Why Rewrite a Boundary as \(x\) in Terms of \(y\)?
When a region is sliced horizontally, each slice runs from a left boundary to a right boundary. To find its width, we need the \(x\)-coordinates of those boundaries at a given height \(y\). A curve given as \(y=f(x)\) may therefore need to be rearranged as \(x\) in terms of \(y\).
Rearranging is not just an algebraic convenience. A curve may fail to have a single \(x\)-value for each height, or an algebraic equation may produce solutions that are outside the part of the curve bounding the region. The correct formula must match the actual boundary and its interval.
When Can a Boundary Be Inverted?
Suppose a boundary is described by \(y=f(x)\) for \(x\) in an interval \(I\). It can be written as a single function \(x=g(y)\) on the corresponding range if each height in that range comes from exactly one \(x\)-value in \(I\). In other words, \(f\) must be one-to-one on \(I\). In standard calculus, a continuous, strictly increasing or strictly decreasing function on an interval is one-to-one there.
The range matters: the inverse formula is valid only for \(y\)-values in \(J\). If the original curve is specified only on part of its full domain, use the range of that restricted part, not the range of an unrestricted formula.
If the curve is not one-to-one on the interval under consideration, there may be several \(x\)-values at the same height. In that case, it cannot be represented there by one inverse function. Often the remedy is to split the curve into intervals on which it is one-to-one and solve for each branch separately. Which branches are needed depends on the region being sliced.
From the Inverse Branch to a Slice Width
At a fixed height \(y\), suppose a horizontal slice of the region runs from \(x=L(y)\) on the left to \(x=R(y)\) on the right. Its width is \(R(y)-L(y)\). If these endpoints vary continuously across \(a\leq y\leq b\), the area is obtained by adding the widths of thin horizontal strips:
Here \(a\) and \(b\) are the bottom and top heights of the region. If a horizontal slice has more than one separate piece, its total width is the sum of the widths of those pieces. The integral must reflect that geometry; a single right-minus-left expression spanning a gap would count points outside the region.
The following result explains why solving for \(x\) is the relevant algebraic step.
Why Inversion Gives the Correct Endpoint
Fix a height \(y\) in the range of the boundary. By definition of the inverse, \(f^{-1}(y)\) is the unique \(x\) in the chosen interval for which \(f(x)=y\). A horizontal line at that height meets this boundary at precisely that \(x\)-coordinate. If another boundary meets the same horizontal line at \(x=h(y)\), the distance between the two intersections is the absolute difference of their coordinates. When the endpoints are identified as left and right, that distance is \(R(y)-L(y)\).
Adding these widths over the full range of heights gives the area, because each thin horizontal strip has area approximately its width times its thickness \(dy\). Taking the limit of the strip sums gives the definite integral of the width. This is why the inverse branches and their valid intervals must be established before setting up the integral.
Worked Example: Inverting a Line for Horizontal Slices
Worked Example: Inverting a Line for Horizontal Slices
Find the area of the region bounded by \(y=2x+1\), \(x=0\), \(y=1\), and \(y=5\), using horizontal slices.
Step 1: Solve the curve equation for \(x\). Starting from \(y=2x+1\), subtract \(1\) and divide by \(2\):
This is valid for every real \(y\), since the line is one-to-one and its range is all real numbers.
Step 2: Identify the endpoints of a slice. The horizontal boundaries show that the region runs from \(y=1\) to \(y=5\). At each such height, the left endpoint is \(x=0\), and the right endpoint is \(x=(y-1)/2\). The width is therefore
Step 3: Integrate the width.
Check: Using vertical slices, \(x\) runs from \(0\) to \(2\), since \(y=5\) meets the line at \(x=(5-1)/2=2\). The vertical height is \((2x+1)-1=2x\), so the area is \(\int_0^2 2x\,dx=[x^2]_0^2=4\), agreeing with the horizontal-slice result.
Worked Example: A Cubic Boundary and Its Valid Range
Worked Example: A Cubic Boundary and Its Valid Range
Find the area enclosed by \(y=(x-1)^3+2\), \(x=3\), and \(y=2\), using horizontal slices.
Step 1: Rearrange the cubic equation. Subtract \(2\), take the real cube root, and add \(1\):
The cubic is strictly increasing, so each height corresponds to exactly one \(x\)-value. The boundary meets \(y=2\) at \(x=1\) and meets \(x=3\) at \(y=(3-1)^3+2=10\). The relevant height interval is \(2\leq y\leq10\).
Step 2: Determine left and right. At the bottom height \(y=2\), the curve is at \(x=1\), while the vertical boundary is at \(x=3\). Across the region, the curve remains to the left of \(x=3\). Thus the width is
Step 3: Integrate. With \(u=y-2\), the limits become \(0\) and \(8\), and \(dy=du\):
Check: Vertically, for \(1\leq x\leq3\), the region lies between \(y=2\) and \(y=(x-1)^3+2\). Its area is \(\int_1^3 (x-1)^3\,dx=[(x-1)^4/4]_1^3=16/4-0=4\). The two slice directions agree.
Worked Example: Choosing Branches on a Restricted Interval
Worked Example: Choosing Branches on a Restricted Interval
Rewrite \(y=(x-2)^2\) as \(x\) in terms of \(y\), when the boundary is restricted to \(0\leq x\leq3\). Identify where each branch is valid.
Step 1: Solve the equation.
Since \(y=(x-2)^2\), the possible heights satisfy \(y\geq0\). The restriction \(0\leq x\leq3\) also limits the range: the minimum height is \(0\), at \(x=2\), and the maximum is \(4\), at \(x=0\). Hence the relevant range is \(0\leq y\leq4\).
Step 2: Test each branch against the original interval. For the minus branch, \(x=2-\sqrt y\). If \(0\leq y\leq4\), then \(0\leq\sqrt y\leq2\), so \(0\leq2-\sqrt y\leq2\). This branch is valid throughout the range.
For the plus branch, \(x=2+\sqrt y\). To satisfy \(x\leq3\), we need \(2+\sqrt y\leq3\), or \(\sqrt y\leq1\), which is equivalent to \(y\leq1\). Thus the plus branch is valid only for \(0\leq y\leq1\).
Step 3: Check the branches. At \(y=1\), they give \(x=1\) and \(x=3\); both lie in the specified interval and satisfy \((x-2)^2=1\). At \(y=4\), the minus branch gives \(x=0\), which satisfies \((0-2)^2=4\). The plus branch gives \(x=4\), which solves the equation but violates \(x\leq3\), so it is not part of the specified boundary.
Therefore, for heights \(0<y<1\), both branches describe points on the restricted curve. For \(1<y\leq4\), only \(x=2-\sqrt y\) describes points on it. This is precisely why solving an equation is not enough: the solutions must also be checked against the boundary interval.
A Common Mistake: Reversing the Horizontal Order
In the cubic example, the curve is the left boundary and \(x=3\) is the right boundary. A common error is to write the width as
which is left minus right. For \(2<y<10\), the curve has \(x<3\), so this expression is negative. Integrating it would produce \(-4\), a signed result that cannot represent the geometric area.
The correction is to identify the order at a representative height, or from the region’s geometry, and subtract left from right:
Another related error is to use every algebraic solution without checking the interval of the boundary. In the branch example, \(x=2+\sqrt y\) is a valid solution of the equation for \(y=4\), but it is not on the specified curve segment because it gives \(x=4\notin[0,3]\).
A Reliable Procedure
Record the \(x\)-interval on which the original curve actually forms part of the region.
Determine the minimum and maximum \(y\)-values on that segment; these give the candidate limits for horizontal integration.
Rearrange the equation and keep every candidate branch until its validity has been checked.
Use the original \(x\)-interval and the region’s geometry to select valid branches and determine which endpoint is left and which is right.
Integrate right minus left over the correct height interval, then check that widths are nonnegative and, where practical, verify the result another way.
Why Careful Branch Selection Matters
Horizontal integration depends on the actual horizontal extent of the region at each height. An inverse formula that is algebraically correct but uses the wrong branch may place an endpoint outside the region. Using a range that is too large may add slices that do not belong; using a range that is too small may omit part of the region. If a boundary changes branch or a slice separates into multiple pieces, those changes must be reflected in the setup.
The practical habit is simple: after solving for \(x\), substitute the proposed branch back into the original equation and check that its \(x\)-values lie on the specified boundary segment. Then compare the endpoints at a representative height to confirm their left-to-right order. These checks turn an algebraic rearrangement into a valid horizontal-slice description.
Check Your Understanding
For each question, pay attention to both the algebraic equation and the interval on which the boundary is specified.
- What condition on a boundary \(y=f(x)\) allows it to be expressed as one inverse function \(x=f^{-1}(y)\) on an interval?
- Rearrange \(y=4x-3\) as \(x\) in terms of \(y\). What is the range of \(y\) if the boundary is restricted to \(1\leq x\leq4\)?
- For \(y=(x-1)^2\) restricted to \(0\leq x\leq2\), find both algebraic branches for \(x\) and state where each is valid.
- At a fixed height, the left and right endpoints are \(x=L(y)\) and \(x=R(y)\). Which difference gives the horizontal width, and why?
- Why can an algebraic solution of the boundary equation fail to describe a point on the boundary segment used for an area calculation?
