Locating Slice Transitions At Boundary Crossings

Why Slice Endpoints Change

When a region is described by curved or piecewise boundaries, the endpoints of its slices need not come from the same curves throughout the region. A vertical slice may have one upper boundary for smaller \(x\)-values and another for larger \(x\)-values. A horizontal slice may likewise change its left or right boundary as \(y\) varies.

The coordinate where the active boundary changes is a slice transition. Finding it is a geometric task: solve for where the relevant boundary curves intersect, then decide whether that intersection actually changes the edge of the region. These coordinates tell us where to split an iterated integral so its bounds describe the region correctly.

Definition: A slice transition is a coordinate value at which an endpoint of a vertical or horizontal slice changes from one boundary curve to another. A crossing is a candidate transition only if both curves form that endpoint on the region near the crossing.

For a vertical slice, compare boundary heights as \(x\) varies. For a horizontal slice, compare boundary \(x\)-coordinates as \(y\) varies. In either case, equate the relevant boundary expressions to find candidate crossings. Then test which curve actually bounds the slice on each side of the candidate.

Why Splitting at a Crossing Works

Suppose two continuous boundary functions \(g(x)\) and \(h(x)\) can form the upper edge of a region. Their order can change only where \(g(x)=h(x)\). Between consecutive intersections, the continuous difference \(g(x)-h(x)\) cannot change sign: if it did, the Intermediate Value Theorem would imply that the difference equals zero somewhere in between. Thus the same curve remains the upper edge throughout each such interval, provided both curves are relevant boundaries there.

Once the interval is split at the crossing, each piece has a consistent upper and lower boundary. The integral over the whole region is the sum of the integrals over the pieces. This follows from additivity of the integral over regions whose interiors do not overlap; the shared slice at the split coordinate has area zero.

\iint_D f(x,y)\,dA = \int_a^c\int_{\ell_1(x)}^{u_1(x)} f(x,y)\,dy\,dx + \int_c^b\int_{\ell_2(x)}^{u_2(x)} f(x,y)\,dy\,dx.

Here the region is divided at \(x=c\), and the indicated bounds describe its two pieces. The formula requires that the pieces together cover \(D\), apart from their shared boundary, and that \(f\) be integrable on \(D\). The same principle applies to horizontal slices, splitting the outer \(y\)-interval at a transition value.

Worked Example: A Vertical Slice Changes Its Upper Boundary

Worked Example: A Switch Between Two Upper Curves

Consider the region \(D\) with \(0\le x\le 3\), \(y\ge 0\), and \(y\) below both lines \(y=x+1\) and \(y=3-x\). Find the vertical-slice transition and write the area integral.

The two possible upper boundaries meet where

x+1=3-x \quad\Longrightarrow\quad 2x=2 \quad\Longrightarrow\quad x=1.

At this coordinate, both lines have height \(y=2\). To determine which line is lower on either side, test \(x=0\): the heights are \(1\) and \(3\), so \(y=x+1\) is the upper edge of the region there. Test \(x=2\): the heights are \(3\) and \(1\), so \(y=3-x\) is the upper edge there. Since the region lies below both lines, its upper boundary is the smaller of the two heights. The switch at \(x=1\) is therefore an actual slice transition.

The lower edge is \(y=0\) throughout. For the area, use \(f(x,y)=1\) and split at \(x=1\):

\iint_D 1\,dA = \int_0^1\int_0^{x+1}1\,dy\,dx + \int_1^3\int_0^{3-x}1\,dy\,dx.

Evaluate the first piece:

\int_0^1\int_0^{x+1}1\,dy\,dx = \int_0^1(x+1)\,dx = \left[\frac{x^2}{2}+x\right]_0^1 = \frac12+1 = \frac32.

Evaluate the second piece:

\int_1^3\int_0^{3-x}1\,dy\,dx = \int_1^3(3-x)\,dx = \left[3x-\frac{x^2}{2}\right]_1^3 = \frac92-\frac52 = 2.

The total area is \(\frac32+2=\frac72\). As a check, the antiderivatives differentiate to the integrands: \(\frac{d}{dx}(\frac{x^2}{2}+x)=x+1\) and \(\frac{d}{dx}(3x-\frac{x^2}{2})=3-x\). The two pieces also use the lower of the two candidate heights on their respective intervals, as required.

Candidate Crossings Must Be Relevant

Solving an equation between two curves gives candidate coordinates, not an automatic instruction to split. A crossing may lie outside the portion of the curves that bounds the region, or the crossing curves may not be the active endpoints on either side. To decide, first identify the region and the possible slice endpoints. Then compare those endpoints on each side of the candidate, using a test value or the sign of their difference.

Identify the slice direction.
For vertical slices, compare \(y\)-values at fixed \(x\). For horizontal slices, compare \(x\)-values at fixed \(y\).

Find candidate crossings.
Set the relevant boundary expressions equal and solve for the coordinate along which the slices vary.

Check the region on both sides.
Determine which curves really form the slice endpoints before and after each candidate.

Split only at actual transitions.
Use one pair of endpoint formulas on each resulting interval.

Worked Example: A Horizontal Slice Changes Its Right Endpoint

Worked Example: A Parabola and a Line

Let \(D\) be the region between \(y=x^2\) and \(y=2-x\). Find the horizontal-slice transition and use horizontal slices to calculate \(\iint_D y\,dA\).

First find where the curves meet:

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

The intersections are \((-2,4)\) and \((1,1)\). In horizontal slices, the parabola gives \(x=-\sqrt y\) on its left branch and \(x=\sqrt y\) on its right branch; the line gives \(x=2-y\). The region runs from \(y=0\) to \(y=4\). The right endpoint changes at the intersection \((1,1)\), so the transition height is \(y=1\).

For \(0\le y\le1\), the right endpoint is \(x=\sqrt y\). For instance, at \(y=\frac14\), \(\sqrt y=\frac12\) while \(2-y=\frac74\); the region between the curves ends at \(\frac12\). For \(1\le y\le4\), the right endpoint is \(x=2-y\): at \(y=2\), it is \(0\), while \(\sqrt y>0\). The left endpoint throughout is \(x=-\sqrt y\). Therefore,

\iint_D y\,dA = \int_0^1\int_{-\sqrt y}^{\sqrt y}y\,dx\,dy + \int_1^4\int_{-\sqrt y}^{2-y}y\,dx\,dy.

Since \(y\) is constant with respect to the inner variable \(x\), the first piece is

\int_0^1 2y\sqrt y\,dy = 2\int_0^1 y^{3/2}\,dy = \left[\frac45y^{5/2}\right]_0^1 = \frac45.

The second piece is

\begin{aligned} \int_1^4 y\bigl((2-y)-(-\sqrt y)\bigr)\,dy &=\int_1^4(2y-y^2+y^{3/2})\,dy\\ &=\left[y^2-\frac{y^3}{3}+\frac25y^{5/2}\right]_1^4\\ &=\left(16-\frac{64}{3}+\frac{64}{5}\right) -\left(1-\frac13+\frac25\right)\\ &=\frac{112}{15}-\frac{16}{15} =\frac{32}{5}. \end{aligned}

The total is \(\frac45+\frac{32}{5}=\frac{36}{5}\). Differentiating the displayed antiderivative gives \(2y-y^2+y^{3/2}\), confirming the second integration. The horizontal bounds also close correctly: at \(y=4\), both endpoints equal \(-2\), the intersection at the top of the region.

Worked Example: A Vertical Slice Changes Its Lower Boundary

Worked Example: The Larger of Two Lower Curves

Consider the region \(D\) defined by \(0\le x\le2\) and \(\max(x,2-x)\le y\le3\). The lower edge is whichever of the lines \(y=x\) and \(y=2-x\) is higher. Find its transition and calculate \(\iint_D x\,dA\).

The lower curves meet where

x=2-x \quad\Longrightarrow\quad x=1,\qquad y=1.

At \(x=\frac12\), the lines have heights \(\frac12\) and \(\frac32\), so \(y=2-x\) is the lower edge of the region there (the higher of the two lower constraints). At \(x=\frac32\), the heights are \(\frac32\) and \(\frac12\), so \(y=x\) is the lower edge. The actual transition is \(x=1\). The upper edge \(y=3\) does not change. Thus

\iint_D x\,dA = \int_0^1\int_{2-x}^3 x\,dy\,dx + \int_1^2\int_x^3 x\,dy\,dx.

For the first piece, integrate with respect to \(y\), then \(x\):

\begin{aligned} \int_0^1\int_{2-x}^3 x\,dy\,dx &=\int_0^1 x\bigl(3-(2-x)\bigr)\,dx\\ &=\int_0^1(x+x^2)\,dx\\ &=\left[\frac{x^2}{2}+\frac{x^3}{3}\right]_0^1\\ &=\frac12+\frac13=\frac56. \end{aligned}

For the second piece,

\begin{aligned} \int_1^2\int_x^3 x\,dy\,dx &=\int_1^2 x(3-x)\,dx\\ &=\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_1^2\\ &=\left(6-\frac83\right)-\left(\frac32-\frac13\right)\\ &=\frac{10}{3}-\frac76=\frac{13}{6}. \end{aligned}

Adding gives \(\frac56+\frac{13}{6}=3\). As a check, the antiderivatives differentiate to \(x+x^2\) and \(3x-x^2\), respectively. The region’s lower edge is continuous at the transition: both formulas give \(y=1\) when \(x=1\).

A Common Mistake: Splitting at the Wrong Crossing

Suppose the upper edge is the lower of \(y=x+1\) and \(y=3-x\) on \(0\le x\le3\). A common error is to use \(y=3-x\) throughout because it is the descending line. That produces the incorrect area calculation

\int_0^3(3-x)\,dx = \left[3x-\frac{x^2}{2}\right]_0^3 = \frac92.

The result is wrong because on \(0\le x<1\), the line \(y=x+1\) lies below \(y=3-x\) and therefore forms the region’s upper edge. Correcting the bounds requires the crossing \(x=1\): use \(x+1\) on \([0,1]\), and \(3-x\) on \([1,3]\). The corrected area is \(\frac32+2=\frac72\), not \(\frac92\). The error is not in evaluating an integral; it is in failing to check which boundary is active on each interval.

Practical Checks Before Integrating

Before writing an iterated integral for a region with competing boundaries, make sure that every split has a geometric reason. An intersection matters when it changes an actual endpoint, not merely because two equations happen to have a common solution. Check the endpoint ordering on each side of the crossing, and verify that the pieces cover the intended coordinate interval without gaps or overlaps in their interiors.

Key takeaway: Solve boundary equations to find candidate transition coordinates, test which curves bound the slices on either side, and split the integral exactly where an active endpoint changes.

Check Your Understanding

Use boundary intersections and endpoint comparisons to answer these questions.

For a vertical slice with possible upper boundaries \(y=2x+1\) and \(y=5-x\), what equation gives the candidate transition coordinate?
Why does an intersection of two curves not necessarily require splitting an integral?
For the region between \(y=x^2\) and \(y=2-x\), at what height does the horizontal right endpoint change, and which curves meet there?
If two continuous candidate endpoints have no intersection on an interval, what can you conclude about their ordering there?
What geometric mistake is made by using one boundary formula across an interval where the active slice endpoint changes?