Multivariable Calculus Quiz: Domains And Level Sets

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Domains And Level SetsQuestion 1 of 16

Consider the function $f(x, y) = e^{y - x^2}$ . Which of the following statements correctly describes the level curves of this function?

A family of parabolas opening upward, with vertices moving up the

y

-axis as the level value increases.

A family of parabolas opening downward, with vertices moving down the

y

-axis as the level value increases.

A family of hyperbolas with vertices on the

y

-axis.

A family of parabolas opening upward, with vertices moving down the

y

-axis as the level value increases.

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Domains And Level Sets

Practice Domains And Level Sets in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Domains And Level Sets, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Consider the function $f(x, y) = e^{y - x^2}$ . Which of the following statements correctly describes the level curves of this function?

A family of parabolas opening upward, with vertices moving up the $y$ -axis as the level value increases. (correct answer)
A family of parabolas opening downward, with vertices moving down the $y$ -axis as the level value increases.
A family of hyperbolas with vertices on the $y$ -axis.
A family of parabolas opening upward, with vertices moving down the $y$ -axis as the level value increases.

Explanation: A level curve is defined by the equation

f(x, y) = c

for some constant

c

. Since the range of the exponential function is positive numbers, we must have

c > 0

. The equation is

e^{y - x^2} = c

. Taking the natural logarithm of both sides gives

y - x^2 = \ln(c)

, which can be rewritten as

y = x^2 + \ln(c)

. This is the equation of a family of parabolas opening upward. The vertex of each parabola is at

(0, \ln(c))

. As the level value

c

increases,

\ln(c)

also increases, so the vertex moves up the

y

-axis.

Question 2

What is the level set of the function $f(x,y) = \frac{(x-y)^2}{x^2+y^2}$ corresponding to the level $c=0$ ?

The line $y=x$ .
The line $y=x$ , with the origin excluded. (correct answer)
The single point $(0,0)$ .
The pair of lines $y=x$ and $y=-x$ .

Explanation: The level set for

c=0

is the set of all points

(x,y)

such that

f(x,y) = 0

. This requires the numerator to be zero and the denominator to be non-zero. The numerator is

(x-y)^2 = 0

, which implies

x-y=0

, or

y=x

. The denominator is

x^2+y^2

. For the function to be defined,

x^2+y^2 \ne 0

, which means

(x,y)

cannot be the origin

(0,0)

. Therefore, the level set is the line

y=x

with the point

(0,0)

removed.

Question 3

The temperature on a rectangular plate defined by $|x| \le 2$ and $|y| \le 3$ is given by the function $T(x,y) = 90 - 5x^2 - 2y^2$ . An isotherm is a curve of constant temperature. Which of the following statements is true?

Isotherms for temperatures less than 90 are hyperbolas.
The isotherm corresponding to the lowest temperature on the plate passes through the point $(0,0)$ .
The isotherm corresponding to the lowest temperature on the plate passes through the points $(2,3), (-2,3), (2,-3),$ and $(-2,-3)$ . (correct answer)
The isotherm for $T=90$ is the line segment from $(0,-3)$ to $(0,3)$ .

Explanation: The isotherms are the level curves of

T(x,y)

, given by

90 - 5x^2 - 2y^2 = c

, or

5x^2 + 2y^2 = 90 - c

. For

c < 90

, these curves are ellipses centered at the origin. The function

T(x,y)

has its maximum value of 90 at

(0,0)

and decreases as

(x,y)

moves away from the origin. The lowest temperature on the plate will therefore occur at the point(s) within the domain

|x| \le 2, |y| \le 3

that are 'farthest' from the origin in the sense of the elliptical level curves. This occurs at the four corners of the rectangular domain:

(\pm 2, \pm 3)

. The temperature at these corners is

T(2,3) = 90 - 5(2^2) - 2(3^2) = 90 - 20 - 18 = 52

. This is the minimum temperature. The isotherm for

T=52

5x^2+2y^2 = 90 - 52 = 38

, and this ellipse passes through all four corner points.

Question 4

Consider the function $f(x,y) = \frac{\ln(x+y-1)}{\sqrt{4-x^2-y^2}}$ . Which of the following best describes the domain of $f$ ?

The region $x^2 + y^2 \leq 4$ where $x + y > 1$ including the boundary circle
The closed disk $x^2 + y^2 \leq 4$ where $x + y \geq 1$
The region $x^2 + y^2 < 4$ where $x + y > 1$ excluding the circle $x^2 + y^2 = 4$
The interior of the disk $x^2 + y^2 < 4$ where $x + y > 1$ (correct answer)

Explanation: When finding the domain of a multivariable function, you need to identify where each component is mathematically defined and then find their intersection. For

f(x,y) = \frac{\ln(x+y-1)}{\sqrt{4-x^2-y^2}}

, examine each part separately. The natural logarithm

\ln(x+y-1)

requires its argument to be strictly positive, so you need

x+y-1 > 0

, which means

x+y > 1

. The denominator

\sqrt{4-x^2-y^2}

requires the expression under the square root to be strictly positive (since it's in the denominator, it cannot equal zero). This gives you

4-x^2-y^2 > 0

, or equivalently

x^2+y^2 < 4

. The domain is where both conditions hold simultaneously: the interior of the disk

x^2+y^2 < 4

intersected with the half-plane

x+y > 1

. Choice A incorrectly includes the boundary circle

x^2+y^2 = 4

, but this would make the denominator zero (undefined). Choice B has two errors: it includes the boundary circle and uses

x+y \geq 1

, but the logarithm requires

x+y > 1

strictly. Choice C correctly excludes the boundary circle but redundantly states this exclusion - the main issue is that it doesn't clearly specify that we want the interior of the disk. Study tip: For domain problems, systematically check each function component's requirements, paying special attention to whether inequalities are strict or non-strict, especially when expressions appear in denominators or inside logarithms.

Question 5

The level sets of $F(x,y,z) = x^2 + y^2 - z^2$ are:

Paraboloids opening upward for $c > 0$ , paraboloids opening downward for $c < 0$ , and a plane for $c = 0$
Ellipsoids for $c > 0$ , empty sets for $c < 0$ , and a point for $c = 0$
Hyperboloids of one sheet for $c > 0$ , hyperboloids of two sheets for $c < 0$ , and a cone for $c = 0$ (correct answer)
Cylinders with elliptical cross-sections for all values of $c$

Explanation: When analyzing level sets of functions in three variables, you're looking at the surfaces formed by setting

F(x,y,z) = c

for different constant values. The function

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

creates a family of quadric surfaces that change character based on the value of

c

. For

c > 0

: Setting

x^2 + y^2 - z^2 = c

gives us

x^2 + y^2 = z^2 + c

. Since the right side is always positive, this describes a hyperboloid of one sheet - a surface that extends infinitely and has a "saddle-like" shape that's connected throughout. For

c < 0

: The equation becomes

x^2 + y^2 = z^2 + c

. Since

c < 0

, we need

z^2 \geq |c|

for real solutions. This creates two separate surfaces (one for

z \geq \sqrt{|c|}

and one for

z \leq -\sqrt{|c|}

), forming a hyperboloid of two sheets. For

c = 0

: We get

x^2 + y^2 = z^2

, which is a circular cone with vertex at the origin. This confirms answer C is correct. Answer A incorrectly identifies these as paraboloids, which would involve linear terms. Answer B describes surfaces like

x^2 + y^2 + z^2 = c

, where all variables are squared with the same sign. Answer D suggests the surfaces don't change character with

c

, missing the fundamental distinction between the three cases. Strategy tip: For quadric surface problems, focus on the signs of the squared terms and whether the constant is positive, negative, or zero - this determines the surface type.

Question 6

Let $f(x,y) = \arcsin(x^2 + y^2 - 1)$ . For which value of $c$ does the level set $f(x,y) = c$ consist of a single point?

$c = \frac{\pi}{2}$
$c = \pi$
$c = 0$
$c = -\frac{\pi}{2}$ (correct answer)

Explanation: The level set is given by the equation

\arcsin(x^2 + y^2 - 1) = c

. Applying the sine function to both sides gives

x^2 + y^2 - 1 = \sin(c)

, or

x^2 + y^2 = 1 + \sin(c)

. This equation describes a circle centered at the origin with radius

r = \sqrt{1 + \sin(c)}

. For the level set to be a single point (the origin), the radius must be zero. This requires

1 + \sin(c) = 0

, which means

\sin(c) = -1

. The range of the

\arcsin

function is

[-\frac{\pi}{2}, \frac{\pi}{2}]

. The only value of

c

in this interval for which

\sin(c) = -1

c = -\frac{\pi}{2}

Question 7

Consider the function $f(x, y) = \ln(x^2 + y^2)$ . Which statement best describes the spacing of its level curves $f(x,y)=c$ for equally spaced values of $c$ ?

The level curves are circles that become more sparsely spaced as $c$ increases. (correct answer)
The level curves are circles that become more densely spaced as $c$ increases.
The level curves are equally spaced concentric circles.
The level curves are parabolas that become more sparsely spaced as $c$ increases.

Explanation: The level curves are given by

\ln(x^2 + y^2) = c

. Exponentiating both sides gives

x^2 + y^2 = e^c

. These are concentric circles centered at the origin with radius

r(c) = \sqrt{e^c} = e^{c/2}

. To analyze the spacing, consider equally spaced values of

c

, such as

c, c+1, c+2, \dots

. The radii of the corresponding circles are

e^{c/2}, e^{(c+1)/2}, e^{(c+2)/2}, \dots

. The distance between two consecutive curves is the difference in their radii:

\Delta r = r(c+1) - r(c) = e^{(c+1)/2} - e^{c/2} = e^{c/2}(e^{1/2} - 1)

. Since

e^{1/2} - 1

is a positive constant and

e^{c/2}

is an increasing function of

c

, the distance

\Delta r

increases as

c

increases. Therefore, the level curves become more sparsely spaced.

Question 8

The domain of the function $f(x, y) = \frac{1}{\cos(\pi x) - 1}$ is the entire $xy$ -plane with a set of points removed. What is the geometric description of the removed set?

A set of horizontal lines of the form $y=2k$ for any integer $k$ .
A set of vertical lines of the form $x=2k$ for any integer $k$ . (correct answer)
A lattice of points $(2k, n)$ where $k$ and $n$ are any integers.
A set of vertical lines of the form $x=k$ for any integer $k$ .

Explanation: The function is defined everywhere except where the denominator is zero. The denominator is

\cos(\pi x) - 1 = 0

, which means

\cos(\pi x) = 1

. The cosine function equals 1 when its argument is an even multiple of

\pi

. So, we must have

\pi x = 2k\pi

for any integer

k

. Dividing by

\pi

gives

x = 2k

. This condition depends only on

x

; the value of

y

can be anything. Therefore, the set of points removed from the domain is the set of all points

(x,y)

where

x

is an even integer. This corresponds to an infinite set of vertical lines of the form

x=..., -4, -2, 0, 2, 4, ...

Question 9

For the function $f(x,y) = e^{-\sqrt{4-x^2-y^2}}$ , the level set $f(x,y)=c$ is a circle for certain values of $c$ . For which of the following values of $c$ is the level set a circle?

$c = \frac{1}{e^3}$
$c = \frac{1}{e^2}$
$c = e$
$c = \frac{1}{e}$ (correct answer)

Explanation: First, determine the range of

f

. The domain is

4-x^2-y^2 \ge 0

, or

x^2+y^2 \le 4

. The value of

u = \sqrt{4-x^2-y^2}

ranges from

0

(when

x^2+y^2=4

) to

2

(when

x^2+y^2=0

). Thus, the range of

f(x,y) = e^{-u}

[e^{-2}, e^0] = [1/e^2, 1]

. A level set can only exist for

c

in this range. Now, set

f(x,y)=c

e^{-\sqrt{4-x^2-y^2}} = c

. This gives

-\sqrt{4-x^2-y^2} = \ln c

, so

\sqrt{4-x^2-y^2} = -\ln c = \ln(1/c)

. Squaring both sides yields

4-x^2-y^2 = (\ln(1/c))^2

, or

x^2+y^2 = 4 - (\ln(1/c))^2

. For this to be a circle, the radius squared must be positive:

4 - (\ln(1/c))^2 > 0

. This means

(\ln(1/c))^2 < 4

, or

|\ln(1/c)| < 2

. This implies

-2 < \ln(1/c) < 2

, so

e^{-2} < 1/c < e^2

, which means

1/e^2 < c < e^2

. We need a value of

c

that is in both the range

[1/e^2, 1]

and the interval

(1/e^2, e^2)

. The intersection is

(1/e^2, 1]

. Of the choices, only

c = 1/e

is in this interval. For

c=1/e^2

, the radius is 0, which is a point, not a circle.

Question 10

Let $D$ be the domain of the function $f(x,y) = \sqrt{1 - x^2} + \sqrt{1 - y^2}$ , and let $C$ be the level set $f(x,y) = 2$ . Which of the following statements is true?

$C$ is the boundary of the domain $D$ .
$C$ consists of the four corner points of the domain $D$ .
$C$ is the center point of the domain $D$ . (correct answer)
$C$ is a circle of radius 1 centered at the origin.

Explanation: The domain

D

requires both

1 - x^2 \ge 0

and

1 - y^2 \ge 0

. These inequalities simplify to

|x| \le 1

and

|y| \le 1

, which define a closed square in the

xy

-plane with vertices at

(\pm 1, \pm 1)

. The center of this square is

(0,0)

. The level set

C

is defined by

\sqrt{1 - x^2} + \sqrt{1 - y^2} = 2

. Within the domain, the maximum value of

\sqrt{1 - x^2}

is 1 (attained only at

x=0

), and the maximum value of

\sqrt{1 - y^2}

is 1 (attained only at

y=0

). The sum can equal 2 only if both terms achieve their maximum value simultaneously. This occurs only when

x=0

and

y=0

. Thus, the level set

C

consists of the single point

(0,0)

, which is the center of the domain

D

Question 11

Which statement accurately describes the level surfaces $f(x, y, z) = c$ of the function $f(x, y, z) = x^2 + y^2 - z^2$ ?

They are hyperboloids of two sheets for $c > 0$ , a cone for $c = 0$ , and hyperboloids of one sheet for $c < 0$ .
They are hyperboloids of one sheet for $c > 0$ , a cone for $c = 0$ , and hyperboloids of two sheets for $c < 0$ . (correct answer)
They are ellipsoids for $c > 0$ , a point for $c = 0$ , and empty for $c < 0$ .
They are elliptic paraboloids for all $c \ne 0$ and a plane for $c=0$ .

Explanation: The level surfaces are given by the equation

x^2 + y^2 - z^2 = c

. This is the standard form for a family of quadric surfaces. When

c > 0

, we have a hyperboloid of one sheet. For example, if

c=1

, the surface is connected and opens along the

z

-axis. When

c = 0

, the equation is

x^2 + y^2 = z^2

, which is the equation of a double cone with its vertex at the origin. When

c < 0

, let

c = -k^2

for some

k>0

. The equation becomes

x^2 + y^2 - z^2 = -k^2

, or

z^2 - x^2 - y^2 = k^2

, which is the equation of a hyperboloid of two sheets.

Question 12

Let $D$ be the domain of the function $f(x, y) = \frac{\ln(y - x^2)}{\sqrt{9 - x^2 - y^2}}$ . Which of the following provides the most accurate geometric description of $D$ ?

The set of points inside the circle $x^2 + y^2 = 9$ and below the parabola $y = x^2$ .
The set of points inside the circle $x^2 + y^2 = 9$ and above the parabola $y = x^2$ . (correct answer)
The set of points outside the circle $x^2 + y^2 = 9$ and above the parabola $y = x^2$ .
The set of points bounded by the circle $x^2 + y^2 = 9$ and the parabola $y = x^2$ .

Explanation: The domain of the function is constrained by two conditions. First, the argument of the natural logarithm must be positive, so

y - x^2 > 0

, which means

y > x^2

. This describes the region above the parabola

y = x^2

. Second, the argument of the square root in the denominator must be positive, so

9 - x^2 - y^2 > 0

, which means

x^2 + y^2 < 9

. This describes the region inside the circle of radius 3 centered at the origin. The domain

D

is the intersection of these two regions: the set of points that are both inside the circle and above the parabola.

Question 13

The function $p(x,y) = \tan^{-1}\left(\frac{y}{x}\right) + \tan^{-1}\left(\frac{x}{y}\right)$ has a domain that excludes:

Only the origin and points where $xy < 0$
Only the coordinate axes where $x = 0$ or $y = 0$ (correct answer)
Only the origin $(0,0)$
Only points in the second and fourth quadrants

Explanation: The function p(x,y) = arctan(y/x) + arctan(x/y) is undefined when either denominator is zero. This occurs when x = 0 or y = 0, which are the coordinate axes. The function is defined in all four quadrants where both x ≠ 0 and y ≠ 0. Note that arctan(y/x) is undefined when x = 0, and arctan(x/y) is undefined when y = 0. Choice A incorrectly restricts to positive products, choice C only excludes the origin, and choice D incorrectly excludes entire quadrants.

Question 14

For the function $g(x,y,z) = \frac{xyz}{x^2+y^2+z^2-1}$ , which statement about its domain is correct?

The domain includes all points except the origin $(0,0,0)$
The domain is the interior of the unit ball where $x^2+y^2+z^2 < 1$
The domain is the exterior of the unit ball where $x^2+y^2+z^2 > 1$
The domain includes all points except those on the unit sphere $x^2+y^2+z^2 = 1$ (correct answer)

Explanation: When finding the domain of a multivariable function, you need to identify where the function is undefined. For rational functions like this one, the critical issue is where the denominator equals zero. The function

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

is undefined when its denominator equals zero. Setting

x^2+y^2+z^2-1 = 0

gives us

x^2+y^2+z^2 = 1

, which is the equation of a unit sphere centered at the origin. The function is defined everywhere except on this sphere, making answer D correct. Let's see why the other options are wrong. Option A suggests the function is undefined only at the origin

(0,0,0)

. However, substituting the origin into the denominator gives

0^2+0^2+0^2-1 = -1 \neq 0

, so the function is actually defined at the origin. Option B claims the domain is only the interior of the unit ball (

x^2+y^2+z^2 < 1

). This excludes points outside the unit sphere where the function is perfectly well-defined. Option C makes the opposite error, claiming the domain is only the exterior (

x^2+y^2+z^2 > 1

), which incorrectly excludes the interior points where the function is also defined. Remember: for rational functions, always set the denominator equal to zero to find excluded points. The domain includes all other points in the appropriate space (here, all of

\mathbb{R}^3

except the unit sphere).

Question 15

Consider the function $r(x,y,z) = \frac{1}{\sqrt{(x-1)^2 + (y-2)^2 + (z-3)^2 - 9}}$ . The boundary of the domain of this function is:

A sphere of radius $3$ centered at the origin
A sphere of radius $9$ centered at $(1,2,3)$
A sphere of radius $3$ centered at $(1,2,3)$ (correct answer)
A sphere of radius $\sqrt{9}$ centered at $(-1,-2,-3)$

Explanation: When analyzing the domain of a function with a square root in the denominator, you need to determine where the expression is both positive (for the square root) and non-zero (to avoid division by zero). The function is undefined where the denominator equals zero or becomes imaginary. For

r(x,y,z)

to be defined, the expression under the square root must be strictly positive:

(x-1)^2 + (y-2)^2 + (z-3)^2 - 9 > 0

. This can be rewritten as

(x-1)^2 + (y-2)^2 + (z-3)^2 > 9

. The boundary of the domain occurs where the inequality becomes an equality:

(x-1)^2 + (y-2)^2 + (z-3)^2 = 9

. This is the standard form of a sphere with center

(1,2,3)

and radius

\sqrt{9} = 3

. The function is defined outside this sphere but undefined on and inside it. Looking at the wrong answers: A incorrectly places the center at the origin instead of

(1,2,3)

. B uses radius

9

instead of

\sqrt{9} = 3

- this confuses the value under the square root with the actual radius. D has the wrong center

(-1,-2,-3)

, likely from a sign error, though it correctly calculates the radius as

\sqrt{9}

. The correct answer is C: a sphere of radius

3

centered at

(1,2,3)

. Study tip: For domain problems involving square roots in denominators, always set the expression equal to zero to find the boundary, and remember that the radius of a sphere equation

(x-h)^2 + (y-k)^2 + (z-l)^2 = r^2

r

, not

r^2

Question 16

For the function $k(x,y) = \sqrt{\ln(xy)}$ , what additional constraint must be satisfied for points in the first quadrant where $xy > 1$ ?

No additional constraint is needed beyond $xy > 1$ in the first quadrant (correct answer)
We must also have $x + y > 2$ in addition to $xy > 1$
We must also have $x^2 + y^2 > 1$ in addition to $xy > 1$
We must also have $|x - y| < 1$ in addition to $xy > 1$

Explanation: For k(x,y) = √(ln(xy)) to be defined, we need: (1) xy > 0 (for the logarithm to be defined), and (2) ln(xy) ≥ 0 (for the square root), which means xy ≥ 1. In the first quadrant, xy > 0 is automatically satisfied. The condition xy > 1 ensures ln(xy) > 0 > 0, so the square root is defined. No additional constraints are needed. Choices B, C, and D introduce unnecessary conditions that aren't required for the function to be defined.