Piecewise Boundaries And Absolute-Value Constraints

Why Boundary Formulas Need to Be Unpacked

An absolute-value constraint can describe a boundary compactly, but it can hide a change in the curve’s equation. To write integral limits, make that change explicit: determine where the expression inside the absolute value changes sign, write the corresponding curve on each interval, and split the limits wherever a boundary formula changes.

The same care is needed for a boundary already given piecewise. Its formula is valid only on its stated interval. Writing a single set of limits without respecting those intervals can describe the wrong region, even if the boundary equation looks familiar.

Key idea: Before constructing limits, identify the valid coordinate ranges, the sign changes or piecewise breakpoints, and the boundary curves on each resulting segment.

Turning Absolute Values into Curve Segments

The absolute value of a real number \(u\) is \(u\) when \(u\ge 0\), and \(-u\) when \(u<0\). Thus, for example, the graph \(y=|x|\) is not one line: it is \(y=-x\) for \(x<0\) and \(y=x\) for \(x\ge 0\). The formulas meet at \(x=0\).

Definition: A piecewise boundary is a boundary specified by different formulas on different parts of its coordinate range. The points where the formula changes are called breakpoints. To use the boundary in integral limits, keep each formula attached to the interval on which it is valid.

A useful general form is \(|y-g(x)|\le h(x)\). Where \(h(x)\ge 0\), this inequality is equivalent to \(g(x)-h(x)\le y\le g(x)+h(x)\). If \(h(x)<0\), there are no values of \(y\) satisfying the inequality. When these lower or upper curves change formula, split the relevant coordinate interval at the change.

For a Type I description, a region is written using vertical slices: \(a\le x\le b\), with a lower \(y\)-boundary and an upper \(y\)-boundary. If either boundary changes formula at \(x=c\), write separate limits on the intervals to the left and right of \(c\). A Type II description instead uses horizontal slices, with \(y\) as the outer variable and left and right \(x\)-boundaries. The same rule applies when a boundary changes formula as \(y\) varies.

A Reliable Procedure

Identify the constraint.
Keep track of which expressions are bounded and whether the inequality specifies points inside or outside a boundary.

Find the sign changes or breakpoints.
Set the expression inside each absolute value equal to zero, or use the stated endpoints of piecewise formulas.

Write explicit boundary formulas.
Use each formula only on its valid interval, and find the coordinate range where the constraints can hold.

Construct and check the limits.
Split at relevant breakpoints, then test shared endpoints and substitute the proposed boundaries into the original constraint.

Worked Examples

Worked Example: The Region Between \(y=|x|\) and a Horizontal Line

Describe the region \(R\) satisfying \(|x|\le y\le 2\) using vertical slices, and then give a horizontal-slice description.

For the region to be nonempty, \(|x|\le 2\), which gives \(-2\le x\le 2\). The lower boundary \(y=|x|\) changes formula where \(x=0\): it is \(y=-x\) for \(-2\le x\le 0\), and \(y=x\) for \(0\le x\le 2\). Thus a Type I description splits at \(x=0\):

\iint_R f(x,y)\,dA = \int_{-2}^{0}\int_{-x}^{2} f(x,y)\,dy\,dx + \int_{0}^{2}\int_{x}^{2} f(x,y)\,dy\,dx.

To describe the same region horizontally, fix \(y\). Since \(y\ge |x|\), we have \(-y\le x\le y\). The region has \(0\le y\le 2\), so the Type II limits are:

\iint_R f(x,y)\,dA = \int_{0}^{2}\int_{-y}^{y} f(x,y)\,dx\,dy.

Check: At \(x=-2\), the lower boundary is \(-(-2)=2\); at \(x=0\), both lower-bound formulas give \(0\); and at \(x=2\), the lower boundary is \(2\). Substituting \(y=|x|\) into the original inequality gives equality in \(|x|\le y\), while the other boundary \(y=2\) supplies the upper limit. For a horizontal slice, \(x=-y\) and \(x=y\) both satisfy \(|x|=y\), and \(y\) ranges from \(0\) to \(2\). These checks confirm that both descriptions use the stated region.

Worked Example: An Absolute-Value Constraint with a Horizontal Breakpoint

Describe \(R=\{(x,y): |y-1|\le x\le 2\}\) using horizontal slices.

The inequality \(|y-1|\le x\) requires \(x\ge 0\). Together with \(x\le 2\), this means \(|y-1|\le 2\), so \(-1\le y\le 3\). The absolute value changes formula at \(y=1\). For \(-1\le y\le 1\), \(|y-1|=1-y\); for \(1\le y\le 3\), \(|y-1|=y-1\). Hence the left boundary for each horizontal slice is different on the two intervals:

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

Check: At \(y=-1\), the left limit is \(1-(-1)=2\); at the breakpoint \(y=1\), both formulas give \(0\); and at \(y=3\), the left limit is \(3-1=2\). Each outer endpoint therefore collapses to the boundary \(x=2\), as required. On the first interval, substituting \(x=1-y=|y-1|\) gives equality in the original absolute-value constraint. On the second, substituting \(x=y-1=|y-1|\) gives the same equality. The right limit \(x=2\) satisfies \(x\le 2\) throughout.

Worked Example: The Diamond \(|x|+|y|\le 2\)

Construct a Type I description of the region \(R=\{(x,y): |x|+|y|\le 2\}\).

First, \(|x|\le 2\), so \(-2\le x\le 2\). For each such \(x\), the constraint on \(y\) is \(|y|\le 2-|x|\). Its lower and upper boundaries are \(y=-(2-|x|)\) and \(y=2-|x|\). Because \(|x|\) changes formula at \(x=0\), the limits split there. When \(-2\le x\le 0\), \(|x|=-x\), so the bounds are \(-2-x\le y\le 2+x\). When \(0\le x\le 2\), \(|x|=x\), so the bounds are \(x-2\le y\le 2-x\). Therefore:

\iint_R f(x,y)\,dA = \int_{-2}^{0}\int_{-2-x}^{2+x} f(x,y)\,dy\,dx + \int_{0}^{2}\int_{x-2}^{2-x} f(x,y)\,dy\,dx.

Check: At \(x=-2\), both \(y\)-limits are \(0\); at \(x=0\), the left interval gives limits \(-2,2\) and the right interval gives the same limits; at \(x=2\), both limits are \(0\). Substituting either boundary into the original constraint confirms it: on the left interval, \(|x|=-x\) and the upper boundary \(y=2+x\) gives \(-x+|2+x|=2\), since \(2+x\ge0\); the lower boundary \(y=-2-x\) gives \(-x+|-2-x|=2\), since \(-2-x\le0\). On the right interval, the upper boundary \(y=2-x\) and lower boundary \(y=x-2\) likewise give \(x+|y|=2\). The limits therefore trace the four edges of the diamond.

Piecewise Boundaries and Breakpoints

Not every change of formula comes from an absolute value. For example, suppose a lower boundary is given by \(y=p(x)\) for \(a\le x\le c\), and by \(y=q(x)\) for \(c\le x\le b\), while the upper boundary is \(y=r(x)\) throughout. A Type I integral over the region must use the applicable lower formula on each interval:

\int_a^c\int_{p(x)}^{r(x)} f(x,y)\,dy\,dx + \int_c^b\int_{q(x)}^{r(x)} f(x,y)\,dy\,dx.

Before using this form, check that each lower boundary stays below the upper boundary on its interval, and verify that the formulas agree at \(x=c\) if the boundary is intended to join there. If the curves meet or exchange their upper-lower roles elsewhere, the limits must also account for that change. A formula’s stated interval is part of the boundary description, not an optional detail.

A Common Mistake

A frequent error is to replace \(|x|\) by \(x\) without restricting to \(x\ge 0\). For example, the region \(|x|\le y\le 2\) does not have lower bound \(y=x\) across the full interval \([-2,2]\). At \(x=-1\), the original lower boundary is \(y=|-1|=1\), but the incorrect formula gives \(y=-1\). It would include points that violate the original constraint.

The correction is to preserve the sign cases: use \(y=-x\) when \(x<0\), and \(y=x\) when \(x\ge0\). Then split the integral at \(x=0\), or choose horizontal slices, for which the same region has the single interval \(-y\le x\le y\).

Takeaway: An absolute-value boundary is a compact way to specify several curve segments. Correct limits come from resolving its sign cases, keeping each explicit formula on its valid interval, and checking the resulting boundaries against the original constraint.

Why This Matters

Iterated integrals are built from the actual edges of a region. When a boundary changes formula, an integral that ignores the breakpoint may use the wrong edge for part of the region. Conversely, splitting at every sign change makes the geometry visible and gives limits that can be checked directly. The same procedure applies whether the final description uses vertical or horizontal slices: choose the slicing direction, identify how the boundary varies along that direction, and split wherever its explicit formula changes.

Check Your Understanding

For each question, identify the relevant breakpoints before writing any integral limits.

Write \(y=|x-2|\) as explicit formulas on the intervals to either side of its breakpoint.
For \(|x|\le y\le 3\), what is the allowed range of \(x\), and at which value must a Type I description split?
For \(|y+1|\le x\le 4\), determine the allowed range of \(y\) and the value where a Type II description splits.
For \(|x|+|y|\le 3\), write the lower and upper \(y\)-boundaries on each side of \(x=0\).
What should you check at a shared endpoint of two piecewise boundary formulas?