Multivariable-Calculus · Unit 7 · Video 1 · Interactive Practice

Slicing a Surface: How to Picture a Function of Two Variables

IKey Formulas

Formula	Name	What it says
$(x, y) \mapsto f(x, y)$	Function of two variables	One number for each point of the domain
$z = f(x, y)$	Graph	The point above $(x, y)$ at height $f(x, y)$ ; together, a surface
$x = 0,\ \ y = 0,\ \ z = 0$	Coordinate slices	Setting one variable constant leaves a plane curve
$r^2 = x^2 + y^2 \Rightarrow z = 1 - r^2$	Surface of revolution	Height depends only on the distance $r$ to the $z$ -axis

Key Insight: A surface is hard to picture whole, but each slice by a plane is a familiar curve in that plane: the slices of $z = 1 - x^2 - y^2$ are two downward parabolas and the unit circle.

IIVisualization 1 — Where the Formula Makes Sense

The domain is the set of points $(x, y)$ where the formula returns a number.

Function x 1 y 2

IIIVisualization 2 — Slicing by the Coordinate Planes

Holding one of $x$ , $y$ , $z$ constant cuts the surface $z = f(x, y)$ in a familiar plane curve.

Surface Slice c 0 View 20°

IVVisualization 3 — The Dome Is a Surface of Revolution

When $f$ depends only on $r = \sqrt{x^2 + y^2}$ , every vertical plane through the $z$ -axis cuts the same curve.

Surface θ 30°

VQuiz Questions

Problem 1 · Height Above a Point

Given: $f(x, y) = 1 - x^2 - y^2$ — find $f(1, 2)$ , the height of the graph above the point $(1, 2)$ .

✅ Correct!

f(1, 2) = 1 - 1 - 4 = -4

, so the graph passes through

(1, 2, -4)

, four units below the

xy

-plane.

❌ Check the signs. Both squares are subtracted:

1 - 1^2 - 2^2

, not

1 + 1^2 + 2^2

❌ Square the $y$ -coordinate. The term is

y^2 = 2^2 = 4

, not

2

❌ The sign matters. A negative value means the graph lies below the

xy

-plane there; height

4

would be above it.

❌ Not quite. Substitute

x = 1

y = 2

into

1 - x^2 - y^2

Show solution

f(1, 2) = 1 - 1^2 - 2^2 = 1 - 1 - 4 = -4

The point of the graph above $(1, 2)$ is $(1, 2, -4)$ . Since $1^2 + 2^2 = 5 > 1$ , the point $(1, 2)$ lies outside the unit circle, where the dome has already dropped below the $xy$ -plane.

Problem 2 · Two Conditions at Once

Given: $f(x, y) = \dfrac{\sqrt{y}}{x}$ — find its domain.

✅ Correct! The root needs

y \geq 0

and the division needs

x \neq 0

: the closed upper half-plane with the

y

-axis removed.

❌ One condition is missing. That handles

\sqrt{y}

, but

x

sits in a denominator: every point with

x = 0

must go.

❌ Too strict.

\sqrt{0} = 0

is fine, so points on the

x

-axis with

x \neq 0

belong: e.g.

f(2, 0) = 0/2 = 0

❌ One condition is missing. That handles the denominator, but

\sqrt{y}

is not a real number when

y < 0

Show solution

Each operation in the formula imposes its own condition, and a point must satisfy all of them.

$\sqrt{y}$ is defined exactly when $y \geq 0$ (the $x$ -axis included).
Dividing by $x$ is defined exactly when $x \neq 0$ .

Domain: $\{(x, y) : y \geq 0,\ x \neq 0\}$ — the upper half-plane with the $x$ -axis kept (except the origin) and the $y$ -axis removed.

Problem 3 · Slicing a Taller Dome

Given: the surface $z = 4 - x^2 - y^2$ — find its slice by the plane $x = 1$ and the curve where it meets the $xy$ -plane.

Slice by the plane x = 1

Where the surface meets the xy-plane

✅ Correct! A downward parabola with top at height

3

in the plane

x = 1

, and a dome of height

4

sitting on the circle of radius

2

❌ Substitute $x = 1$ .

z = 4 - y^2

is the slice by

x = 0

; at

x = 1

the term

x^2

removes one more unit.

❌ Check the substitution.

4 - 1^2 - y^2 = 3 - y^2

: the

x^2

term is subtracted and

-y^2

keeps its sign.

❌ Radius, not radius squared.

x^2 + y^2 = 4

says the distance to the origin squared is

4

, so the radius is

\sqrt{4} = 2

❌ That is a vertical slice.

z = 4 - y^2

is the slice by

x = 0

; the

xy

-plane is

z = 0

❌ Set $z = 0$ .

0 = 4 - x^2 - y^2

gives

x^2 + y^2 = 4

Show solution

Slice $x = 1$ :

z = 4 - 1^2 - y^2 = 3 - y^2

a downward parabola with its top at $(1, 0, 3)$ .

Slice $z = 0$ :

0 = 4 - x^2 - y^2 \iff x^2 + y^2 = 4

the circle of radius $\sqrt{4} = 2$ about the origin. The part of the surface above the $xy$ -plane is a dome of height $4$ sitting on that circle.

Problem 4 · Spotting a Surface of Revolution

Given: four surfaces — find the one that is a surface of revolution about the $z$ -axis (every vertical plane through the $z$ -axis cuts the same curve).

✅ Correct! With

r^2 = x^2 + y^2

it reads

z = e^{-r^2}

: the height depends only on

r

, so it is a bell-shaped hill symmetric about the

z

-axis.

❌ Unequal coefficients. Along the

x

-axis

z = s^2

but along the

y

-axis

z = 2s^2

: different parabolas, so the cross-sections

z = c

are ellipses, not circles.

❌ That is the saddle. Along the

y

-axis it curves up (

z = s^2

) and along the

x

-axis it curves down (

z = -s^2

❌ A tilted plane. Its slice along the

x

-axis is flat (

z = 0

) but along the

y

-axis it has slope

-1

Show solution

A surface $z = f(x, y)$ is a surface of revolution about the $z$ -axis exactly when $f$ depends on $x$ and $y$ only through $r^2 = x^2 + y^2$ . In the vertical plane at angle $\theta$ , put $x = s\cos\theta$ , $y = s\sin\theta$ :

$e^{-(x^2 + y^2)} = e^{-s^2}$ for every $\theta$ — the same curve. ✓
$x^2 + 2y^2 = s^2(1 + \sin^2\theta)$ — depends on $\theta$ .
$y^2 - x^2 = -s^2\cos 2\theta$ — depends on $\theta$ .
$-y = -s\sin\theta$ — depends on $\theta$ .

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