Reversing Order For A Region With One Slice Interval

Why Reverse the Order?

A double integral can describe the same region using vertical slices or horizontal slices. Reversing the order means describing those same points with the other kind of slice, then changing the integration limits to match. This can make a calculation easier, but the first task is geometric: the new limits must cover exactly the original region.

Here we focus on regions for which each horizontal slice is either empty or one interval. If the original bounds describe vertical slices, the goal is to find the bottom and top of the region and, at each height, the leftmost and rightmost points of the horizontal slice.

Key idea: Reversing the order does not change the region or the integrand. It changes which variable is integrated first, so the limits must be redrawn from the region’s geometry.

From Vertical Slices to Horizontal Slices

Suppose a region is given by vertical slices as

R=\{(x,y):a\leq x\leq b,\ L(x)\leq y\leq U(x)\}.

For a fixed \(x\), the values of \(y\) run from the lower boundary \(L(x)\) to the upper boundary \(U(x)\). To reverse the order, first find the full range of \(y\) values attained in \(R\). Then fix a value of \(y\) in that range and find where the horizontal line at that height enters and leaves the region.

If those two points have \(x\)-coordinates \(\alpha(y)\) and \(\beta(y)\), and the slice is one interval, then the horizontal description is

R=\{(x,y):c\leq y\leq d,\ \alpha(y)\leq x\leq\beta(y)\}.

The corresponding iterated integrals are

\int_a^b\int_{L(x)}^{U(x)} f(x,y)\,dy\,dx = \int_c^d\int_{\alpha(y)}^{\beta(y)} f(x,y)\,dx\,dy,

provided the region is bounded, the boundary curves are sufficiently regular for the integrals to exist, and \(f\) is continuous on the region. These conditions ensure that both iterated integrals represent the same double integral.

A Reliable Sketch-Based Method

Read the original region.
Translate the nested limits into inequalities and sketch the boundary curves and endpoints.

Find the new outer range.
Read the lowest and highest \(y\)-values attained anywhere in the region.

Draw a horizontal slice.
At a representative height \(y\), identify the left and right boundary intersections.

Write the new limits.
Express the left and right \(x\)-coordinates as functions of \(y\), using the appropriate boundary equations.

Check the description.
Confirm that the limits include every point of the region exactly once and that each horizontal slice is one interval.

Boundary equations often need to be solved for \(x\) to obtain horizontal-slice limits. Use the sketch to choose the correct solution branch and to set the range where that branch applies. Inverting an equation without checking the geometry can select the wrong endpoint.

Worked Examples

Worked Example: Reversing the Order Under a Parabola

Reverse the order of integration for the area of the region described by \(-\sqrt{2}\leq x\leq\sqrt{2}\) and \(x^2\leq y\leq2\), then check the result.

The region lies above \(y=x^2\) and below \(y=2\). The parabola meets the line \(y=2\) when \(x^2=2\), so \(x=-\sqrt{2}\) and \(x=\sqrt{2}\). Its lowest point is \(y=0\), and the top of the region is \(y=2\). Thus \(0\leq y\leq2\).

At a fixed height \(y\), the boundary equation \(y=x^2\) gives \(x=-\sqrt{y}\) on the left and \(x=\sqrt{y}\) on the right. The horizontal slice is the interval between them. Therefore the reversed integral for the area is

\int_0^2\int_{-\sqrt{y}}^{\sqrt{y}}1\,dx\,dy.

Evaluate it:

\begin{aligned} \int_0^2\int_{-\sqrt{y}}^{\sqrt{y}}1\,dx\,dy &=\int_0^2\bigl(\sqrt{y}-(-\sqrt{y})\bigr)\,dy\\ &=\int_0^2 2\sqrt{y}\,dy\\ &=\left[\frac{4}{3}y^{3/2}\right]_0^2\\ &=\frac{4}{3}(2\sqrt{2})-0\\ &=\frac{8\sqrt{2}}{3}. \end{aligned}

Check using the original order:

\begin{aligned} \int_{-\sqrt{2}}^{\sqrt{2}}\int_{x^2}^{2}1\,dy\,dx &=\int_{-\sqrt{2}}^{\sqrt{2}}(2-x^2)\,dx\\ &=\left[2x-\frac{x^3}{3}\right]_{-\sqrt{2}}^{\sqrt{2}}\\ &=\left(2\sqrt{2}-\frac{2\sqrt{2}}{3}\right) -\left(-2\sqrt{2}+\frac{2\sqrt{2}}{3}\right)\\ &=\frac{8\sqrt{2}}{3}. \end{aligned}

Both orders give the same area. The horizontal bounds also include the full slice on both sides of the \(y\)-axis.

Worked Example: Reversing a Region Bounded by a Logarithmic Curve

Reverse the order in

\int_1^2\int_0^{\ln x}1\,dy\,dx.

The inequalities are \(1\leq x\leq2\) and \(0\leq y\leq\ln x\). The region runs from \(y=0\) up to \(y=\ln 2\). For a fixed \(y\), the inequality \(y\leq\ln x\) is equivalent to \(x\geq e^y\). Together with \(x\leq2\), this gives the horizontal interval \(e^y\leq x\leq2\). Thus the reversed integral is

\int_0^{\ln 2}\int_{e^y}^{2}1\,dx\,dy.

Evaluate in the reversed order:

\begin{aligned} \int_0^{\ln 2}\int_{e^y}^{2}1\,dx\,dy &=\int_0^{\ln 2}(2-e^y)\,dy\\ &=\left[2y-e^y\right]_0^{\ln 2}\\ &=(2\ln 2-2)-(0-1)\\ &=2\ln 2-1. \end{aligned}

Check with the original order:

\begin{aligned} \int_1^2\int_0^{\ln x}1\,dy\,dx &=\int_1^2\ln x\,dx\\ &=\left[x\ln x-x\right]_1^2\\ &=(2\ln 2-2)-(0-1)\\ &=2\ln 2-1. \end{aligned}

The two evaluations agree. At \(y=0\), the new slice runs from \(x=e^0=1\) to \(x=2\); at \(y=\ln 2\), it collapses to \(x=2\), as the sketch requires.

Worked Example: Reversing the Order Below a Sine Curve

Describe the region below \(y=\sin x\) for \(0\leq x\leq\frac{\pi}{2}\), above \(y=0\), using horizontal slices. Then compare the two area integrals.

The region’s \(y\)-values range from \(0\) to \(1\), since \(\sin 0=0\) and \(\sin(\frac{\pi}{2})=1\). For a fixed \(y\) in this range, the curve intersection satisfies \(y=\sin x\). On \(0\leq x\leq\frac{\pi}{2}\), this gives \(x=\arcsin y\). The slice extends from the curve to the right edge \(x=\frac{\pi}{2}\), so the reversed limits are

\int_0^1\int_{\arcsin y}^{\pi/2}1\,dx\,dy.

Evaluate the reversed integral. Since \(\frac{d}{dy}\bigl(y\arcsin y+\sqrt{1-y^2}\bigr)=\arcsin y\) for \(0<y<1\),

\begin{aligned} \int_0^1\int_{\arcsin y}^{\pi/2}1\,dx\,dy &=\int_0^1\left(\frac{\pi}{2}-\arcsin y\right)\,dy\\ &=\left[\frac{\pi y}{2}-y\arcsin y-\sqrt{1-y^2}\right]_0^1\\ &=\left(\frac{\pi}{2}-\frac{\pi}{2}-0\right)-(0-0-1)\\ &=1. \end{aligned}

In the original order,

\begin{aligned} \int_0^{\pi/2}\int_0^{\sin x}1\,dy\,dx &=\int_0^{\pi/2}\sin x\,dx\\ &=\left[-\cos x\right]_0^{\pi/2}\\ &=0-(-1)\\ &=1. \end{aligned}

Both orders give area \(1\). The inverse curve is used only on the stated interval, where \(\arcsin y\) is the correct left endpoint.

Why the Reversal Works

The change of limits can be justified by describing the same set of points in two ways. For example, suppose the original region is \(a\leq x\leq b\), \(L(x)\leq y\leq U(x)\). At a fixed height \(y\), the points in the region are exactly the \(x\)-values satisfying all three conditions

a\leq x\leq b,\qquad L(x)\leq y,\qquad y\leq U(x).

If solving these conditions gives one interval \(\alpha(y)\leq x\leq\beta(y)\), then the set of points has the horizontal description \(c\leq y\leq d\), \(\alpha(y)\leq x\leq\beta(y)\). The bounds have not changed the region; they have only changed how its points are grouped into slices.

For a continuous function on a bounded region with regular boundary, the Fubini theorem for double integrals permits the two iterated descriptions of that same region. Consequently, changing the order gives the same integral when the new bounds describe precisely the same points. The geometric check is essential: the theorem does not make incorrect limits valid.

A Common Mistake: Reversing the Boundary Inequality

In the logarithmic example, the condition \(y\leq\ln x\) implies \(x\geq e^y\), not \(x\leq e^y\). If the endpoints are accidentally reversed, one might write

\int_0^{\ln 2}\int_1^{e^y}1\,dx\,dy.

This describes the interval from \(1\) to \(e^y\), which lies on the other side of the curve. Its value is

\begin{aligned} \int_0^{\ln 2}(e^y-1)\,dy &=\left[e^y-y\right]_0^{\ln 2}\\ &=(2-\ln 2)-(1-0)\\ &=1-\ln 2, \end{aligned}

which does not match the correct area \(2\ln 2-1\). A quick geometric check catches the error: at \(y=0\), the original region extends from \(x=1\) to \(x=2\), whereas the incorrect bounds collapse to \(x=1\). When deriving new limits, test a boundary height and confirm which side of the boundary belongs to the region.

Check Your Understanding

For each question, begin by identifying the region and sketching a representative horizontal slice.

For \(\int_{-1}^{1}\int_{x^2}^{1}1\,dy\,dx\), what are the reversed limits?
In a reversal, what do the outer limits describe geometrically?
For the region \(1\leq x\leq e\), \(0\leq y\leq\ln x\), what are the left and right endpoints of a horizontal slice?
Why is solving a boundary equation for \(x\) not enough, by itself, to determine the correct horizontal limits?
What geometric check can reveal that new bounds describe the wrong side of a boundary?