From Inequalities to Slice Bounds
A planar region is often given by several inequalities rather than by a picture or a set of integral limits. To describe its slices, fix one coordinate and determine which values of the other coordinate satisfy all the inequalities. The result is an interval for each slice, together with the range of fixed-coordinate values for which that interval exists.
For a vertical slice, \(x\) is fixed and we find the allowed \(y\)-values. For a horizontal slice, \(y\) is fixed and we find the allowed \(x\)-values. The key algebraic task is to isolate the coordinate that varies. The key logical task is to combine the restrictions rather than treating each inequality separately.
A Reliable Translation Method
Suppose a region is defined by several inequalities. To find vertical-slice bounds, hold \(x\) fixed and solve each inequality for \(y\). Inequalities that say \(y\) is at least some expression provide lower bounds; inequalities that say \(y\) is at most an expression provide upper bounds. The allowed interval begins at the greatest lower bound and ends at the least upper bound.
For horizontal slices, hold \(y\) fixed and solve for \(x\). Again, combine all lower bounds and all upper bounds. A proposed interval is valid only when its lower endpoint does not exceed its upper endpoint. That condition may restrict the outer coordinate range or require the description to be split into pieces.
Decide whether to fix \(x\) and vary \(y\), or fix \(y\) and vary \(x\).
Rearrange every inequality. When dividing by a negative number, reverse the inequality sign.
Use the strongest lower bound and strongest upper bound to form the allowed interval.
Require the lower bound to be no greater than the upper bound, then verify that the resulting limits satisfy the original constraints.
Why the Bounds Must Be Combined
Fix \(x\) at some value. If one inequality requires \(y\ge L_1(x)\) and another requires \(y\ge L_2(x)\), then \(y\) must satisfy both. It must therefore be at least the larger of the two lower limits. Similarly, if \(y\le U_1(x)\) and \(y\le U_2(x)\), it must be no greater than the smaller upper limit.
This gives a direct justification for the slice procedure: a point belongs to the region precisely when its varying coordinate lies between every required lower and upper limit. The same reasoning applies to horizontal slices after exchanging the roles of \(x\) and \(y\). If the resulting lower bound exceeds the upper bound, there are no allowed points on that slice.
Worked Example: Vertical Slices Between a Line and a Horizontal Boundary
Worked Example: Vertical Slices Between a Line and a Horizontal Boundary
Describe the region defined by \(-1\le x\le 2\), \(y\ge 2x-1\), and \(y\le 5\) using vertical slices.
Here \(x\) is already restricted to the interval \([-1,2]\), so fix any \(x\) in that interval. The inequality \(y\ge 2x-1\) gives the lower limit \(2x-1\). The inequality \(y\le 5\) gives the upper limit \(5\). Thus the candidate interval is \(2x-1\le y\le 5\).
Check that this interval is nonempty throughout the stated \(x\)-range. The lower expression \(2x-1\) increases as \(x\) increases, so its largest value on \([-1,2]\) is \(2(2)-1=3\). Since \(3\le 5\), the lower limit never exceeds the upper limit. At the two endpoints, the lower values are \(2(-1)-1=-3\) and \(2(2)-1=3\), both at most \(5\).
To check the description, take any point satisfying these slice bounds. Its \(x\)-coordinate is in \([-1,2]\), and its \(y\)-coordinate satisfies both \(y\ge 2x-1\) and \(y\le 5\), exactly the original conditions. Conversely, any point satisfying the original inequalities has \(x\in[-1,2]\) and \(y\) between those same two limits. The descriptions agree.
Worked Example: Horizontal Slices Between Two Slanted Boundaries
Worked Example: Horizontal Slices Between Two Slanted Boundaries
Describe the region \( -1\le y\le 2\), \(x\ge y-2\), and \(x\le 2y+4\) using horizontal slices.
Fix \(y\) in \([-1,2]\). The first slanted inequality already gives a lower bound on \(x\): \(x\ge y-2\). The second gives the upper bound \(x\le 2y+4\). Before accepting these as an interval, check their order:
Every \(y\) in the specified range satisfies \(y\ge-1\), and hence \(y\ge-6\). The interval is therefore nonempty for the entire range. At \(y=-1\), the limits are \(x=-1-2=-3\) and \(x=2(-1)+4=2\). At \(y=2\), they are \(x=2-2=0\) and \(x=2(2)+4=8\).
For any point in this description, the outer bounds give \(-1\le y\le2\), while the inner bounds give \(x\ge y-2\) and \(x\le2y+4\). These are exactly the original inequalities. The endpoint checks also confirm that the interval has the correct order at the ends of the stated \(y\)-range.
When a Bound Changes
Sometimes several inequalities supply lower bounds, or several supply upper bounds. The actual endpoint of a slice is then the most restrictive one: the greater lower limit or the lesser upper limit. If different candidate limits are most restrictive for different fixed-coordinate values, the slice description may need to be split into ranges.
For example, with \(0\le x\le2\), \(y\ge0\), and \(x+y\le4\), a horizontal slice has \(0\le x\le4-y\), but it must also obey \(x\le2\). Its upper endpoint is therefore the lesser of \(2\) and \(4-y\). For \(0\le y\le2\), \(4-y\ge2\), so the upper endpoint is \(2\). For \(2\le y\le4\), \(4-y\le2\), so the upper endpoint is \(4-y\). The description is piecewise because the more restrictive upper bound changes.
The change occurs where the candidate limits agree: \(2=4-y\), which gives \(y=2\). This comparison is not an extra condition on the points; it identifies which of the original restrictions supplies the active slice endpoint on each range.
Worked Example: Slices of a Quarter-Disk
Worked Example: Slices of a Quarter-Disk
Write the region defined by \(x^2+y^2\le4\), \(x\ge0\), and \(y\ge0\) using both vertical and horizontal slices.
For vertical slices, fix \(x\). The constraints \(x\ge0\) and \(x^2+y^2\le4\) imply \(0\le x\le2\): since \(y^2\ge0\), we have \(x^2\le4\), and nonnegative \(x\) then gives \(x\le2\). For each such \(x\), the constraint \(y\ge0\) supplies the lower limit. From the disk inequality, \(y^2\le4-x^2\); with \(y\ge0\), this becomes \(y\le\sqrt{4-x^2}\).
For horizontal slices, fix \(y\). The nonnegativity condition gives \(y\ge0\), and the disk inequality implies \(y^2\le4\), so \(0\le y\le2\). Solving \(x^2+y^2\le4\) for nonnegative \(x\) gives \(0\le x\le\sqrt{4-y^2}\).
To check either description, the square-root upper limit is nonnegative on its stated outer range. If \(0\le y\le\sqrt{4-x^2}\), then \(y\ge0\) and \(y^2\le4-x^2\), so \(x^2+y^2\le4\). The horizontal version gives the same inequality by interchanging \(x\) and \(y\). Conversely, every point in the original region has nonnegative coordinates and satisfies the corresponding square-root bound. Both descriptions therefore specify exactly the original quarter-disk.
A Common Sign Error
When isolating a coordinate, division by a negative number reverses the inequality. For instance, start with \(y\le -2x+4\). Subtracting \(4\) gives \(y-4\le-2x\). Dividing by \(-2\) reverses the sign, so the correct result is \(x\le(4-y)/2\).
A common incorrect calculation is \(y-4\le-2x\), followed by \(x\ge(4-y)/2\), with no reversal. The point \((0,0)\) shows the error: it satisfies the original inequality because \(0\le4\), but the incorrect bound would require \(0\ge2\), which is false. Testing a point against the original conditions is a useful way to detect a sign mistake.
Why Slice Bounds Matter
Slice bounds turn a geometric description into a precise algebraic one. For a vertical slice, they tell you exactly which \(y\)-values occur at a fixed \(x\); for a horizontal slice, they tell you which \(x\)-values occur at a fixed \(y\). This is the step that makes an inequality-defined region usable in an iterated integral. Correctness depends on including every constraint, choosing the restrictive bounds, and stating the full range over which each slice exists.
Check Your Understanding
For each question, identify the varying coordinate and keep every original inequality in view.
- For \(0\le x\le1\), \(y\ge x-2\), and \(y\le3\), what are the vertical-slice bounds?
- For \(0\le y\le2\), \(x\ge -y\), and \(x\le y+1\), check that the horizontal-slice interval is nonempty throughout the stated range.
- When two inequalities both give lower bounds for \(y\), which one determines the actual lower endpoint?
- Starting from \(y\ge -3x+6\), isolate \(x\) and state whether the inequality sign changes.
- For the quarter-disk \(x^2+y^2\le9\), \(x\ge0\), \(y\ge0\), write the vertical-slice bounds.