Worked Workshop: Integrating Over Irregular Planar Regions

Why Irregular Regions Need Careful Slicing

A double integral over a rectangle has constant outer limits and constant inner limits. For a region bounded by intersecting curves, however, the boundary of a slice can change as the slice moves. A correct setup must account for those changes: an omitted piece, an incorrectly ordered pair of bounds, or a missed crossing changes the region being integrated over.

The main task is therefore geometric before it is computational. Find the boundary intersections, determine which boundary is active on each interval, and split the integral wherever that description changes. A second description, using the other slicing direction, can provide an independent check.

Definition: A vertical slice of a planar region fixes \(x\) and varies \(y\); a horizontal slice fixes \(y\) and varies \(x\). A region is described by vertical slices if it can be written using a range of \(x\)-values and, for each \(x\), a single interval of \(y\)-values. A horizontal-slice description is defined in the same way with \(x\) and \(y\) interchanged. If the boundary of a slice changes formula, divide the region into pieces so each piece has valid limits.

For a continuous function \(f\) on a bounded region \(R\), a valid vertical description gives an iterated integral of the form \(\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx\). A valid horizontal description gives \(\int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy\). If a description needs multiple pieces, add the integrals over those pieces. The lower and upper limits must describe the actual slice throughout each stated interval.

Worked Example: Area of a Region with Two Upper Boundaries

Consider the region above the \(x\)-axis and below both \(y=x^2\) and \(y=2-x\), with \(x\ge 0\). Find its area using both slicing directions.

Worked Example: Splitting at the Curves’ Intersection

First find where the two curves meet:

x^2=2-x \quad\Longleftrightarrow\quad (x+2)(x-1)=0.

The intersection with \(x\ge 0\) is \((1,1)\). The line meets the \(x\)-axis at \((2,0)\), and the parabola meets it at \((0,0)\). For \(0\le x\le1\), the parabola is the lower of the two upper boundaries; for \(1\le x\le2\), the line is lower. Thus a vertical description must split at \(x=1\):

\begin{aligned} \operatorname{Area}(R) &=\int_0^1\int_0^{x^2}1\,dy\,dx +\int_1^2\int_0^{2-x}1\,dy\,dx\\ &=\int_0^1 x^2\,dx+\int_1^2(2-x)\,dx\\ &=\frac13+\left[2x-\frac{x^2}{2}\right]_1^2\\ &=\frac13+\left(2-\frac32\right) =\frac13+\frac12 =\frac56. \end{aligned}

For horizontal slices, the region’s height ranges from \(y=0\) to \(y=1\). At a fixed height \(y\), the parabola gives \(x=\sqrt y\), and the line gives \(x=2-y\). The slice lies between these boundaries, so the bounds are:

\operatorname{Area}(R) =\int_0^1\int_{\sqrt y}^{2-y}1\,dx\,dy =\int_0^1(2-y-\sqrt y)\,dy =2-\frac12-\frac23 =\frac56.

The two descriptions give the same area. In particular, the horizontal endpoints meet at \(y=1\), the top of the region; they do not switch order before that height.

How to Find the Pieces and Their Limits

A boundary crossing is a candidate location for a split because it can change which curve bounds a slice. A split is needed when that change affects the slice description in the chosen direction. Intersections alone do not determine the limits: the inequalities defining the region and a test point on each interval tell you which boundary lies inside the region.

List the boundaries.
Translate the description of the region into equations or inequalities, including any straight edges such as coordinate axes.

Find relevant intersections.
Solve the boundary equations and keep the points that lie on the region’s boundary. Their coordinates may determine split values in either slicing direction.

Choose a slicing direction.
For that direction, determine the outer range and inspect a typical slice on each side of every candidate split.

Write and check the bounds.
Put the smaller endpoint first, check that each slice stays inside the region, and add the integrals if the description has multiple pieces.

The reason a split works is that it divides the region into pieces on which the same boundary formulas apply. Each piece is then a region with an unchanging lower and upper boundary formula for its slices. Integrating over every piece and adding the results accounts for the full region without counting any interior points twice; the shared edges have zero area.

Worked Example: Checking an Integral with the Other Slicing Direction

Use the region from the first example, and evaluate \(\iint_R (x+y)\,dA\). The integrand is continuous, and either of the region descriptions provides valid limits.

Worked Example: Evaluating with Horizontal Slices

The horizontal bounds are \(0\le y\le1\) and \(\sqrt y\le x\le2-y\). Integrate first with respect to \(x\):

\begin{aligned} \iint_R(x+y)\,dA &=\int_0^1\int_{\sqrt y}^{2-y}(x+y)\,dx\,dy\\ &=\int_0^1\left[\frac{x^2}{2}+yx\right]_{\sqrt y}^{2-y}\,dy\\ &=\int_0^1\left( \frac{(2-y)^2}{2}+y(2-y) -\frac{y}{2}-y\sqrt y \right)dy\\ &=\int_0^1\left(2-\frac{y}{2}-\frac{y^2}{2}-y^{3/2}\right)dy\\ &=\left[2y-\frac{y^2}{4}-\frac{y^3}{6}-\frac{2}{5}y^{5/2}\right]_0^1\\ &=2-\frac14-\frac16-\frac25 =\frac{71}{60}. \end{aligned}

Check the value using vertical slices, which require a split at \(x=1\):

\begin{aligned} \iint_R(x+y)\,dA &=\int_0^1\int_0^{x^2}(x+y)\,dy\,dx +\int_1^2\int_0^{2-x}(x+y)\,dy\,dx\\ &=\int_0^1\left(x^3+\frac{x^4}{2}\right)dx +\int_1^2\left(2-\frac{x^2}{2}\right)dx\\ &=\left[\frac{x^4}{4}+\frac{x^5}{10}\right]_0^1 +\left[2x-\frac{x^3}{6}\right]_1^2\\ &=\left(\frac14+\frac1{10}\right) +\left(\frac83-\frac{11}{6}\right)\\ &=\frac7{20}+\frac56 =\frac{21}{60}+\frac{50}{60} =\frac{71}{60}. \end{aligned}

The independent calculation agrees. The intermediate endpoint values are \(\left[2x-x^3/6\right]_{x=2}=8/3\) and \(\left[2x-x^3/6\right]_{x=1}=11/6\), so their difference is \(5/6\).

Worked Example: A Horizontal Description That Needs Two Pieces

Let \(R\) be the region between the parabola \(y=3-x^2\) and the line \(y=1+x\). Find its area in both slicing directions. The crossings satisfy \(3-x^2=1+x\), or \((x+2)(x-1)=0\), so they occur at \((-2,-1)\) and \((1,2)\). On \([-2,1]\), the parabola lies above the line because \(3-x^2-(1+x)=(1-x)(x+2)\ge0\).

Worked Example: Splitting at a Change in Horizontal Bounds

Vertical slices use the same upper and lower curves for \(-2\le x\le1\):

\begin{aligned} \operatorname{Area}(R) &=\int_{-2}^1\int_{1+x}^{3-x^2}1\,dy\,dx\\ &=\int_{-2}^1(2-x-x^2)\,dx\\ &=\left[2x-\frac{x^2}{2}-\frac{x^3}{3}\right]_{-2}^1\\ &=\frac76-\left(-\frac{10}{3}\right) =\frac76+\frac{20}{6} =\frac92. \end{aligned}

For horizontal slices, the parabola gives two possible \(x\)-values, \(x=-\sqrt{3-y}\) and \(x=\sqrt{3-y}\). When \(-1\le y\le2\), the line \(x=y-1\) is the right boundary, while the left boundary is the left branch of the parabola. When \(2\le y\le3\), a horizontal slice lies between both branches of the parabola. Therefore, the horizontal description changes at \(y=2\):

\begin{aligned} \operatorname{Area}(R) &=\int_{-1}^2\int_{-\sqrt{3-y}}^{y-1}1\,dx\,dy +\int_2^3\int_{-\sqrt{3-y}}^{\sqrt{3-y}}1\,dx\,dy\\ &=\int_{-1}^2\left(y-1+\sqrt{3-y}\right)dy +\int_2^3 2\sqrt{3-y}\,dy\\ &=-\frac32+\frac{14}{3}+\frac43\\ &=-\frac{9}{6}+\frac{28}{6}+\frac{8}{6} =\frac{27}{6} =\frac92. \end{aligned}

For the square-root integrals, the substitution \(u=3-y\) gives \(\int_{-1}^2\sqrt{3-y}\,dy=\int_1^4\sqrt u\,du=14/3\), and \(\int_2^3 2\sqrt{3-y}\,dy=2\int_0^1\sqrt u\,du=4/3\). The horizontal and vertical computations agree.

A Common Mistake: Reversing the Horizontal Endpoints

For the first region, a tempting but incorrect horizontal description is \(0\le x\le\sqrt y\) for \(0\le y\le1\). At \(y=1/4\), this would place the slice between \(x=0\) and \(x=1/2\). But the region is below \(y=x^2\), so at this height its points must satisfy \(x^2\ge1/4\), hence \(x\ge1/2\). The line boundary also requires \(x\le2-1/4=7/4\). The actual slice is \(\frac12\le x\le\frac74\).

The correction is to derive both endpoints from the inequalities defining the region, not just solve for \(x\) and assume that the result is the left endpoint. Here \(y\le x^2\) gives \(x\ge\sqrt y\) because \(x\ge0\), while \(y\le2-x\) gives \(x\le2-y\). Therefore the correct horizontal description is \(\sqrt y\le x\le2-y\) for \(0\le y\le1\). A quick test at an interior height can expose reversed or misplaced bounds before integration begins.

Why the Decomposition Matters

Choosing a valid decomposition is not bookkeeping added after the mathematics. It is what ensures that the integral measures the intended region. For a region with intersecting boundaries, the split values record where a slice changes which edge it meets. Once those values are identified, each iterated integral follows directly from the geometry.

When possible, describe the same region in both orders. Agreement of the limits and values is a useful check; disagreement often signals a missed piece, a reversed endpoint, or an intersection assigned to the wrong interval. A second description is not always simpler, but it can make an error visible.

Check Your Understanding

For each question, identify the boundary information that determines the slice limits before attempting any integration.

For the region above the \(x\)-axis and below \(y=x^2\) and \(y=2-x\), why does a vertical description split at \(x=1\)?
For that same region, what are the horizontal bounds, and at what height do the endpoints meet?
In the region between \(y=3-x^2\) and \(y=1+x\), why does a horizontal description split at \(y=2\)?
At \(y=1/4\) in the first region, what are the actual horizontal endpoints, and which inequalities determine them?
When can calculating the same double integral in the other slicing direction help check a setup?