Why Slice Direction Matters
A bounded region can often be described in two ways: by vertical slices, which lead to integration with respect to \(x\), or by horizontal slices, which lead to integration with respect to \(y\). Both can give the same area, but the boundary functions and limits are different. A useful setup must describe the region accurately at every slice, not just use the right curves somewhere in the calculation.
Choosing a direction is therefore a geometric decision. Sketch or inspect the region, identify what each slice crosses first and last, and check whether those boundaries stay the same over the full range. If they do not, the setup may need to be split into separate integrals.
Describing an Area with Slices
A vertical slice at a particular \(x\) runs from a lower boundary \(y=g(x)\) to an upper boundary \(y=f(x)\). If those boundaries remain in that order for \(a\le x\le b\), the area is
A horizontal slice at a particular \(y\) runs from a left boundary \(x=u(y)\) to a right boundary \(x=v(y)\). If those boundaries remain in that order for \(c\le y\le d\), the area is
The word “top” or “right” refers to the position along the slice, not to which curve seems larger in a different coordinate direction. In particular, horizontal slices require boundary equations solved for \(x\) as functions of \(y\), when possible. A curve written as \(y=f(x)\) may need to be inverted to use it as a horizontal boundary.
A Justification for the Two Setups
The slice formulas come from approximating the region with thin strips. For vertical slices, divide the interval from \(a\) to \(b\) into narrow pieces of width \(\Delta x\). Over one such piece, the strip’s height is approximately \(f(x)-g(x)\), so its area is approximately \((f(x)-g(x))\Delta x\). Adding the strips gives an approximation to the region’s area:
As the widths shrink toward zero, these sums approach the definite integral. The same reasoning applies to horizontal strips: a strip of thickness \(\Delta y\) has approximate width \(v(y)-u(y)\), giving the sum
Its limit is the integral with respect to \(y\). This also explains why changing slice direction does not change the area: each method adds the areas of thinner and thinner strips that fill the same region. The functions and limits differ because the strips cross the region in different directions.
Worked Example: Both Directions Give One Integral
Worked Example: Both Directions Give One Integral
Find the area bounded by \(y=x^2\) and \(y=2x\). First find where the curves intersect:
For \(0<x<2\), \(2x-x^2=x(2-x)>0\), so the line is above the parabola. Vertical slices give
For horizontal slices, the curves must be written as \(x\) in terms of \(y\). From \(y=2x\), \(x=y/2\); from \(y=x^2\), the relevant branch is \(x=\sqrt y\), since the region lies at \(x\ge0\). The \(y\)-values run from \(0\) to \(4\). For \(0<y<4\), \(y/2<\sqrt y\), so the line is the left boundary and the parabola is the right boundary. Thus
The two answers agree. As a direct check, the vertical integrand is positive between the intersection points, and both methods describe the region between the same two curves.
Choosing a Direction When One Setup Splits
A setup using one direction may require multiple integrals even when the other direction uses just one. The reason is that a slice can encounter different boundary curves in different parts of the region. The following example shows why checking the boundary at each slice matters.
Worked Example: Horizontal Slices Avoid a Split
Find the area enclosed by \(x=y^2\) and \(x=2-y\). Intersections satisfy
so \(y=-2\) or \(y=1\). For \(-2<y<1\), the horizontal slice runs from the parabola \(x=y^2\) on the left to the line \(x=2-y\) on the right. The area is one integral:
For vertical slices, solve \(x=y^2\) as \(y=\pm\sqrt{x}\). The region runs from \(x=0\) to \(x=4\). Its lower boundary is \(y=-\sqrt{x}\), but its upper boundary changes: it is \(y=\sqrt{x}\) from \(x=0\) to \(x=1\), then \(y=2-x\) from \(x=1\) to \(x=4\). To locate the change, solve \(\sqrt{x}=2-x\), with \(x\le2\); squaring gives \(x=(2-x)^2\), or \((x-1)(x-4)=0\), and only \(x=1\) satisfies \(x\le2\). Therefore
Both setups give \(9/2\). The horizontal setup is shorter because its left and right boundaries remain the same throughout the full \(y\)-range. For a check on the vertical setup, its first piece contributes \(4/3\), its second contributes \(19/6\), and their sum is \(9/2\), matching the horizontal result.
Check Which Region Is Intended
The curves alone do not always specify a requested portion of the plane. A prompt may ask for one half of a region, or for all the area enclosed by the curves. Make that choice explicit before setting limits. In particular, a line such as \(x=0\) does not automatically restrict a region to \(x\ge0\); that restriction must be part of the specified region.
Worked Example: A Specified Half and the Entire Region
The curves \(y=x^2\) and \(y=4\) enclose the region between \(x=-2\) and \(x=2\), since \(x^2=4\) gives \(x=-2\) and \(x=2\). If the question asks for the right-hand half, meaning the part with \(x\ge0\), vertical slices give
For the same half-region, horizontal slices run from \(x=0\) to \(x=\sqrt y\), with \(0\le y\le4\). Therefore
If the question instead asks for the entire region enclosed by the two curves, the \(x\)-limits must include both sides. By symmetry, its area is
Thus the area of the specified right-hand half is \(16/3\), while the area of the entire enclosed region is \(32/3\). The boundary \(x=0\) is a cut dividing the full region into halves; it is not one of the two curves that enclose the full region.
A Practical Choice Strategy
When both directions are possible, the shorter integral is not automatically the correct one. First confirm that it represents the intended region; then compare the amount of boundary solving and splitting required. A reliable setup follows these checks:
Find relevant intersections and identify any extra restriction, such as \(x\ge0\), that specifies only part of a larger region.
Identify its lower and upper boundaries, then check whether those boundaries stay the same from the leftmost to the rightmost point.
Identify its left and right boundaries and check whether they stay the same from the bottom to the top.
Prefer a setup with clear boundaries and fewer pieces, then check that the limits and integrand describe every slice of the intended region.
A Common Mistake: Using the Wrong “Top” or “Right” Boundary
A common error is to subtract the functions in the order they were given instead of checking their position along the chosen slice. For example, on \(0<x<2\), \(y=x^2\) is below \(y=2x\). Writing
does not give a negative area; it shows that the subtraction order is reversed. The correction is to use upper minus lower:
Similarly, for horizontal slices the order is right minus left, not necessarily the same order as the curves were listed. If the proposed area is negative, check the order of the boundaries before proceeding.
Check Your Understanding
For each question, focus on how the slices match the region and whether the boundary functions stay consistent.
- For the region between \(y=x^2\) and \(y=2x\), which curve is the right boundary for horizontal slices?
- For the region enclosed by \(x=y^2\) and \(x=2-y\), why does a vertical-slice setup need two integrals?
- What information does the \(x\)-range provide in a vertical-slice area integral?
- If a prompt specifies the part of a region with \(x\ge0\), what should you check before using that restriction as an integration limit?
- For horizontal slices, which boundary is subtracted from which?
