Why Curve Order Matters
When two curves bound a region, an area expression depends on which curve is on which side. For vertical slices, the relevant comparison is between the curves' \(y\)-values at the same \(x\). For horizontal slices, it is between their \(x\)-values at the same \(y\). In either case, the difference should be taken in the order that makes it nonnegative throughout the interval being considered.
A graph can suggest the order, but a comparison at a carefully chosen test point gives a precise check. To extend that comparison across an interval, we also need to know that the curves do not intersect inside that interval. This tutorial focuses on determining and using that order; it does not calculate the resulting area.
Comparing Vertical Curves
Suppose two continuous functions \(y=f(x)\) and \(y=g(x)\) are defined on an interval. At a chosen \(x\)-coordinate, evaluate both functions using that same \(x\). The one with the greater output has the greater \(y\)-value there. If the order is fixed on the interval, the vertical difference from the lower curve to the upper curve is
For example, if \(f(x)>g(x)\) throughout the interval, then \(f\) is above \(g\), and the vertical difference is \(f(x)-g(x)\). Reversing that subtraction would make the difference negative.
Comparing Horizontal Boundaries
For horizontal slices, represent the boundaries as \(x=p(y)\) and \(x=q(y)\). At a chosen \(y\)-coordinate, compare the two \(x\)-values. The greater \(x\)-value is farther to the right. If \(p(y)>q(y)\) throughout the interval, then \(p\) is the right boundary and \(q\) is the left boundary. The horizontal difference is \(p(y)-q(y)\).
Why One Test Point Can Determine the Order
A test point gives a comparison at one location, not automatically across an entire interval. The reason that one comparison can be enough is that continuous curves cannot change order without meeting. The following argument makes that condition explicit.
Here is a proof. Define the difference \(h(x)=f(x)-g(x)\). Since \(f\) and \(g\) are continuous, \(h\) is continuous. The no-intersection condition says \(h(x)\ne 0\) at every interior point. Suppose \(h\) were positive at one interior point and negative at another. The Intermediate Value Theorem would then imply that \(h\) equals zero at some point between them. That would mean \(f=g\) at that point, contradicting the no-intersection condition. So \(h\) cannot change sign in the interval. Its sign at a test point therefore gives its sign throughout.
The same reasoning applies to horizontal boundaries: use the continuous difference \(p(y)-q(y)\) and verify that it does not equal zero in the interval. If the boundaries meet at interval endpoints, that is compatible with the result; the comparison concerns interior points.
A Reliable Test-Point Procedure
Use an interval on which the boundaries have no intersection in its interior. If an intersection divides the range into separate intervals, determine the order separately on each one.
Pick a value that makes the function evaluations straightforward, while ensuring it actually lies in the interval.
For vertical slices, evaluate both \(y\)-functions at the same \(x\). For horizontal slices, evaluate both \(x\)-functions at the same \(y\).
Subtract lower from upper, or left from right. Use the continuity and no-intersection condition to justify that this order applies throughout the interval.
Worked Examples: Using Test Points
Worked Example: Identifying the Upper Curve
Consider \(y=x+2\) and \(y=x^2\) on the interval \((-1,2)\). Determine which curve is above and write the vertical difference.
Choose \(x=0\), which lies inside \((-1,2)\). The line has value \(0+2=2\), while the parabola has value \(0^2=0\). Since \(2>0\), the line is above the parabola at this test point.
To verify that the order holds throughout the interval, subtract the parabola from the line and factor:
For \(-1<x<2\), both \(2-x>0\) and \(x+1>0\), so the difference is positive throughout. The curves meet at the interval endpoints, since \(x^2=x+2\) when \(x=-1\) or \(x=2\). Thus the line is above the parabola in the interior, and the nonnegative vertical difference is \(x+2-x^2\).
Check: At the test point \(x=0\), the difference is \(0+2-0^2=2\), matching the direct comparison \(2-0=2\).
Worked Example: Identifying the Right Boundary
For horizontal slices, compare \(x=4-y^2\) and \(x=y+2\) on \((-2,1)\). Determine the right boundary and write the horizontal difference.
Choose \(y=0\), which is in the interval. The first boundary has \(x\)-value \(4-0^2=4\), and the second has \(x\)-value \(0+2=2\). Since \(4>2\), \(x=4-y^2\) is farther right at the test point.
Factor the difference to check the order across the interval:
When \(-2<y<1\), both \(1-y>0\) and \(y+2>0\). Therefore the difference is positive throughout. The boundaries meet at \(y=-2\) and \(y=1\), and \(4-y^2\) is the right boundary between those values.
Check: At \(y=0\), the difference evaluates to \(2-0-0=2\), which equals the directly compared horizontal distance \(4-2=2\).
Worked Example: Rechecking the Order on Separate Intervals
Compare \(y=x^2\) and \(y=1\) over \([-2,2]\). Their intersections occur at \(x=-1\) and \(x=1\), so use the three intervals separated by those points. Determine the upper curve on each interval and write the corresponding vertical difference.
On \((-2,-1)\), choose \(x=-\frac{3}{2}\). Then \(x^2=\frac{9}{4}\), and \(\frac{9}{4}>1\). Thus \(y=x^2\) is above \(y=1\), and the difference is \(x^2-1\).
On \((-1,1)\), choose \(x=0\). Then \(x^2=0<1\). Thus \(y=1\) is above \(y=x^2\), and the difference is \(1-x^2\).
On \((1,2)\), choose \(x=\frac{3}{2}\). Again \(x^2=\frac{9}{4}>1\). Thus \(y=x^2\) is above \(y=1\), and the difference is \(x^2-1\).
Each comparison is made at a point inside its own interval. The differences confirm the order there: \(x^2-1>0\) when \(|x|>1\), while \(1-x^2>0\) when \(|x|<1\). At \(x=-1\) and \(x=1\), the curves meet, so the order changes across those points.
Check: Substituting the three test points into the stated differences gives \(\frac{9}{4}-1=\frac{5}{4}>0\), \(1-0=1>0\), and \(\frac{9}{4}-1=\frac{5}{4}>0\), respectively, matching each direct comparison.
A Common Mistake: Subtracting in the Wrong Order
Suppose that at a test point the upper curve has value \(5\) and the lower curve has value \(2\). A tempting but incorrect difference is \(2-5=-3\). This is lower minus upper, so it gives a negative value, even though the vertical separation is positive.
The correction is to name the boundaries before subtracting: identify the upper curve and subtract the lower one. For horizontal slices, identify the right and left boundaries first, then subtract left from right. Do not choose subtraction order based on which function is written first.
When a Test Point Is Not Enough
A single comparison does not establish the order on an interval if the curves might intersect inside it. For instance, \(y=x\) and \(y=-x\) have opposite orders on the two sides of \(x=0\). A test point on one side cannot determine the order on the other. The interval must be divided at the intersection, and a test point chosen within each resulting interval.
Likewise, a point outside the interval cannot serve as its test point. Its comparison may describe a different part of the graph, where the curves have a different order. Always check that the test value lies strictly inside the interval being analyzed.
Once the order is established, the resulting difference is the appropriate nonnegative integrand for slices in that direction. That sign check is useful even before any integration: it catches reversed boundaries and helps ensure that the expression represents a geometric separation rather than a negative quantity.
Check Your Understanding
For each question, state which values should be compared and explain why the chosen subtraction gives a nonnegative difference.
- For \(y=f(x)\) and \(y=g(x)\), suppose \(f(2)=7\) and \(g(2)=4\). Which curve is above at \(x=2\), and what is the vertical difference there?
- For horizontal boundaries \(x=p(y)\) and \(x=q(y)\), suppose \(p(-1)=3\) and \(q(-1)=6\). Which boundary is to the right at \(y=-1\), and what is the horizontal difference?
- Why does continuity, together with no intersections inside an interval, allow one test point to determine the ordering throughout the interval?
- The curves \(y=x^2\) and \(y=4\) meet at \(x=-2\) and \(x=2\). Which test point could determine their order on \((-2,2)\), and which curve is above?
- What is wrong with using a test point outside the interval where the order is being determined?
