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Integrals over General Regions · Tutorial 542 of 999

Nonconvex Regions With Disconnected Slices

Identify disconnected slices and write the region as a union of simpler pieces so its double integral includes each part without counting gaps.

Intermediate 9 min read

What You'll Learn

  • Recognize when a horizontal slice consists of more than one interval.
  • Distinguish a disconnected slice from a single interval spanning its outer endpoints.
  • Partition a region into simpler pieces and express its integral as a sum.
  • Handle slice bounds that vary with the slice level.
  • Check an integral representation by comparing it with the actual slices.

Why Disconnected Slices Need Special Care

A planar region can be connected as a whole and still have horizontal slices that split into separate pieces. A U-shaped region is a familiar example: a low horizontal slice may cross the bottom of the U in one interval, while a higher slice meets only its two arms. If we describe that higher slice by one interval from its leftmost to its rightmost point, we accidentally include the empty gap between the arms.

This matters when setting up a double integral. The inner bounds must describe the points that actually lie in the region, not merely the smallest interval containing them. When a slice has several components, integrate over each component and add the results, or divide the region into simpler pieces and add their integrals.

Definition: For a planar region \(R\), its horizontal slice at height \(y\) is the set \(R_y=\{x:(x,y)\in R\}\). A slice is disconnected when it consists of two or more separated intervals rather than one interval. If \(R_y\) is a union of intervals, its integral along that slice is the sum of the integrals over those intervals.

Representing the Region by Its Slices

A region has disconnected horizontal slices if, for some value of \(y\), the points of the region at that height occur in separated intervals of \(x\). For instance, if \(R_y=[a,b]\cup[c,d]\) with \(b<c\), then the gap \((b,c)\) is not part of the slice. The notation \([a,d]\) would incorrectly fill that gap.

For a bounded region with piecewise-continuous boundaries and a continuous integrand \(f\), the double integral can be computed by integrating across each slice and then integrating over the slice levels. If the slice consists of finitely many intervals, the inner integral is a sum over those intervals. The number of intervals may change as \(y\) changes, so it can be necessary to divide the range of \(y\) into bands where the slice description stays consistent.

Key rule: If \(R_y=[a_1(y),b_1(y)]\cup\cdots\cup[a_k(y),b_k(y)]\), with disjoint interval interiors, then the contribution from that slice is \(\sum_{j=1}^{k}\int_{a_j(y)}^{b_j(y)} f(x,y)\,dx\). Integrate this sum over the applicable values of \(y\).

Often it is clearer to partition the entire region into simple pieces. If the pieces have disjoint interiors and cover \(R\), then the integral over \(R\) is the sum of the integrals over those pieces. Shared edges do not affect the double integral: they have area zero. The following examples show both approaches.

Worked Examples

Worked Example: A U-Shaped Region

Let \(R\) consist of a bottom rectangle \(0\leq x\leq4,\ 0\leq y\leq1\), together with two arms \(0\leq x\leq1,\ 1\leq y\leq3\) and \(3\leq x\leq4,\ 1\leq y\leq3\). Set up and evaluate \(\iint_R(1+x)\,dA\).

For \(0\leq y\leq1\), the horizontal slice is the single interval \([0,4]\). For \(1\leq y\leq3\), it consists of \([0,1]\) and \([3,4]\). Therefore the integral is

$$ \int_0^1\int_0^4(1+x)\,dx\,dy + \int_1^3\left( \int_0^1(1+x)\,dx+ \int_3^4(1+x)\,dx \right)dy. $$

An antiderivative of \(1+x\) with respect to \(x\) is \(x+x^2/2\). The bottom slice contributes

$$ \int_0^4(1+x)\,dx = \left[x+\frac{x^2}{2}\right]_0^4 = 4+\frac{16}{2} = 12. $$

The left arm contributes \(1+1/2=3/2\) at each height. The right arm contributes

$$ \int_3^4(1+x)\,dx = \left[x+\frac{x^2}{2}\right]_3^4 = (4+8)-\left(3+\frac{9}{2}\right) = 12-\frac{15}{2} = \frac{9}{2}. $$

The two upper contributions sum to \(3/2+9/2=6\) at each height. Since that band has height \(3-1=2\), it contributes \(2\cdot6=12\). Thus the total is \(12+12=24\).

As a check, partitioning the region into its three rectangles gives bottom contribution \(12\), left-arm contribution \(2(3/2)=3\), and right-arm contribution \(2(9/2)=9\). Their sum is \(12+3+9=24\), as required.

Worked Example: Two Lobes with Slanted Edges

Consider the region whose horizontal slice, for \(0\leq y\leq2\), is \([0,y/2]\cup[3-y/2,3]\). Find its area and evaluate \(\iint_R x\,dA\). At each height, the intervals are separate: the right endpoint of the left interval is \(y/2\), and the left endpoint of the right interval is \(3-y/2\). Their gap has width \(3-y\), which is at least \(1\) throughout this range.

For the area, the total slice length is \(y/2+y/2=y\). Hence

$$ \operatorname{Area}(R)=\int_0^2 y\,dy =\left[\frac{y^2}{2}\right]_0^2 =2. $$

For the integral of \(x\), the left interval contributes

$$ \int_0^{y/2}x\,dx = \left[\frac{x^2}{2}\right]_0^{y/2} = \frac{y^2}{8}. $$

The right interval contributes

$$ \int_{3-y/2}^{3}x\,dx = \left[\frac{x^2}{2}\right]_{3-y/2}^{3} = \frac{9-(3-y/2)^2}{2} = \frac{3y}{2}-\frac{y^2}{8}. $$

Adding the two interval contributions cancels the quadratic terms, leaving \(3y/2\). Therefore

$$ \iint_R x\,dA = \int_0^2\frac{3y}{2}\,dy = \left[\frac{3y^2}{4}\right]_0^2 = 3. $$

As a check, integrating over the two lobes separately gives \(\int_0^2 y^2/8\,dy=1/3\) for the left lobe and \(\int_0^2(3y/2-y^2/8)\,dy=8/3\) for the right lobe. Their sum is \(1/3+8/3=3\).

Worked Example: Three Disconnected Intervals in a Slice

Let \(R\) contain the rectangle \(0\leq x\leq6,\ 0\leq y\leq1\), and, above it, the three rectangles \(0\leq x\leq1,\ 1\leq y\leq3\), \(2\leq x\leq3,\ 1\leq y\leq3\), and \(5\leq x\leq6,\ 1\leq y\leq3\). Evaluate \(\iint_R x\,dA\). The lower band has one interval in each slice; the upper band has three.

In the lower band, the integral across each slice is

$$ \int_0^6 x\,dx = \left[\frac{x^2}{2}\right]_0^6 = \frac{36}{2} = 18. $$

The band has height \(1\), so its contribution is \(1\cdot18=18\). In the upper band, the three interval integrals at each height are

$$ \int_0^1x\,dx=\frac{1}{2},\qquad \int_2^3x\,dx=\frac{9-4}{2}=\frac{5}{2},\qquad \int_5^6x\,dx=\frac{36-25}{2}=\frac{11}{2}. $$

Their sum is \(1/2+5/2+11/2=17/2\). The upper band has height \(3-1=2\), so its contribution is \(2(17/2)=17\). The total is \(18+17=35\).

Check by integrating over the four rectangles separately: the lower rectangle contributes \(18\). The three upper rectangles have height \(2\), so their contributions are \(2(1/2)=1\), \(2(5/2)=5\), and \(2(11/2)=11\). These upper contributions sum to \(1+5+11=17\); adding the lower contribution gives \(18+17=35\).

Why Adding the Pieces Works

The method follows from additivity of integrals. For any fixed height \(y\) at which the slice is a union of intervals with disjoint interiors, the one-dimensional integral across the slice is the sum of the integrals across its component intervals. Integrating that equality over a band of heights gives the corresponding equality of double integrals.

1
Describe the slice.
At a fixed height, list every interval of \(x\)-values that belongs to the region, leaving any gaps out.
2
Identify changes.
If the slice endpoints or the number of intervals change with height, divide the height range at the relevant transition levels.
3
Add the contributions.
Integrate over each interval in each band, then add the results. Equivalently, partition the region into simpler pieces and add their integrals.

This reasoning requires that the integrals exist; in the usual setting of a bounded region with piecewise-continuous boundary and a continuous integrand, they do. The pieces must cover the region, and their interiors must not overlap. Shared boundary curves do not cause double counting because they have area zero.

A Common Mistake: Filling the Gap

For the U-shaped region in the first example, a slice at any height \(1<y<3\) is \([0,1]\cup[3,4]\), not \([0,4]\). Replacing the two intervals with the single interval between their outer endpoints includes the gap from \(x=1\) to \(x=3\).

For example, if the integrand is \(1\), the actual upper slice contributes \(1+1=2\), its total interval length. The incorrect bound gives \(\int_0^4 1\,dx=4\), counting the gap as though it were part of the region. Across the upper band, the correct area contribution is \(2\cdot2=4\); the incorrect one is \(2\cdot4=8\). The lower rectangle has area \(4\), so the correct total area is \(8\), not the \(12\) obtained by filling in the U.

Practical check: Before writing one inner interval, ask whether every point between its endpoints actually lies in the slice. If not, write separate intervals or divide the region into pieces.

Check Your Understanding

For each question, focus on whether the proposed bounds include exactly the points in the region.

  1. A slice is \([0,2]\cup[5,7]\). What interval must not be included in the integral across that slice?
  2. Why can a connected region have disconnected horizontal slices?
  3. For a band of height \(3\) whose slice is \([0,1]\cup[4,6]\), write the inner integral for a continuous integrand \(f(x,y)\).
  4. When a region is divided into pieces sharing edges, why does adding the piece integrals not count positive area twice?
  5. In the three-interval example, what upper-band contribution results from mistakenly replacing the intervals by \([0,6]\) when integrating \(x\)?

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