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Integrals over General Regions · Tutorial 549 of 999

Worked Workshop: Reversing Polynomial Integrals

Practice turning polynomial double integrals into correctly reversed, piecewise iterated integrals, then evaluate and verify each result.

Intermediate 10 min read

What You'll Learn

  • Sketch a region from its original iterated bounds before reversing the order.
  • Identify when horizontal slices are disconnected and require separate integrals.
  • Find where a reversed boundary changes and split the outer interval there.
  • Evaluate polynomial double integrals using reversed bounds.
  • Check reversed calculations against the original order of integration.
  • Recognize and correct a common error in reversing bounds.

Why Work Through the Region First?

A double integral is tied to both its integrand and its region. Reversing the order changes which variable is integrated first, so the bounds must describe the same points in a different way. For polynomial integrands, the arithmetic may be straightforward; the main work is often finding the correct reversed bounds.

A reliable approach is to read the original bounds as inequalities, sketch the resulting region, and then describe its horizontal slices. Some horizontal slices may split into separate intervals. In other regions, a boundary changes partway through the range of the new outer variable, so the reversed integral must be split at that transition.

Working principle: Reverse the description of the region, not just the order of the differentials. The reversed bounds must include every point in the original region exactly once.

A Slice-Based Justification

Suppose a bounded planar region \(R\) has a description using vertical slices and another description using horizontal slices. If \(f\) is continuous on \(R\), the iterated integrals in either order give the same double integral, provided each description uses bounds that cover the region correctly. The boundaries in the examples below are polynomial curves, and the regions are bounded, so this standard change-of-order result applies.

The reason the bounds must sometimes be split is geometric. At a fixed height \(y\), the points of the region form a horizontal slice. If that slice has two separated pieces, one interval of \(x\)-values cannot describe it. If the slice has one interval but its endpoint is determined by different curves at different heights, the formula for that endpoint changes at the height where the curves meet.

1
Read the original bounds.
Translate them into inequalities that specify which points belong to the region.
2
Sketch the boundary and the region.
Mark intersections, endpoints, and any gaps that can occur in a horizontal slice.
3
Describe the horizontal slices.
Find the full range of \(y\), then determine the allowed \(x\)-values at each height.
4
Split where the description changes.
Use separate integrals for disconnected intervals or for different boundary formulas.
5
Evaluate and check.
Compute the reversed integral and, when useful, compare it with the original order.

Worked Example: A Region with Disconnected Horizontal Slices

Worked Example: A Region with Disconnected Horizontal Slices

Reverse and evaluate the integral

$$ \int_0^2\int_0^{(x-1)^2} x\,dy\,dx. $$

The bounds describe \(0\le x\le 2\) and \(0\le y\le (x-1)^2\). The boundary curve \(y=(x-1)^2\) has height \(1\) at \(x=0\) and \(x=2\), and reaches height \(0\) at \(x=1\). Thus \(0\le y\le1\). At a fixed \(y\), the inequality \(y\le(x-1)^2\) becomes \(|x-1|\ge\sqrt y\). Together with \(0\le x\le2\), this gives two intervals: \(0\le x\le1-\sqrt y\) and \(1+\sqrt y\le x\le2\).

The reversed integral is therefore

$$ \int_0^1\left( \int_0^{1-\sqrt y}x\,dx+ \int_{1+\sqrt y}^2x\,dx \right)dy. $$

Evaluate the two inner integrals separately. The first is

$$ \int_0^{1-\sqrt y}x\,dx =\left[\frac{x^2}{2}\right]_0^{1-\sqrt y} =\frac{(1-\sqrt y)^2}{2}. $$

The second is

$$ \int_{1+\sqrt y}^2x\,dx =\left[\frac{x^2}{2}\right]_{1+\sqrt y}^2 =2-\frac{(1+\sqrt y)^2}{2}. $$

Adding and simplifying, with \(s=\sqrt y\), gives

$$ \frac{(1-s)^2}{2}+2-\frac{(1+s)^2}{2} =\frac{1-2s+s^2+4-1-2s-s^2}{2} =2-2\sqrt y. $$

Hence the value is

$$ \int_0^1(2-2\sqrt y)\,dy =\left[2y-\frac{4}{3}y^{3/2}\right]_0^1 =2-\frac43 =\frac23. $$

Check in the original order:

$$ \int_0^2\int_0^{(x-1)^2}x\,dy\,dx =\int_0^2x(x-1)^2\,dx =\int_0^2(x^3-2x^2+x)\,dx =\left[\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}\right]_0^2 =4-\frac{16}{3}+2 =\frac23. $$

The check agrees. A single interval from \(x=0\) to \(x=2\) would be wrong for most heights: the region has a gap around \(x=1\) whenever \(y>0\).

Worked Example: A Boundary Changes at a Transition Height

Worked Example: A Boundary Changes at a Transition Height

Reverse and evaluate

$$ \int_0^1\int_{x^2}^{2-x}x\,dy\,dx. $$

The region is \(0\le x\le1\), between \(y=x^2\) and \(y=2-x\). The curves meet at \(x=1\), where both have height \(1\). The smallest height in the region is \(0\), and the largest is \(2\). To find the horizontal slices, use \(x^2\le y\) and \(y\le2-x\), along with \(0\le x\le1\).

For \(0\le y\le1\), the parabola gives \(x\le\sqrt y\), while the line gives \(x\le2-y\). In this range, \(\sqrt y\le2-y\), so the slice runs from \(0\) to \(\sqrt y\). For \(1\le y\le2\), the parabola condition allows all \(x\) from \(0\) to \(1\), but the line requires \(x\le2-y\). The slice then runs from \(0\) to \(2-y\). The transition occurs at \(y=1\).

Thus the reversed integral is

$$ \int_0^1\int_0^{\sqrt y}x\,dx\,dy +\int_1^2\int_0^{2-y}x\,dx\,dy. $$

Evaluate the first part:

$$ \int_0^1\left[\frac{x^2}{2}\right]_0^{\sqrt y}dy =\int_0^1\frac y2\,dy =\left[\frac{y^2}{4}\right]_0^1 =\frac14. $$

For the second part,

$$ \int_1^2\left[\frac{x^2}{2}\right]_0^{2-y}dy =\frac12\int_1^2(2-y)^2\,dy. $$

Using \(u=2-y\), the limits change from \(y=1,2\) to \(u=1,0\), and \(dy=-du\). Therefore

$$ \frac12\int_1^2(2-y)^2\,dy =\frac12\int_0^1u^2\,du =\frac12\left[\frac{u^3}{3}\right]_0^1 =\frac16. $$

The total is \(\frac14+\frac16=\frac5{12}\). Check directly in the original order:

$$ \int_0^1\int_{x^2}^{2-x}x\,dy\,dx =\int_0^1x(2-x-x^2)\,dx =\left[x^2-\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1 =1-\frac13-\frac14 =\frac5{12}. $$

The two descriptions agree. The split is essential because the upper endpoint of a horizontal slice is \(\sqrt y\) below height \(1\), but \(2-y\) above height \(1\).

Worked Example: Reversing a Symmetric Region

Worked Example: Reversing a Symmetric Region

Reverse and evaluate

$$ \int_{-2}^{2}\int_{x^2}^{4}(x^2+y)\,dy\,dx. $$

The region lies above \(y=x^2\) and below \(y=4\), with \(-2\le x\le2\). Its horizontal slices have heights \(0\le y\le4\). At a fixed height, \(x^2\le y\) means \(-\sqrt y\le x\le\sqrt y\). The reversed integral is

$$ \int_0^4\int_{-\sqrt y}^{\sqrt y}(x^2+y)\,dx\,dy. $$

For the inner integral, an antiderivative with respect to \(x\) is \(x^3/3+yx\). Substitution at the two endpoints gives

$$ \left[\frac{x^3}{3}+yx\right]_{-\sqrt y}^{\sqrt y} =\left(\frac{y^{3/2}}{3}+y^{3/2}\right) -\left(-\frac{y^{3/2}}{3}-y^{3/2}\right) =\frac{8}{3}y^{3/2}. $$

Therefore

$$ \int_0^4\frac83y^{3/2}\,dy =\frac83\left[\frac25y^{5/2}\right]_0^4 =\frac{16}{15}\cdot 32 =\frac{512}{15}, $$

since \(4^{5/2}=32\). To check, integrate in the original order:

$$ \begin{aligned} \int_{-2}^{2}\int_{x^2}^{4}(x^2+y)\,dy\,dx &=\int_{-2}^{2}\left[x^2y+\frac{y^2}{2}\right]_{x^2}^{4}dx\\ &=\int_{-2}^{2}\left(4x^2-\!x^4+8-\frac{x^4}{2}\right)dx\\ &=\int_{-2}^{2}\left(8+4x^2-\frac32x^4\right)dx\\ &=32+\frac{64}{3}-\frac{96}{5} =\frac{512}{15}. \end{aligned} $$

Here the horizontal slices remain a single interval throughout, so no split is needed. The reversed bounds also make the symmetry of each slice explicit.

A Common Reversal Error

A frequent mistake is to reverse the differentials but keep the original outer endpoints. In the first example, that error might produce

$$ \int_0^1\int_0^2x\,dx\,dy =\int_0^1 2\,dy =2. $$

This is not the same region: at a height such as \(y=1/4\), for example, the original region allows \(0\le x\le1/2\) and \(3/2\le x\le2\), not every \(x\) between \(0\) and \(2\). The correct reversed bounds preserve those two intervals, and the integral is \(\frac23\), not \(2\).

Correction: After reversing, test a typical horizontal slice against the original inequalities. If the slice has a gap or a changing endpoint, reflect that in the bounds before evaluating.

Why the Sketch Matters

The examples illustrate three distinct features that a sketch can reveal: disconnected slices, a change in the curve defining a slice endpoint, and a region whose slices need no splitting. The same habits apply even when the integrand is more complicated. Bounds come from the region, not from the integrand, so a polynomial integrand cannot compensate for a geometric error.

Before doing the antiderivatives, check that the new outer interval covers all heights in the region and that the inner bounds describe the entire horizontal slice at each height. Then check the final value by evaluating in the original order when that calculation is manageable. Agreement is a useful verification; it does not replace correct bounds.

Check Your Understanding

For each question, identify the geometry before attempting the integrations.

  1. For the region \(0\le x\le2\), \(0\le y\le(x-1)^2\), why are the horizontal slices disconnected for \(0<y<1\)?
  2. For the region \(0\le x\le1\), \(x^2\le y\le2-x\), at what height does the upper \(x\)-bound change, and what are its formulas on either side?
  3. What inequalities describe the horizontal slices of the region \(x^2\le y\le4\), \(-2\le x\le2\)?
  4. Why is it generally incorrect to retain the original outer endpoints after reversing the order of integration?
  5. What geometric checks can you make to decide whether a reversed integral needs to be split?

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