Domain Restrictions for Functions of Several Variables
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.1: “Functions of Several Variables” (domain and range objective) |
| 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 (Examples 1 and 2) |
Both texts open the multivariable course by asking the same first question about any new function: where is it defined? OpenStax Calculus Volume 3 is free and openly licensed.
Key idea
You already do this in one variable. When you meet $f(x) = \sqrt{x-2}$, you do not panic. You think “the inside of a square root cannot be negative,” write $x - 2 \ge 0$, and read off $x \ge 2$. That instinct is the entire skill. Nothing changes except that the input is now a pair $(x, y)$ instead of a single number $x$.
Picture the input space as a flat sheet of paper, the $xy$-plane. Each point on the sheet is one allowed or forbidden input. Finding the domain of $f(x, y)$ means shading the region of the sheet where the formula returns a real number and leaving the forbidden part blank. The same three rules that governed single-variable domains govern this shading: a square root needs a nonnegative inside, a logarithm needs a positive argument, and a denominator cannot be zero. The work in this course is reading those rules off a two-variable formula and turning each one into a sentence about the plane.
So the picture comes first. A domain is not a list of symbols; it is a shaded region. The symbols are how you describe the region precisely.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
If the inequality step is shaky, review Solving Inequalities. If the logarithm rule is shaky, review Logarithm Rules.
Quick Reference
Definition (domain of a two-variable function). If $f$ is given by a formula and no domain is stated, the domain of $f$ is the set of all pairs $(x, y)$ for which the formula produces a real number (in words: every input the formula can actually accept).
The three risky operations and their rules.
| Operation in the formula | Condition you must impose | Why |
|---|---|---|
| $\sqrt{u}$ (even root) | $u \ge 0$ | A real square root of a negative number does not exist |
| $\ln u$ or $\log u$ | $u > 0$ | The logarithm is defined only for positive inputs |
| $\dfrac{\text{something}}{u}$ | $u \neq 0$ | Division by zero is undefined |
The procedure (flashcard size).
- Read the formula and circle every square root, logarithm, and denominator.
- For each circled piece, write the matching condition ($\ge 0$, $> 0$, or $\neq 0$).
- Solve each condition for the region of the plane it describes.
- Combine all conditions with “and”: a point is in the domain only if it satisfies every one.
- State the domain in set-builder form $\{(x, y) \mid \dots\}$ and, when you can, name the shape.
Keyword clues. When you see a root sign, reach for “inside $\ge 0$.” When you see $\ln$ or $\log$, reach for “argument $> 0$.” When you see a fraction bar, reach for “bottom $\neq 0$.”
Key Concepts
1. The single-variable rules carry over unchanged
The rules that decide a domain do not know how many variables the input has. $\sqrt{x-2}$ needs $x - 2 \ge 0$; $\sqrt{x + y + 1}$ needs $x + y + 1 \ge 0$. The only new step is that “solve the condition” now produces a region in the plane instead of an interval on a line.
Inline self-check. A function contains the piece $\ln(x - y)$. What single condition must hold for that piece to be defined?
Show answer
The argument of a logarithm must be positive, so $x - y > 0$, that is, $x > y$. This is the set of points below the line $y = x$ (where the $x$-coordinate exceeds the $y$-coordinate).
2. A square root produces a region bounded by a curve
Example 1. Find the domain of $g(x, y) = \sqrt{9 - x^2 - y^2}$.
Goal. Decide which pairs $(x, y)$ make the inside of the root nonnegative, then name the shape.
Step 1. The only risky operation is the square root. Impose \[ 9 - x^2 - y^2 \ge 0. \]
Step 2. Rearrange to put the variables together: \[ x^2 + y^2 \le 9. \]
Step 3. Recognize the shape. The equation $x^2 + y^2 = 9$ is a circle of radius $3$ centered at the origin, so $x^2 + y^2 \le 9$ is the filled-in disk of radius $3$, boundary included.
\[ \boxed{D = \{(x, y) \mid x^2 + y^2 \le 9\}, \text{ the closed disk of radius } 3 \text{ centered at the origin.}} \]
Recap. One square root gave one inequality, and the inequality named a familiar shape. This is the source of the value $g$ takes on across that disk, which the graphs lesson will lift into three dimensions. (Source: Stewart 14.1, Example 2, p. 973.)
3. A logarithm produces an open region; a denominator removes a curve
Example 2. Find the domain of $f(x, y) = \dfrac{\sqrt{x + y + 1}}{x - 1}$.
Goal. Two risky operations appear at once: a square root and a denominator. Handle each, then combine.
Step 1. Square root: $x + y + 1 \ge 0$, which rearranges to $y \ge -x - 1$. This is the set of points on or above the line $y = -x - 1$.
Step 2. Denominator: $x - 1 \neq 0$, so $x \neq 1$. This removes the vertical line $x = 1$ from the plane.
Step 3. Combine with “and.” A point qualifies only if it is on or above the line $y = -x - 1$ AND not on the line $x = 1$.
\[ \boxed{D = \{(x, y) \mid x + y + 1 \ge 0, \; x \neq 1\}.} \]
Recap. The square root supplied a closed half-plane; the denominator punched a line out of it. (A common error here is to forget the denominator entirely and report only the half-plane. Circle the fraction bar in Step 1 so it is not lost.) (Source: Stewart 14.1, Example 1, p. 972.)
4. Three or more variables: the same rules, one more coordinate
Example 3. Find the domain of $f(x, y, z) = \ln(z - y) + xy\sin z$.
Goal. Decide which triples $(x, y, z)$ are allowed.
Step 1. The term $xy \sin z$ is defined for every input, so it imposes no condition. The only risky operation is the logarithm.
Step 2. Logarithm: $z - y > 0$, that is, $z > y$.
Step 3. Name the region. In three-dimensional space this is everything strictly above the plane $z = y$, a half-space.
\[ \boxed{D = \{(x, y, z) \mid z > y\}, \text{ the half-space above the plane } z = y.} \]
Recap. Adding a third variable did not add a new rule. It only meant the region lives in space rather than in the plane. (Source: Stewart 14.1, Example 14, p. 982.)
5. The step-by-step domain procedure in full
When a formula mixes several risky operations, follow the procedure without shortcuts:
- Identify the formula’s expression.
- Locate every risky operation: denominators (watch for zero), even roots (inside must be $\ge 0$), logarithms (argument must be $> 0$).
- Translate each into an inequality or a “not equal” condition.
- Rewrite each condition in a standard geometric form when possible (a disk, a half-plane, a line removed).
- Double-check for leftover constraints, such as a denominator that is easy to overlook.
- Visualize or describe the region: interior of a circle, a half-plane, or a combination.
This is the same checklist used throughout Stewart 14.1 to find domains; carrying it out in order is what prevents missed conditions.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting to flip the inequality sign | from $-y \ge 3$ writing $y \ge -3$ | Multiplying by $-1$ flips the sign: $y \le -3$ |
| Reporting only one of several conditions | giving $y \ge -x-1$ but dropping $x \neq 1$ | Every risky operation contributes a condition; combine all with “and” |
| Using $\ge 0$ for a logarithm | writing $x - y \ge 0$ for $\ln(x-y)$ | A logarithm needs strictly $> 0$: $x - y > 0$ |
| Confusing the domain (a region) with the range (output values) | calling $0 \le z \le 3$ the domain of $\sqrt{9-x^2-y^2}$ | The domain is the disk of inputs; $[0,3]$ is the range of outputs |
| Leaving the answer as raw symbols | stopping at $9 - x^2 - y^2 \ge 0$ | Finish: rearrange to $x^2 + y^2 \le 9$ and name the disk |
Leveled Practice
Work each problem fully before revealing the answer.
Level 1 -- Direct Application
Problem 1. Find the domain of $f(x, y) = \sqrt{x + 3y - 4}$.
Show answer
The inside of the root must be nonnegative: $x + 3y - 4 \ge 0$, that is, $x + 3y \ge 4$.
This is the set of points on or above the line $x + 3y = 4$ (equivalently $y \ge \tfrac{4 - x}{3}$).
\[ D = \{(x, y) \mid x + 3y \ge 4\}. \]
Problem 2. Find the domain of $g(x, y) = \ln(4 - x^2 - y^2)$.
Show answer
The argument of the logarithm must be positive: $4 - x^2 - y^2 > 0$, that is, $x^2 + y^2 < 4$.
This is the open disk of radius $2$ centered at the origin, boundary excluded.
\[ D = \{(x, y) \mid x^2 + y^2 < 4\}. \]
(The boundary circle is excluded because the logarithm condition is strict.)
Problem 3. Find the domain of $h(x, y) = \dfrac{1}{x - y}$.
Show answer
The denominator may not be zero: $x - y \neq 0$, that is, $x \neq y$.
The domain is the entire plane with the line $y = x$ removed.
\[ D = \{(x, y) \mid x \neq y\}. \]
Level 2 -- Combined Conditions
Problem 4. Find the domain of $f(x, y) = \dfrac{\sqrt{y - x^2}}{x}$.
Show answer
Two risky operations.
Square root: $y - x^2 \ge 0$, that is, $y \ge x^2$. This is the region on or above the parabola $y = x^2$.
Denominator: $x \neq 0$. This removes the $y$-axis.
Combine with “and”: \[ D = \{(x, y) \mid y \ge x^2, \; x \neq 0\}. \]
The region sits on or inside the upward parabola, with the vertical line $x = 0$ taken out.
Problem 5. Find the domain of $g(x, y) = \ln(x) + \sqrt{y}$.
Show answer
Logarithm: $x > 0$.
Square root: $y \ge 0$.
Both must hold: \[ D = \{(x, y) \mid x > 0, \; y \ge 0\}. \]
This is the part of the first quadrant with the positive $y$-axis removed (since $x > 0$ is strict) but the positive $x$-axis included (since $y \ge 0$ allows $y = 0$).
Level 3 -- Three Variables and Interpretation
Problem 6. Find and describe the domain of $f(x, y, z) = \dfrac{1}{\sqrt{z - x^2 - y^2}}$.
Show answer
Two conditions act on the same expression. The square root needs its inside nonnegative, and that same square root sits in a denominator, so it may not be zero. A nonnegative quantity that is also nonzero is strictly positive: \[ z - x^2 - y^2 > 0, \quad \text{that is,} \quad z > x^2 + y^2. \]
In space this is the set of points strictly above the paraboloid $z = x^2 + y^2$ (the bowl that opens upward).
\[ D = \{(x, y, z) \mid z > x^2 + y^2\}. \]
This problem shows why the “denominator that contains a root” case collapses two conditions into one strict inequality.
Mastery Checklist
You have mastered this skill when you can, without notes:
Mental Model
Think of finding a domain as drawing a fence around the inputs the formula can accept.
The formula is a machine. Some inputs make it break: a negative under a square root, a zero in a denominator, a non-positive argument in a logarithm. Each risky operation is a rule about where the fence goes. A square root draws a fence along a curve and keeps the side where the inside stays nonnegative. A logarithm draws the same fence but stands one step back, refusing the boundary itself. A denominator does not enclose a region; it snips a single curve out of the middle.
When several risky operations appear, you draw several fences and keep only the land inside all of them at once. That intersection, the part of the plane (or of space) that survives every fence, is the domain. Reading the domain back is then just describing the shape of the land that is left.
Connections
Built from
- Solving Inequalities: Every domain condition is an inequality, and the sign-flip rule decides which side of a boundary survives.
- Logarithm Rules: The fact that $\ln u$ exists only for $u > 0$ is the source of the strict-boundary logarithm condition.
- Factoring Techniques: Rearranging a condition such as $9 - x^2 - y^2 \ge 0$ into $x^2 + y^2 \le 9$ relies on comfortable algebraic manipulation.
Leads to
- Functions of Two Variables: The domain is half of the definition of a two-variable function; the other half is the rule that assigns each allowed input its output.
- Functions of Three or More Variables: The identical procedure produces a region in space, a half-space or the inside of a surface.
Where this shows up
Finding a domain is the first move in nearly every multivariable problem. Before you can take a partial derivative, set up a double integral over a region, or check continuity at a point, you must know where the function lives. In Chapter 15 the domain of a function becomes the very region you integrate over, so a misread domain becomes a wrong integral. The habit you build here, reading a formula and naming the region where it is defined, is used in physics whenever a model is valid only inside a boundary, and in any computing task where an input must be validated before it is used.
Audience Notes
For students who find math intimidating: You already know how to do this with one variable. Say the three rules out loud, circle the risky pieces, and write one inequality for each. The two-variable case adds nothing new except that the answer is a shaded region.
For students who want depth: The set where a continuous formula is defined is open or closed depending on whether each boundary is strict. The logarithm and the open denominator give open conditions; the square root gives a closed one. This distinction between open and closed domains becomes important when you study limits and continuity of multivariable functions in the next section.
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