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Integrals over General Regions · Tutorial 540 of 999

Regions Between Two Curves With Switching Order

Use intersections and sign tests to identify which curve is above the other, split vertical slices where needed, and write bounds that describe the intended region.

Intermediate 10 min read

What You'll Learn

  • Find all intersections of two boundary curves on a specified interval.
  • Use a test point in each interval to determine which curve is upper and which is lower.
  • Split vertical-slice bounds at crossings where the order changes.
  • Distinguish crossings that change order from tangencies that do not.
  • Write and check iterated integrals over regions between curves.

Why the Order of the Curves Matters

To describe a region between two curves using vertical slices, each slice needs a lower boundary and an upper boundary. If the curves switch order as \(x\) changes, one pair of bounds cannot describe the whole region correctly. The interval must be split wherever the order changes.

The key is not to guess which curve is on top from its equation or from a single sketch. Find their intersections, compare their values between successive intersections, and use the resulting order to write the bounds.

Definition: Suppose a region lies between the graphs \(y=f(x)\) and \(y=g(x)\) for \(a\leq x\leq b\). A vertical slice at \(x\) runs from the curve with the smaller \(y\)-value to the curve with the larger \(y\)-value. If that order changes within \([a,b]\), split the interval at the points where it changes.

Intersections and Order Tests

The curves intersect where \(f(x)=g(x)\). Solving this equation gives candidate \(x\)-values at which the upper-lower order might change. Sort the intersection values that lie in the region’s \(x\)-interval. They divide the interval into smaller intervals. On each smaller interval, choose a test value and compare \(f(x)\) and \(g(x)\): the larger value is the upper boundary, and the smaller is the lower boundary.

A crossing often changes the order, but an intersection does not automatically mean that it does. A curve can touch another curve and turn back without passing through it. Therefore, check the order on both sides of each intersection instead of assuming that it switches.

Useful test: Let \(h(x)=f(x)-g(x)\). Where \(h(x)>0\), \(f\) is above \(g\); where \(h(x)<0\), \(f\) is below \(g\). A change in the sign of \(h\) means the order changes.

Why the Order Stays Fixed Between Intersections

This sign test works because continuous curves cannot change order without meeting. Here is the precise argument. Suppose \(f\) and \(g\) are continuous on an interval, and suppose they do not intersect there. Then \(h=f-g\) is continuous and has no zero on that interval. If \(h\) were positive at one point and negative at another, the Intermediate Value Theorem would imply that \(h\) equals zero somewhere between them. That would mean \(f=g\) at that point, contradicting the assumption that the curves do not intersect. Thus the sign of \(h\)—and therefore the order of the curves—stays fixed between consecutive intersections.

Consequence: For continuous boundary curves, once all intersections in the \(x\)-interval have been found, one test point in each resulting open interval determines the order throughout that interval.

Worked Example: Two Curves with One Internal Change

Worked Example: Two Curves with One Internal Change

Describe the region between \(y=x^2\) and \(y=x\) for \(-1\leq x\leq 1\) using vertical slices, then find its area.

First find the intersections:

$$ x^2=x \quad\Longrightarrow\quad x(x-1)=0 \quad\Longrightarrow\quad x=0\ \text{or}\ x=1. $$

The interval \([-1,1]\) is divided at \(x=0\) and \(x=1\). Test \(x=-\frac12\) on \([-1,0]\): \(x^2=\frac14\) and \(x=-\frac12\), so \(x^2\) is above \(x\). Test \(x=\frac12\) on \((0,1)\): \(x=\frac12\) and \(x^2=\frac14\), so \(x\) is above \(x^2\). The curves meet at the endpoint \(x=1\).

Thus the vertical-slice bounds are:

$$ \begin{array}{ll} -1\leq x\leq 0: & x\leq y\leq x^2,\\[2pt] 0\leq x\leq 1: & x^2\leq y\leq x. \end{array} $$

For the area, integrate upper minus lower on each interval:

$$ \begin{aligned} A &=\int_{-1}^{0}(x^2-x)\,dx+\int_{0}^{1}(x-x^2)\,dx\\ &=\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_{-1}^{0} +\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_{0}^{1}\\ &=0-\left(-\frac13-\frac12\right) +\left(\frac12-\frac13\right)-0\\ &=\frac56+\frac16=1. \end{aligned} $$

Check the order at the test points: on the left, \(x^2-x=\frac14-(-\frac12)=\frac34>0\); on the right, \(x-x^2=\frac12-\frac14=\frac14>0\). Both integrals therefore use a nonnegative slice height. The intersections also satisfy \(x^2=x\) at \(x=0\) and \(x=1\), as required.

Worked Example: Several Order Changes

Worked Example: Several Order Changes

For \(-2\leq x\leq 2\), find vertical-slice bounds and the area between \(y=x^3\) and \(y=x\).

Find every intersection:

$$ x^3=x \quad\Longrightarrow\quad x(x-1)(x+1)=0 \quad\Longrightarrow\quad x=-1,\ 0,\ 1. $$

These points divide the interval into four parts. Compare the curves at \(x=-\frac32,-\frac12,\frac12,\frac32\), respectively:

$$ \begin{array}{c|c|c|c} \text{Test }x & x^3 & x & \text{Upper curve}\\ \hline -\frac32 & -\frac{27}{8} & -\frac32 & y=x\\ -\frac12 & -\frac18 & -\frac12 & y=x^3\\ \frac12 & \frac18 & \frac12 & y=x\\ \frac32 & \frac{27}{8} & \frac32 & y=x^3 \end{array} $$

The bounds are consequently:

$$ \begin{array}{ll} -2\leq x\leq -1: & x^3\leq y\leq x,\\ -1\leq x\leq 0: & x\leq y\leq x^3,\\ 0\leq x\leq 1: & x^3\leq y\leq x,\\ 1\leq x\leq 2: & x\leq y\leq x^3. \end{array} $$

Calculate the area by integrating the upper curve minus the lower curve in each interval:

$$ \begin{aligned} A={}&\int_{-2}^{-1}(x-x^3)\,dx +\int_{-1}^{0}(x^3-x)\,dx\\ &+\int_{0}^{1}(x-x^3)\,dx +\int_{1}^{2}(x^3-x)\,dx. \end{aligned} $$

For the first integral, an antiderivative is \(\frac{x^2}{2}-\frac{x^4}{4}\): its values at \(-1\) and \(-2\) are \(\frac14\) and \(-2\), so the result is \(\frac14-(-2)=\frac94\). For the second, an antiderivative is \(\frac{x^4}{4}-\frac{x^2}{2}\): its values at \(0\) and \(-1\) are \(0\) and \(-\frac14\), giving \(\frac14\). The third integral is \(\frac12-\frac14=\frac14\). For the fourth, the same antiderivative \(\frac{x^4}{4}-\frac{x^2}{2}\) has values \(2\) at \(2\) and \(-\frac14\) at \(1\), giving \(\frac94\). Therefore:

$$ A=\frac94+\frac14+\frac14+\frac94=5. $$

Each computed area is positive, consistent with the upper-lower order in its interval. Substitution into \(x^3=x\) confirms all three split points: \((-1)^3=-1\), \(0^3=0\), and \(1^3=1\).

Writing the Integral for a General Integrand

The same bounds apply when integrating a function over the region, not just when calculating its area. For the region in the second example, let the integrand be a continuous function \(q(x,y)\). The vertical-slice description gives the iterated integral below. The split is necessary because the inner limits change when the curves exchange order.

$$ \begin{aligned} \iint_R q(x,y)\,dA ={}&\int_{-2}^{-1}\int_{x^3}^{x}q(x,y)\,dy\,dx +\int_{-1}^{0}\int_x^{x^3}q(x,y)\,dy\,dx\\ &+\int_{0}^{1}\int_{x^3}^{x}q(x,y)\,dy\,dx +\int_{1}^{2}\int_x^{x^3}q(x,y)\,dy\,dx. \end{aligned} $$

To check the bounds, fix any \(x\) strictly inside one of the four intervals. The inner integral starts at the smaller of \(x^3\) and \(x\), and ends at the larger. At a split point where the curves meet, the slice height is zero. Thus adjoining intervals meet without creating a gap or overlapping vertical slices.

Worked Example: Testing Whether Order Actually Changes

Worked Example: Testing Whether Order Actually Changes

Consider \(y=x^2\) and \(y=0\) for \(-1\leq x\leq 1\). Their only intersection is \(x=0\). Determine the slice bounds and area.

For \(x=-\frac12\), the curve values are \(x^2=\frac14\) and \(0\), so \(y=x^2\) is above \(y=0\). For \(x=\frac12\), they are again \(\frac14\) and \(0\). The order is the same on both sides: the curves touch at \(0\), but do not switch order. Therefore a valid description uses one set of bounds across the whole interval:

$$ -1\leq x\leq 1, \qquad 0\leq y\leq x^2. $$

The area is:

$$ A=\int_{-1}^{1}(x^2-0)\,dx =\left[\frac{x^3}{3}\right]_{-1}^{1} =\frac13-\left(-\frac13\right) =\frac23. $$

Check: \(x^2\geq0\) for every \(x\), so the stated lower and upper boundaries are correctly ordered on the whole interval. The curves agree at \(x=0\), as expected, and the computed area is positive.

A Common Mistake: Assuming Every Intersection Reverses the Order

A common error is to treat every intersection as an order switch. For \(y=x^2\) and \(y=0\), that would lead someone to put \(y=x^2\) below \(y=0\) on \([-1,0]\), producing the alleged slice height \(0-x^2\). At \(x=-\frac12\), this height is \(0-\frac14=-\frac14\), which cannot be the height of a vertical slice. The correction is to compare the curves on both sides of the intersection: \(x^2\) is positive on both sides, so it stays above zero. An intersection is a place to check for a change, not proof that a change occurred.

A Reliable Procedure

For two continuous curves over a specified \(x\)-interval, the following process gives valid vertical bounds:

1
Find the intersections.
Solve \(f(x)=g(x)\), and keep the solutions that lie in the interval describing the region.
2
Split at the candidate points.
List the intersection \(x\)-values in order, together with the interval’s endpoints.
3
Test each interval.
Choose an interior \(x\)-value and compare the two function values to identify the upper and lower curves.
4
Write the bounds in order.
For each interval, put the lower curve first and the upper curve second in the inner \(y\)-limits.
5
Check the description.
Substitute intersection values into both curves and verify that each chosen slice has nonnegative height.

Check Your Understanding

Use intersections and test points to decide whether the bounds need to be split, and identify the lower and upper curve on each interval.

  1. For \(y=x^2\) and \(y=2x\) on \(0\leq x\leq 3\), find the intersections and state the upper curve on each resulting interval.
  2. For \(y=x^2-1\) and \(y=0\) on \(-2\leq x\leq 2\), determine whether the order changes at every intersection.
  3. For \(y=x^3\) and \(y=2x\) on \(-2\leq x\leq 2\), find all intersection values and describe how you would test the order between them.
  4. Explain why two continuous curves that do not intersect on an interval cannot exchange upper-lower order there.
  5. For a region between \(y=f(x)\) and \(y=g(x)\), what does a negative value of \(f(x)-g(x)\) tell you about their order at that \(x\)?

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