Why Three Curves Require Careful Boundary Selection
When three graphs appear in an area problem, it is tempting to assume that all three form the boundary, or to choose the curves that look closest together. Neither approach is reliable. A curve may intersect the others without bounding the particular region of interest. The correct boundary depends on which enclosed region is being considered and on how its edges are traced.
A systematic approach is to find the relevant intersections, identify which region they enclose, and compare the curves at a test point inside that region. The comparison tells us which curve supplies each edge of a slice, so that the area integral has the right limits and the right difference.
Intersections, Order, and the Region of Interest
Suppose three curves are given by \(y=f(x)\), \(y=g(x)\), and \(y=h(x)\). A pairwise intersection is a point where two of these functions have equal values. For example, intersections of \(f\) and \(g\) satisfy \(f(x)=g(x)\), subject to both functions being defined.
Those solutions are candidates for endpoints or boundary junctions, but they do not by themselves tell us which curve bounds a chosen region. We must also establish which region is intended. If the region is described in a diagram or in words, use that information. If the curves form several enclosed regions, identify the particular component before writing an integral.
For vertical slices, a slice at \(x\) runs from a lower \(y\)-value to an upper \(y\)-value. If the lower and upper boundaries are \(y=L(x)\) and \(y=U(x)\), respectively, the area is
Here \(a\) and \(b\) are the left and right endpoints of the region. For horizontal slices, the analogous formula uses the right boundary minus the left boundary:
where \(L(y)\) and \(R(y)\) are the left and right \(x\)-coordinates, and \(c\) and \(d\) are the bottom and top \(y\)-coordinates. These formulas require the stated boundaries to describe the region throughout the interval of integration.
Why a Test Point Determines the Order
Assume two functions \(f\) and \(g\) are continuous on an interval and do not intersect there. Then \(f(x)-g(x)\) is also continuous and never equals zero on that interval. Its sign cannot change: if it changed from positive to negative, the Intermediate Value Theorem would force it to equal zero somewhere in between, which would be an intersection.
Therefore, one test point in the interval is enough to determine the order everywhere on that interval. If \(f(x_0)>g(x_0)\) at the test point \(x_0\), then \(f(x)>g(x)\) throughout the interval, provided the functions remain continuous and have no intersection there. For horizontal slices, apply the same reasoning to the \(x\)-coordinates of the boundaries as functions of \(y\).
A Reliable Boundary-Finding Procedure
Determine which enclosed component the problem asks about. Do not assume every curve shown is part of its boundary.
Solve the equality for each relevant pair of curves. Use the intersection points to locate possible endpoints and junctions.
Decide whether vertical or horizontal slices make the region’s edges easiest to express as functions.
Choose a point between the relevant intersections. Compare \(y\)-values for vertical slices or \(x\)-values for horizontal slices.
Write the positive slice width—top minus bottom or right minus left—over the interval covering the chosen region.
Confirm that the endpoints are on the boundary, the width is nonnegative inside the interval, and the proposed edges actually trace the specified region.
Worked Example: A Third Curve Lies Outside the Lens
Worked Example: A Third Curve Lies Outside the Lens
Consider \(y=x^2\), \(y=2x\), and \(y=0\). Find the area of the lens enclosed by the parabola and the line, and determine whether the \(x\)-axis forms part of its boundary.
First find the intersections of the parabola and the line:
The two curves meet at \((0,0)\) and \((2,4)\). Between these \(x\)-values, test \(x=1\). The three \(y\)-values are
Thus the line is above the parabola, and the parabola is above the \(x\)-axis at this test point. The line and parabola enclose the lens from \(x=0\) to \(x=2\); the \(x\)-axis lies below that lens and is not one of its edges. The area is therefore
Check the integrand: at \(x=1\), it gives \(2-1=1\), which is the vertical distance between the line and parabola there. At the endpoints it is \(0\), as expected where the two boundaries meet. The antiderivative differentiates to \(2x-x^2\), confirming the integration.
Worked Example: Three Curves Form a Triangle
Worked Example: Three Curves Form a Triangle
Find the area of the region in the first quadrant bounded by \(y=0\), \(y=x\), and \(y=4-x\).
The line \(y=x\) meets the \(x\)-axis at \((0,0)\), and \(y=4-x\) meets it at \((4,0)\). The two sloping lines meet where
These intersections identify the triangular region. Horizontal slices are convenient: at height \(y\), the left edge \(y=x\) gives \(x=y\), while the right edge \(y=4-x\) gives \(x=4-y\). For example, at \(y=1\) the edge positions are \(x=1\) and \(x=3\), so the order is left to right as stated. The region runs from \(y=0\) to \(y=2\). Hence
The width is \(4-2y\), which is nonnegative for \(0\le y\le2\), and it becomes zero at the top vertex \(y=2\). Differentiating \(4y-y^2\) gives \(4-2y\). As a geometric check, the triangle has base \(4\) and height \(2\), so its area is \(\frac12(4)(2)=4\), agreeing with the integral.
Worked Example: Choosing Horizontal Slices for a Curved Boundary
Worked Example: Choosing Horizontal Slices for a Curved Boundary
Find the area of the region bounded by \(y=x^2\), \(y=2x\), and \(y=1\) whose boundary includes the segment of \(y=1\) from \((\frac12,1)\) to \((1,1)\).
The intersections with \(y=1\) are found separately. For the line,
For the parabola,
The specified boundary uses the point \((1,1)\), not \((-1,1)\). The line and parabola meet where \(x^2=2x\), so \(x=0\) or \(x=2\); the relevant meeting point for this region is \((2,4)\). Thus the region is enclosed by the horizontal segment at \(y=1\), the line from \((\frac12,1)\) to \((2,4)\), and the right-hand branch of the parabola from \((1,1)\) to \((2,4)\).
Horizontal slices make the curved boundary easy to describe. Rewrite the line as \(x=y/2\). On the relevant right-hand branch, rewrite the parabola as \(x=\sqrt{y}\). Test \(y=2\): the line gives \(x=1\), while the parabola gives \(x=\sqrt{2}\), and \(1<\sqrt2\). Therefore the line is left of the parabola for the interior heights, and the limits are \(1\le y\le4\). The area is
At \(y=4\), the antiderivative is \(\frac{16}{3}-4=\frac{4}{3}\). At \(y=1\), it is \(\frac{2}{3}-\frac14=\frac{8}{12}-\frac{3}{12}=\frac{5}{12}\). Thus
Differentiating the antiderivative gives \(\sqrt y-y/2\), the right-minus-left width. That width is zero at \(y=4\), where the curves meet, and is positive at \(y=2\), since \(\sqrt2-1>0\). These checks agree with the identified boundaries.
A Common Boundary-Selection Mistake
In the first example, a tempting but incorrect calculation is to treat the \(x\)-axis as the lower boundary of the lens and write
The arithmetic is correct, but the boundary choice is not. At \(x=1\), the lens lies between \(y=1\) and \(y=2\), not between \(y=0\) and \(y=2\). The proposed integral adds the area below the lens, and its right endpoint does not close against the \(x\)-axis and line at a shared intersection. For the specified lens, the lower curve is \(y=x^2\), giving \(\int_0^2(2x-x^2)\,dx=4/3\).
The general lesson is that an intersection belonging to some pair of curves does not automatically make those curves the boundaries of the region under discussion. Check the actual slice edges at an interior point, and verify that the proposed edges connect at the limits.
Why Boundary Identification Matters
The definite integral records the widths of slices across a region. If a curve is mistakenly included or excluded, the integral may still be easy to evaluate, but it represents a different geometric set. Correct limits alone are not enough; the integrand must also describe the actual distance between the region’s edges.
Before evaluating, ask three questions:
- Do the endpoints correspond to boundary intersections of the region I selected?
- At an interior test point, are my proposed upper and lower, or right and left, curves in the correct order?
- Does the slice width remain nonnegative and close to zero where the relevant boundaries meet?
If the order of curves changes inside the interval, a single difference may no longer describe every slice. That situation requires locating the change and treating the intervals appropriately. For the regions in the examples above, the selected slice boundaries keep a consistent order throughout the stated limits.
Check Your Understanding
For each question, identify the boundary logic before calculating.
- For \(y=x^2\), \(y=2x\), and \(y=0\), what test point verifies that the \(x\)-axis is not the lower boundary of the lens between the parabola and line?
- Why does one test point determine the order of two continuous, nonintersecting curves on an interval?
- For the triangle bounded by \(y=0\), \(y=x\), and \(y=4-x\), what are the left and right \(x\)-coordinates of a horizontal slice at height \(y\)?
- For \(y=x^2\), \(y=2x\), and \(y=1\), why is \(x=\sqrt y\), rather than \(x=-\sqrt y\), the relevant parabola branch for the region specified in the example?
- What two checks can reveal that an area integral uses a curve that is not actually part of the chosen region’s boundary?
