Why a Three-Curve Area May Need More Than One Integral
When a region is bounded by three curves, its boundary may follow one curve for part of the region and another curve for the rest. A vertical slice still has a height: the upper boundary value minus the lower boundary value. But if either boundary changes at an intersection, one formula for the height may not apply across the whole region.
The essential task is therefore not simply to list the curves or find every intersection. You must identify the particular boundary arcs of the chosen region, determine where those arcs change, and keep the upper and lower values in the correct order on every interval.
Intersections and Boundary Arcs
For curves written as functions of \(x\), an intersection occurs at an \(x\)-value where the functions have equal values. For three functions \(f\), \(g\), and \(h\), the possible intersection values come from solving the three pairwise equations \(f(x)=g(x)\), \(f(x)=h(x)\), and \(g(x)=h(x)\). Only intersections relevant to the chosen region can serve as its partition points.
An important subtlety is that a curve may pass through an \(x\)-interval without its entire graph being part of the region's boundary there. For example, two lines may both have values at \(x=\tfrac12\), even though only one of their line segments lies along the selected region. The boundary consists of particular arcs joining the region's vertices, not automatically every point on every curve.
Once the boundary arcs are identified, find where their formulas meet and sort those \(x\)-values from left to right. Between consecutive partition points, the identity of each boundary arc is fixed. A test value inside each interval helps verify which boundary is above the other.
The Piecewise Area Rule
Suppose the region extends from \(x=a\) to \(x=b\), and its boundary changes at \(c\), where \(a<c<b\). If the upper and lower boundaries are \(u_1,l_1\) from \(a\) to \(c\), and \(u_2,l_2\) from \(c\) to \(b\), then its area is
The functions in each difference must describe the actual boundary arcs on that interval. The formula is not a license to choose whichever of the three curve values happens to be largest or smallest at every \(x\): the selected region and its boundary arcs determine which curves apply.
Why Partitioning Works
Here is the justification for the piecewise rule. On an interval where the upper boundary is \(u(x)\) and the lower boundary is \(l(x)\), a thin vertical slice has height \(u(x)-l(x)\). If its width is \(\Delta x\), its area is approximately \((u(x)-l(x))\Delta x\). Adding these approximate slice areas and taking the limit as the widths shrink gives the definite integral of the slice height.
When the boundary changes at \(c\), the same reasoning applies separately on \([a,c]\) and \([c,b]\). The two parts meet at the partition point and their areas add, so the total area is the sum of the two integrals. More generally, with partition points \(a=x_0<x_1<\cdots<x_n=b\), the area is the sum of the correctly ordered slice-height integrals on all the subintervals.
This setup assumes the stated boundary functions describe the selected region on each interval and that the resulting slice heights are nonnegative. If an intersection merely touches without changing the boundary arc, it may be included as a partition point, but it need not create a new formula.
Worked Example: A Triangle with a Changing Upper Boundary
Worked Example: A Triangle with a Changing Upper Boundary
Find the area of the triangular region bounded by \(y=0\), \(y=x\), and \(y=2-x\).
Step 1: Find the vertices and the relevant partition. The line \(y=x\) meets \(y=0\) at \(x=0\). The line \(y=2-x\) meets \(y=0\) at \(x=2\). The two sloping lines meet where \(x=2-x\), so \(2x=2\), \(x=1\), and \(y=1\). The boundary runs from \((0,0)\) to \((1,1)\) along \(y=x\), then from \((1,1)\) to \((2,0)\) along \(y=2-x\), with \(y=0\) as the lower boundary.
Step 2: Write the area on each interval. The upper boundary is \(y=x\) on \(0\le x\le1\), and \(y=2-x\) on \(1\le x\le2\). Thus
Step 3: Evaluate.
Check: The first contribution is the area of a right triangle with base \(1\) and height \(1\), namely \(\tfrac12\). The second has the same base and height, so it also has area \(\tfrac12\). Their total agrees with the integral calculation.
Worked Example: A Parabola and a Line Form the Upper Boundary
Worked Example: A Parabola and a Line Form the Upper Boundary
Find the area of the region bounded by \(y=0\), \(y=x^2\), and \(y=2-x\).
Step 1: Find the boundary intersections. The parabola meets \(y=0\) at \(x=0\). The line meets \(y=0\) at \(2-x=0\), so \(x=2\). The parabola and line meet where \(x^2=2-x\), or \(x^2+x-2=0\). Factoring gives \((x+2)(x-1)=0\), so their intersection values are \(x=-2\) and \(x=1\). The boundary of the region in the first quadrant uses the intersection at \(x=1\).
Step 2: Identify the boundary arcs. The region is enclosed by the \(x\)-axis from \(x=0\) to \(x=2\), the parabola from \((0,0)\) to \((1,1)\), and the line from \((1,1)\) to \((2,0)\). Thus the upper boundary is \(x^2\) on \([0,1]\), then \(2-x\) on \([1,2]\). For instance, at \(x=\tfrac12\), the boundary arc is the parabola at \(y=\tfrac14\); the line's value there is not part of this region's boundary.
Step 3: Set up and evaluate.
Check: The antiderivatives differentiate back to \(x^2\) and \(2-x\), respectively. The two boundary arcs meet at \((1,1)\), and each is above \(y=0\) on its stated interval, so both contributions are positive. The resulting area is \(\tfrac56\) square units.
Worked Example: The Upper Boundary Changes to a Horizontal Line
Worked Example: The Upper Boundary Changes to a Horizontal Line
Find the area enclosed by \(y=x^2\), \(y=2x\), and \(y=2\).
Step 1: Find the relevant intersections. The parabola and the line \(y=2x\) meet where \(x^2=2x\), so \(x(x-2)=0\), giving \(x=0\) and \(x=2\). The line \(y=2x\) meets \(y=2\) at \(x=1\). The parabola meets \(y=2\) at \(x^2=2\), so the positive intersection is \(x=\sqrt2\). The enclosed region has vertices \((0,0)\), \((1,2)\), and \((\sqrt2,2)\), with the parabola joining \((\sqrt2,2)\) back to \((0,0)\).
Step 2: Determine the upper and lower boundaries. On \(0<x<1\), \(2x>x^2\), so the line is above the parabola. On \(1<x<\sqrt2\), the boundary at the top is \(y=2\), while the parabola remains below it. The upper boundary therefore changes at \(x=1\); the lower boundary is \(x^2\) on both intervals.
Step 3: Integrate the two slice heights.
Check: The first antiderivative differentiates to \(2x-x^2\), and the second to \(2-x^2\). The first height is positive for \(0<x<1\), and the second is positive for \(1<x<\sqrt2\), since \(x^2<2\) there. The final value is positive because \(4\sqrt2>3\).
A Common Mistake: Choosing a Curve That Is Not on the Boundary
In the triangular example, it is tempting to treat \(y=2-x\) as the upper boundary all the way from \(x=0\) to \(x=1\), and \(y=x\) as the upper boundary from \(x=1\) to \(x=2\). That reverses the actual boundary arcs. It gives
which is not the area of the triangular region. At \(x=\tfrac12\), for example, the chosen region extends from \(y=0\) to \(y=\tfrac12\), not to \(y=\tfrac32\). The line \(y=2-x\) has value \(\tfrac32\) there, but that point on the line is not on the triangle's boundary. The correct boundary arcs give
Before integrating, check both questions on every interval: Is this curve actually part of the selected region's boundary here? And, if so, is it the upper or lower boundary?
A Reliable Setup Procedure
For a region bounded by three curves, use this sequence before evaluating any integral:
Solve the pairwise equations and keep the intersection values that belong to the chosen region.
Identify which curve segment joins each pair of relevant vertices. Do not assume every part of every curve borders the region.
Sort the \(x\)-coordinates where those boundary arcs change, and split the interval there.
On each interval, identify the upper and lower boundary and confirm the ordering with a test value if useful.
Integrate upper minus lower on every interval, then verify that the contributions are nonnegative and the boundary assignments match the region.
Why This Skill Matters
A correct area calculation depends on both calculus and geometry. The definite integral adds slice heights, but it cannot correct a mistaken choice of boundary. In a three-curve problem, intersections tell you where a change might occur; tracing the boundary tells you whether that change actually belongs to the region being measured.
Once the boundary arcs and partition points are clear, the integration is usually routine. Careful setup prevents a common error: integrating a curve value that is present in the picture but lies outside the selected region. Keep the geometry, the interval, and the upper-minus-lower order connected at every step.
Check Your Understanding
For each question, focus first on the region's boundary arcs before writing an integral.
- Why can a curve have a value at a particular \(x\) without being part of the selected region's boundary at that slice?
- What equations should be solved to find possible intersections of three curves \(f\), \(g\), and \(h\)?
- For the region bounded by \(y=0\), \(y=x\), and \(y=2-x\), which upper boundary applies on \([0,1]\), and which applies on \([1,2]\)?
- In the region bounded by \(y=x^2\), \(y=2x\), and \(y=2\), at what \(x\)-value does the upper boundary change?
- What two checks can help detect an incorrectly ordered area integral?
