Tutorials › Calculus III › Verifying Reversed Bounds Describe The Same Region

Home  /  Tutorials  /  Integrals over General Regions

Integrals over General Regions · Tutorial 537 of 999

Verifying Reversed Bounds Describe The Same Region

Learn a systematic way to decide whether proposed reversed bounds describe exactly the original planar region.

Intermediate 10 min read

What You'll Learn

  • Test whether the reversed outer interval covers every horizontal slice of the region.
  • Check that inner endpoints are the correct left and right boundaries.
  • Use the original inequalities to verify that proposed bounds include no extra points.
  • Identify when a change in the active boundary requires piecewise bounds.
  • Use an area calculation as a consistency check without treating it as proof of matching regions.

Why Verify Reversed Bounds?

A reversed iterated integral is useful only if its bounds describe the same region as the original integral. A plausible-looking pair of limits can still omit part of the region, include points outside it, or miss a change in boundary partway through the outer interval.

The reliable check is geometric and algebraic: identify the region from the original bounds, then test the proposed outer interval, inner endpoints, and boundary curves against that region. A sketch can make the checks visible; inequalities can make them exact.

Key idea: Matching the names or equations of boundary curves is not enough. At each value of the proposed outer variable, the inner interval must contain exactly the points of the original region on that slice.

A Precise Region Check

For bounds of the form \(a\le x\le b\), \(g_1(x)\le y\le g_2(x)\), the region consists of all points satisfying those four inequalities. After reversing the order, bounds \(c\le y\le d\), \(h_1(y)\le x\le h_2(y)\) describe the same region only if they give exactly those same points.

This leads to three checks. First, the outer interval must cover every value of the new outer variable that occurs in the region, and no values outside its range. Second, for each value in that interval, the inner endpoints must be the leftmost and rightmost points of the region on that horizontal slice. Third, each endpoint must lie on the appropriate boundary curve, and the interval between the endpoints must not cross outside the region.

1
Check the outer interval.
Find the smallest and largest values of the new outer variable that occur anywhere in the original region.
2
Check each horizontal slice.
At a fixed outer-variable value, identify the full interval of inner-variable values belonging to the region.
3
Check the boundary curves.
Substitute the proposed endpoints into the original boundary equations or inequalities and verify that the slice lies inside the region.

These checks concern the region, not just a computed integral. For a continuous integrand on a bounded region, equality of the regions justifies the corresponding change in order of integration. Computing both integrals can provide a useful check, but equal answers alone do not prove that the bounds describe the same region.

Worked Example: Checking a Single Horizontal Slice Interval

Worked Example: Checking a Single Horizontal Slice Interval

Consider the region described by \(0\le x\le 1\) and \(x^2\le y\le 1\). A proposed reversal is \(0\le y\le 1\), \(0\le x\le\sqrt{y}\). Verify the bounds.

The original inequalities show that \(y\) ranges from \(0\) to \(1\): the smallest value \(x^2\) can take is \(0\), and the upper boundary is \(y=1\). Fix a value \(y\) in this interval. The lower-bound condition \(x^2\le y\), together with \(x\ge0\), gives \(0\le x\le\sqrt{y}\). The original condition \(x\le1\) adds no further restriction because \(\sqrt{y}\le1\) when \(0\le y\le1\).

Thus the proposed interval gives exactly the points in the region on every horizontal slice. The lower endpoint \(x=0\) is the vertical boundary, and the upper endpoint \(x=\sqrt{y}\) is the curve \(y=x^2\) solved for \(x\). The outer endpoints \(y=0\) and \(y=1\) are the bottom and top levels of the region.

An area calculation provides an additional consistency check. Integrating the constant function \(1\) in the original order gives:

$$ \int_0^1\int_{x^2}^{1}1\,dy\,dx =\int_0^1(1-x^2)\,dx =\left[x-\frac{x^3}{3}\right]_0^1 =1-\frac13 =\frac23. $$

Using the proposed reversed bounds gives:

$$ \int_0^1\int_0^{\sqrt{y}}1\,dx\,dy =\int_0^1\sqrt{y}\,dy =\left[\frac23y^{3/2}\right]_0^1 =\frac23. $$

The slice checks establish that the bounds match; the equal areas support that conclusion.

A Complete Inequality Check

Here is the underlying reason the slice method works. The original bounds are a compact way to state inequalities defining membership in a region. Reversing the order does not change those inequalities; it reorganizes them so that, for each fixed value of the new outer variable, the allowed values of the inner variable are explicit.

Region-equivalence principle: Proposed reversed bounds describe the original region if, for every outer-variable value in their interval, their inner interval consists exactly of the points satisfying the original region's inequalities.

This is a direct test, not an inference from a picture alone. For example, suppose the original region is described by \(0\le x\le1\) and \(x^2\le y\le1\). Holding \(y\) fixed, the conditions become \(0\le y\le1\), \(x\ge0\), and \(x^2\le y\). Since \(x\ge0\), the last inequality is equivalent to \(x\le\sqrt{y}\). Therefore the original inequalities are equivalent to \(0\le y\le1\) and \(0\le x\le\sqrt{y}\), which proves the proposed bounds describe exactly the same points.

Worked Example: Detecting a Missing Half of the Region

Worked Example: Detecting a Missing Half of the Region

Now consider \(-1\le x\le1\) and \(x^2\le y\le1\). Suppose someone proposes the reversed bounds \(0\le y\le1\), \(0\le x\le\sqrt{y}\). Check the proposal and correct it if needed.

The outer interval \(0\le y\le1\) is correct: the region reaches \(y=0\) at \(x=0\) and has upper boundary \(y=1\). But for any fixed \(y\) between \(0\) and \(1\), the inequality \(x^2\le y\) is equivalent to \(-\sqrt{y}\le x\le\sqrt{y}\). The proposed lower endpoint \(x=0\) excludes the part of the slice with negative \(x\). For example, at \(y=\frac14\), the original slice includes \(-\frac12\le x\le\frac12\), whereas the proposal includes only \(0\le x\le\frac12\).

The area calculation makes the omission visible. The original bounds give:

$$ \int_{-1}^{1}\int_{x^2}^{1}1\,dy\,dx =\int_{-1}^{1}(1-x^2)\,dx =\left[x-\frac{x^3}{3}\right]_{-1}^{1} =\frac23-\left(-\frac23\right) =\frac43. $$

The incorrect proposal gives only:

$$ \int_0^1\int_0^{\sqrt{y}}1\,dx\,dy =\int_0^1\sqrt{y}\,dy =\left[\frac23y^{3/2}\right]_0^1 =\frac23. $$

The corrected reversed bounds are \(0\le y\le1\), \(-\sqrt{y}\le x\le\sqrt{y}\). They give:

$$ \int_0^1\int_{-\sqrt{y}}^{\sqrt{y}}1\,dx\,dy =\int_0^1 2\sqrt{y}\,dy =\left[\frac43y^{3/2}\right]_0^1 =\frac43. $$

The correction includes both sides of every horizontal slice. The matching area is a check; the inequalities identify precisely why the original proposal failed.

When the Boundary Changes Along the Outer Interval

Not every region can be described by one unchanging pair of inner endpoints. As the outer variable moves, a different boundary may become the limiting left or right endpoint. In that case, the reversed description needs separate intervals, with the transition value included as an endpoint of the relevant pieces.

Worked Example: Verifying a Piecewise Reversal

Worked Example: Verifying a Piecewise Reversal

Consider the region \(-1\le x\le1\), \(x^2\le y\le2-x^2\). Verify the piecewise reversed bounds:

$$ \begin{cases} 0\le y\le1, & -\sqrt{y}\le x\le\sqrt{y},\\ 1\le y\le2, & -\sqrt{2-y}\le x\le\sqrt{2-y}. \end{cases} $$

The lower boundary \(y=x^2\) is nonnegative, and the upper boundary \(y=2-x^2\) is at most \(2\). The region therefore has \(0\le y\le2\). To find the horizontal slice at a fixed \(y\), rewrite both boundary conditions:

$$ x^2\le y \quad\Longleftrightarrow\quad -\sqrt{y}\le x\le\sqrt{y}, \qquad y\le2-x^2 \quad\Longleftrightarrow\quad -\sqrt{2-y}\le x\le\sqrt{2-y}. $$

The slice must satisfy both conditions, so its endpoints are determined by the more restrictive of the two symmetric intervals. When \(0\le y\le1\), \(y\le2-y\), hence \(\sqrt{y}\le\sqrt{2-y}\); the first interval is narrower. When \(1\le y\le2\), \(2-y\le y\), hence \(\sqrt{2-y}\le\sqrt{y}\); the second interval is narrower. The transition occurs at \(y=1\), where the two endpoint magnitudes are both \(1\). This proves that the proposed pieces give exactly the horizontal slices of the region.

As a check, the area in the original order is:

$$ \begin{aligned} \int_{-1}^{1}\int_{x^2}^{2-x^2}1\,dy\,dx &=\int_{-1}^{1}(2-2x^2)\,dx\\ &=\left[2x-\frac{2x^3}{3}\right]_{-1}^{1}\\ &=\left(2-\frac23\right)-\left(-2+\frac23\right) =\frac83. \end{aligned} $$

The proposed piecewise bounds give:

$$ \begin{aligned} \int_0^1\int_{-\sqrt{y}}^{\sqrt{y}}1\,dx\,dy +\int_1^2\int_{-\sqrt{2-y}}^{\sqrt{2-y}}1\,dx\,dy &=\int_0^1 2\sqrt{y}\,dy+\int_1^2 2\sqrt{2-y}\,dy\\ &=\left[\frac43y^{3/2}\right]_0^1 +\left[-\frac43(2-y)^{3/2}\right]_1^2\\ &=\frac43+\frac43 =\frac83. \end{aligned} $$

A single proposed interval \(-\sqrt{y}\le x\le\sqrt{y}\) for all \(0\le y\le2\) would be wrong: for \(y>\!1\), it can extend beyond the upper boundary \(y=2-x^2\). For instance, \(y=\frac32\) and \(x=\frac9{10}\) satisfy \(x^2\le y\), but \(y\le2-x^2\) fails because \(\frac32>\frac{119}{100}\). The transition at \(y=1\) is essential.

A Practical Audit

Before using reversed limits, carry out the following checks. If the bounds are piecewise, perform the slice and boundary checks on every piece. A quick sketch can help locate the relevant edges, but the inequalities determine exactly which points are included.

  • Outer range: Find the minimum and maximum values of the new outer variable over the original region.
  • Slice interval: Fix a value in that range and determine all allowed values of the inner variable, not just one side of the region.
  • Endpoint source: Check that each endpoint comes from a boundary of the region and that the boundary is solved for the correct variable.
  • Transition check: If two candidate boundaries can exchange roles, find where they meet and test which is active on each side.
  • Final consistency check: Compare areas or integrals when convenient, while remembering that a matching value alone does not certify matching regions.

A common error is to solve one boundary equation correctly and assume it gives the whole reversed interval. It may give only one endpoint, or only one side of a slice. Another is to use correct inner endpoints over an outer interval that is too wide. In either case, substitute a fixed outer value into the original inequalities: that simple test reveals whether the proposed slice includes exactly the required points.

Check Your Understanding

For each question, use the original inequalities to verify the proposed bounds rather than relying only on a computed integral.

  1. For \(0\le x\le2\), \(x^2\le y\le4\), what outer interval and horizontal slice endpoints describe the region?
  2. For \(-2\le x\le2\), \(x^2\le y\le4\), why would a proposed inner interval \(0\le x\le\sqrt{y}\) omit part of the region?
  3. What three features should be checked when auditing reversed bounds?
  4. Why can matching area values support a proposed reversal but not, by themselves, prove that the regions are identical?
  5. In the piecewise example, what changes at \(y=1\), and how does that affect the inner endpoints?

Previous

Next

Back to Tutorials