Why Rewriting Boundaries Takes Care
A horizontal slice has a fixed \(y\)-coordinate and extends from a left boundary to a right boundary. To describe its width, the boundary equations must give \(x\) in terms of \(y\). But solving an equation for \(x\) can produce multiple answers, and not every answer belongs to the region being measured.
The essential task is therefore twofold: rewrite each boundary as one or more functions of \(y\), then use the region’s location to decide which branch supplies each edge of the slice. A correct algebraic solution is not automatically the correct boundary.
Solving for \(x\) and Restricting the Branch
Suppose a curve is given by an equation involving \(x\) and \(y\). To use it as a horizontal boundary, solve the equation for \(x\). For instance, \(y=x^2\) gives \(x=\pm\sqrt{y}\) when \(y\geq 0\). These two expressions describe different parts of the same curve: \(x=\sqrt{y}\) is its right branch, while \(x=-\sqrt{y}\) is its left branch.
More generally, solving an equation may not produce a single function on the whole curve. If the curve doubles back in the horizontal direction, it may need to be described by separate branches. The region determines which branch applies: inspect a horizontal slice and ask which point on the curve is its left endpoint and which is its right endpoint.
For a region whose horizontal slice at each height \(y\) runs from \(x_L(y)\) to \(x_R(y)\), with \(x_L(y)\leq x_R(y)\), its area is found by integrating the width \(x_R(y)-x_L(y)\) over the region’s vertical extent. This is the horizontal-slice area formula. The work in this tutorial is identifying the correct functions for those two boundaries.
A Reliable Branch-Selection Process
Keep every algebraic branch that can describe the curve; do not discard a sign before checking the region.
Use the region’s endpoints, intersections, and other horizontal boundaries to determine the heights covered.
At a representative height, identify which boundary lies to the left and which lies to the right. Confirm the ordering throughout the interval.
Use the selected \(x\)-expressions in that order, then integrate over the \(y\)-interval.
Worked Examples
Worked Example: A Parabola and a Vertical Line
Find the area enclosed by \(x=y^2\) and \(x=4\), using horizontal slices.
The parabolic boundary is already expressed as \(x=y^2\). The two curves meet when \(y^2=4\), so \(y=-2\) or \(y=2\). For each \(y\) between these values, \(y^2\leq 4\). Thus the parabola is the left boundary and the line is the right boundary.
An antiderivative is \(4y-\frac{y^3}{3}\), since its derivative is \(4-y^2\). Therefore,
The result is positive, as an area must be. As a check, describe the same region using vertical slices: for \(0\leq x\leq 4\), the parabola has \(y\)-values from \(-\sqrt{x}\) to \(\sqrt{x}\). Its vertical-slice area is
This agrees with the horizontal-slice result.
Worked Example: Choosing Both Branches of a Parabola
Find the area enclosed by \(y=x^2\) and the horizontal line \(y=2\), using horizontal slices.
Solving \(y=x^2\) for \(x\) gives \(x=\pm\sqrt{y}\). The region extends from \(y=0\), where the parabola meets itself at its vertex, to \(y=2\). At any height in this interval, the left edge is the negative solution and the right edge is the positive solution:
Consequently, the width is \(\sqrt{y}-(-\sqrt{y})=2\sqrt{y}\), not zero. Integrating,
The antiderivative checks because \(\frac{d}{dy}\left(\frac{4}{3}y^{3/2}\right)=2\sqrt{y}\) for \(y>0\). A second check uses vertical slices: \(x\) ranges from \(-\sqrt{2}\) to \(\sqrt{2}\), and the slice height is \(2-x^2\). Thus
The matching values confirm that both branches were included and ordered correctly.
Worked Example: Selecting the Branch That Bounds an Enclosed Region
Find the area enclosed by \(y=x^2\) and \(y=2x\), using horizontal slices.
The curves intersect where \(x^2=2x\), so \(x(x-2)=0\), giving \(x=0\) and \(x=2\). Their corresponding \(y\)-coordinates are \(0\) and \(4\). Solving the parabola for \(x\) gives \(x=\pm\sqrt{y}\); solving the line gives \(x=y/2\).
The enclosed region lies to the right of the \(y\)-axis. At a height such as \(y=1\), the line gives \(x=1/2\), while the two parabola branches give \(x=-1\) and \(x=1\). The enclosed part runs from the line at \(x=1/2\) to the positive parabola branch at \(x=1\). Thus, throughout \(0\leq y\leq 4\), use \(x_L(y)=y/2\) and \(x_R(y)=\sqrt{y}\).
The antiderivative is valid because its derivative is \(\sqrt{y}-y/2\). To check the area independently, vertical slices run from \(y=x^2\) to \(y=2x\) for \(0\leq x\leq2\):
The negative square-root branch is part of the full parabola, but it is not part of this enclosed region. Including it would describe a different slice.
Why the Branch Check Works
At a fixed height \(y\), points on a horizontal slice have the same \(y\)-coordinate and different \(x\)-coordinates. The left edge is the smallest \(x\)-coordinate in the region at that height; the right edge is the largest. Therefore, once the boundary equations have been rewritten as \(x\)-values, their geometric positions determine which expressions belong in the width.
This also explains why solving an equation alone is insufficient. The equation \(y=x^2\) has two solutions for \(x\) when \(y>0\), but those solutions are not interchangeable. If the region is on the right side of the \(y\)-axis, the positive branch may bound it; if it spans both sides, the negative and positive branches may be its left and right edges. The region, not the algebraic sign convention, decides.
Common Mistake: Keeping the Wrong Root
For the region between \(y=x^2\) and \(y=2x\), a tempting but incorrect setup is to use the negative parabola branch as the left edge and the line as the right edge:
This does not describe the enclosed region. At \(y=1\), the proposed endpoints are \(x=-1\) and \(x=1/2\), but the enclosed region lies between \(x=1/2\) and \(x=1\). The negative branch is outside it. The corrected expression is
A useful safeguard is to test a height strictly inside the interval, not just at an intersection. Substitute that \(y\)-value into each candidate boundary and compare the resulting \(x\)-coordinates with the visible region.
What to Verify Before Integrating
- Does each boundary expression actually solve its original curve equation?
- Is the \(y\)-interval the full vertical extent of the region?
- Are the chosen branches part of the enclosed region, rather than another part of the same curve?
- Does the proposed right boundary have an \(x\)-value at least as large as the left boundary throughout the interval?
- Does the resulting area have the expected sign and agree with the geometry?
Branch selection is not an extra algebraic trick. It is the step that connects the equations of the curves to the particular region being measured. Once that connection is correct, the horizontal-slice integrand follows as right minus left.
Check Your Understanding
For each question, identify the relevant branch or explain how you would decide which branch belongs in the area integral.
- Solving \(y=x^2\) gives two expressions for \(x\). What are they, and what do their signs tell you geometrically?
- For the region enclosed by \(y=x^2\) and \(y=2\), what are the left and right boundaries as functions of \(y\)?
- For \(y=x^2\) and \(y=2x\), why is the negative square-root branch not part of the enclosed region?
- What inequality must hold between \(x_L(y)\) and \(x_R(y)\) for the width \(x_R(y)-x_L(y)\) to be nonnegative?
- Why should you test a height inside the interval when choosing between algebraic branches?
