Why Horizontal Slices Need Their Own Bounds
A horizontal slice is taken at a fixed height \(y\). Its length is measured from a left endpoint to a right endpoint, so describing it requires two things: the range of \(y\)-values that meet the region, and the \(x\)-coordinates of the slice’s left and right endpoints at each height.
This is different from simply reading off the \(x\)-interval of a region. For horizontal slices, the limits of the slicing variable are \(y\)-values, and the boundary curves must be written as \(x\) in terms of \(y\), when possible. These bounds and functions are the geometric setup for cross sections built on horizontal base segments.
The Vertical Range and the Boundary Functions
Suppose a bounded region \(D\) lies in the \(xy\)-plane. Its vertical range is the set of heights \(y\) for which a horizontal line at height \(y\) intersects \(D\). If every height between the minimum and maximum occurs, we can write this range as \(a\le y\le b\).
At a chosen height in that range, the horizontal slice consists of the \(x\)-values for which \((x,y)\) belongs to the region. When the slice is a single segment, its endpoints can be written \(L(y)\) and \(R(y)\), with \(L(y)\le R(y)\).
To find these functions from equations given as \(y=f(x)\), fix \(y\) and solve \(y=f(x)\) for \(x\). This may produce more than one branch. The correct branch depends on which part of the boundary forms the left or right endpoint. A graph, the region’s restrictions, and a test value of \(y\) help identify the correct choices.
A Reliable Setup Method
Record its boundary curves and any stated restrictions. Locate where the boundaries meet or where the region reaches a highest or lowest point.
Determine the minimum and maximum \(y\)-values in the region. These are the outer limits for horizontal slices.
Rewrite the boundary equations to express their relevant \(x\)-coordinates in terms of \(y\).
At an interior height, compare the candidate \(x\)-values. The smaller one is left; the larger one is right.
Test the endpoints and an interior value. Confirm that the boundary functions lie on the stated curves and that \(L(y)\le R(y)\).
Why the Slice Length Is Right Minus Left
At a fixed height \(y\), the horizontal segment contains all \(x\)-values from \(L(y)\) to \(R(y)\). On a number line, the distance between a smaller coordinate and a larger coordinate is the larger minus the smaller. Therefore the segment’s length is \(R(y)-L(y)\), not the reverse.
The same reasoning explains why the vertical bounds are the lowest and highest heights reached by the region: horizontal slices are taken at fixed \(y\), so the range of those heights determines which slices exist. In the examples below, endpoint checks and interior checks verify both parts of the setup.
Worked Examples
Worked Example: A Region Between a Parabola and a Horizontal Line
Find the vertical range, left and right boundary functions, and horizontal slice length for the region bounded by \(y=x^2\) and \(y=4\).
First find where the curves meet. Set their \(y\)-values equal:
The intersection points are \((-2,4)\) and \((2,4)\). The lowest point in the region is the parabola’s vertex \((0,0)\), and the highest height is \(4\). Thus the vertical range is \(0\le y\le4\).
At a fixed height \(y\), solve the parabola equation for \(x\):
The negative branch is the left boundary and the positive branch is the right boundary. Therefore \(L(y)=-\sqrt{y}\) and \(R(y)=\sqrt{y}\), giving slice length
Check the endpoints: at \(y=0\), both boundaries are \(x=0\); at \(y=4\), they are \(x=-2\) and \(x=2\), where the parabola meets the line \(y=4\). At the interior height \(y=1\), the boundaries are \(-1\) and \(1\), and \(-1\le1\). The setup is consistent.
Worked Example: A Sideways Parabola and a Vertical Line
Find the vertical range, left and right boundary functions, and slice length for the region bounded by \(x=y^2\) and \(x=4\).
The curves meet when \(y^2=4\), so \(y=-2\) or \(y=2\). These intersections give the lowest and highest heights of the bounded region: \(-2\le y\le2\).
At each allowed height, the sideways parabola gives the left endpoint \(x=y^2\), while the vertical line gives the right endpoint \(x=4\). Thus
The horizontal slice length is
Check the endpoints: when \(y=-2\) or \(y=2\), \(L(y)=4=R(y)\), so the slice collapses to an intersection point. At \(y=0\), \(L(0)=0\) and \(R(0)=4\), so \(0\le4\), and the slice length is \(4-0=4\). This confirms the boundary order throughout the interval.
Worked Example: Selecting the Correct Branches from Two Curves
Find the vertical range, left and right boundary functions, and slice length for the region enclosed by \(y=x^2\) and \(y=2x\).
Find the intersections by setting the equations equal:
The corresponding heights are \(y=0\) and \(y=4\). The region lies between these heights, so \(0\le y\le4\).
For a horizontal slice, rewrite both boundary equations as \(x\)-functions. The line gives \(y=2x\), so \(x=y/2\). The parabola gives \(y=x^2\), so \(x=\pm\sqrt{y}\). However, this enclosed region lies to the right of the \(y\)-axis: its \(x\)-values run from \(0\) to \(2\). Therefore the relevant parabola branch is \(x=\sqrt{y}\), not \(x=-\sqrt{y}\).
At \(y=1\), for example, the line gives \(x=1/2\), while the parabola’s positive branch gives \(x=1\). Hence the line is left and the parabola is right. The functions and slice length are
Check the order for every \(y\) in the range: both sides are nonnegative, and squaring the comparison \(y/2\le\sqrt{y}\) gives \(y^2/4\le y\). This is equivalent to \(y(y-4)\le0\), which holds for \(0\le y\le4\). At \(y=0\) and \(y=4\), the boundaries coincide, as required at the intersections.
Common Mistake: Using the Wrong Bounds or Branch
A frequent error is to use the region’s \(x\)-intersection values as the bounds for horizontal slices. In the previous example, the intersections have \(x\)-coordinates \(0\) and \(2\), but horizontal slices are indexed by \(y\). The corresponding heights are \(0\) and \(4\), so the correct range is \(0\le y\le4\).
Another error is to select an inverse branch without checking the region. From \(y=x^2\), both \(x=-\sqrt{y}\) and \(x=\sqrt{y}\) solve the equation. Using \(-\sqrt{y}\) as the parabola boundary in the region between \(y=x^2\) and \(y=2x\) would place the endpoint to the left of the \(y\)-axis, outside that region. The restriction \(0\le x\le2\) selects \(+\sqrt{y}\). Always use the actual region, not just the equation, to decide which branch applies.
Check Your Understanding
For each question, identify the vertical range and the horizontal slice boundaries. Check your answers at the endpoints or at an interior height.
- For the region bounded by \(y=x^2\) and \(y=9\), what are the vertical bounds and the left and right boundary functions?
- For the region bounded by \(x=y^2\) and \(x=9\), what are the vertical bounds and slice length?
- For the region enclosed by \(y=x^2\) and \(y=4x\), which branch of the parabola is used for the right boundary of a horizontal slice?
- Why are the intersection \(x\)-coordinates not automatically the bounds for horizontal slices?
- What inequality should be true between the left and right boundary functions throughout the vertical range?
