Graphing Inequalities and Planar Regions

Core Definition

A two-variable inequality in the Cartesian plane defines a subset $D \subset \mathbb{R}^2$ consisting of all ordered pairs $(x, y)$ satisfying a relation such as $g(x, y) \le c$ , $g(x, y) < c$ , $g(x, y) \ge c$ , or $g(x, y) > c$ . Graphing the inequality involves plotting the boundary curve $g(x, y) = c$ , rendering it as either solid (boundary included) or dashed (boundary excluded), and shading the half-plane or subregion containing points that satisfy the inequality.

Concept Explanation

Graphing regions in $\mathbb{R}^2$ is a four-step systematic algorithm:

Step 1: Identify and Graph the Boundary Curve

Replace the inequality symbol ( $<, \le, >, \ge$ ) with an equals sign ( $=$ ) to obtain the governing boundary equation:

g(x, y) = c

Classify and sketch this 2D curve using standard Cartesian models:

Linear boundary: $ax + by = c$ (straight line)
Parabolic boundary: $y = ax^2 + bx + c$ or $x = ay^2 + by + c$
Conic boundary: $x^2 + y^2 = r^2$ (circle) or $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (ellipse)

Step 2: Choose Boundary Line Type (Solid vs. Dashed)

Solid Line ( $\le$ or $\ge$ ): The boundary points satisfy the relation and belong to the set. In topology, this contributes toward a closed region.
Dashed Line ( $<$ or $>$ ): The boundary points do not satisfy the relation and are strictly excluded. In topology, this produces an open region.
Punctured / Excluded Curves ( $\ne$ ): When an expression forbids equality (e.g., a denominator $Q(x, y) \ne 0$ ), draw the curve $Q(x, y) = 0$ as a dashed line cutting through the region.

Step 3: Test-Point Method for Shading

A continuous boundary curve divides the Cartesian plane into two or more non-overlapping test regions.

Select an arbitrary test point $(x_0, y_0)$ not lying on the boundary curve. The origin $(0, 0)$ is usually the fastest algebraic choice.
Substitute $(x_0, y_0)$ into the original inequality.
If the resulting statement is TRUE, shade the entire region containing $(x_0, y_0)$ .
If the resulting statement is FALSE, shade the opposite region across the boundary curve.

Step 4: Topological Classification of the Planar Region

Open Region: Contains none of its boundary points (all boundaries dashed; every point in $D$ has an open disk entirely within $D$ ).
Closed Region: Contains all of its boundary points (all boundaries solid).
Neither Open Nor Closed: Contains some, but not all, boundary curves (e.g., $x^2 + y^2 \le 4$ with $x > 0$ ).
Bounded Region: The entire region can be enclosed inside a circle or disk of finite radius $R$ centered at the origin ( $x^2 + y^2 \le R^2$ ).
Unbounded Region: The region extends infinitely far in at least one direction across $\mathbb{R}^2$ .

When to Use It / Method Comparison

Inequality Type	Boundary Formatting	Topological Nature	Sample Calculus III Context
Non-strict ( $\le, \ge$ )	Solid boundary line	Closed set (if all boundaries solid)	Domain of radical $\sqrt{9 - x^2 - y^2} \implies x^2 + y^2 \le 9$
Strict ( $<, >$ )	Dashed boundary line	Open set (if all boundaries dashed)	Domain of logarithm $\ln(y - x^2) \implies y > x^2$
Non-zero denominator ( $\ne$ )	Dashed exclusion curve	Excluded locus / puncture line	Domain of fraction $\frac{1}{x - y} \implies y \ne x$
Compound / Intersecting	Combined dashed & solid	Often neither open nor closed	Domain of $\frac{\sqrt{16 - x^2 - y^2}}{\ln(y)} \implies x^2 + y^2 \le 16$ , $y > 0$ , $y \ne 1$

Why We Use It in Calculus (Calculus III Context)

Visualizing Multivariable Function Domains: In Session 08, the natural domain of $z = f(x, y)$ is a subset $D \subset \mathbb{R}^2$ . Visualizing this domain as a shaded 2D planar region clarifies where the function is defined, continuous, and differentiable.
Extreme Value Theorem in Higher Dimensions: Finding global extrema of $z = f(x, y)$ on a set $D$ requires that $D$ be both closed and bounded (compact). If the region is open or unbounded, absolute maxima/minima are not guaranteed to exist.
Setting Up Limits of Integration for Multiple Integrals: In double integrals $\iint_D f(x, y)\,dA$ , identifying top/bottom curves ( $y = g_2(x)$ to $y = g_1(x)$ ) or left/right curves requires accurately graphing planar inequality regions.

Engineering Applications

Mechanical Machine Design & Allowable Stress: Failure criteria (such as Tresca or von Mises yield criteria) define safe operating envelopes in principal stress space $(\sigma_1, \sigma_2)$ as closed, bounded geometric regions bounded by ellipses or hexagons.
Control Engineering & Stability Margins: Nyquist and root-locus stability envelopes map system gain and phase margins to open half-planes or disks in the complex coordinate plane.
Operations Research & Resource Allocation: Linear programming establishes feasible operational regions via intersections of linear half-spaces ( $Ax \le b$ ), where optimal cost vertices lie on polygonal boundaries.

Example Problems

Example Problem 1: Circular Disk with Linear Exclusion

Graph and classify the planar region $D \subset \mathbb{R}^2$ satisfying the compound inequality system:

x^2 + y^2 \le 16 \quad \text{and} \quad y > x

Solution

Step 1: Analyze the first inequality $x^2 + y^2 \le 16$

Boundary: $x^2 + y^2 = 16$ , which is a circle centered at $(0, 0)$ with radius $r = 4$ .
Line Type: Solid (non-strict inequality $\le$ ).
Test Point: Test $(0, 0)$ : $0^2 + 0^2 = 0 \le 16$ (TRUE). The shaded region is the interior and boundary of the disk of radius $4$ .

Step 2: Analyze the second inequality $y > x$

Boundary: $y = x$ , which is a straight line passing through the origin at a $45^\circ$ angle.
Line Type: Dashed (strict inequality $>$ ).
Test Point: The line passes through $(0, 0)$ , so choose a point off the line: $(0, 2)$ . Substitute: $2 > 0$ (TRUE). The shaded region is the upper half-plane strictly above the line $y = x$ .

Step 3: Intersect the two regions

The set $D$ is the upper-left circular segment of the disk of radius $4$ bounded on the bottom by the line segment $y = x$ between $(-2\sqrt{2}, -2\sqrt{2})$ and $(2\sqrt{2}, 2\sqrt{2})$ .
Boundary curves: The circular arc $x^2 + y^2 = 16$ with $y > x$ is solid. The diagonal chord $y = x$ with $x^2 + y^2 < 16$ is dashed.

Step 4: Topological Classification

Open / Closed: Neither open nor closed, because it includes its circular boundary arc but excludes its diagonal chord boundary.
Bounded / Unbounded: Bounded, because the entire set lies completely within the disk $x^2 + y^2 \le 16$ .

Example Problem 2: Domain of a Multivariable Function with Parabolic Boundary

Determine and sketch the domain of the function:

f(x, y) = \frac{\sqrt{y - x^2 + 1}}{x}

Classify the domain as open, closed, or neither, and bounded or unbounded.

Solution

Step 1: State the algebraic domain restrictions

Radicand must be non-negative: $y - x^2 + 1 \ge 0 \implies y \ge x^2 - 1$
Denominator cannot equal zero: $x \ne 0$

Step 2: Graph boundary curves and line styles

Parabolic boundary: $y = x^2 - 1$ . This is an upward-opening parabola with vertex at $(0, -1)$ and $x$ -intercepts at $(\pm 1, 0)$ . Since the inequality is non-strict ( $\ge$ ), this boundary is solid (except where $x = 0$ ).
Vertical axis boundary: $x = 0$ (the $y$ -axis). Since $x \ne 0$ , the $y$ -axis is an excluded boundary, plotted as a dashed vertical line.

Step 3: Determine the shaded interior region

Test point $(0, 1)$ lies inside the upward bowl of the parabola: $1 \ge 0^2 - 1 = -1 \quad (\text{TRUE})$
Shade the interior bowl of the parabola $y \ge x^2 - 1$ , but puncture/exclude the entire $y$ -axis ( $x = 0$ ) running straight up through the vertex $(0, -1)$ .

Step 4: Classify the region

Open / Closed: Neither open nor closed, because it includes the parabolic boundary points (where $x \ne 0$ ) but excludes the line $x = 0$ (including the vertex $(0, -1)$ ).
Bounded / Unbounded: Unbounded, because the parabolic bowl extends infinitely upward ( $y \to \infty$ ) and outward ( $x \to \pm \infty$ ).

Connections & References

Parent Concepts: Calculus Foundations
Related Geometric Concept: Conic Sections in the Cartesian Plane
Downstream Calculus Applications:

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