One Region, Two Ways to Slice
A bounded region in the plane can often be described using vertical slices or horizontal slices. A vertical slice has thickness \(dx\), so its length is measured from a lower \(y\)-boundary to an upper \(y\)-boundary. A horizontal slice has thickness \(dy\), so its length is measured from a left \(x\)-boundary to a right \(x\)-boundary.
Both descriptions can give the area of the same region, but they do not necessarily have the same bounds or integrands. In particular, a boundary that is easy to write as \(y=f(x)\) may require solving for \(x\) to use horizontal slices. Also, the left or right boundary of a horizontal slice can change as the slice moves through the region.
What the Two Integrals Measure
Suppose a region has one vertical slice for each \(x\) from \(a\) to \(b\), with upper boundary \(y=u(x)\) and lower boundary \(y=\ell(x)\). Its area is
The integrand is the vertical slice length: top minus bottom. Alternatively, if horizontal slices run from a left boundary \(x=L(y)\) to a right boundary \(x=R(y)\), for \(y\) from \(c\) to \(d\), then
The integrand is now the horizontal slice length: right minus left. To find horizontal boundaries from equations written as \(y=f(x)\), solve those equations for \(x\) in terms of \(y\), keeping track of which branch applies.
Why the Two Setups Give the Same Area
Each setup adds the areas of thin slices that cover the region. With vertical slices, one approximates the area by adding thin rectangles whose widths are changes in \(x\) and whose heights are vertical slice lengths. With horizontal slices, one adds thin rectangles whose heights are changes in \(y\) and whose widths are horizontal slice lengths.
As the slices become thinner, each sum approaches the area of the same region. The two descriptions may divide the region differently, and one may require more pieces, but neither changes which points belong to the region. When the boundaries are continuous or piecewise continuous and the region is bounded, the corresponding area integrals therefore have the same value, provided each setup covers the region exactly once.
Worked Example: A Parabolic Region with Simple Horizontal Bounds
Worked Example: Reversing the Slices Between Two Curves
Find the area enclosed by \(y=x\) and \(y=x^2\), and set it up using both \(dx\) and \(dy\).
Step 1: Find the intersections. Set the equations equal:
At both intersection points, \(y=x\), so the intersection points are \((0,0)\) and \((1,1)\).
Step 2: Use vertical slices. On \(0\le x\le1\), \(x\ge x^2\), so the line is above the parabola. The vertical slice length is \(x-x^2\):
The antiderivative checks because \(\frac{d}{dx}\left(\frac{x^2}{2}-\frac{x^3}{3}\right)=x-x^2\).
Step 3: Use horizontal slices. Solve each boundary equation for \(x\). The line \(y=x\) gives \(x=y\); the parabola \(y=x^2\) gives \(x=\sqrt y\) on this region. For \(0\le y\le1\), \(y\le\sqrt y\), so the horizontal slice runs from \(x=y\) to \(x=\sqrt y\):
The derivative of the displayed antiderivative is \(\sqrt y-y\). Both descriptions use the full ranges \(0\le x\le1\) and \(0\le y\le1\), and both give area \(\frac16\).
When Horizontal Slices Need More Than One Integral
A vertical-slice description may use one upper curve and one lower curve throughout, while the horizontal slices change their left or right boundaries. In that situation, a single horizontal integral with one pair of boundary formulas can miss part of the region. Split the \(y\)-range at the height where the boundary changes.
Worked Example: A Region Whose Horizontal Slice Changes Boundary
Find the area enclosed by \(y=x^2\) and \(y=2-x\), using both vertical and horizontal slices. Verify that the horizontal description covers the whole region.
Step 1: Find the intersections and vertical description. Solve \(x^2=2-x\):
The corresponding \(y\)-values are \(4\) and \(1\). Between \(x=-2\) and \(x=1\), the line is above the parabola; for example, at \(x=0\), the line has value \(2\) and the parabola has value \(0\). Thus
At \(x=1\), the antiderivative is \(2-\frac12-\frac13=\frac76\). At \(x=-2\), it is \(-4-2+\frac83=-\frac{10}{3}\). Therefore
The antiderivative checks: its derivative is \(2-x-x^2\), the vertical slice length.
Step 2: Identify the horizontal boundaries. The parabola \(y=x^2\) gives two branches \(x=-\sqrt y\) and \(x=\sqrt y\). The line gives \(x=2-y\). The lowest height in the region is \(y=0\), at the parabola's vertex, and the greatest is \(y=4\), at the left intersection.
For \(0\le y\le1\), the horizontal slice runs between the two branches of the parabola, from \(x=-\sqrt y\) to \(x=\sqrt y\). For \(1\le y\le4\), its left endpoint remains \(x=-\sqrt y\), but its right endpoint is the line \(x=2-y\). The change occurs at \(y=1\), where the right parabola branch and the line meet at \(x=1\).
Step 3: Integrate both horizontal pieces.
The first integral is \(\frac43\). For the second, the antiderivative at \(4\) is \(8-8+\frac{16}{3}=\frac{16}{3}\), and at \(1\) it is \(2-\frac12+\frac23=\frac{13}{6}\). Hence
The derivatives of the two antiderivatives are \(2\sqrt y\) and \(2-y+\sqrt y\), as required. The horizontal result agrees with the vertical result. The split is essential: below \(y=1\), the right endpoint is the parabola; above \(y=1\), it is the line.
A Second Check: A Symmetric Parabolic Region
Worked Example: Comparing Slices Across a Full Parabola
Find the area enclosed by \(y=x^2\) and \(y=4\), using both slice directions.
Vertical slices: The intersections satisfy \(x^2=4\), so \(x=-2\) or \(x=2\). The line \(y=4\) is the upper boundary and \(y=x^2\) is the lower boundary:
The antiderivative differentiates to \(4-x^2\). For horizontal slices, \(y\) ranges from \(0\) to \(4\), and solving \(y=x^2\) gives \(x=-\sqrt y\) and \(x=\sqrt y\). Thus
The derivative of \(\frac43y^{3/2}\) is \(2\sqrt y\). At every height from \(0\) to \(4\), the horizontal interval includes both parabola branches, so this one integral covers the full region.
A Reliable Comparison Process
When checking whether \(dx\) and \(dy\) setups describe the same enclosure, use a short sequence of geometric checks before evaluating either integral.
These determine where the curves meet and often provide the outer limits of one or both slicing variables.
Identify the lower and upper boundaries throughout the \(x\)-range, and use upper minus lower.
Invert the boundary equations and choose the correct branches for the region, not merely all algebraic solutions.
Check near the bottom, top, and any height where a boundary may change. Split the integral at each change.
Evaluate exactly when possible. Agreement is a useful check, but it does not replace verifying the geometry of each setup.
Common Mistake: Using One Right Boundary at Every Height
In the region between \(y=x^2\) and \(y=2-x\), it is incorrect to use the line as the right boundary for every horizontal slice from \(y=0\) to \(y=4\). For \(0<y<1\), the line lies to the right of the region; the right boundary is \(x=\sqrt y\). Using the line throughout gives the wrong setup
This calculation produces \(\left[2y-\frac{y^2}{2}+\frac23y^{3/2}\right]_0^4=\frac{16}{3}\), not the region's area \(\frac92\). The problem is visible geometrically: below \(y=1\), the proposed right endpoint extends beyond the parabola and includes points outside the enclosure. The correction is to use the two pieces
Another frequent error is reversing the horizontal difference. The horizontal slice length is right minus left, not left minus right. If a proposed slice length is negative within the region, recheck the endpoint order or the branch choices.
Why Comparing Setups Is Useful
Sometimes one direction gives a shorter integral, while the other makes the shape or boundary changes easier to see. Comparing them helps reveal whether bounds span the entire region and whether each integrand measures an actual slice length. It also provides an independent check on an area calculation: the two exact values should agree when both descriptions are correct.
Do not expect the two integrals to look alike. A \(dx\) integrand measures vertical length, and a \(dy\) integrand measures horizontal length. Their limits and formulas can be quite different even though both accumulate the same area.
Check Your Understanding
For each question, explain how the slice endpoints determine the bounds and integrand.
- For the region between \(y=x\) and \(y=x^2\) from their intersections, what are the horizontal boundaries in terms of \(y\)?
- In the region between \(y=x^2\) and \(y=2-x\), at what height does the horizontal right boundary change, and which boundary applies on either side?
- Why is the horizontal slice length written as right boundary minus left boundary?
- What geometric error occurs if the line \(x=2-y\) is used as the right boundary for every horizontal slice of that region?
- State one reason the \(dx\) and \(dy\) integrals for the same region can have different numbers of terms.
