One Region, Two Ways to Slice
A bounded region can often be described by thin vertical slices or by thin horizontal slices. Each description leads to a definite integral for the same geometric area, but the bounds and the expression for slice length generally change when the slicing direction changes.
For a vertical slice, the length is measured from a lower curve to an upper curve, so it is a difference of \(y\)-values. For a horizontal slice, the length is measured from a left boundary to a right boundary, so it is a difference of \(x\)-values. Keeping track of which direction a slice runs is the key to setting up the correct integral.
The two formulas use the same principle: add up slice lengths over the range of slices. The variable of integration identifies the slicing direction. Integrating with respect to \(x\) uses vertical slices; integrating with respect to \(y\) uses horizontal slices.
Worked Example: A Parabolic Region
Worked Example: Comparing Slices Between Two Curves
Find the area enclosed by \(y=x^2\) and \(y=2x\), and express it using both vertical and horizontal slices.
Step 1: Find the intersections. At an intersection, the two \(y\)-values are equal:
The corresponding points are \((0,0)\) and \((2,4)\). For \(0\leq x\leq 2\), the line is above the parabola: \(2x-x^2=x(2-x)\geq 0\). Thus a vertical slice has length \(2x-x^2\).
Step 2: Integrate using vertical slices.
Step 3: Describe the same region horizontally. The \(y\)-values run from \(0\) to \(4\). Solving each boundary equation for \(x\), with \(x\geq 0\), gives \(y=x^2\Rightarrow x=\sqrt y\) and \(y=2x\Rightarrow x=y/2\). For \(0\leq y\leq4\), \(y/2\leq\sqrt y\), so the horizontal slice runs from \(x=y/2\) to \(x=\sqrt y\).
Step 4: Integrate using horizontal slices.
Both descriptions give area \(4/3\). As a check on the horizontal setup, at \(y=1\) the slice goes from \(x=1/2\) to \(x=1\), with length \(1-1/2=1/2\), agreeing with \(\sqrt{1}-1/2\).
Why the Two Integrals Agree
The agreement is not a coincidence specific to the example. Both integrals add the lengths of slices that fill the same region. The direction of the slices changes which coordinate is allowed to vary, not the region being measured.
Here the boundary functions may be piecewise-defined, with the intervals divided where their formulas change. The essential conditions are that the slice lengths describe the same region and that each integral adds nonnegative lengths over the full projection of that region.
Justification. Enclose the region in a rectangle and divide that rectangle into a fine grid of small rectangles. Grid rectangles lying wholly inside the region contribute their full area; those lying wholly outside contribute zero. Only rectangles meeting the boundary can create a difference between this grid estimate and the true area. Since the boundary is piecewise smooth, the total area of these boundary rectangles tends to zero as the grid is refined.
Now group the same grid rectangles by their vertical columns. In each column, the included lengths approximate the vertical slice length, and adding column areas gives the limit \(\int_a^b [f(x)-g(x)]\,dx\). Group them instead by horizontal rows. The included lengths approximate the horizontal slice length, and adding row areas gives \(\int_c^d [v(y)-u(y)]\,dy\). Both groupings estimate the area of the same region, and their boundary errors tend to zero. Their limits must therefore be equal, which proves the formula.
This reasoning also explains why the bounds matter. The vertical bounds cover the region’s full horizontal projection, while the horizontal bounds cover its full vertical projection. Integrating a correct slice length over only part of a projection, or over a range that includes points outside the region, will not compute its area.
Worked Example: A Triangular Region
Worked Example: A Line and a Vertical Boundary
Find the area bounded by \(y=0\), \(y=2x\), and \(x=2\), using both slicing directions.
Vertical slices: The region runs from \(x=0\), where the line meets \(y=0\), to \(x=2\). The lower boundary is \(0\) and the upper boundary is \(2x\). Therefore,
Horizontal slices: The \(y\)-values run from \(0\) to \(4\), since \(y=2x=4\) at \(x=2\). Solving \(y=2x\) for \(x\) gives the left boundary \(x=y/2\); the right boundary is \(x=2\). Hence,
The two values agree. At the top edge, \(y=4\), the horizontal slice has length \(2-4/2=0\), as expected where the region narrows to a point.
Worked Example: The Region Under a Parabola
Worked Example: A Parabolic Cap
Find the area bounded by \(y=x^2\) and \(y=4\), using both vertical and horizontal slices.
Vertical slices: The curves intersect where \(x^2=4\), so \(x=-2\) or \(x=2\). The line \(y=4\) is above \(y=x^2\) throughout this interval. Thus,
Horizontal slices: The \(y\)-values run from \(0\) to \(4\). For a fixed \(y\), the equation \(y=x^2\) gives \(x=-\sqrt y\) on the left and \(x=\sqrt y\) on the right. The slice length is therefore \(\sqrt y-(-\sqrt y)=2\sqrt y\). Hence,
Both methods give \(32/3\). For a check on the horizontal width, at \(y=1\) the slice extends from \(x=-1\) to \(x=1\), so its length is \(2\), exactly the value \(2\sqrt{1}\).
Choosing and Checking a Slicing Direction
One direction may lead to a simpler integral than the other. A useful setup process is to trace the region in that direction, identify the endpoints of a representative slice, and then find the range over which the slice moves. For horizontal slices, this often requires solving boundary equations for \(x\); use the appropriate branch when a curve has more than one inverse branch.
Decide whether the slices will be vertical or horizontal.
Use the region’s projection onto the axis along which the slices are spaced.
Use upper \(y\)-value minus lower \(y\)-value vertically, or right \(x\)-value minus left \(x\)-value horizontally.
At a convenient coordinate, verify that the stated endpoints and length match the region.
A frequent mistake is to subtract horizontal endpoints in the wrong order. For the region between \(y=x^2\) and \(y=2x\), the horizontal endpoints are \(x=y/2\) on the left and \(x=\sqrt y\) on the right. Reversing them gives
This negative value is a warning: the expression is left endpoint minus right endpoint, not a nonnegative slice length. Correcting the order gives \(\sqrt y-y/2\) and area \(4/3\). A negative result from a purported slice-length integral should prompt a check of the boundary order and bounds.
Comparing vertical and horizontal integrals is a useful way to check a setup, but the two integrals need not look alike. They use different variables and can involve different boundary expressions. What must agree is their value, provided both integrals describe the same region completely.
Check Your Understanding
For each question, identify the slicing direction and check that the stated length is nonnegative over the given range.
- For the region between \(y=x^2\) and \(y=2x\), what are the horizontal bounds in \(x\) at a height \(y\) between \(0\) and \(4\)?
- For the triangular region bounded by \(y=0\), \(y=2x\), and \(x=2\), what is the horizontal slice length at height \(y\)?
- For the region between \(y=x^2\) and \(y=4\), why does a horizontal slice have length \(2\sqrt y\) rather than \(\sqrt y\)?
- If a proposed horizontal slice length is negative, what should you check first?
- In the parabolic region between \(y=x^2\) and \(y=2x\), what area does each slicing direction give?
