Regions, Lines, and Circles in the Plane
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.1: “Functions of Several Variables” (domain sketching) |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-1-functions-of-several-variables |
| Supplementary | Stewart Calculus (9th ed.), Section 14.1: “Functions of Several Variables,” pp. 972-973 (domains sketched in Figures 2-4) |
OpenStax Calculus Volume 3 is free and openly licensed. Both texts sketch every domain as a region in the plane, which is the skill this lesson rehearses.
Equations Draw Curves, Inequalities Shade Regions
You already read shapes off equations. You know that $y = 2x + 1$ is a line and that $x^2 + y^2 = 9$ is a circle. The new habit in this course is small: an equation draws a curve, but an inequality shades a region. The line $x + y = 1$ is one curve; the inequality $x + y \le 1$ is the half-plane on one side of that curve.
Picture the plane as a sheet you can shade. Every domain you will meet in Chapter 14 is a shaded patch of that sheet: a half-plane on one side of a line, the inside or the outside of a circle, or the overlap of a few such patches. The previous lesson found those patches from a formula’s square roots and logarithms. Once you have a patch, name it precisely: which curve is the boundary, which side is shaded, whether the boundary itself is included, and how to write all of that in symbols so another reader sees the same shaded sheet you do.
So the boundary curve and the shaded side are two separate decisions. Equations give you the boundary; inequalities tell you the side.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
If lines and circles are shaky, review Basic Geometric Formulas and Graphing Functions.
Quick Reference
Lines. \[ y = mx + b \quad (\text{slope } m, \; y\text{-intercept } b), \qquad Ax + By + C = 0 \quad (\text{general form}). \]
Circles. \[ (x - h)^2 + (y - k)^2 = r^2 \quad (\text{center } (h, k), \; \text{radius } r). \] When the center is the origin this is $x^2 + y^2 = r^2$.
From equation to region.
| Symbol | Boundary | Region described |
|---|---|---|
| $=$ | drawn solid | the curve itself only |
| $\le$ or $\ge$ | drawn solid (included) | the curve plus one side |
| $<$ or $>$ | drawn dashed (excluded) | one side only, boundary not included |
Interval notation. $(a, b)$ means $a < x < b$; $[a, b]$ means $a \le x \le b$; $(-\infty, \infty)$ means no restriction.
Set-builder notation. $\{(x, y) \mid \text{conditions on } x \text{ and } y\}$ reads “the set of all pairs $(x, y)$ such that the conditions hold.” Example: $\{(x, y) \mid x^2 + y^2 \le 9\}$ is the closed disk of radius $3$.
The procedure (flashcard size).
- Write the boundary as an equation and identify its shape (line or circle).
- Decide solid or dashed from the inequality symbol.
- Test one easy point (often the origin) to learn which side is shaded.
- State the region in set-builder form and name the shape.
Key Concepts
1. An inequality in $x$ and $y$ shades a half-plane
A linear inequality such as $x + y + 1 \ge 0$ describes everything on one side of a line, together with the line itself.
Example 1. Describe the region $x + y + 1 \ge 0$.
Goal. Find the boundary line, then decide which side is shaded.
Step 1. Boundary: set $x + y + 1 = 0$, that is, $y = -x - 1$. This is a line of slope $-1$ through $(0, -1)$.
Step 2. The symbol is $\ge$, so the boundary line is solid and included.
Step 3. Test the origin $(0, 0)$: $0 + 0 + 1 = 1 \ge 0$ is true, so the origin is in the region. The shaded side is the side containing the origin.
\[ \boxed{\{(x, y) \mid x + y + 1 \ge 0\}, \text{ the closed half-plane on and above the line } y = -x - 1.} \]
Recap. The line was the boundary; the test point told me which half to keep. (Source: Stewart 14.1, Example 1 and Figure 2, pp. 972-973.)
Inline self-check. Is the point $(-2, -2)$ inside the region $x + y + 1 \ge 0$?
Show answer
Substitute: $-2 + (-2) + 1 = -3$, which is not $\ge 0$. So $(-2, -2)$ is not in the region. It lies on the far side of the line from the origin.
2. A circle inequality shades a disk
Example 2. Describe the region $x^2 + y^2 \le 9$.
Goal. Identify the boundary circle, then decide inside or outside.
Step 1. Boundary: $x^2 + y^2 = 9$ is the circle of radius $3$ centered at the origin.
Step 2. The symbol is $\le$, so the circle is solid and included.
Step 3. Test the origin $(0, 0)$: $0 \le 9$ is true, so the origin is in the region. The shaded part is the inside.
\[ \boxed{\{(x, y) \mid x^2 + y^2 \le 9\}, \text{ the closed disk of radius } 3 \text{ centered at the origin.}} \]
Recap. “Inside the circle” is exactly the inequality $\le$; “outside” would be $\ge$. (A common error is to read $x^2 + y^2 \le 9$ as a circle rather than as the filled disk; the equation is the circle, the inequality is the region.) (Source: Stewart 14.1, Example 2 and Figure 4, p. 973.)
3. A strict inequality excludes its boundary
Example 3. Describe the region $y > x^2$.
Goal. Find the boundary curve and decide which side, watching the strictness.
Step 1. Boundary: $y = x^2$ is the standard upward parabola through the origin.
Step 2. The symbol is $>$ (strict), so the parabola is dashed and excluded.
Step 3. Test the point $(0, 1)$, which is above the vertex: $1 > 0^2 = 0$ is true, so the inside (above the parabola) is shaded.
\[ \boxed{\{(x, y) \mid y > x^2\}, \text{ the open region above the parabola, boundary excluded.}} \]
Recap. The only difference from Example 2 is the strictness: a strict inequality leaves its boundary out, which is why we say “open region.” This is the same distinction that separated the logarithm domain (strict) from the square-root domain (non-strict) in the previous lesson.
4. Several conditions describe the overlap
When a domain carries more than one condition, the region is the set of points satisfying all of them at once.
Example 4. Describe the region $\{(x, y) \mid x^2 + y^2 < 4, \; y \ge 0\}$.
Goal. Intersect a disk condition with a half-plane condition.
Step 1. $x^2 + y^2 < 4$ is the open disk of radius $2$ (boundary circle excluded).
Step 2. $y \ge 0$ is the closed upper half-plane (the $x$-axis included).
Step 3. The overlap is the upper half of the open disk: the points inside the radius-$2$ circle with nonnegative height. The flat bottom edge along the $x$-axis is included (from $y \ge 0$), but the curved circular edge is not (from the strict $<$).
\[ \boxed{\text{The upper half-disk of radius } 2, \text{ flat edge included, curved edge excluded.}} \]
Recap. “And” means intersection. Each condition shades a region; the domain is what they have in common. Describing which parts of the boundary are in or out is part of a complete answer.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Treating an inequality as a curve | calling $x^2 + y^2 \le 9$ “a circle” | The equation is the circle; the inequality is the filled disk |
| Shading the wrong side | assuming $\ge$ always means “above” | Test one point; the truth of the inequality there picks the side |
| Drawing a strict boundary as solid | a solid line for $x + y < 1$ | A strict inequality ($<$ or $>$) excludes its boundary; draw it dashed |
| Mixing up center and radius | reading $(x-2)^2 + (y+3)^2 = 16$ as radius $16$ | Radius is $\sqrt{16} = 4$, center $(2, -3)$ |
| Reading the center sign backward | center $(-2, 3)$ for $(x-2)^2 + (y+3)^2$ | Center is $(h, k)$ where the form is $(x-h)^2 + (y-k)^2$, so $(2, -3)$ |
Leveled Practice
Work each problem fully before revealing the answer.
Level 1 -- Direct Application
Problem 1. Describe in words the region $\{(x, y) \mid x^2 + y^2 < 1\}$.
Show answer
The boundary $x^2 + y^2 = 1$ is the unit circle (radius $1$, centered at the origin). The strict $<$ excludes the boundary and the origin satisfies $0 < 1$, so the region is the open disk of radius $1$ centered at the origin, boundary circle not included.
Problem 2. Write in set-builder form: all points $(x, y)$ where $x > y$ and $y \ge -1$.
Show answer
\[ \{(x, y) \mid x > y, \; y \ge -1\}. \]
This is the part of the plane below the line $y = x$ (where $x$ exceeds $y$) and on or above the horizontal line $y = -1$.
Problem 3. Identify the shape of $x^2 + y^2 \le 16$ and state its boundary.
Show answer
The closed disk of radius $4$ centered at the origin. Its boundary is the circle $x^2 + y^2 = 16$, which is included because the inequality is non-strict.
Level 2 -- Translate and Sketch
Problem 4. Describe the region $2x + y \le 4$.
Show answer
Boundary: $2x + y = 4$, that is, $y = 4 - 2x$, a line of slope $-2$ through $(0, 4)$. The symbol $\le$ makes the line solid and included.
Test the origin: $2(0) + 0 = 0 \le 4$ is true, so the origin is in the region.
The region is the closed half-plane on and below the line $y = 4 - 2x$.
Problem 5. Describe the region $\{(x, y) \mid 1 \le x^2 + y^2 \le 4\}$.
Show answer
Two conditions on the same expression. $x^2 + y^2 \ge 1$ is the outside of the circle of radius $1$ (boundary included). $x^2 + y^2 \le 4$ is the inside of the circle of radius $2$ (boundary included).
The overlap is the closed annulus (ring) between radius $1$ and radius $2$, both boundary circles included.
Level 3 -- Combined Regions and Interpretation
Problem 6. A function has domain $\{(x, y) \mid y \ge x^2, \; x^2 + y^2 \le 4\}$. Describe the region in words.
Show answer
$y \ge x^2$ is the closed region on and above the upward parabola through the origin. $x^2 + y^2 \le 4$ is the closed disk of radius $2$ centered at the origin.
The overlap is the part of the disk that sits on or above the parabola: a closed region bounded below by the parabola $y = x^2$ and above by the upper arc of the circle of radius $2$, with both boundary pieces included.
This kind of region, bounded by a parabola underneath and a circle on top, is exactly the sort of domain that becomes the region of integration in Chapter 15.
Mastery Checklist
You have mastered this skill when you can, without notes:
Mental Model
Think of an inequality in two variables as a line drawn on a sheet of paper, with one half of the sheet colored in.
The equation tells you where to draw the line or the curve. The inequality symbol tells you two more things: which half of the sheet to color, and whether to draw the boundary in solid ink (it counts) or as a dashed pencil line (it does not). A circle inequality colors the inside or the outside of a ring drawn on the sheet.
When a domain has several conditions, you color one sheet for each condition, then stack the sheets on a light table and keep only the part that is colored on every sheet. That common region, lit through all the layers, is the domain. Naming the region is just describing the shape of the lit-up patch and saying which edges are solid ink and which are dashed.
Connections
Built from
- Solving Inequalities: The test-a-point method and the solid-versus-dashed boundary rule come straight from one-variable inequalities.
- Basic Geometric Formulas: Recognizing the line and circle equations and reading center and radius is the geometric half of this skill.
- Graphing Functions: Sketching the boundary curve is the same graphing you did for single-variable functions.
Leads to
- Functions of Two Variables: Every domain you sketch for a two-variable function is one of these regions.
- Level Curves and Contour Maps: A level curve is the boundary curve $f(x, y) = k$; reading which side has higher values is the same side-of-a-curve reasoning used here.
Where this is used
Sketching a region precisely is what makes the rest of multivariable calculus possible to set up. In Chapter 15, a double integral is computed over exactly one of these regions, and the limits of integration are read off the boundary curves. A region described carelessly leads directly to wrong integration limits. The same skill of describing a feasible region by a set of inequalities is the heart of linear programming in operations research and of constraint regions in optimization, which you meet again in the Lagrange-multiplier section at the end of Chapter 14.
Audience Notes
For students who find math intimidating: You only need two decisions for each inequality: which side, and whether the edge counts. Test the origin (it is the easiest point) and look at the symbol. Those two checks are all the method requires.
For students who want depth: The distinction between a region that includes its boundary (closed) and one that excludes it (open) is the start of point-set topology, and it controls whether a continuous function on the region is guaranteed to attain a maximum. The Extreme Value Theorem for two variables, used in the optimization section, requires a closed and bounded region precisely for this reason.
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