MATH-283 (Calculus III) Curriculum

Session 08 - Multivariable Functions

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Determine the algebraic domain and range of two-variable functions $z = f(x, y)$ , classify domain boundaries (open, closed, bounded, unbounded), sketch 2D input regions in $\mathbb{R}^2$ , and evaluate multi-input function values.
Analyze functions of three variables $w = f(x, y, z)$ , identify excluded surfaces (planes, cylinders, spheres), define 3D input domains as solid volumes or open half-spaces in $\mathbb{R}^3$ , and determine real output ranges.
Construct and interpret horizontal slices (level curves $f(x,y) = c$ in 2D and level surfaces $f(x,y,z) = c$ in 3D), analyze vertical cross-sectional traces ( $x = k, y = k, z = k$ ), and reconstruct 3D surface geometry (paraboloids, saddles, ellipsoids) from 2D contour maps and slice curves.

Micro-Lecture

1. Functions of Several Variables & Higher-Dimensional Input Spaces

Mathematical Foundation: In single-variable calculus, functions $y = f(x)$ map a 1D real input to a 1D output, producing a planar curve. In multivariable calculus, a function of two variables $z = f(x, y)$ assigns to each ordered pair $(x, y) \in \mathbb{R}^2$ in its domain a unique real scalar $z \in \mathbb{R}$ . The graph of $f(x, y)$ forms a two-dimensional surface embedded in 3D space $\mathbb{R}^3$ . For three variables, $w = f(x, y, z)$ maps spatial coordinates in $\mathbb{R}^3$ to a scalar output $w \in \mathbb{R}$ .
Engineering Context: Physical engineering systems inherently depend on multiple independent variables: thermodynamic pressure $P(V, T)$ depends on chamber volume and temperature; microchip power dissipation $P(f, V)$ depends on clock frequency and supply voltage; aerodynamic drag $F_D(\rho, v)$ depends on fluid density and velocity.

2. Multivariable Domains & Boundary Classifications

Mathematical Foundation: The domain of $f(x, y)$ is the subset of $\mathbb{R}^2$ where the function is mathematically defined. As detailed in Domain and Range of Multivariable Functions, restrictions arise from non-zero denominators ( $Q(x,y) \ne 0$ ), non-negative even radicals ( $u(x,y) \ge 0$ ), strictly positive logarithmic arguments ( $u(x,y) > 0$ ), and trigonometric asymptotes. Domains in $\mathbb{R}^2$ form planar regions categorized as open (dashed boundary, excluded points) or closed (solid boundary, included points), and bounded (contained in a finite disk) or unbounded. In $\mathbb{R}^3$ , domains form solid volumes, shells, or open half-spaces.
Engineering Context: Domain calculations define the safe operating envelope of engineering systems, identifying structural boundary limits for gas containment vessels and physical material constraints in load-bearing joints.

3. Visualizing Surfaces via Slices: Level Curves, Level Surfaces, and Traces

Mathematical Foundation: Because 3D surfaces and 4D scalar fields cannot be fully visualized in a single 2D projection, multivariable calculus uses dimensional slicing. Slicing a surface $z = f(x, y)$ horizontally with planes $z = c$ yields Level Curves and Contour Maps $f(x, y) = c$ . Closely spaced contour lines indicate steep gradients, while widely spaced lines indicate gradual slopes. Slicing vertically with coordinate planes yields Traces and Cross-Sections of Surfaces ( $z = f(x, k)$ or $z = f(k, y)$ ). For three variables, setting $f(x, y, z) = c$ creates 3D Level Surfaces (isosurfaces), such as isothermal shells or equipotential surfaces.
Engineering Context: Civil engineers use topographical contour maps to route highway grades; aerospace propulsion engineers inspect 3D thermal isosurfaces inside rocket nozzles to identify hot-spots requiring active regenerative cooling.

Skill Block 1

Worked Example A: Radical Function Domain and Range

Given the two-variable function:

f(x, y) = \sqrt{9 - x^2 - y^2}

Determine the mathematical domain in set-builder notation. Describe its geometric boundary and classify the region as open or closed, and bounded or unbounded.
Determine the mathematical range in interval notation.
Evaluate the function at the origin $(0, 0)$ and at the coordinate point $(1, 2)$ .

Worked Solution:

Analyze Domain Condition: State the Radicand Non-Negativity Condition:
$9 - x^2 - y^2 \ge 0$
Rearrange the terms:
$x^2 + y^2 \le 9$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 \le 9\}$
- Geometric Description: The domain consists of all points lying on or inside a circle centered at the origin with radius $r = \sqrt{9} = 3$ .
- Classification: Because the region contains its entire boundary circle $x^2 + y^2 = 9$ (non-strict inequality $\le$ ), it is closed. Because it is contained within radius $3$ , it is bounded. (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Extremal Value Formulation for Quadratic Radicands: The sum of squares is bounded on the domain: $0 \le x^2 + y^2 \le 9$ .
- The maximum radicand occurs at the origin $(0, 0)$ where $x^2 + y^2 = 0$ : $z_{\max} = f(0, 0) = \sqrt{9 - 0} = 3$
- The minimum radicand occurs along the boundary where $x^2 + y^2 = 9$ : $z_{\min} = \sqrt{9 - 9} = 0$
- Range in Interval Notation: $R = [0, 3]$
Evaluate at Points:
- At $(0, 0)$ : $f(0, 0) = \sqrt{9 - 0^2 - 0^2} = \sqrt{9} = 3$
- At $(1, 2)$ : Check domain: $1^2 + 2^2 = 1 + 4 = 5 \le 9$ (valid interior point). $f(1, 2) = \sqrt{9 - 1^2 - 2^2} = \sqrt{9 - 5} = \sqrt{4} = 2$

Worked Example B: Rational Function with Excluded Boundary Lines

Given the rational function:

g(x, y) = \frac{1}{x^2 - y^2}

Find the mathematical domain and describe the geometric curves excluded from the $xy$ -plane.
Determine the mathematical range in interval notation.
Evaluate $g(3, 1)$ and explain why $g(2, 2)$ is undefined.

Worked Solution:

Analyze Domain Condition: State the Non-Zero Denominator Condition for Rational Functions:
$x^2 - y^2 \ne 0$
Factor the difference of squares:
$(x - y)(x + y) \ne 0 \implies y \ne x \quad \text{and} \quad y \ne -x$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid y \ne \pm x\}$
- Geometric Description: The domain consists of all points in the $xy$ -plane except for the two intersecting diagonal lines $y = x$ and $y = -x$ .
- Classification: The region is open (boundary lines are excluded) and unbounded (spans infinitely across $\mathbb{R}^2$ ). (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Range Formulation for Rational Inversion Functions: Let $u = x^2 - y^2$ . The expression $u$ can take on any non-zero real value: $u \in (-\infty, 0) \cup (0, \infty)$ .
- When $u > 0$ , as $u \to 0^+$ , $g \to +\infty$ , and as $u \to \infty$ , $g \to 0^+$ . Output spans $(0, \infty)$ .
- When $u < 0$ , as $u \to 0^-$ , $g \to -\infty$ , and as $u \to -\infty$ , $g \to 0^-$ . Output spans $(-\infty, 0)$ .
- Because the numerator is constant ( $1 \ne 0$ ), $g(x, y)$ can never equal zero.
- Range in Interval Notation: $R = (-\infty, 0) \cup (0, \infty)$
Evaluate at Points:
- At $(3, 1)$ : $g(3, 1) = \frac{1}{3^2 - 1^2} = \frac{1}{9 - 1} = \frac{1}{8}$
- At $(2, 2)$ : Denominator evaluates to $2^2 - 2^2 = 0$ . Division by zero is undefined; $(2, 2)$ lies on the excluded boundary line $y = x$ .

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain the fundamental difference between the domain of a single-variable function $y = f(x)$ and the domain of a two-variable function $z = f(x, y)$ in terms of geometric dimensionality.
(b) In Worked Example A, explain why the range of $f(x, y) = \sqrt{9 - x^2 - y^2}$ is bounded by $[0, 3]$ rather than $(-\infty, 3]$ .
(c) In the linked application Terrain Surveying & Road Grade Safety (Civil Engineering), why is evaluating whether coordinates satisfy the domain condition critical before heavy equipment is dispatched to a mountain site?

Solution Key:

(a) For a single-variable function $f: \mathbb{R} \to \mathbb{R}$ , the input is a single real scalar $x$ , making the domain a one-dimensional subset of the number line. For a function of two variables $f: \mathbb{R}^2 \to \mathbb{R}$ , an input is an ordered pair $(x, y)$ , meaning the domain represents a two-dimensional geometric region in the $xy$ -plane.
(b) The radical expression $\sqrt{u}$ is defined exclusively as the principal (non-negative) root, establishing an absolute lower bound of $0$ . Within the domain disk $x^2 + y^2 \le 9$ , the maximum radicand is $9 - 0 = 9$ (at the origin), yielding $z = 3$ , while the minimum radicand is $9 - 9 = 0$ (on the boundary circle), yielding $z = 0$ . Thus, $R = [0, 3]$ .
(c) In the civil engineering scenario, the radical model $z = \sqrt{16 - x^2 - y^2}$ models an surveyed hillside. Points where $x^2 + y^2 > 16$ produce imaginary roots, indicating points outside the surveyed physical embankment. Verifying that a coordinate lies in the interior of the domain ensures the road grade exists and can bear heavy construction vehicle loads.

Show your steps and calculations here.

Practice Problem 2

Given the radical function with a linear boundary:

f(x, y) = \sqrt{4 - 2x - y}

Find the mathematical domain in set-builder notation and describe the boundary line in the $xy$ -plane.
Classify the domain as open or closed, and bounded or unbounded.
Evaluate $f(1, 1)$ and determine if the point $(3, 1)$ lies in the domain.

Solution Key:

Analyze Domain Condition: State the Radicand Non-Negativity Condition:
$4 - 2x - y \ge 0 \implies y \le 4 - 2x$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid y \le 4 - 2x\}$
- Geometric Description: The domain consists of all points lying on and below the straight line $y = -2x + 4$ in the $xy$ -plane (a closed half-plane).
Classify Domain: Because the inequality is non-strict ( $\le$ ), the boundary line $y = -2x + 4$ is included in the domain, making the region closed. Because the half-plane extends infinitely in the $x$ and negative $y$ directions, it is unbounded.
Evaluate at Points:
- At $(1, 1)$ : $4 - 2(1) - 1 = 4 - 2 - 1 = 1 \ge 0$ $f(1, 1) = \sqrt{1} = 1$
- At $(3, 1)$ : $4 - 2(3) - 1 = 4 - 6 - 1 = -3 < 0$ Because the radicand is negative, $(3, 1)$ is not in the domain ( $f(3, 1) \notin \mathbb{R}$ ).

Show your steps and calculations here.

Practice Problem 3

Given the rational radical function:

g(x, y) = \frac{1}{\sqrt{16 - x^2 - y^2}}

Find the mathematical domain in set-builder notation.
Explain why the boundary curve $x^2 + y^2 = 16$ is excluded, and classify the domain as open or closed, and bounded or unbounded.
Determine the mathematical range in interval notation.

Solution Key:

Analyze Domain Condition: State the Strict Positivity Condition for Denominator Radicals: The radicand must be non-negative ( $\ge 0$ ) for the square root, and the denominator cannot equal zero:
$16 - x^2 - y^2 > 0 \implies x^2 + y^2 < 16$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 + y^2 < 16\}$
Classify Domain: Setting $x^2 + y^2 = 16$ yields $\sqrt{16 - 16} = \sqrt{0} = 0$ in the denominator, which is undefined. Therefore, the boundary circle is drawn with a dashed line and excluded.
- The domain is an open region (contains none of its boundary points).
- Because the disk is completely contained within radius $4$ , it is bounded.
Analyze Range: State the Range Formulation for Inverted Radicals: Within the open disk $x^2 + y^2 < 16$ , the term $x^2 + y^2$ takes on values in $[0, 16)$ .
- At the origin $(0, 0)$ , $x^2 + y^2 = 0$ , giving the minimum value of $g$ : $g(0, 0) = \frac{1}{\sqrt{16 - 0}} = \frac{1}{4}$
- As points approach the boundary circle ( $x^2 + y^2 \to 16^-$ ), the denominator approaches $0$ from the positive side: $\lim_{x^2+y^2 \to 16^-} \frac{1}{\sqrt{16 - x^2 - y^2}} = +\infty$
- Range in Interval Notation: $R = [1/4, \infty)$

Show your steps and calculations here.

Practice Problem 4

Given the logarithmic multivariable function:

k(x, y) = \ln(x^2 - y^2)

State the algebraic domain condition and express the domain in set-builder notation.
Graphically describe the regions in the $xy$ -plane where the function is defined.
Evaluate $k(3, 1)$ in exact form and round to three decimal places.

Solution Key:

Analyze Domain Condition: State the Logarithmic Positivity Rule: The argument of the natural logarithm must be strictly greater than zero:
$x^2 - y^2 > 0 \implies x^2 > y^2 \implies |x| > |y|$
- Set-Builder Notation: $D = \{(x, y) \in \mathbb{R}^2 \mid x^2 > y^2\}$
Geometric Description: The boundary curves are the lines $x^2 - y^2 = 0 \implies y = \pm x$ . Testing sample test points in the four regions divided by these diagonal lines:
- On the $x$ -axis at $(2, 0)$ : $2^2 - 0^2 = 4 > 0$ (Valid $\implies$ Left and Right regions are inside domain).
- On the $y$ -axis at $(0, 2)$ : $0^2 - 2^2 = -4 < 0$ (Invalid $\implies$ Top and Bottom regions are excluded). The domain consists of the two open horizontal conical wedges containing the $x$ -axis where $|x| > |y|$ , with the boundary lines $y = \pm x$ dashed and excluded.
Evaluate at $(3, 1)$ : Substitute $x = 3$ and $y = 1$ :
$k(3, 1) = \ln(3^2 - 1^2) = \ln(9 - 1) = \ln(8)$ In simplified logarithmic form: $k(3, 1) = \ln(2^3) = 3\ln(2) \approx 2.079$

Show your steps and calculations here.

Practice Problem 5

Given the trigonometric multivariable function:

h(x, y) = \tan(x - y)

State the condition required for the tangent function to be defined.
Determine the mathematical domain in set-builder notation and describe the excluded geometry in the $xy$ -plane.
Evaluate $h\left(\frac{\pi}{4}, 0\right)$ and determine if $\left(\frac{\pi}{2}, 0\right)$ lies in the domain.

Solution Key:

Analyze Domain Condition: State the Trigonometric Tangent Pole Condition:
$\tan(x - y) = \frac{\sin(x - y)}{\cos(x - y)}$ The function is undefined wherever $\cos(x - y) = 0$ , which occurs at odd integer multiples of $\frac{\pi}{2}$ : $x - y \ne \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$
Formulate Domain: Solve for $y$ :
$y \ne x - \frac{\pi}{2} - n\pi, \quad n \in \mathbb{Z}$
- Set-Builder Notation: $D = \left\{(x, y) \in \mathbb{R}^2 \;\middle|\; y \ne x - \frac{\pi}{2} - n\pi, \, n \in \mathbb{Z}\right\}$
- Geometric Description: The domain is the entire $xy$ -plane excluding an infinite sequence of evenly spaced parallel lines with slope $m = 1$ , each separated by vertical distance $\pi$ .
Evaluate at Points:
- At $(\pi/4, 0)$ : $x - y = \frac{\pi}{4} - 0 = \frac{\pi}{4}$ $h\left(\frac{\pi}{4}, 0\right) = \tan\left(\frac{\pi}{4}\right) = 1$
- At $(\pi/2, 0)$ : $x - y = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ This matches the condition for $n = 0$ . $\cos(\pi/2) = 0$ , so $\tan(\pi/2)$ is undefined. The point $(\pi/2, 0)$ lies directly on an excluded boundary line and is not in the domain.

Show your steps and calculations here.

Regroup 1

Emphasize that multivariable domains are 2D geometric regions, not 1D intervals. Always sketch the boundary curve first, then use a test point (like $(0,0)$ ) to decide which side to shade.
Clarify the difference between non-strict inequalities ( $\le$ , closed solid boundary) and strict inequalities ( $<$ , open dashed boundary).

Skill Block 2

Worked Example A: 3D Radical Function and Solid Ball Domain

Given the three-variable function:

h(x, y, z) = \sqrt{9 - x^2 - y^2 - z^2}

Find the mathematical domain in set-builder notation and describe the 3D solid region.
Determine the mathematical range in interval notation.
Evaluate $h(1, 2, 0)$ and determine if $(2, 2, 2)$ lies in the domain.

Worked Solution:

Analyze Domain Condition: State the Three-Variable Radicand Non-Negativity Condition:
$9 - x^2 - y^2 - z^2 \ge 0$
Rearrange terms:
$x^2 + y^2 + z^2 \le 9$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \le 9\}$
- Geometric Description: The domain is a solid 3D sphere (closed ball) centered at the origin $(0, 0, 0)$ with a radius of $R = \sqrt{9} = 3$ .
- Topological Classification: Closed and bounded in $\mathbb{R}^3$ . (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Radial Extremum Analysis for 3D Radicals: Let $\rho^2 = x^2 + y^2 + z^2 \in [0, 9]$ .
- At the origin $(0, 0, 0)$ , $\rho = 0$ , giving the maximum value: $h_{\max} = \sqrt{9 - 0} = 3$
- On the boundary shell $\rho^2 = 9$ : $h_{\min} = \sqrt{9 - 9} = 0$
- Range in Interval Notation: $R = [0, 3]$
Evaluate at Points:
- At $(1, 2, 0)$ : Compute $\rho^2 = 1^2 + 2^2 + 0^2 = 1 + 4 + 0 = 5 \le 9$ (valid interior point). $h(1, 2, 0) = \sqrt{9 - 5} = \sqrt{4} = 2$
- At $(2, 2, 2)$ : Compute $\rho^2 = 2^2 + 2^2 + 2^2 = 4 + 4 + 4 = 12$ . Since $12 > 9$ , the point lies outside the solid ball; $h(2, 2, 2) = \sqrt{-3} \notin \mathbb{R}$ is not in the domain.

Worked Example B: 3D Rational Function with Excluded Spherical Shell

Given the three-variable rational function:

g(x, y, z) = \frac{1}{x^2 + y^2 + z^2 - 1}

Find the mathematical domain and describe the excluded 3D surface.
Determine the mathematical range in interval notation.
Evaluate $g(0, 0, 0)$ and $g(2, 0, 0)$ , and explain why points with $x^2 + y^2 + z^2 = 1$ are undefined.

Worked Solution:

Analyze Domain Condition: State the Three-Variable Non-Zero Denominator Condition:
$x^2 + y^2 + z^2 - 1 \ne 0 \implies x^2 + y^2 + z^2 \ne 1$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \ne 1\}$
- Geometric Description: The domain consists of the entirety of three-dimensional space $\mathbb{R}^3$ with the spherical shell of radius $R = 1$ centered at the origin removed. (Defined in Domain and Range of Multivariable Functions.)
Analyze Range: State the Range Formulation for 3D Inverted Fields: Let $\rho^2 = x^2 + y^2 + z^2 \ge 0$ . Denominator is $d = \rho^2 - 1$ .
- Inside the unit sphere ( $\rho^2 \in [0, 1)$ ): $d \in [-1, 0)$ . At the origin $\rho = 0$ , $g(0,0,0) = \frac{1}{-1} = -1$ . As $\rho \to 1^-$ , $d \to 0^- \implies g \to -\infty$ . This covers $(-\infty, -1]$ .
- Outside the unit sphere ( $\rho^2 \in (1, \infty)$ ): $d \in (0, \infty)$ . As $\rho \to 1^+$ , $d \to 0^+ \implies g \to +\infty$ . As $\rho \to \infty$ , $d \to \infty \implies g \to 0^+$ . This covers $(0, \infty)$ .
- Range in Interval Notation: $R = (-\infty, -1] \cup (0, \infty)$
Evaluate at Points:
- At $(0, 0, 0)$ : $g(0, 0, 0) = \frac{1}{0 - 1} = -1$
- At $(2, 0, 0)$ : $g(2, 0, 0) = \frac{1}{2^2 + 0^2 + 0^2 - 1} = \frac{1}{4 - 1} = \frac{1}{3}$
- Points satisfying $x^2 + y^2 + z^2 = 1$ lie on the unit spherical shell, making the denominator zero and causing an infinite division-by-zero discontinuity.

Practice Problem 6: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Why can we not graph a three-variable function $w = f(x, y, z)$ directly using the standard 3D coordinate system used for two-variable functions $z = f(x, y)$ ?
(b) In Worked Example B, explain why the interval $(-1, 0]$ is excluded from the range of $g(x, y, z) = \frac{1}{x^2 + y^2 + z^2 - 1}$ .
(c) In the linked application Structural Stress Analysis in Spherical Joints (Civil & Mechanical Engineering), why was the inspection point $(3, 4, 2)$ physically invalid for ultrasonic evaluation?

Solution Key:

(a) For a function $z = f(x, y)$ , there are two input variables and one output variable, totaling three dimensions $(x, y, z)$ which match physical 3D space. For $w = f(x, y, z)$ , there are three input variables and one output variable, which requires a four-dimensional coordinate space $(x, y, z, w)$ . As a result, functions of three variables are visualized using 3D Level Surfaces rather than direct 4D surface graphs.
(b) For $\rho^2 \in [0, 1)$ , the denominator $\rho^2 - 1$ varies from $-1$ to $0^-$ . Inverting this gives $\frac{1}{\rho^2 - 1} \in (-\infty, -1]$ . The value $-1$ is achieved at the origin, and values closer to zero cannot be formed because $\rho^2 \ge 0$ . For $\rho^2 > 1$ , the denominator is strictly positive, giving outputs in $(0, \infty)$ . Therefore, no inputs can ever produce outputs in the interval $(-1, 0]$ .
(c) The steel joint has a physical radius of $5\text{ cm}$ , defined by the domain ball $x^2 + y^2 + z^2 \le 25$ . Calculating the distance of the probe at $(3, 4, 2)$ yields $\sqrt{9 + 16 + 4} = \sqrt{29} \approx 5.385\text{ cm}$ . Substituting this into the stress formula produces $\sqrt{25 - 29} = \sqrt{-4} = 2i$ , which is non-real. Ultrasonic waves cannot measure steel stress in ambient air outside the joint boundary.

Show your steps and calculations here.

Practice Problem 7

Given the 3D polynomial function:

f(x, y, z) = x^2 + y^2 + z^2

State the mathematical domain and range in appropriate notation.
Evaluate $f(1, 2, 2)$ and $f(-2, 3, -1)$ .
Explain the geometric meaning of the output value in terms of distance from the origin.

Solution Key:

Analyze Domain and Range:
- The expression is a polynomial with no denominators, radicals, or logarithms. The domain is all of 3D space: $D = \{(x, y, z) \in \mathbb{R}^3\} = \mathbb{R}^3$
- Each squared term is non-negative ( $x^2 \ge 0, y^2 \ge 0, z^2 \ge 0$ ). The minimum output occurs at $(0, 0, 0)$ , where $f(0, 0, 0) = 0$ . As points move arbitrarily far from the origin, $f \to \infty$ . $R = [0, \infty)$
Evaluate at Points:
- At $(1, 2, 2)$ : $f(1, 2, 2) = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9$
- At $(-2, 3, -1)$ : $f(-2, 3, -1) = (-2)^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14$
Geometric Meaning: By the 3D Pythagorean theorem, the distance from the origin to $(x, y, z)$ is $\rho = \sqrt{x^2 + y^2 + z^2}$ . Therefore, $f(x, y, z) = \rho^2$ , representing the square of the radial distance from the origin.

Show your steps and calculations here.

Practice Problem 8

Given the 3D rational function:

g(x, y, z) = \frac{1}{x^2 + y^2 + z^2 - 9}

Find the mathematical domain in set-builder notation and describe the excluded boundary surface.
Determine the mathematical range in interval notation.
Evaluate $g(2, 1, 2)$ and explain the result.

Solution Key:

Analyze Domain Condition: State the Three-Variable Non-Zero Denominator Condition:
$x^2 + y^2 + z^2 - 9 \ne 0 \implies x^2 + y^2 + z^2 \ne 9$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \ne 9\}$
- Geometric Description: The domain consists of all points in $\mathbb{R}^3$ except those lying on the sphere of radius $R = \sqrt{9} = 3$ centered at the origin.
Analyze Range: Let $\rho^2 = x^2 + y^2 + z^2 \ge 0$ .
- Inside the sphere ( $\rho^2 < 9$ ), $\rho^2 - 9 \in [-9, 0)$ . At the origin $\rho = 0$ , $g = -1/9$ . As $\rho \to 3^-$ , $g \to -\infty$ . Range component: $(-\infty, -1/9]$ .
- Outside the sphere ( $\rho^2 > 9$ ), $\rho^2 - 9 \in (0, \infty)$ . As $\rho \to 3^+$ , $g \to +\infty$ . As $\rho \to \infty$ , $g \to 0^+$ . Range component: $(0, \infty)$ .
- Range in Interval Notation: $R = (-\infty, -1/9] \cup (0, \infty)$
Evaluate at $(2, 1, 2)$ : Compute the denominator:
$x^2 + y^2 + z^2 - 9 = 2^2 + 1^2 + 2^2 - 9 = 4 + 1 + 4 - 9 = 9 - 9 = 0$ $g(2, 1, 2) = \frac{1}{0} \implies \textbf{Undefined}$ The point $(2, 1, 2)$ is at distance $\rho = \sqrt{9} = 3$ from the origin, lying exactly on the excluded boundary sphere.

Show your steps and calculations here.

Practice Problem 9

Given the 3D solid sphere function:

h(x, y, z) = \sqrt{25 - (x^2 + y^2 + z^2)}

Determine the mathematical domain in set-builder notation and state its radius.
Determine the mathematical range in interval notation.
Evaluate $h(3, 4, 0)$ and explain why $h(3, 4, 2)$ is undefined.

Solution Key:

Analyze Domain Condition: State the Three-Variable Radicand Non-Negativity Condition:
$25 - (x^2 + y^2 + z^2) \ge 0 \implies x^2 + y^2 + z^2 \le 25$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 \le 25\}$
- Geometric Description: A solid 3D ball of radius $R = \sqrt{25} = 5$ centered at $(0, 0, 0)$ .
Analyze Range: The sum $\rho^2 = x^2 + y^2 + z^2$ is bounded between $0$ (at the origin) and $25$ (on the outer boundary shell).
- Maximum output at $(0, 0, 0)$ : $\sqrt{25 - 0} = 5$ .
- Minimum output on the shell: $\sqrt{25 - 25} = 0$ .
- Range: $R = [0, 5]$ .
Evaluate at Points:
- At $(3, 4, 0)$ : $\rho^2 = 3^2 + 4^2 + 0^2 = 9 + 16 + 0 = 25$ $h(3, 4, 0) = \sqrt{25 - 25} = \sqrt{0} = 0$ This point rests directly on the outer spherical boundary shell.
- At $(3, 4, 2)$ : $\rho^2 = 3^2 + 4^2 + 2^2 = 9 + 16 + 4 = 29$ $h(3, 4, 2) = \sqrt{25 - 29} = \sqrt{-4} = 2i \notin \mathbb{R}$ Because $29 > 25$ , the point lies outside the solid ball, so the function is undefined in the real numbers.

Show your steps and calculations here.

Practice Problem 10

Given the logarithmic function of three variables:

m(x, y, z) = \ln(x + 2y + 3z)

State the domain condition and write the domain in set-builder notation.
Geometrically describe the domain in $\mathbb{R}^3$ and name the boundary surface.
Evaluate $m(1, 1, 1)$ in exact form and round to three decimal places.

Solution Key:

Analyze Domain Condition: State the Logarithmic Positivity Condition: The argument of the logarithm must be strictly positive:
$x + 2y + 3z > 0$
- Set-Builder Notation: $D = \{(x, y, z) \in \mathbb{R}^3 \mid x + 2y + 3z > 0\}$
Geometric Description: The boundary surface is the flat plane $x + 2y + 3z = 0$ passing through the origin $(0, 0, 0)$ with normal vector $\vec{n} = \langle 1, 2, 3 \rangle$ . The condition $x + 2y + 3z > 0$ defines an open half-space extending infinitely on the positive side of this boundary plane (in the direction of the normal vector), with the plane itself excluded.
Evaluate at $(1, 1, 1)$ : Substitute $x = 1, y = 1, z = 1$ :
$x + 2y + 3z = 1 + 2(1) + 3(1) = 1 + 2 + 3 = 6 > 0$ $m(1, 1, 1) = \ln(6) \approx 1.792$

Show your steps and calculations here.

Regroup 2

Review the geometric leap from 2D planar domains to 3D solid volumetric domains. Excluded curves in 2D (like $x^2 + y^2 = 9$ ) become excluded 3D surfaces (like the sphere $x^2 + y^2 + z^2 = 9$ ).
Remind students that three-variable functions $w = f(x, y, z)$ cannot be graphed directly in 3D; their behavior is visualized through 3D level surfaces.

Skill Block 3

Worked Example A: Level Curves, Vertical Traces, and Paraboloid Reconstruction

Let $f(x, y) = x^2 + y^2$ .

Find the equations of the level curves for $c = 1, 4, 9$ and describe them in words.
Find the vertical traces in the planes $x = 0$ and $y = 0$ .
Synthesize the 2D slices to name and describe the 3D surface $z = f(x, y)$ .

Worked Solution:

Find Level Curves: State the Level Curve Equation:
$f(x, y) = c \implies x^2 + y^2 = c$
- For $c = 1$ : $x^2 + y^2 = 1$ (circle centered at $(0, 0)$ with radius $r = 1$ ).
- For $c = 4$ : $x^2 + y^2 = 4$ (circle centered at $(0, 0)$ with radius $r = 2$ ).
- For $c = 9$ : $x^2 + y^2 = 9$ (circle centered at $(0, 0)$ with radius $r = 3$ ).
- Geometric Description: Concentric circles centered at the origin whose radii increase as $r = \sqrt{c}$ . (Defined in Level Curves and Contour Maps.)
Find Vertical Traces: State the Vertical Trace Formulation:
- In the plane $x = 0$ ( $yz$ -plane): Substitute $x = 0$ into $z = x^2 + y^2$ : $z = 0^2 + y^2 \implies z = y^2$ This is an upward-opening parabola with its vertex at $(y, z) = (0, 0)$ .
- In the plane $y = 0$ ( $xz$ -plane): Substitute $y = 0$ into $z = x^2 + y^2$ : $z = x^2 + 0^2 \implies z = x^2$ This is an upward-opening parabola with its vertex at $(x, z) = (0, 0)$ . (Defined in Traces and Cross-Sections of Surfaces.)
3D Surface Synthesis: Because the horizontal cross-sections are circles and the orthogonal vertical cross-sections are identical upward-opening parabolas, the 3D surface is a circular paraboloid (a smooth, rotational bowl shape resting at the origin $(0, 0, 0)$ and opening upwards along the positive $z$ -axis).

Worked Example B: Saddle Surface Analysis via Level Curves and Traces

Let $f(x, y) = x^2 - y^2$ .

Find the level curves for $c = -4, -1, 0, 1, 4$ and sketch their qualitative behavior.
Find the vertical traces in the planes $x = 0$ and $y = 0$ .
Describe the 3D surface $z = f(x, y)$ and explain why the origin is called a saddle point.

Worked Solution:

Find Level Curves: State the Level Curve Equation:
$x^2 - y^2 = c$
- For $c = 0$ : $x^2 - y^2 = 0 \implies (x - y)(x + y) = 0 \implies y = x$ and $y = -x$ . The level set consists of two perpendicular straight lines intersecting at the origin.
- For $c = 1$ and $c = 4$ : $x^2 - y^2 = c \implies \frac{x^2}{c} - \frac{y^2}{c} = 1$ . These are horizontal hyperbolas opening to the left and right along the $x$ -axis, with vertices at $(\pm \sqrt{c}, 0)$ .
- For $c = -1$ and $c = -4$ : $x^2 - y^2 = c \implies y^2 - x^2 = |c| \implies \frac{y^2}{|c|} - \frac{x^2}{|c|} = 1$ . These are vertical hyperbolas opening upward and downward along the $y$ -axis, with vertices at $(0, \pm \sqrt{|c|})$ . (Defined in Level Curves and Contour Maps.)
Find Vertical Traces:
- In the plane $x = 0$ ( $yz$ -plane): $z = 0^2 - y^2 \implies z = -y^2$ This is a downward-opening parabola with its maximum at $(y, z) = (0, 0)$ .
- In the plane $y = 0$ ( $xz$ -plane): $z = x^2 - 0^2 \implies z = x^2$ This is an upward-opening parabola with its minimum at $(x, z) = (0, 0)$ . (Defined in Traces and Cross-Sections of Surfaces.)
3D Surface Synthesis: The surface is a hyperbolic paraboloid (commonly called a saddle surface or Pringles chip). Along the $x$ -direction, the origin is a local minimum ( $z$ curves upward like $x^2$ ); along the perpendicular $y$ -direction, the origin is a local maximum ( $z$ curves downward like $-y^2$ ). Because the surface bends up in one direction and down in the other, $(0, 0, 0)$ is a saddle point.

Practice Problem 11: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain the geometric distinction between a horizontal level curve $f(x, y) = c$ and a vertical trace $z = f(x, k)$ .
(b) In Worked Example A, explain why the spacing between concentric circular level curves decreases as elevation $c$ increases, and what this reveals about the slope of a paraboloid.
(c) In the linked application Aerospace Structural Thermal Isosurfaces (Aerospace & Mechanical Engineering), how does a 3D level surface help propulsion engineers identify cooling channel placement?

Solution Key:

(a) Slicing the graph of $z = f(x, y)$ with a horizontal plane $z = c$ freezes the dependent variable (output), producing a curve in $\mathbb{R}^2$ where every point has the identical elevation. Slicing with a vertical plane $y = k$ freezes an independent input variable, producing a 2D profile curve in the $xz$ -plane showing how elevation changes as only $x$ varies.
(b) For $z = r^2$ , we have $r = \sqrt{z}$ . The derivative is $\frac{dr}{dz} = \frac{1}{2\sqrt{z}}$ . As elevation $z$ increases, the horizontal radial expansion needed to gain another unit of elevation shrinks towards zero. On a contour map with equal elevation intervals ( $\Delta z = 1$ ), the spacing between successive contour rings becomes tighter and tighter, visually demonstrating that the slope ( $\frac{dz}{dr} = 2r$ ) is increasing.
(c) In solid rocket nozzles, temperature cannot be modeled as a flat 2D map because heat conducts three-dimensionally through the metal injector flange. By computing the level surface $T(x, y, z) = 400^{\circ}\text{C}$ (which forms a sphere of radius $R = 2\text{ cm}$ ), propulsion engineers pinpoint the exact physical boundary where copper loses structural tensile strength, directing where regenerative coolant channels must circulate to prevent catastrophic nozzle burnout.

Show your steps and calculations here.

Practice Problem 12

Given the elliptic paraboloid function:

f(x, y) = 4x^2 + y^2

Find the equations of the level curves for $c = 4$ and $c = 16$ . Describe their geometric shape in the $xy$ -plane, identifying the semi-major and semi-minor axes.
Find the vertical traces in the coordinate planes $x = 0$ and $y = 0$ .
Describe the 3D surface $z = f(x, y)$ and explain how it differs from a circular paraboloid.

Solution Key:

Find Level Curves: State the Level Curve Equation:
$4x^2 + y^2 = c$
- For $c = 4$ : $\frac{4x^2}{4} + \frac{y^2}{4} = 1 \implies \frac{x^2}{1^2} + \frac{y^2}{2^2} = 1$ This is an ellipse centered at $(0, 0)$ with semi-axis $1$ along the $x$ -axis and semi-axis $2$ along the $y$ -axis.
- For $c = 16$ : $\frac{4x^2}{16} + \frac{y^2}{16} = 1 \implies \frac{x^2}{2^2} + \frac{y^2}{4^2} = 1$ This is an expanded concentric ellipse with semi-axis $2$ along the $x$ -axis and semi-axis $4$ along the $y$ -axis.
Find Vertical Traces:
- Trace in $x = 0$ ( $yz$ -plane): $z = 4(0)^2 + y^2 \implies z = y^2$ This is a standard upward-opening parabola with vertex at $(0, 0)$ .
- Trace in $y = 0$ ( $xz$ -plane): $z = 4x^2 + 0^2 \implies z = 4x^2$ This is a much steeper upward-opening parabola with vertex at $(0, 0)$ .
3D Surface Synthesis: The surface is an elliptic paraboloid. Because the coefficient of $x^2$ ( $4$ ) is greater than the coefficient of $y^2$ ( $1$ ), the surface rises much faster along the $x$ -direction than the $y$ -direction, creating an oval, trough-like bowl rather than a radially symmetric circular bowl.

Show your steps and calculations here.

Practice Problem 13

Given the linear multivariable function:

f(x, y) = 2x + 3y

Find the equations of the level curves for $c = 0, 6, 12$ . Describe their geometric shape in the $xy$ -plane and state their common slope.
Find the vertical traces in the coordinate planes $x = 0$ and $y = 0$ .
Describe the 3D surface $z = f(x, y)$ .

Solution Key:

Find Level Curves: Set $2x + 3y = c$ and solve for $y$ :
$3y = -2x + c \implies y = -\frac{2}{3}x + \frac{c}{3}$
- For $c = 0$ : $y = -\frac{2}{3}x$ (line through origin).
- For $c = 6$ : $y = -\frac{2}{3}x + 2$ ( $y$ -intercept at $2$ ).
- For $c = 12$ : $y = -\frac{2}{3}x + 4$ ( $y$ -intercept at $4$ ).
- Geometric Description: A family of equally spaced, parallel straight lines across the $xy$ -plane, all possessing a constant slope of $m = -2/3$ .
Find Vertical Traces:
- In $x = 0$ ( $yz$ -plane): $z = 2(0) + 3y \implies z = 3y$ A straight line through the origin with slope $3$ .
- In $y = 0$ ( $xz$ -plane): $z = 2x + 3(0) \implies z = 2x$ A straight line through the origin with slope $2$ .
3D Surface Synthesis: Because all horizontal level curves are parallel straight lines and all vertical traces are straight lines, the graph $z = 2x + 3y \implies 2x + 3y - z = 0$ is a flat tilted plane in $\mathbb{R}^3$ passing through the origin with constant normal vector $\vec{n} = \langle 2, 3, -1 \rangle$ .

Show your steps and calculations here.

Practice Problem 14

Given the 3D spherical field function:

g(x, y, z) = x^2 + y^2 + z^2

Find the equations of the level surfaces for $c = 1, 4, 9$ . Describe their geometric shape in $\mathbb{R}^3$ and state their radii.
Determine the algebraic equation of the planar cross-section along the equatorial plane $z = 0$ .
Describe the physical behavior of this scalar field in space.

Solution Key:

Find Level Surfaces: State the Level Surface Formulation:
$x^2 + y^2 + z^2 = c$
- For $c = 1$ : $x^2 + y^2 + z^2 = 1$ (spherical shell of radius $r = 1$ ).
- For $c = 4$ : $x^2 + y^2 + z^2 = 4$ (spherical shell of radius $r = 2$ ).
- For $c = 9$ : $x^2 + y^2 + z^2 = 9$ (spherical shell of radius $r = 3$ ).
- Geometric Description: A family of concentric, nested 3D spherical shells expanding outward from the coordinate origin.
Equatorial Cross-Section ( $z = 0$ ): Substitute $z = 0$ into the level surface equation:
$x^2 + y^2 + 0^2 = c \implies x^2 + y^2 = c$ In the horizontal cutting plane $z = 0$ , the cross-sections are concentric circles centered at the origin of radius $\sqrt{c}$ .
Field Interpretation: The function describes a point-source or centrally symmetric field (such as gravitational potential, electrostatic potential around a point charge, or thermal radiation from a central point heater) where output values depend strictly on spatial distance $\rho = \sqrt{x^2 + y^2 + z^2}$ from the core.

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Practice Problem 15

Given the 3D scalar function where the coordinate $z$ is absent:

F(x, y, z) = x^2 + y^2

Determine the equations of the level surfaces for $c = 1, 4, 9$ .
Geometrically describe these level surfaces in $\mathbb{R}^3$ .
Find the vertical cross-sectional trace produced by slicing the field with the plane $x = 0$ .

Solution Key:

Find Level Surfaces: State the Level Surface Equation:
$F(x, y, z) = c \implies x^2 + y^2 = c$
- For $c = 1$ : $x^2 + y^2 = 1$ .
- For $c = 4$ : $x^2 + y^2 = 4$ .
- For $c = 9$ : $x^2 + y^2 = 9$ .
Geometric Description in $\mathbb{R}^3$ : In 2D, $x^2 + y^2 = c$ is a circle of radius $\sqrt{c}$ . However, because the coordinate $z$ does not appear in the equation, $z$ can take on any real value: $-\infty < z < \infty$ . Each level surface is a circular cylinder centered coaxial with the $z$ -axis with radius $r = \sqrt{c}$ , extending infinitely along the $z$ -axis. As $c$ increases, the cylinders expand in radius, forming nested concentric pipes.
Cross-Section in $x = 0$ ( $yz$ -plane): Substitute $x = 0$ into the level surface equation:
$0^2 + y^2 = c \implies y^2 = c \implies y = \pm \sqrt{c}$ In 3D space, $y = \sqrt{c}$ and $y = -\sqrt{c}$ are two flat vertical planes parallel to the $xz$ -plane, located at distance $\sqrt{c}$ on either side of the $z$ -axis.

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Regroup 3

Review how 2D contour maps translate into 3D topography. Remember: closely spaced level curves mean steep slopes; parallel linear level curves mean tilted planes; concentric circles mean paraboloids or cones.
Emphasize the distinction between level curves ( $f(x, y) = c$ , horizontal slices in $\mathbb{R}^2$ ) and level surfaces ( $f(x, y, z) = c$ , 2D shells in $\mathbb{R}^3$ ).

Synthesis Wrap-up

Core Takeaways from Session 08:
1. Multivariable functions $z = f(x, y)$ and $w = f(x, y, z)$ map higher-dimensional coordinates to single real outputs, modeling physical systems with multiple coupled parameters.
2. The domain of a two-variable function is a planar region in $\mathbb{R}^2$ (classified as open/closed, bounded/unbounded); the domain of a three-variable function is a solid volume or open half-space in $\mathbb{R}^3$ .
3. Slicing surfaces horizontally yields level curves (contour lines); slicing vertically yields traces. Combining orthogonal slices allows engineers to reconstruct complex quadric surfaces.
4. In three dimensions, level sets $f(x, y, z) = c$ form level surfaces (isosurfaces), allowing engineers to visualize volumetric fields such as thermal and pressure distributions.
Knewton Alta Homework Alignment:
- Functions of Two Variables
- Functions of More Than Two Variables and Applications
- Level Curves and Traces of Functions of Two Variables