Why Vertical-Slice Bounds Need Checking
A vertical slice fixes an \(x\)-value and looks at the points of the region directly above or below it. To describe those slices correctly, you need to know two different things: the interval of \(x\)-values where the region exists, and the lower and upper boundaries of the region at each such \(x\).
Solving equations where boundary curves meet gives important candidate endpoints. But an intersection calculation alone does not guarantee that the curves have been assigned to the correct sides of the region. A quick check between the intersections is essential, especially when the region is described as lying below more than one curve.
A Reliable Method
Suppose a region is bounded by curves \(y=f(x)\) and \(y=g(x)\). Intersections satisfy \(f(x)=g(x)\), so solving this equation identifies candidate \(x\)-values where one curve may meet or cross the other. These values often provide the endpoints of the region’s \(x\)-interval. They do not, by themselves, establish which curve is above the other inside that interval.
List the curves or lines that enclose the region and determine which intersections could be endpoints of its \(x\)-extent.
Set the relevant boundary equations equal and solve. Include intersections with any other boundary, such as the \(x\)-axis.
At a value strictly between consecutive intersections, compare the boundary heights or test whether a point lies inside the stated region.
Give the full \(x\)-interval and, for each \(x\) in it, identify the lower and upper \(y\)-boundaries.
If the same two continuous curves form the slice boundaries throughout an interval, their order cannot switch inside that interval unless they intersect there. This fact helps make a test at one interior value useful, provided all possible intersections have been found.
Worked Example: A Region Below Two Lines
Worked Example: A Triangular Region with a Changing Upper Boundary
Find the \(x\)-bounds and describe the vertical slices of the region bounded by \(y=0\), \(y=x\), and \(y=2-x\).
Step 1: Find the boundary intersections. The line \(y=x\) meets \(y=0\) when \(x=0\). The line \(y=2-x\) meets \(y=0\) when \(2-x=0\), so \(x=2\). The two sloping lines meet when
Their intersection has \(y=1\). The vertices therefore have \(x\)-coordinates \(0\), \(1\), and \(2\), so the region’s \(x\)-extent is \(0\le x\le2\).
Step 2: Check which sloping line forms the upper boundary. The region lies above \(y=0\) and below both sloping lines. Thus, for any fixed \(x\), its upper boundary is the smaller of \(x\) and \(2-x\). For \(0\le x\le1\), \(x\le2-x\), so the upper boundary is \(y=x\). For \(1\le x\le2\), \(2-x\le x\), so the upper boundary is \(y=2-x\).
For instance, at \(x=\frac12\), the sloping-line heights are \(\frac12\) and \(\frac32\), and the region extends from \(y=0\) to \(y=\frac12\). At \(x=\frac32\), the heights are \(\frac32\) and \(\frac12\), and the region extends from \(y=0\) to \(y=\frac12\).
Check: At \(x=0\) and \(x=2\), the slice has zero height because the boundary meets \(y=0\). At \(x=1\), both sloping boundaries give \(y=1\), as required at their intersection.
Worked Example: Two Curves Enclosing a Region
Worked Example: A Parabola and a Line
Find the \(x\)-bounds and the lower and upper boundaries for the region enclosed by \(y=x^2\) and \(y=2x\).
Step 1: Solve for intersections.
Thus, \(x=0\) or \(x=2\). The corresponding intersection points are \((0,0)\) and \((2,4)\). These give the candidate \(x\)-interval \(0\le x\le2\).
Step 2: Verify the order between the intersections. Subtract the parabola’s height from the line’s height:
For \(0\le x\le2\), both \(x\) and \(2-x\) are nonnegative, so \(x(2-x)\ge0\). Therefore \(2x\ge x^2\) throughout the interval: the line is the upper boundary and the parabola is the lower boundary.
For a sample check at \(x=1\), the line has height \(2(1)=2\) and the parabola has height \(1^2=1\), confirming the order.
Check: At \(x=0\), both curves give \(y=0\). At \(x=2\), both give \(y=4\). These endpoint checks agree with the intersection calculation.
Worked Example: A Horizontal Line and a Parabola
Worked Example: A Region Between \(y=x^2\) and \(y=4\)
Find the \(x\)-bounds and describe the vertical slices of the region enclosed by \(y=x^2\) and \(y=4\).
Step 1: Find the intersections.
The intersection points are \((-2,4)\) and \((2,4)\), giving the \(x\)-interval \([-2,2]\).
Step 2: Verify the boundary order throughout the interval. For \(-2\le x\le2\), \(-2\le x\le2\) implies \(0\le x^2\le4\). Hence the parabola is the lower boundary and the horizontal line is the upper boundary throughout the interval.
For example, at \(x=-1\), \(x^2=1\), so the slice runs from \(y=1\) to \(y=4\). At \(x=0\), it runs from \(y=0\) to \(y=4\).
Check: At both endpoints, \(x^2=4\), matching the line \(y=4\). The slice height is zero there, consistent with the curves meeting.
Why the Boundary Order Can Be Checked
Here is a useful justification for the test-value method. Let \(f\) and \(g\) be continuous on an interval, and suppose they do not intersect inside that interval. Define their difference by \(h(x)=f(x)-g(x)\). If the order changed, there would be two points \(a\) and \(b\) in the interval where \(h(a)>0\) and \(h(b)<0\), or the reverse.
Because \(h\) is continuous, the Intermediate Value Theorem says that it must take every value between \(h(a)\) and \(h(b)\). In particular, it must equal \(0\) at some point between \(a\) and \(b\). At that point, \(f(x)-g(x)=0\), so \(f(x)=g(x)\): the curves intersect. This contradicts the assumption that they do not intersect inside the interval. Therefore the order cannot change without an intersection.
Common Mistake: Choosing the Higher of Two Upper Limits
When a region lies below two curves, a frequent error is to choose the higher curve as the upper boundary. But a point in the region must lie below both curves. It cannot extend above the lower one.
In the triangular example, at \(x=\frac12\), the two sloping lines have heights \(y=\frac12\) and \(y=\frac32\). Choosing \(y=\frac32\) as the upper boundary would include points such as \(\left(\frac12,1\right)\). That point is above \(y=x\), since \(1>\frac12\), so it is outside the region. The correct upper boundary there is \(y=x\), and the slice runs from \(y=0\) to \(y=\frac12\).
At \(x=\frac32\), choosing \(y=x=\frac32\) would similarly include points above \(y=2-x=\frac12\). The correct upper boundary is \(y=2-x\), and the slice runs from \(y=0\) to \(y=\frac12\). These checks expose the error: being below one boundary does not suffice when the region must also be below another.
Keep the \(x\)-Interval and Slice Boundaries Distinct
The \(x\)-bounds answer, “Where does the region extend from left to right?” The lower and upper boundaries answer, “For this particular \(x\), which \(y\)-values belong to the slice?” These are related but different pieces of information. An intersection may set an endpoint of the \(x\)-interval, while the curves’ relative heights determine the slice limits in between.
Before using vertical slices in an application, check that the proposed interval covers the region and that the proposed boundaries contain every slice. Solve for intersections, compare boundary values between them, and test endpoints. When a third boundary is present, such as the \(x\)-axis in the triangular example, check it as well; intersections between the other two curves alone do not describe the whole region.
Check Your Understanding
For each question, identify the \(x\)-bounds and verify the boundary order rather than relying only on a sketch.
- Find the \(x\)-bounds for the region enclosed by \(y=x^2\) and \(y=3x\). Which curve is above the other between the intersections?
- A region lies above \(y=0\) and below both \(y=2x\) and \(y=4-x\). Find the \(x\)-coordinates where the relevant boundaries meet.
- For the region between \(y=1\) and \(y=x^2\), solve for the intersection \(x\)-values and determine which curve is the upper boundary between them.
- In a region below two curves, why must the upper boundary be the lower of the two curve heights at a fixed \(x\)?
- Explain why two continuous curves cannot change order on an interval without intersecting there.
