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Integrals over General Regions · Tutorial 548 of 999

Common Errors In Bounds For General Regions

Compare each integration bound with the slice of the region it is meant to describe, then correct common errors and verify the resulting integral.

Intermediate 10 min read

What You'll Learn

  • Interpret inner limits as endpoints of slices parallel to the inner variable’s axis.
  • Distinguish a reversed integral from the same region with correctly ordered endpoints.
  • Detect when a single pair of bounds omits part of a region or includes points outside it.
  • Check a proposed change of integration order by deriving the new slice intervals.
  • Use test points, sketches, and exact integral values to verify corrections.

Why Bounds Need to Be Checked

Nested bounds are a compact way to describe a planar region, but a plausible-looking integral can describe the wrong set of points. An inner limit might use the wrong curve, run in the wrong direction, or cover only part of the region. A change in integration order can also fail if its new limits are guessed rather than derived.

The reliable check is geometric: fix the outer variable and ask which values of the inner variable belong to the region. The inner limits must be the endpoints of exactly that slice. Then check that the outer limits cover every required slice, and no others.

Key idea: Bounds describe a region before they describe an integral. Check the set of points encoded by the bounds first; only then evaluate the integral.

What a Pair of Nested Bounds Describes

Suppose \(a\le x\le b\), and for each such \(x\), the variable \(y\) ranges from \(g(x)\) to \(h(x)\), where \(g(x)\le h(x)\). The bounds describe the region

$$ D=\{(x,y):a\le x\le b,\ g(x)\le y\le h(x)\}. $$

For each fixed \(x\), the points in the region lie on a vertical slice, running from \(y=g(x)\) to \(y=h(x)\). Correspondingly, the integral \(\int_a^b\int_{g(x)}^{h(x)} f(x,y)\,dy\,dx\) accumulates the values of \(f\) along those slices and then across the allowed \(x\)-values. If the bounds instead put \(x\) inside and \(y\) outside, the inner slices are horizontal.

Definition: A set described by \(a\le x\le b\) and \(g(x)\le y\le h(x)\) is vertically simple over \([a,b]\): each vertical line with \(x\) in that interval meets the region in one interval, possibly a single point.

These descriptions impose specific checks. The lower endpoint must not exceed the upper endpoint for the stated outer-variable range. The endpoints must be the boundaries of the intended slice, and the outer interval must include every slice needed to cover the region. If a slice splits into separate intervals, one inner interval does not describe it completely.

Why Correct Limits Represent the Region

The connection between the bounds and the region follows directly from the definition of an iterated integral. Consider a continuous function \(f\) on a bounded region \(D\) described by the bounds above. The indicator function \(\mathbf{1}_D(x,y)\) equals \(1\) at points in \(D\) and \(0\) elsewhere. Thus integrating \(f\mathbf{1}_D\) over the enclosing rectangle includes \(f\) precisely on the desired region.

$$ \int_a^b\int_{g(x)}^{h(x)} f(x,y)\,dy\,dx = \int_a^b\int_{-\infty}^{\infty} f(x,y)\mathbf{1}_D(x,y)\,dy\,dx = \iint_D f(x,y)\,dA. $$

Here the indicator vanishes outside the slice from \(g(x)\) to \(h(x)\), so the inner integral over all \(y\) reduces to the stated inner limits. The same reasoning applies to horizontal slices when the order is \(dx\,dy\). For a bounded region and a continuous integrand, the integral can be expressed using either order, provided the new bounds describe the same region. The validity of changing order does not justify simply exchanging variable names or copying the old endpoint formulas.

Worked Example: An Incorrect Upper Boundary

Worked Example: An Incorrect Upper Boundary

Consider the region between \(y=x^2\) and \(y=x\), for \(0\le x\le1\). A proposed area integral is \(\int_0^1\int_{x^2}^{1}1\,dy\,dx\). Does it describe the stated region?

At a fixed \(x\), the intended slice runs from the parabola \(y=x^2\) up to the line \(y=x\). For \(0<x<1\), the value \(1\) is above \(x\), so the proposed upper limit includes points outside the region. For example, at \(x=\tfrac12\), the intended slice is \(y\in[\tfrac14,\tfrac12]\), whereas the proposed bounds give \(y\in[\tfrac14,1]\).

The correct bounds are therefore

$$ \int_0^1\int_{x^2}^{x}1\,dy\,dx = \int_0^1(x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac12-\frac13 = \frac16. $$

As a check, the slice height is \(x-x^2\), which is nonnegative on \([0,1]\) and vanishes at both intersections \(x=0\) and \(x=1\). The proposed integral instead gives \(\int_0^1(1-x^2)\,dx=1-\tfrac13=\tfrac23\), confirming it represents a larger region.

Reversed Endpoints and the Sign of an Integral

For a one-variable definite integral, reversing the endpoints changes the sign: \(\int_u^v q(t)\,dt=-\int_v^u q(t)\,dt\). This rule also applies to an inner integral. It does not mean that the region has changed into a region with negative area; it means that the iterated integral is oriented in the opposite direction along each inner slice.

Worked Example: Reversed Endpoints on a Triangular Region

Let \(D=\{(x,y):0\le x\le1,\ 0\le y\le x\}\). Its vertical slices begin at \(y=0\) and end at \(y=x\). The area is

$$ \iint_D 1\,dA = \int_0^1\int_0^x 1\,dy\,dx = \int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac12. $$

If the inner endpoints are mistakenly written in reverse order, the calculation becomes

$$ \int_0^1\int_x^0 1\,dy\,dx = \int_0^1(0-x)\,dx = -\int_0^1x\,dx = -\frac12. $$

The negative answer is the warning: it is not the area of \(D\). Correcting the endpoint order gives \(+\tfrac12\), and the check is \(\int_0^1\int_0^x1\,dy\,dx=\tfrac12\), as required. If an application intentionally uses reversed endpoints, the sign must be retained; one must not silently treat that integral as an ordinary positive area.

When One Pair of Bounds Misses a Piece

A region may have different upper or lower boundaries in different parts of its projection onto the outer-variable axis. In that case, using one boundary formula throughout can include points that are not in the region or omit points that are. A change in the boundary often occurs where two curves cross, so checking their values on either side of the crossing is essential.

Worked Example: Splitting at a Change of Boundary

Consider the region \(D\) defined by \(0\le x\le2\), \(y\ge0\), \(y\le x\), and \(y\le2-x\). Its upper edge is whichever of \(x\) and \(2-x\) is smaller. These two functions meet when \(x=2-x\), so \(x=1\). For \(0\le x\le1\), \(x\le2-x\); for \(1\le x\le2\), \(2-x\le x\).

Thus the region needs two pieces, and its area is

$$ \begin{aligned} \iint_D1\,dA &=\int_0^1\int_0^x1\,dy\,dx+\int_1^2\int_0^{2-x}1\,dy\,dx\\ &=\int_0^1x\,dx+\int_1^2(2-x)\,dx\\ &=\left[\frac{x^2}{2}\right]_0^1+\left[2x-\frac{x^2}{2}\right]_1^2\\ &=\frac12+\left(2-\frac32\right)\\ &=1. \end{aligned} $$

To check the split, test one point on each side: at \(x=\tfrac12\), the smaller upper bound is \(x=\tfrac12\); at \(x=\tfrac32\), it is \(2-x=\tfrac12\). Using \(y\le x\) on the entire interval would give area \(\int_0^2 x\,dx=2\), counting points above the sloping edge \(y=2-x\) when \(x>1\). The two-piece bounds follow the actual upper edge on both sides of the crossing.

A Change of Order Must Preserve the Region

Changing the order means describing the same set of points with slices in the other direction. The new inner limits therefore need to be found from the region’s boundary inequalities, not obtained by swapping symbols in the old formulas. This is especially important when a curve must be solved for the other variable: the resulting endpoint functions may be different.

Worked Example: Why Swapping Bound Formulas Can Fail

Take \(D=\{(x,y):0\le x\le1,\ x^2\le y\le x\}\) and integrate \(f(x,y)=x\). For horizontal slices, the condition \(y\le x\) gives \(x\ge y\), while \(x^2\le y\), with \(x\ge0\), gives \(x\le\sqrt y\). The range of \(y\) is from \(0\) to \(1\). Hence the actual reversed bounds are \(y\le x\le\sqrt y\).

Using the original bounds first gives

$$ \int_0^1\int_{x^2}^{x}x\,dy\,dx = \int_0^1x(x-x^2)\,dx = \int_0^1(x^2-x^3)\,dx = \frac13-\frac14 = \frac1{12}. $$

A tempting but incorrect “swap” is \(\int_0^1\int_{y^2}^{y}x\,dx\,dy\). It gives \(\int_0^1\tfrac12(y^2-y^4)\,dy=\tfrac12(\tfrac13-\tfrac15)=\tfrac1{15}\), not \(\tfrac1{12}\). The discrepancy signals that the bounds need checking. The correct reversed integral is

$$ \int_0^1\int_y^{\sqrt y}x\,dx\,dy = \int_0^1\left[\frac{x^2}{2}\right]_{y}^{\sqrt y}\,dy = \frac12\int_0^1(y-y^2)\,dy = \frac12\left(\frac12-\frac13\right) = \frac1{12}. $$

Both correctly bounded integrals agree. In the mistaken version, the proposed horizontal slice runs from \(x=y^2\) to \(x=y\), but the region requires \(x\) to run from \(y\) to \(\sqrt y\). For instance, at \(y=\tfrac14\), the correct slice is \(x\in[\tfrac14,\tfrac12]\), not \(x\in[\tfrac1{16},\tfrac14]\).

A Practical Error-Checking Routine

When a set of bounds looks uncertain, check each part against the geometry it claims to encode. A short, systematic test often catches a mistake before any integration begins.

1
Read the outer interval.
Identify the full range of the outer variable and check that it covers the region’s projection onto that axis.
2
Check a typical inner slice.
Fix an interior value of the outer variable. Confirm that the inner limits are the correct endpoints and are ordered correctly.
3
Test boundary transitions.
If two boundary curves might switch roles, solve for their intersection and check which one is upper or lower on each side.
4
For a new order, derive new endpoints.
Describe horizontal or vertical slices in the new direction directly from the region’s inequalities. Split the range if a slice boundary changes.
5
Use a final consistency check.
For an area integral, confirm the result is nonnegative. If evaluating the same continuous integrand in two orders, compare the results as a check that both sets of bounds describe the same region.

Common Mistake: Trusting the Algebra Without Testing the Set

One common error is to manipulate endpoint expressions while never checking what points the result includes. For example, from \(x^2\le y\le x\), a mistaken order change might be written as \(y^2\le x\le y\). At \(y=\tfrac14\), those limits prescribe \(\tfrac1{16}\le x\le\tfrac14\). But the original inequalities require \(y\le x\le\sqrt y\), or \(\tfrac14\le x\le\tfrac12\). The intervals differ, so the proposed change of order is wrong even though the new expressions look related to the old ones.

The correction is to start with the original inequalities and solve them for the inner variable in the new order. Then test at least one ordinary interior slice and, where needed, one slice on each side of a boundary crossing. A sketch can help, but the inequalities and slice checks establish exactly what the bounds mean.

Takeaway: A correct integral requires correct endpoints, correct endpoint order, complete coverage of the region, and a valid description of each slice. Never accept a proposed set of bounds solely because its formulas resemble familiar ones.

Check Your Understanding

For each question, first identify the region or slice that the bounds describe.

  1. For \(0\le x\le1\), what region is described by \(x^2\le y\le1\), and how does it differ from the region between \(y=x^2\) and \(y=x\)?
  2. What sign results from reversing the inner limits in \(\int_0^1\int_0^x1\,dy\,dx\)? What are the correctly ordered limits for area?
  3. For \(0\le x\le2\), \(0\le y\le x\), and \(y\le2-x\), at what value of \(x\) does the upper boundary switch?
  4. Starting from \(0\le x\le1\) and \(x^2\le y\le x\), derive the horizontal slice limits for \(x\).
  5. Why does a change in integration order require checking that the new bounds describe the same region?

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