From Inequalities to Regions
A system of Cartesian inequalities describes a set of points in three-dimensional space. Each inequality imposes a condition on the coordinates \(x\), \(y\), and \(z\); the region described by the system consists exactly of the points that satisfy every condition.
The geometric interpretation requires care. An inequality does not always select the points on one side of a single surface. For instance, \(x^2\leq 1\) is equivalent to \(-1\leq x\leq 1\), so it selects the slab between two planes. Other inequalities can select several disconnected pieces. The expression in the inequality, its sign, and any factorizations or equivalent forms determine the geometry.
To describe a solid, identify what each condition allows, then find the points satisfying all conditions together. Replacing an inequality by equality reveals a candidate surface; the other inequalities determine which portions of that surface are relevant to the solid.
Equality Surfaces and Active Boundaries
For an inequality \(g(x,y,z)\leq 0\), points satisfying \(g(x,y,z)=0\) lie on its equality surface. The equality surface is a candidate boundary, not automatically a boundary of the full region. Some of it may be excluded by another inequality, or the inequality may be redundant because the other conditions already imply it.
For example, the inequalities \(x^2\leq 4\) and \(x^2\leq 9\) together describe just \(-2\leq x\leq 2\). The equality \(x^2=9\) gives the planes \(x=-3\) and \(x=3\), but neither bounds the resulting region. In contrast, \(x^2=4\) gives its two bounding planes.
Why Strict Inequalities Give Interior Points
Here is the justification for that result. Let \(p\) be a point where \(g_i(p)<0\) for every inequality. Continuity means that for each \(i\), values of \(g_i\) remain close to \(g_i(p)\) when the point is moved a sufficiently small distance. Since \(g_i(p)\) is strictly negative, there is a neighborhood of \(p\) in which \(g_i\) remains negative. There are finitely many inequalities, so we can choose a neighborhood small enough to work for all of them at once. Every point in this neighborhood satisfies every inequality; thus the neighborhood lies in the region and \(p\) is an interior point.
Consequently, a point on the boundary cannot satisfy all inequalities strictly: at least one must be an equality there. But an equality point may still be an interior point if its condition is redundant. This is why candidate surfaces must be checked against the rest of the system.
Put each inequality in a form that makes its sign or range easier to interpret.
Set each defining expression equal to zero, or replace the inequality symbol with equality.
Test each candidate surface against the other inequalities and discard portions that cannot belong to the region.
Solve the surface equations together, then check that the resulting points or curves satisfy the full system.
Worked Example: A Circular Cylinder with Two Caps
Worked Example: A Circular Cylinder with Two Caps
Describe the region \(x^2+y^2\leq 4,\;0\leq z\leq 3\), and identify its bounding surfaces and their intersections.
The first inequality contains no \(z\). In the \(xy\)-plane, \(x^2+y^2\leq 4\) describes the disk of radius \(2\) centered at the origin. Allowing \(z\) to vary produces the solid circular cylinder above that disk.
The other condition can be written as two inequalities:
These restrict the cylinder to the space between the horizontal planes \(z=0\) and \(z=3\). The candidate bounding surfaces are therefore the lateral cylinder and the two planes. On the lateral surface, \(x^2+y^2=4\); the caps are the disks in \(z=0\) and \(z=3\).
To find the intersections of the lateral surface with each cap, substitute the cap equation into the cylinder equation. At \(z=0\), the intersection is \(x^2+y^2=4,\;z=0\). At \(z=3\), it is \(x^2+y^2=4,\;z=3\). These are the two circular rim curves.
Check an interior point, \((1,0,1)\): \(1^2+0^2=1<4\), and \(0<1<3\), so it satisfies all conditions strictly. Check a point outside, \((3,0,1)\): \(3^2+0^2=9>4\), so it fails the cylinder condition. Finally, a point on the top rim, \((2,0,3)\), gives \(2^2+0^2=4\) and \(z=3\), confirming that it lies on both the cylinder and the top cap.
Worked Example: A Paraboloid Capped by a Plane
Worked Example: A Paraboloid Capped by a Plane
Describe the region \(x^2+y^2\leq z\leq 4\), and find where its two bounding surfaces meet.
The double inequality means both \(z\geq x^2+y^2\) and \(z\leq 4\). The first condition places the points on or above the upward-opening paraboloid \(z=x^2+y^2\); the second places them on or below the horizontal plane \(z=4\). The region is the portion between these surfaces.
At an intersection, both inequalities are equalities. Thus \(z=x^2+y^2\) and \(z=4\), which give
The intersection is a circle of radius \(2\) in the plane \(z=4\). The paraboloid portion in the boundary runs from its vertex \((0,0,0)\) to this circle; the planar portion is the disk \(x^2+y^2\leq 4\) at height \(4\).
The bounds \(0\leq z\leq 4\) follow from the original inequalities: \(z\geq x^2+y^2\geq 0\) and \(z\leq 4\). The point \((1,1,2)\) lies on the paraboloid because \(1^2+1^2=2\), and it satisfies \(2\leq 4\). The point \((0,0,5)\) fails the upper condition \(z\leq 4\). At the intersection point \((2,0,4)\), both surface equations hold: \(2^2+0^2=4=z\). This verifies the stated circle equation at one of its points.
When One Inequality Produces Multiple Pieces
Expressions involving squares can produce regions between two surfaces or regions on both outer sides of them. For instance, \(x^2\leq 1\) is equivalent to \(-1\leq x\leq 1\), whereas \(x^2\geq 1\) is equivalent to \(x\leq -1\) or \(x\geq 1\). The latter selects two separate ranges of \(x\), not a single half-space.
Worked Example: Two Disconnected Cylindrical Pieces
Describe the set \(1\leq x^2\leq 4,\;y^2+z^2\leq 1\), and identify its boundary surfaces and their intersections.
The first condition combines \(x^2\geq 1\) and \(x^2\leq 4\). Taking square roots with both signs accounted for gives \(-2\leq x\leq -1\) or \(1\leq x\leq 2\). The second condition describes a disk of radius \(1\) in the \(yz\)-plane, extended along the \(x\)-direction. Together the conditions give two separate pieces of this cylinder, one with \(-2\leq x\leq -1\), the other with \(1\leq x\leq 2\).
The boundary surfaces are the circular cylinder \(y^2+z^2=1\) and the four planes \(x=-2,\;x=-1,\;x=1,\;x=2\). Each plane meets the cylinder in a circle. For example, at \(x=1\), the intersection is \(y^2+z^2=1,\;x=1\); the same equation in \(y,z\) holds at each of the other three plane values of \(x\).
Check \((1.5,0,0)\): \(1.5^2=2.25\), which is between \(1\) and \(4\), and \(0^2+0^2=0\leq 1\), so it belongs to the positive-\(x\) piece. The point \((-1.5,0,0)\) also belongs, since its squared \(x\)-coordinate is again \(2.25\). The point \((0,0,0)\) fails \(1\leq x^2\), because \(1\leq 0\) is false. Thus the two pieces are separated by a gap; the condition \(x^2\geq 1\) does not mean “on one side” of a single plane.
Intersections and Geometric Checks
When two candidate surfaces meet, their intersection is found by imposing both equations at once. The result may be a curve, a finite collection of points, or no points. The algebraic intersection must still be checked against every inequality: a curve solving two surface equations might lie entirely outside the region because it violates a third condition.
For a reliable description, check more than the surface equations. Test a point believed to be inside against every original inequality. Test a point believed to be outside and identify which condition it violates. For proposed edges or rims, substitute into the equations of both meeting surfaces and verify the remaining inequalities. These checks help distinguish the boundary of the actual solid from a collection of surfaces that merely appeared in the equations.
Why This Description Matters
A precise geometric description makes clear what a system of inequalities includes, which surfaces enclose it, and where those surfaces meet. This is essential before representing a solid in a later calculation: an incorrect interpretation of a sign, an omitted boundary, or an overlooked disconnected piece changes the region itself.
The key discipline is to move between algebra and geometry in both directions. Use the algebra to identify candidate surfaces and their intersections; use geometric reasoning to interpret the signs and distinguish the relevant portions. Then check representative points in the original inequalities rather than relying on a sketch alone.
Check Your Understanding
For each question, identify the relevant sign conditions and explain how they determine the geometric region.
- What region is described by \(x^2\leq 9\), and which two planes form its boundary?
- Rewrite \(x^2\geq 4\) as a union of intervals for \(x\). Why can the resulting set have two pieces?
- For \(x^2+y^2\leq z\leq 9\), find the equations of the two bounding surfaces and their intersection.
- Why does a point satisfying equality in one defining inequality not necessarily lie on the boundary of the full region?
- What must be checked after solving two candidate surface equations to find their intersection within a solid?