Graphing Inequalities and Planar Regions

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:

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.

1. 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.
2. Substitute $(x_0, y_0)$ into the original inequality.
3. If the resulting statement is TRUE, shade the entire region containing $(x_0, y_0)$.
4. 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 $\mathbb{R}^2$.

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When to Use It / Method Comparison

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Why We Use It in Calculus (Calculus III Context)

1. 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.
2. 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.
3. 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.

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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.

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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:

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 $x^2 + y^2 \le 16$.

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

Determine and sketch the domain of the function:

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

Solution

Step 1: State the algebraic domain restrictions

1. Radicand must be non-negative:
   $$y - x^2 + 1 \ge 0 \implies y \ge x^2 - 1$$
2. 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 ($x \to \pm \infty$).

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Connections & References

• Parent Concepts: Calculus Foundations
• Related Geometric Concept: Conic Sections in the Cartesian Plane
• Downstream Calculus Applications:
  • Domain and Range of Multivariable Functions
  • Double Integrals
  • Calc-3 Discussion 04