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Bounds For Regions Clipped By Multiple Inequalities

Learn to combine multiple inequalities into slice bounds, identify which boundary is active, and split the integral wherever those bounds change or cease to be feasible.

Intermediate 11 min read

What You'll Learn

  • Convert each inequality into a candidate lower or upper boundary on a vertical slice.
  • Use the maximum lower bound and minimum upper bound to describe the slice.
  • Find boundary intersections that can change which candidate is active.
  • Restrict the bounds to slices where the lower boundary does not exceed the upper boundary.
  • Split an iterated integral at active-boundary transitions and feasibility endpoints.
  • Check the resulting bounds against every original inequality.

Why Multiple Inequalities Require Care

Many planar regions are described not by a single pair of curves, but by several inequalities imposed at once. Each inequality restricts the region, so the actual region is their intersection. On a vertical slice, one inequality may supply the lower edge, another the upper edge, and the active edge can change as the slice moves across the plane.

The central task is therefore not to list every boundary as an integration limit. It is to determine which boundary is actually active on each part of the region. Once those active boundaries are identified, an iterated integral can be written with the correct limits.

Key idea: On a vertical slice, combine all lower-bound candidates by taking their maximum, and all upper-bound candidates by taking their minimum. Keep the slice only where the resulting lower bound is no greater than the upper bound.

Turning Inequalities into Slice Bounds

Fix a value of \(x\) and consider the possible \(y\)-values. An inequality of the form \(y\ge g(x)\) gives a candidate lower boundary \(g(x)\); an inequality \(y\le h(x)\) gives a candidate upper boundary \(h(x)\). If the region has several such constraints, write the lower candidates as \(g_1(x),\ldots,g_m(x)\) and the upper candidates as \(h_1(x),\ldots,h_n(x)\).

$$ L(x)=\max\{g_1(x),\ldots,g_m(x)\}, \qquad U(x)=\min\{h_1(x),\ldots,h_n(x)\}. $$

The constraints on that slice are equivalent to \(L(x)\le y\le U(x)\). If \(L(x)>U(x)\), there is no \(y\) satisfying all the inequalities, so that value of \(x\) contributes no points to the region.

Some inequalities constrain \(x\) directly rather than bounding \(y\). For example, \(a\le x\le b\) restricts the interval of slices to examine. An inequality involving \(y\) can also yield an \(x\)-only restriction after the candidate bounds are compared: feasibility requires \(L(x)\le U(x)\). When an inequality is rearranged to isolate \(y\), its direction must be handled correctly. Dividing by a negative coefficient reverses the inequality; if the coefficient of \(y\) is zero, that inequality restricts \(x\) instead.

Definition: For a fixed \(x\), the vertical slice of a region is the set of \(y\)-values satisfying all the region’s inequalities at that \(x\). When it is nonempty and described by a single interval, it is \([L(x),U(x)]\), with \(L(x)\le U(x)\).

Finding Where the Active Boundary Changes

To determine which lower candidate is largest, or which upper candidate is smallest, compare their values. For continuous candidate functions on an interval, a change in which candidate is largest or smallest can occur only where candidates are equal. Thus, solve equations such as \(g_i(x)=g_j(x)\) and \(h_i(x)=h_j(x)\), then test the candidate order between the resulting transition points.

Continuity is an important condition in that statement. If candidates are not continuous, the active bound can change at a discontinuity without the candidates ever being equal. In that case, discontinuities must also be checked as possible transition points. In the examples below, the boundary functions are continuous on the intervals under consideration.

Intersections between lower and upper candidates matter for a different reason: they can mark where slices start or stop being feasible. After identifying possible transitions, check \(L(x)\le U(x)\) on each interval. A crossing of two curves is not automatically an active-boundary switch, and a switch between active candidates is not automatically an endpoint of the region.

Practical rule: Find candidate intersections and feasibility endpoints, sort them along the \(x\)-axis, and test one value from each resulting interval. The test determines the active maximum and minimum there; checking \(L(x)\le U(x)\) determines whether that interval contains slices of the region.

A Justification for the Maximum–Minimum Rule

Suppose the region is specified by \(y\ge g_i(x)\) for every \(i\), and \(y\le h_j(x)\) for every \(j\). For a fixed \(x\), satisfying every lower constraint means that \(y\) must be at least as large as the largest lower candidate. Satisfying every upper constraint means that \(y\) must be no larger than the smallest upper candidate. These two requirements together give exactly the interval from \(L(x)\) to \(U(x)\), provided it is nonempty.

$$ \bigl(y\ge g_i(x)\ \text{for every }i\bigr) \ \text{and}\ \bigl(y\le h_j(x)\ \text{for every }j\bigr) \quad\Longleftrightarrow\quad L(x)\le y\le U(x). $$

For continuous candidates, the transition test follows from the intermediate value theorem. For two lower candidates \(g_1\) and \(g_2\), the difference \(g_1-g_2\) is continuous. If its sign changes between two \(x\)-values, it must equal zero somewhere between them. Hence the candidates tie at a point where their ordering can switch. The same reasoning applies to two upper candidates. This argument establishes why intersections locate possible switches; testing the intervals between intersections establishes which candidate is active.

Worked Example: Four Linear Constraints

Worked Example: Four Linear Constraints

Describe the vertical bounds and find the area of the region defined by \(0\le x\le3\), \(y\ge0\), \(y\ge2x-2\), \(y\le4-x\), and \(y\le x+1\).

The lower candidates are \(0\) and \(2x-2\). They are equal when \(0=2x-2\), so \(x=1\). Testing \(x=0\) gives \(0>-2\), so the active lower bound is \(0\) before \(1\), and \(2x-2\) after \(1\).

The upper candidates \(4-x\) and \(x+1\) are equal when \(4-x=x+1\), giving \(x=\frac32\). At \(x=0\), \(x+1=1<4\); at \(x=2\), \(4-x=2<3\). Thus the active upper bound is \(x+1\) before \(\frac32\), and \(4-x\) after it.

Now check feasibility. On \(0\le x\le1\), the lower bound \(0\) does not exceed \(x+1\). On \(1\le x\le\frac32\), feasibility is \(2x-2\le x+1\), or \(x\le3\), which holds throughout. On \(\frac32\le x\le3\), feasibility is \(2x-2\le4-x\), or \(3x\le6\), so \(x\le2\). The feasible \(x\)-interval is therefore \(0\le x\le2\).

The area is obtained by integrating the slice height, splitting at both active-boundary changes and the feasibility endpoint:

$$ \begin{aligned} A &=\int_0^1\bigl((x+1)-0\bigr)\,dx +\int_1^{3/2}\bigl((x+1)-(2x-2)\bigr)\,dx\\ &\qquad+\int_{3/2}^2\bigl((4-x)-(2x-2)\bigr)\,dx\\ &=\left[\frac{x^2}{2}+x\right]_0^1 +\left[3x-\frac{x^2}{2}\right]_1^{3/2} +\left[6x-\frac{3x^2}{2}\right]_{3/2}^2\\ &=\frac32+\frac78+\frac38 =\frac{22}{8} =\frac{11}{4}. \end{aligned} $$

For a check, the three slice heights are \(x+1\), \(3-x\), and \(6-3x\) on the stated intervals. They are nonnegative throughout their intervals, vanish at the terminal point \(x=2\), and give total area \(\frac{11}{4}\) when integrated. The final upper bound at \(x=2\) equals the lower bound: \(4-2=2\) and \(2(2)-2=2\).

Worked Example: A Parabola and Several Clipping Lines

Worked Example: A Parabola and Several Clipping Lines

Find the area of the region defined by \(-1\le x\le2\), \(y\ge x^2\), \(y\ge3x-2\), \(y\le x+3\), and \(y\le4\).

The lower candidates meet where \(x^2=3x-2\), or \((x-1)(x-2)=0\). At \(x=0\), \(x^2=0>-2=3x-2\), so the parabola is the active lower bound on \([-1,1]\). At \(x=\frac32\), \(x^2=\frac94<\frac52=3x-2\), so the line is active on \([1,2]\). The upper candidates meet where \(x+3=4\), namely \(x=1\). The line \(x+3\) is smaller for \(x<1\), and \(4\) is smaller for \(x>1\).

On \([-1,1]\), feasibility holds: \(x^2\le x+3\), since \(x^2-x-3\le0\) on that interval. On \([1,2]\), feasibility requires \(3x-2\le4\), which holds for \(x\le2\). The upper and lower bounds agree at \(x=2\). Therefore the area is

$$ \begin{aligned} A &=\int_{-1}^{1}\bigl((x+3)-x^2\bigr)\,dx +\int_{1}^{2}\bigl(4-(3x-2)\bigr)\,dx\\ &=\left[\frac{x^2}{2}+3x-\frac{x^3}{3}\right]_{-1}^{1} +\left[6x-\frac{3x^2}{2}\right]_{1}^{2}\\ &=\left(\frac{19}{6}-\left(-\frac{13}{6}\right)\right) +\left(6-\frac92\right)\\ &=\frac{32}{6}+\frac32 =\frac{16}{3}+\frac32 =\frac{41}{6}. \end{aligned} $$

As a check on the transition, at \(x=1\) the lower candidates both equal \(1\), and the upper candidates both equal \(4\). The slices from the left and right therefore meet at the same vertical interval \(1\le y\le4\); the split does not add or omit a strip of area.

Worked Example: Feasibility Determines the Final Slice

Worked Example: Feasibility Determines the Final Slice

Find the area of the region satisfying \(y\ge0\), \(y\ge x-1\), \(y\le2\), and \(y\le4-x^2\). No separate \(x\)-interval is given, so it must be determined from feasibility.

The lower candidates \(0\) and \(x-1\) switch at \(x=1\): the active lower bound is \(0\) for \(x\le1\), and \(x-1\) for \(x\ge1\). The upper candidates \(2\) and \(4-x^2\) meet when \(x^2=2\), at \(x=\pm\sqrt2\). The active upper bound is \(4-x^2\) for \(x<-\sqrt2\) and \(x>\sqrt2\), and \(2\) between those values.

Where the lower bound is \(0\), feasibility requires \(4-x^2\ge0\) whenever the parabola is the active upper bound. On the negative side this gives \(x\ge-2\). On the positive side, where \(x\ge1\), the active lower bound is \(x-1\), and the final feasibility condition is \(x-1\le4-x^2\). Equivalently, \(x^2+x-5\le0\). Its positive root is \(\alpha=\frac{-1+\sqrt{21}}{2}\), so the region ends at \(x=\alpha\). The slice intervals are consequently \([-2,-\sqrt2]\), \([-\sqrt2,1]\), \([1,\sqrt2]\), and \([\sqrt2,\alpha]\).

$$ \begin{aligned} A &=\int_{-2}^{-\sqrt2}(4-x^2)\,dx +\int_{-\sqrt2}^{1}2\,dx +\int_{1}^{\sqrt2}\bigl(2-(x-1)\bigr)\,dx\\ &\qquad+\int_{\sqrt2}^{\alpha}\bigl((4-x^2)-(x-1)\bigr)\,dx\\ &=\left[4x-\frac{x^3}{3}\right]_{-2}^{-\sqrt2} +2(1+\sqrt2) +\left[3x-\frac{x^2}{2}\right]_1^{\sqrt2}\\ &\qquad+\left[5x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{\sqrt2}^{\alpha}\\ &=\left(\frac{16}{3}-\frac{10\sqrt2}{3}\right) +(2+2\sqrt2) +\left(3\sqrt2-\frac72\right) +\left(\frac{7\alpha}{2}+\frac16-\frac{13\sqrt2}{3}\right)\\ &=4-\frac{8\sqrt2}{3}+\frac{7\alpha}{2} =4-\frac{8\sqrt2}{3}+\frac{7(\sqrt{21}-1)}{4}. \end{aligned} $$

In simplifying the last integral, use \(\alpha^2=5-\alpha\), which follows from \(\alpha^2+\alpha-5=0\). Multiplying by \(\alpha\) gives \(\alpha^3=6\alpha-5\), so the antiderivative at \(\alpha\) is \(5\alpha-\frac{5-\alpha}{2}-\frac{6\alpha-5}{3}=\frac{7\alpha}{2}-\frac56\). At \(x=-2\) and \(x=\alpha\), the active upper and lower bounds agree: \(4-(-2)^2=0\), and \(4-\alpha^2=\alpha-1\). These endpoint checks confirm that the feasibility limits are consistent with the original inequalities.

Using the Bounds in an Iterated Integral

Once the active boundaries and feasible intervals have been established, the same splits can be used for an iterated double integral. For a continuous function \(f\) over a region whose vertical slices are the intervals just found, integrate \(f(x,y)\) from \(L(x)\) to \(U(x)\) on each interval, then add the pieces. For example, the region in the first worked example gives

$$ \iint_R f(x,y)\,dA = \int_0^1\int_0^{x+1}f(x,y)\,dy\,dx +\int_1^{3/2}\int_{2x-2}^{x+1}f(x,y)\,dy\,dx +\int_{3/2}^2\int_{2x-2}^{4-x}f(x,y)\,dy\,dx. $$

The splitting points are necessary because a single formula for the lower or upper limit would not describe every slice. The endpoints themselves do not cause a problem: adjoining intervals share their boundary slice, which has zero area and is not counted twice in the double integral.

A Common Pitfall: Treating Every Boundary as Active

A frequent error is to place all candidate lower and upper curves directly into separate integrals without comparing them. For instance, in the first example it would be incorrect to use \(0\) as the lower limit all the way to \(x=2\). At \(x=2\), the constraint \(y\ge2x-2\) requires \(y\ge2\), so the proposed lower bound \(0\) would include points that violate an original inequality.

The correction is to compare candidates first: \(2x-2>0\) for \(x>1\), so the active lower limit changes at \(x=1\). Likewise, using the lower bound \(2x-2\) beyond \(x=2\) in that example would not make the region continue indefinitely: the active upper bound there is \(4-x\), and \(2x-2\le4-x\) fails for \(x>2\). Both active-bound changes and feasibility must be checked.

Why This Method Matters

Regions clipped by multiple inequalities occur naturally when several geometric or physical constraints apply at the same time. The maximum–minimum description gives a systematic way to translate those constraints into iterated-integral limits. It also separates three questions that should not be conflated: what the candidate boundaries are, which candidates are active, and where the slices are feasible.

For a region with complicated boundaries, a clear sketch can help identify likely transitions, but the final bounds should be supported by equations and test values. Verify every proposed interval against the original inequalities, not only against the boundary pair that happened to determine one of its limits.

Check Your Understanding

For each question, identify the active candidates and check feasibility before writing an integral.

  1. If the lower candidates are \(x\) and \(2-x\), at what value of \(x\) do they meet, and which is larger on either side?
  2. If \(y\ge x^2\), \(y\ge3x-2\), \(y\le4\), and \(y\le x+3\), which equations identify possible lower- and upper-bound transitions?
  3. Why must the condition \(L(x)\le U(x)\) be checked even after all candidate intersections have been found?
  4. What additional points should be checked for active-bound changes if candidate functions are not continuous?
  5. For the region \(y\ge0\), \(y\ge x-1\), \(y\le2\), and \(y\le4-x^2\), what equation determines the positive endpoint of the feasible \(x\)-range?

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