Why Slice Length Depends on Direction
A slice through a planar base region meets the region along a line segment. The segment’s length is measured along the slice, so its endpoints depend on whether the slice is vertical or horizontal. Getting this length right is essential when describing the base of a cross section.
For a vertical slice, the endpoints differ in their \(y\)-coordinates, so the length is a top \(y\)-value minus a bottom \(y\)-value. For a horizontal slice, they differ in their \(x\)-coordinates, so the length is a right \(x\)-value minus a left \(x\)-value. The subtraction order is determined by position, not by which curve seems more important.
Vertical Slices: Top Minus Bottom
A vertical slice is parallel to the \(y\)-axis. Fix an \(x\)-value, then identify the top and bottom points where the vertical line at that \(x\) meets the region. If the upper boundary is \(y=f(x)\) and the lower boundary is \(y=g(x)\), with \(f(x)\ge g(x)\) throughout the interval considered, then the vertical slice length is
Both boundary values are \(y\)-coordinates. Subtracting their \(x\)-coordinates would measure nothing along the vertical segment, because both endpoints have the same \(x\)-coordinate.
Worked Example: A Vertical Slice Between Two Curves
Consider the region between \(y=x^2\) and \(y=2x\), for \(0\le x\le 2\). Find the vertical slice length at \(x=1\), and give the length function on the interval.
At a fixed \(x\), compare the two \(y\)-values. Their difference is
For \(0\le x\le2\), both \(x\) and \(2-x\) are nonnegative, so \(2x-x^2\ge0\). Thus \(y=2x\) is the top boundary and \(y=x^2\) is the bottom boundary on this interval. The slice length is
At \(x=1\), the top coordinate is \(2(1)=2\), and the bottom coordinate is \(1^2=1\). Therefore,
Check: the slice runs vertically from \((1,1)\) to \((1,2)\), and the distance between those points is \(\sqrt{(1-1)^2+(2-1)^2}=1\). At the interval endpoints, \(L(0)=0\) and \(L(2)=0\), consistent with the curves meeting there.
Horizontal Slices: Right Minus Left
A horizontal slice is parallel to the \(x\)-axis. Fix a \(y\)-value, then identify the left and right endpoints of the segment in the region. If the right boundary is \(x=r(y)\) and the left boundary is \(x=\ell(y)\), with \(r(y)\ge\ell(y)\), then the horizontal slice length is
These boundaries must be written as \(x\)-coordinates in terms of \(y\). If the region’s curves were originally given as \(y\) in terms of \(x\), finding the horizontal endpoints may require solving those equations for \(x\). When an equation has two branches, choose the branch that actually forms the left or right edge of the region.
Worked Example: A Horizontal Slice with a Curved Left Boundary
Suppose a region is bounded by \(y=0\), \(x=2\), and the curve \(x=y^2\). Find its horizontal slice length for \(0\le y\le\sqrt{2}\).
At a fixed height \(y\), the curved boundary gives the left endpoint \(x=y^2\), and the vertical line gives the right endpoint \(x=2\). The length is therefore
For the stated range, \(0\le y\le\sqrt{2}\), so \(y^2\le2\). Thus the right coordinate is at least the left coordinate and the length is nonnegative. At \(y=1\), the endpoints have \(x\)-coordinates \(1^2=1\) and \(2\), giving
Check: the horizontal segment goes from \((1,1)\) to \((2,1)\); its length is \(\sqrt{(2-1)^2+(1-1)^2}=1\). At \(y=0\), the length is \(2\). At \(y=\sqrt{2}\), the endpoints both have \(x=2\), so the length is \(2-(\sqrt{2})^2=0\), as expected where the boundaries meet.
Why the Subtraction Rule Works
The rule follows directly from the distance formula. For a vertical segment, both endpoints have the same \(x\)-coordinate. For a horizontal segment, both have the same \(y\)-coordinate. In either case, the distance formula reduces to the absolute difference between the other coordinates.
For a vertical segment, let the endpoints be \((x,y_{\text{bottom}})\) and \((x,y_{\text{top}})\), with \(y_{\text{top}}\ge y_{\text{bottom}}\).
The distance is \(\sqrt{(x-x)^2+(y_{\text{top}}-y_{\text{bottom}})^2}=|y_{\text{top}}-y_{\text{bottom}}|\).
Since \(y_{\text{top}}\ge y_{\text{bottom}}\), the absolute value is \(y_{\text{top}}-y_{\text{bottom}}\). The horizontal case gives \(x_{\text{right}}-x_{\text{left}}\) in the same way.
For completeness, the horizontal calculation uses endpoints \((x_{\text{left}},y)\) and \((x_{\text{right}},y)\), where \(x_{\text{right}}\ge x_{\text{left}}\). Their distance is
So “top minus bottom” and “right minus left” are not separate memorized tricks. They are the distance formula with the endpoint order made explicit.
Choosing the Correct Branch and Checking the Order
A boundary curve may not give just one \(x\)-coordinate for each \(y\). For example, the circle \(x^2+y^2=25\) has a left branch and a right branch. Solving for \(x\) gives \(x=\pm\sqrt{25-y^2}\). A horizontal slice of a region on the right side of the circle uses the positive branch as its right endpoint; the negative branch describes the opposite side of the circle.
Worked Example: A Horizontal Slice Between a Line and a Circle
Take the part of the disk \(x^2+y^2\le25\) lying to the right of the line \(x=3\). For \(-4\le y\le4\), find the horizontal slice length.
The circle’s right boundary is found by solving its equation for the positive \(x\)-value:
The left boundary of this part of the disk is the line \(x=3\). The circle branch is indeed to the right of the line on the stated range: since \(-4\le y\le4\), \(y^2\le16\), so \(25-y^2\ge9\), and therefore \(\sqrt{25-y^2}\ge3\). The slice length is
At \(y=0\), the endpoints have \(x\)-coordinates \(3\) and \(\sqrt{25}=5\), so \(L(0)=5-3=2\). At \(y=3\), the circle coordinate is \(\sqrt{25-9}=\sqrt{16}=4\), giving \(L(3)=4-3=1\). Check the point \((4,3)\): \(4^2+3^2=16+9=25\), so it lies on the circle, and its horizontal distance from \(x=3\) is \(4-3=1\). At \(y=4\), the circle coordinate is \(\sqrt{25-16}=3\), so the slice length is \(3-3=0\); the same happens at \(y=-4\).
Checking the order is part of finding the length, not an optional final detail. A proposed length should be nonnegative over the interval where it is used. If a comparison shows it becomes negative, either the endpoints were subtracted in the wrong order or the boundary ordering changes on that interval.
Common Mistake: Reversing the Difference
A frequent error is to use the right boundary minus the left boundary for a vertical slice, or the top boundary minus the bottom boundary for a horizontal slice. Those formulas mix up the direction of the segment. Another error is to subtract in the right direction but fail to check which curve is actually above or farther right.
Worked Example: Correcting a Reversed Vertical Difference
For the region between \(y=x^2\) and \(y=2x\) at \(x=1\), suppose someone writes
A length cannot be negative. At \(x=1\), the two points are \((1,1)\) and \((1,2)\). The point with larger \(y\)-coordinate is on \(y=2x\), so that is the top boundary. Correcting the subtraction gives
Check: the distance between \((1,1)\) and \((1,2)\) is \(1\). The negative result came from subtracting the top coordinate from the bottom coordinate, not from an unusual geometric length.
When a Slice Is Not One Segment
The simple formulas assume that the part of the slice inside the region is one continuous segment with identifiable endpoints. If a horizontal or vertical line meets the region in two separate pieces, there is no single interval from left to right or bottom to top that describes just the region: that interval would include points outside it. In that situation, identify each separate segment and its own endpoints before finding its length.
Likewise, if the boundary order changes, one difference may not stay nonnegative throughout the interval. The response is to use the correct ordering on the portion being considered and keep each slice within its stated range. This careful identification of endpoints prevents both negative lengths and lengths that include points outside the base.
Check Your Understanding
For each question, identify the endpoints before writing a length.
- For a vertical slice between \(y=3x\) and \(y=x^2\) at \(x=2\), which curve is on top, and what is the slice length?
- A horizontal slice has left endpoint \(x=y+1\) and right endpoint \(x=5\). Write its length and state the condition on \(y\) that makes this ordering valid.
- For the circle \(x^2+y^2=16\), write the \(x\)-coordinate of its right branch in terms of \(y\). What horizontal length results between that branch and \(x=2\)?
- Why does subtracting bottom from top give a vertical length, while subtracting left from right gives a horizontal length?
- If a proposed slice-length formula is negative at a point in its stated interval, what should you check first?
