When One Slice Meets More Than One Piece
A vertical slice of a region does not always form one continuous segment. A vertical line might pass through a lower part of the region, then through a gap outside the region, and then through an upper part. The area calculation must count both pieces without counting the gap between them.
The key is to describe each vertical slice by all the intervals it actually contains. For each interval, identify its lower and upper boundary, and add the lengths. Integrating those lengths over the appropriate x-ranges gives the area.
Vertical Slices With Multiple Intervals
For a fixed value of \(x\), a vertical slice consists of the points in the region with that x-coordinate. If the slice has separate intervals from \(y=b_1(x)\) to \(y=t_1(x)\), and from \(y=b_2(x)\) to \(y=t_2(x)\), their total length is
Here \(b_i(x)\) is the bottom boundary and \(t_i(x)\) is the top boundary of interval \(i\). Each interval must have its own top and bottom. If the number of intervals or their boundary curves changes as \(x\) varies, split the x-range at the points where that change occurs.
The inner sum is taken over the intervals present at that x-value. The outer sum accounts for different x-ranges. Endpoints shared by adjacent ranges do not affect the area, so an interval endpoint can be included in either neighboring range.
Why the Rule Works
The definite integral of a slice length accumulates area across the x-direction. For one interval, the slice length is top minus bottom. If a slice has two or more disjoint intervals, its length is the sum of their separate lengths. Additivity of area means the total area of those pieces is the sum of their areas. Therefore, integrating the total slice length gives the sum of the integrals of the individual lengths.
More explicitly, suppose throughout an x-range the slice contains two intervals. Write their lengths as \(L_1(x)=t_1(x)-b_1(x)\) and \(L_2(x)=t_2(x)-b_2(x)\). The total slice length is \(L_1(x)+L_2(x)\), so the area over that range is
This is the additivity property of the definite integral. It justifies adding the separate top-minus-bottom integrals. It also explains why filling a gap between intervals is not allowed: that gap is not part of the slice length.
Worked Example: The Annulus
Worked Example: The Annulus
Find the area of the annulus between the circles \(x^2+y^2=4\) and \(x^2+y^2=1\), using vertical slices.
The outer circle has radius \(2\), and the inner circle has radius \(1\). Solving each circle equation for \(y\) gives the upper and lower branches
For \(-1<x<1\), a vertical slice meets the annulus in two intervals. The lower interval runs from the lower outer circle to the lower inner circle, and the upper interval runs from the upper inner circle to the upper outer circle. For \(1\le |x|\le2\), the inner circle is absent at that x-value, so the slice is one interval through the outer disk. By symmetry about the y-axis, the area is
Equivalently, the annulus is the outer disk with the inner disk removed. This gives a shorter evaluation of the same slice lengths. For a circle of radius \(r\), the full vertical-slice integral is
The antiderivative is valid for \(-r<x<r\); its endpoint values are obtained by continuity. Substituting \(r=2\) gives outer-disk area \(4\pi\), and substituting \(r=1\) gives inner-disk area \(\pi\). Thus
As a check, the annulus area is also \(\pi(2^2-1^2)=3\pi\). The two-interval slices in the central x-range are essential: they exclude the inner disk instead of filling its hole.
Worked Example: Two Curved Bands
Worked Example: Two Curved Bands
Let a region consist of the two parts described by
Find its area using vertical slices, then check the setup with horizontal slices.
For every \(x\in[-1,1]\), the vertical slice contains two disjoint intervals: one from \(x^2\) to \(1\), and one from \(2\) to \(3-x^2\). Their lengths are \(1-x^2\) and \(1-x^2\), respectively. Therefore,
For a horizontal check, the lower part occupies \(0\le y\le1\). At height \(y\) in that range, \(x^2\le y\), so \(-\sqrt y\le x\le\sqrt y\). Its horizontal width is \(2\sqrt y\), and its area is
The upper part occupies \(2\le y\le3\). From \(y\le3-x^2\), we obtain \(-\sqrt{3-y}\le x\le\sqrt{3-y}\). Its area is
where \(u=3-y\), so the bounds change from \(y=2,3\) to \(u=1,0\), with the reversed bounds accounted for in the displayed integral. The horizontal setup gives \(4/3+4/3=8/3\), agreeing with the vertical calculation.
When the Number of Intervals Changes
A region may have one vertical interval over part of its x-range and two over another part. In that case, list the intervals separately on each range. Do not force one formula to cover the entire region if the slice structure changes.
Worked Example: One Component Appears Partway Across
Consider the union of the regions
For \(-1\le x<0\), there is only the first interval. For \(0\le x\le1\), there are two intervals. For \(1<x\le2\), there is only the second interval. The areas of the two components can be integrated separately because their vertical ranges are disjoint:
For a horizontal check, the first component has width \(2\sqrt y\) for \(0\le y\le1\), then width \(2\) for \(1\le y\le2\). Its area is
The second component has \(3\le y\le4\), with \((x-1)^2\le4-y\), giving width \(2\sqrt{4-y}\). Its area is
where \(u=4-y\). The horizontal setup also gives \(10/3+4/3=14/3\). In the vertical setup, the component that starts at \(x=0\) must be included only for \(0\le x\le2\), its actual x-range.
A Common Overcounting Error
In the two-band example, one might take the highest boundary \(3-x^2\) and the lowest boundary \(x^2\), then integrate their difference over \([-1,1]\). That calculation is
This is not the area of the two bands. It includes the empty gap between \(y=1\) and \(y=2\), whose area is \(1\cdot2=2\). Subtracting that gap gives \(14/3-2=8/3\), the correct area. The direct correction is to add the lengths of the two actual intervals, rather than treating the entire distance from the lowest to the highest boundary as part of the region.
A Reliable Setup
Mark x-values where a component begins or ends, boundaries meet, or the number of vertical intervals changes.
For each interval, name its lower boundary and upper boundary. Do not bridge a gap outside the region.
On each range, sum top minus bottom for all present intervals, integrate, and then add the results across ranges.
A horizontal setup can confirm the area, but it too must account for every separate interval at each height.
Horizontal slices are not automatically simpler. Their advantage is that the region may form a single interval at each height even when vertical slices are disconnected. Conversely, horizontal slices can also split into multiple intervals. In either direction, the same rule applies: count only the pieces of the region that the slice actually meets.
Check Your Understanding
Use the interval lengths in each question; do not assume a slice is connected.
- A vertical slice meets a region from \(y=0\) to \(y=1\), and again from \(y=3\) to \(y=5\). What is its total length?
- Why is the expression “highest boundary minus lowest boundary” incorrect for a slice with a gap?
- What x-values should prompt you to split an area integral into separate ranges?
- For a region made of two components with disjoint y-ranges, explain how a horizontal-slice setup can help check a vertical-slice calculation.
- In the annulus example, what part of the region would be counted incorrectly by using only the outer circle’s upper and lower boundaries?
