their endpoints.
Checking a Projection Carefully
For a complicated region, drawing a rough sketch can help reveal whether the projection has gaps, but the sketch should be supported by equations or inequalities. The most direct question is: for this \(x\), does at least one \(y\) place \((x,y)\) in the region? This is the existence test in the definition of projection.
If the region is specified by inequalities, use them to decide whether a suitable \(y\) exists. The disk example used a particularly simple test: the squared \(y\)-term is nonnegative, and choosing \(y=0\) establishes sufficiency once the remaining inequality is met. In other regions, finding an \(x\)-range may require checking the allowed \(y\)-values against more than one boundary.
When the region is made up of several components, determine the projection of each component and then take their union. Do not fill a gap simply because the leftmost and rightmost \(x\)-coordinates are known. A vertical slice through that gap would miss the base entirely.
A Common Mistake: Using a \(y\)-Value as an \(x\)-Bound
For \(y=x^2\) and \(y=2\), a mistaken calculation might say: “The horizontal boundary is \(y=2\), so the right endpoint is \(x=2\).” This confuses a \(y\)-coordinate with an \(x\)-coordinate. The endpoint occurs where the two boundary equations have equal \(y\)-values, so the required equation is \(x^2=2\), not \(x=2\).
The wrong value \(x=2\) fails a direct check: at \(x=2\), the parabola has \(y=4\), while the line has \(y=2\). These boundaries do not meet there. The correct endpoints are \(x=-\sqrt2\) and \(x=\sqrt2\), where both equations give \(y=2\).
Why These Bounds Matter
In a cross-section volume problem, each valid \(x\)-value identifies a position at which the base can be sliced vertically. The bounds tell us where those slices occur. If the base has disconnected horizontal extent, treating it as one continuous interval would include positions with no slice through the base. That changes the region being described, even before any volume calculation is attempted.
It is equally important not to confuse the bounds with the slice’s vertical dimensions. In the region between \(y=x^2\) and \(y=2\), for instance, the \(x\)-bounds are \(-\sqrt2\) and \(\sqrt2\), while the vertical boundaries at a particular \(x\) are \(y=x^2\) and \(y=2\). The first pair gives the range of slice positions; the second pair describes the slice at one position.
Check Your Understanding
For each question, identify the projection onto the \(x\)-axis and explain how you checked it.
- For the region between \(y=x^2\) and \(y=9\), what are the \(x\)-bounds?
- Find the \(x\)-interval of the bounded region between \(y=x^2\) and \(y=2x\).
- A disk is described by \((x-3)^2+y^2\le4\). For which \(x\)-values can a vertical slice meet it?
- Why do two disks with separated projections require a union of intervals rather than one interval between the leftmost and rightmost endpoints?
- In the region between two graphs, what is the difference between the \(x\)-bounds and the \(y\)-values at the top and bottom of a vertical slice?
