How Three Curves Can Enclose a Region
A region enclosed by three curves is not necessarily bounded by all three curves over their entire graphs. Usually, only particular segments of each curve form the boundary. The challenge is to trace those segments in order, find where the boundary changes from one curve to another, and then represent the area without counting regions that are not part of the enclosure.
For vertical slices, the area at a given \(x\)-value is the vertical distance from the lower boundary to the upper boundary. If the curve forming either boundary changes as \(x\) varies, the integral must be split at the change. Pairwise intersections help identify possible change points, but a point of intersection is relevant only if it lies on the boundary of the region being considered.
Finding the Active Boundary
Suppose three curves are given as graphs \(y=f(x)\), \(y=g(x)\), and \(y=h(x)\). To identify a region using vertical slices, first locate the pairwise intersections that could serve as endpoints or boundary changes. Next, determine which parts of the curves connect to form a closed loop. Between consecutive relevant intersection \(x\)-values, use a test point to compare the curve heights. This identifies the curve that forms the top of the chosen region and the curve that forms its bottom.
There can be several bounded regions in the same graph. A single intersection may belong to one region but not another. Also, two curves can intersect without changing their order: for example, they may touch and turn away. For that reason, the intersection list gives candidate locations, not an automatic instruction to split every integral there.
Why the Integral Must Be Split at a Boundary Change
The ordering of two continuous curves cannot switch on an interval without the curves meeting somewhere in that interval. Indeed, if \(f(x)-g(x)\) is positive at one point and negative at another, the Intermediate Value Theorem implies that \(f(x)-g(x)=0\) between those points. Thus, after finding the relevant intersections, a test point in each resulting interval identifies the ordering throughout that interval, provided there are no other intersections inside it.
Here is why this gives the correct area setup. On an interval where \(U\) is the upper boundary and \(L\) is the lower boundary, the vertical-slice formula gives area \(\int (U(x)-L(x))\,dx\). If the boundary changes at \(c\), the part to the left of \(c\) and the part to the right of \(c\) have their own upper and lower boundary functions. Area is additive across adjacent regions, and definite integrals are additive across adjacent intervals. Therefore the total area is the sum of the integrals for the boundary pieces.
This is not merely a matter of making the integrand easier to integrate. The pieces must match the actual boundary of the region. If one curve stops bounding the region at \(x=c\), continuing to use it beyond \(c\) would calculate the area of a different set.
Worked Example: A Three-Sided Region with Two Sloping Curves
Worked Example: A Three-Sided Region with Two Sloping Curves
Find the area of the region bounded by \(y=0\), \(y=x\), and \(y=2-x\).
Step 1: Find the pairwise intersections. The line \(y=x\) meets \(y=0\) when \(x=0\). The line \(y=2-x\) meets \(y=0\) when \(x=2\). The two sloping lines meet when \(x=2-x\), so \(2x=2\) and \(x=1\). Their common height there is \(y=1\).
These points connect the three segments into a closed triangular region. Its horizontal extent is from \(x=0\) to \(x=2\), and the upper boundary changes at \(x=1\).
Step 2: Test the upper boundary on each interval. At \(x=\frac12\), the sloping lines have heights \(\frac12\) and \(\frac32\), so the upper boundary of the region above \(y=0\) is \(y=x\). At \(x=\frac32\), their heights are \(\frac32\) and \(\frac12\), so the upper boundary is \(y=2-x\). In both intervals, the lower boundary is \(y=0\).
Step 3: Set up and evaluate the area.
Check: The first triangular piece has base \(1\) and height \(1\), so its area is \(\frac12\). The second has the same base and height, so it also has area \(\frac12\). Together they give \(1\) square unit, agreeing with the integral.
Worked Example: A Curved Boundary That Changes at an Intersection
Worked Example: A Curved Boundary That Changes at an Intersection
Find the area of the region bounded by \(y=0\), \(y=x^2\), and \(y=2-x\) in the enclosure that lies above the \(x\)-axis.
Step 1: Find the intersections relevant to the enclosure. The parabola \(y=x^2\) meets the \(x\)-axis when \(x^2=0\), giving \(x=0\). The line \(y=2-x\) meets the \(x\)-axis when \(2-x=0\), giving \(x=2\). The parabola and line meet when \(x^2=2-x\), or \(x^2+x-2=0\). Factoring gives \((x+2)(x-1)=0\), so \(x=-2\) or \(x=1\). The intersection at \(x=1\) is on the boundary of this enclosure; \(x=-2\) lies outside its horizontal extent.
Step 2: Determine the upper boundary. The region runs from \(x=0\) to \(x=2\), with lower boundary \(y=0\). At \(x=\frac12\), the parabola has height \(\frac14\) and the line has height \(\frac32\); the parabola is the upper edge of this particular region. At \(x=\frac32\), the heights are \(\frac94\) and \(\frac12\); the line is the upper edge. The boundary switches at their intersection \(x=1\).
Step 3: Split and evaluate the area.
Check: The first integral is the area under \(y=x^2\) from \(0\) to \(1\), and the second is the area under \(y=2-x\) from \(1\) to \(2\); both lie above \(y=0\). Their positive values add to \(\frac56\) square unit.
Worked Example: Choosing the Boundary When Both Curves Are Above a Base
Worked Example: Choosing the Boundary When Both Curves Are Above a Base
Find the area of the region bounded by \(y=1\), \(y=x^2\), and \(y=10-x^2\) on the right side of the \(y\)-axis.
Step 1: Identify the intersections that form the closed boundary. The curve \(y=x^2\) meets \(y=1\) at \(x^2=1\), so the right-side intersection is \(x=1\). The curve \(y=10-x^2\) meets \(y=1\) at \(10-x^2=1\), giving \(x^2=9\), so the right-side intersection is \(x=3\). The two parabolas meet when \(x^2=10-x^2\), so \(2x^2=10\), \(x^2=5\), and the right-side intersection is \(x=\sqrt{5}\). These three points join the curve segments into the selected closed region, which extends from \(x=1\) to \(x=3\).
Step 2: Determine which parabola forms the upper edge. At \(x=2\), the parabola heights are \(4\) and \(6\), so \(y=x^2\) is the upper boundary of the region above \(y=1\). At \(x=\frac52\), the heights are \(\frac{25}{4}\) and \(\frac{15}{4}\), so \(y=10-x^2\) is the upper boundary. The switch occurs at \(x=\sqrt{5}\), where both parabolas have height \(5\). The lower boundary on both intervals is \(y=1\).
Step 3: Set up and evaluate both pieces.
Check: The derivative of \(\frac{x^3}{3}-x\) is \(x^2-1\), and the derivative of \(9x-\frac{x^3}{3}\) is \(9-x^2\), the two slice heights used. The area is approximately \(3.760\) square units, a positive value. The two integrals meet at \(x=\sqrt{5}\), exactly where the upper boundary changes.
A Common Mistake: Using the Same Curve Across the Whole Interval
In the second example, the line and parabola cross at \(x=1\). A tempting but incorrect setup is to calculate the signed difference between those two curves over the entire interval from \(0\) to \(2\):
This is not the area of the specified enclosure. It uses the difference between the line and parabola rather than the distance from the \(x\)-axis to the active upper boundary. It also carries a negative sign on part of the interval, because the parabola rises above the line after \(x=1\).
A Boundary-Tracking Checklist
When three curves create a possible enclosure, organize the setup before integrating:
- Find the pairwise intersections that could be endpoints or boundary changes.
- Trace the curve segments that connect to make the particular closed region you want.
- Record the region's leftmost and rightmost \(x\)-values.
- Between relevant intersection values, test which curves actually form the top and bottom of that region.
- Split the integral where either boundary changes, and use upper minus lower on each interval.
- Check that every selected slice lies inside the traced region and that the resulting area is positive.
The central task is not simply solving three pairs of equations. It is deciding which solutions belong to the same closed boundary and which curve is active on each part of that boundary. Once that is clear, the integral follows directly from the vertical-slice area formula.
Check Your Understanding
For each question, identify the boundary pieces before deciding how to set up the area.
- For \(y=0\), \(y=x\), and \(y=4-x\), find the intersections that form the closed region above the \(x\)-axis and state where its upper boundary changes.
- For \(y=0\), \(y=x^2\), and \(y=3-x\), find the relevant boundary-change point for the enclosure above the \(x\)-axis.
- Why does finding every pairwise intersection not by itself determine which integral represents a chosen enclosed region?
- What information does a test point provide when checking the curves on an interval between intersections?
- Suppose the upper boundary changes at \(x=c\), while the lower boundary stays the same. What must change in the area setup at \(c\)?
