Multivariable Calculus Quiz: Units Orientation And Signs

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Units Orientation And SignsQuestion 1 of 19

Let $\vec{J}$ represent the heat flux density vector in a material, measured in watts per square meter ( $W/m^2$ ). Let $S$ be a surface within the material, with area measured in square meters. The total rate of heat flow across the surface $S$ is given by the flux integral $\Phi = \iint_S \vec{J} \cdot d\vec{S}$ . What are the physical units of $\Phi$ ?

Joules (

J

), representing total energy.

Watts (

W

), representing the rate of energy transfer.

Watts per square meter (

W/m^2

), representing heat flux density.

Joules per meter (

J/m

), representing thermal force.

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Units Orientation And Signs

Practice Units Orientation And Signs 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 Units Orientation And Signs, 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

Joules ( $J$ ), representing total energy.
Watts ( $W$ ), representing the rate of energy transfer. (correct answer)
Watts per square meter ( $W/m^2$ ), representing heat flux density.
Joules per meter ( $J/m$ ), representing thermal force.

Explanation: The integral calculates the flux of the vector field

\vec{J}

over the surface

S

. The units of the integrand are

W/m^2

, and the units of the differential surface area element

d\vec{S}

are

m^2

. Therefore, the units of the integral are

(W/m^2) \cdot m^2 = W

(watts). A watt is a joule per second (

J/s

), which represents power, or the rate of energy transfer.

Question 2

Let $S$ be the surface of a sphere centered at the origin, oriented with the outward-pointing normal vector. Consider the vector field $\vec{F} = \langle x^3, y^3, z^3 \rangle$ . What can be concluded about the sign of the total flux $\Phi = \oiint_S \vec{F} \cdot d\vec{S}$ ?

The flux is positive. (correct answer)
The flux is negative.
The flux is zero.
The sign of the flux depends on the radius of the sphere.

Explanation: By the Divergence Theorem, the flux is equal to the triple integral of the divergence of

\vec{F}

over the volume

V

enclosed by the sphere:

\Phi = \iiint_V (\nabla \cdot \vec{F}) dV

. The divergence is

\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(z^3) = 3x^2 + 3y^2 + 3z^2

. This expression is non-negative everywhere within the sphere and strictly positive everywhere except at the origin. Therefore, its integral over the volume of the sphere must be positive.

Question 3

Let $\vec{E}$ be an electric field measured in Newtons per Coulomb ( $N/C$ ). A student evaluates the line integral $I = \int_C \vec{E} \cdot d\vec{r}$ along a path $C$ , where distance is measured in meters ( $m$ ). What physical quantity is represented by the absolute value of $I$ ?

Work done on a charge, in Joules ( $J$ ).
Electric potential difference, in Volts ( $V$ ). (correct answer)
Magnitude of the electric force, in Newtons ( $N$ ).
Magnitude of the electric field, in Newtons per Coulomb ( $N/C$ ).

Explanation: The line integral of an electric field

\vec{E}

along a path represents the negative of the electric potential difference (voltage) between the endpoints of the path. The units are

(N/C) \cdot m

. Since a Joule is a Newton-meter (

J = N \cdot m

), the units are

J/C

. A Joule per Coulomb is the definition of a Volt (

V

). Work would be the integral of force,

\int_C (q\vec{E}) \cdot d\vec{r}

, which would have units of Joules.

Question 4

Let $S$ be the portion of the plane $z = 4 - 2x - 4y$ in the first octant, oriented with a downward-pointing normal vector. A fluid flows with a constant velocity $\vec{v} = \langle 1, 1, -2 \rangle$ . Without performing a full calculation, determine the sign of the flux of $\vec{v}$ through $S$ .

The flux is positive.
The flux is zero.
The flux is negative. (correct answer)
The sign cannot be determined without the specific bounds of the surface.

Explanation: When analyzing flux through a surface, you're measuring how much fluid flows through that surface. The key insight is that flux depends on the relationship between the velocity field and the surface's orientation—specifically, the dot product of the velocity vector with the normal vector. The plane

z = 4 - 2x - 4y

can be rewritten as

2x + 4y + z = 4

, giving us a normal vector

\vec{n} = \langle 2, 4, 1 \rangle

. However, since the surface is oriented with a downward-pointing normal, we use

\vec{n} = \langle -2, -4, -1 \rangle

. Now examine the dot product:

\vec{v} \cdot \vec{n} = \langle 1, 1, -2 \rangle \cdot \langle -2, -4, -1 \rangle = (1)(-2) + (1)(-4) + (-2)(-1) = -2 - 4 + 2 = -4

. Since this dot product is negative, the flux is negative. Looking at the wrong answers: (A) suggests positive flux, but our dot product calculation shows the velocity and normal vectors point in generally opposite directions. (B) claims zero flux, which would only occur if the velocity were parallel to the surface (perpendicular to the normal), making the dot product zero—clearly not the case here. (D) suggests we need specific bounds, but the sign of flux depends only on the direction relationship between velocity and normal vectors, which we can determine from the given information. Strategy tip: For flux sign questions, focus on the dot product

\vec{v} \cdot \vec{n}

. Positive means flow in the normal direction, negative means flow opposite to the normal, and zero means flow parallel to the surface.

Question 5

Consider a cube with vertices at $(\pm 1, \pm 1, \pm 1)$ . Let $S$ be the face of the cube on the plane $y = 1$ , oriented with an outward-pointing normal vector. For the vector field $\vec{F} = \langle x, y^2, z \rangle$ , is the net flux through the surface $S$ positive, negative, or zero?

Positive (net flow is outward). (correct answer)
Negative (net flow is inward).
Zero.
Positive on one half of the face and negative on the other, with zero net flux.

Explanation: The face

S

lies on the plane

y=1

. The outward-pointing normal vector for this face is

\vec{n} = \langle 0, 1, 0 \rangle

. The flux is given by the integral

\iint_S \vec{F} \cdot \vec{n} \, dA

. The dot product is

\vec{F} \cdot \vec{n} = \langle x, y^2, z \rangle \cdot \langle 0, 1, 0 \rangle = y^2

. On the surface

S

, the value of

y

is fixed at

1

. Therefore, the integrand is constant:

y^2 = 1^2 = 1

. The flux is

\iint_S 1 \, dA = \text{Area}(S)

. Since the face is a square with side length 2, its area is 4. The flux is 4, which is positive, indicating a net outward flow.

Question 6

A surface $S$ is given by the parameterization $\vec{r}(u, v) = \langle u, v, u^2 + v^2 \rangle$ . The normal vector for calculating flux is determined by the cross product $\vec{N} = \vec{r}_u \times \vec{r}_v$ . What is the orientation of the surface corresponding to this normal vector?

The orientation changes from upward to downward depending on $u$ and $v$ .
Downward-pointing.
Inward-pointing toward the $z$ -axis.
Upward-pointing. (correct answer)

Explanation: When determining surface orientation from a parameterization, you need to examine the direction of the normal vector

\vec{N} = \vec{r}_u \times \vec{r}_v

. The key insight is that the

z

-component of this normal vector tells you whether the surface is oriented upward (positive

z

-component) or downward (negative

z

-component). Let's calculate the normal vector. First, find the partial derivatives:

$\vec{r}_u = \langle 1, 0, 2u \rangle$
$\vec{r}_v = \langle 0, 1, 2v \rangle$

Now compute the cross product:

\vec{N} = \vec{r}_u \times \vec{r}_v = \langle (0)(2v) - (2u)(1), (2u)(0) - (1)(2v), (1)(1) - (0)(0) \rangle = \langle -2u, -2v, 1 \rangle

The crucial observation is that the

z

-component equals 1, which is always positive regardless of the values of

u

and

v

. This means the normal vector consistently points in the positive

z

-direction, giving an upward orientation. Choice A is incorrect because the orientation doesn't change—the

z

-component is constant. Choice B is wrong because a downward orientation would require a negative

z

-component. Choice C misinterprets the geometry; this parameterization describes a paraboloid opening upward, not a surface wrapping around the

z

-axis. Study tip: For parameterized surfaces, always check the

z

-component of

\vec{r}_u \times \vec{r}_v

to determine orientation. If it's consistently positive, the surface is upward-pointing; if consistently negative, it's downward-pointing.

Question 7

Let $S$ be the boundary surface of the cube defined by $0 \le x \le 2$ , $0 \le y \le 2$ , and $0 \le z \le 2$ , with the outward-pointing normal. Consider the vector field $\vec{F} = \langle x-1, y-1, z-1 \rangle$ . Without computing the full surface integral, determine the sign of the total flux of $\vec{F}$ out of the cube.

The sign cannot be determined without explicitly calculating six surface integrals.
The total flux is negative.
The total flux is zero.
The total flux is positive. (correct answer)

Explanation: When you encounter a flux problem like this, look for opportunities to use the Divergence Theorem rather than computing six separate surface integrals. The Divergence Theorem states that the flux through a closed surface equals the triple integral of the divergence over the enclosed volume. For

\vec{F} = \langle x-1, y-1, z-1 \rangle

, the divergence is:

\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x-1) + \frac{\partial}{\partial y}(y-1) + \frac{\partial}{\partial z}(z-1) = 1 + 1 + 1 = 3

Since the divergence is the positive constant 3 throughout the cube, the total flux is:

\iiint_V 3 \, dV = 3 \times \text{(volume of cube)} = 3 \times 8 = 24

The positive flux makes physical sense:

\vec{F}

points away from the center point (1,1,1) of the cube. At every point on the cube's boundary, the vector field has a positive component in the outward normal direction. Answer A is wrong because the Divergence Theorem provides an elegant shortcut that avoids computing six integrals. Answer B is incorrect since we calculated a positive value of 24. Answer C misses that the constant positive divergence creates net outflow. Answer D correctly identifies the positive flux. Strategy tip: When you see flux through a closed surface, immediately check if you can use the Divergence Theorem by computing

\nabla \cdot \vec{F}

. If the divergence is constant (especially positive or negative), you can determine the flux sign without lengthy calculations.

Question 8

The work done by a force field $\vec{F}$ on a particle moving along a path $C$ is calculated to be $W = \int_C \vec{F} \cdot d\vec{r} = -25$ Joules. Which of the following is a correct physical interpretation of this result?

The particle gained 25 J of kinetic energy from the force field.
The path $C$ must have been traversed in a clockwise direction.
The force field, on average, acted against the direction of motion. (correct answer)
The force field $\vec{F}$ is not a conservative field.

Explanation: When you encounter work calculations using line integrals, remember that the sign of the result tells you about the energy transfer between the force field and the particle. Work is defined as

W = \int_C \vec{F} \cdot d\vec{r}

, where the dot product measures how much the force aligns with the direction of motion. Since

W = -25

J (negative), the force field removed energy from the particle. This happens when the force generally opposes the particle's motion. Think of it like friction or gravity acting against a moving object - the dot product

\vec{F} \cdot d\vec{r}

is predominantly negative along the path, meaning the force field acted against the direction of motion on average. This makes (C) correct. (A) is wrong because negative work means the particle lost kinetic energy (25 J), not gained it. The work-energy theorem tells us that work done equals the change in kinetic energy. (B) is incorrect because the sign of work has nothing to do with the geometric direction (clockwise vs. counterclockwise) of the path. The sign depends on the relationship between force and motion directions. (D) is false because you cannot determine whether a field is conservative from a single work calculation. Conservative fields have path-independent work, but this single negative value doesn't tell you anything about path dependence. Study tip: Remember that the sign of work in line integrals is all about energy flow - positive means the field gives energy to the particle, negative means the field takes energy away.

Question 9

A particle moves along a closed, counter-clockwise path $C$ that encloses the origin in the $xy$ -plane. The particle is subject to a force field $\vec{F} = \langle -y, x \rangle$ . Let $W$ be the work done by the field on the particle. Which of the following statements about $W$ is correct?

$W > 0$ (correct answer)
$W < 0$
$W = 0$
The sign of $W$ depends on the specific shape and size of the path $C$ .

Explanation: The work done is

W = \oint_C \vec{F} \cdot d\vec{r}

. The vector field

\vec{F} = \langle -y, x \rangle

represents a rotation counter-clockwise around the origin. Since the path

C

is also counter-clockwise, the force vector

\vec{F}

is everywhere tangent to the path and points in the same direction as the motion. Therefore, the dot product

\vec{F} \cdot d\vec{r}

is always positive, and the integral

W

must be positive. Alternatively, using Green's Theorem,

W = \iint_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA = \iint_R (1 - (-1)) dA = 2 \iint_R dA = 2 \cdot \text{Area}(R)

, which is positive.

Question 10

Let $\vec{F} = \langle 0, x \rangle$ . Let $W_1$ be the work done by $\vec{F}$ along the line segment $C_1$ from $(0,0)$ to $(1,1)$ , and let $W_2$ be the work done by $\vec{F}$ along the parabola $C_2$ given by $y=x^2$ from $(0,0)$ to $(1,1)$ . Which statement is true?

$W_1 = W_2$ and both are positive.
$W_1 > 0$ and $W_2 < 0$ .
$W_1 \neq W_2$ and both are positive. (correct answer)
$W_1 = W_2 = 0$ .

Explanation: When you encounter work problems with vector fields, you need to evaluate line integrals to see if the field is path-independent (conservative) or path-dependent. The key insight is checking whether the vector field is conservative by testing if

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

. For

\vec{F} = \langle 0, x \rangle

, we have

P = 0

and

Q = x

. Since

\frac{\partial P}{\partial y} = 0

but

\frac{\partial Q}{\partial x} = 1

, this field is not conservative, so work depends on the path taken. Let's calculate both work values. For the line segment

C_1

, parametrize as

\vec{r}(t) = \langle t, t \rangle

where

t \in [0,1]

. Then

\vec{r}'(t) = \langle 1, 1 \rangle

and

\vec{F}(\vec{r}(t)) = \langle 0, t \rangle

. So

W_1 = \int_0^1 \langle 0, t \rangle \cdot \langle 1, 1 \rangle \, dt = \int_0^1 t \, dt = \frac{1}{2}

. For the parabola

C_2

, parametrize as

\vec{r}(t) = \langle t, t^2 \rangle

where

t \in [0,1]

. Then

\vec{r}'(t) = \langle 1, 2t \rangle

and

\vec{F}(\vec{r}(t)) = \langle 0, t \rangle

. So

W_2 = \int_0^1 \langle 0, t \rangle \cdot \langle 1, 2t \rangle \, dt = \int_0^1 2t^2 \, dt = \frac{2}{3}

. Since

\frac{1}{2} \neq \frac{2}{3}

and both are positive, answer C is correct. Answer A fails because the values aren't equal. Answer B incorrectly claims

W_2 < 0

. Answer D wrongly assumes both works are zero. Remember: always check if a vector field is conservative first—if not, you must compute line integrals along each specific path.

Question 11

Let $S$ be the portion of the paraboloid $z = 9 - x^2 - y^2$ that lies above the $xy$ -plane, and let $C$ be its boundary curve in the $xy$ -plane. According to Stokes' Theorem, the line integral of a vector field $\vec{F}$ around $C$ is equal to the flux of its curl through $S$ . If the surface integral $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$ is calculated using an upward-pointing normal vector for $S$ , what is the required orientation of the boundary curve $C$ ?

Clockwise as viewed from above.
Counter-clockwise as viewed from above. (correct answer)
The orientation of $C$ does not affect the value of the line integral.
The orientation of $C$ depends on the specific vector field $\vec{F}$ .

Explanation: Stokes' Theorem requires compatible orientations between the surface and its boundary curve, determined by the right-hand rule. If you point the thumb of your right hand in the direction of the surface's normal vector (upward), your fingers curl in the direction of the positive orientation for the boundary curve. For a surface above the

xy

-plane with an upward normal, this corresponds to a counter-clockwise orientation of the boundary curve when viewed from above.

Question 12

Consider flux $\iint_S \vec{F} \cdot \vec{n} \, dS$ where $\vec{F} = (x^2, y^2, z^2)$ represents heat flux in W/m² and $S$ is the boundary of a region $R$ oriented outward. If the computed flux is $-120$ W, what does this indicate, and what should be checked?

Heat flows into the region; check that the normal orientation matches the intended physical direction (correct answer)
Heat flows out of the region; check for computational errors since flux should be positive
The surface area calculation is incorrect; flux density cannot yield negative total flux
Heat flows into the region; check that the divergence of $\vec{F}$ is properly calculated

Explanation: Negative flux with outward normal means the vector field points predominantly inward, so heat flows into the region. This is physically reasonable and the sign should be verified against the intended orientation. Choice B incorrectly assumes negative flux indicates an error. Choice C misunderstands that heat flux can be negative (inward flow). Choice D confuses flux calculation with divergence theorem application.

Question 13

A magnetic field $\vec{B}$ produces flux $\Phi = \iint_S \vec{B} \cdot \vec{n} \, dS$ through surface $S$ . Given that $\vec{B}$ is measured in tesla (T) and the surface area is in m², if $\Phi = 0.05$ , a student concludes there is no net magnetic field through the surface. What error in reasoning occurred?

Assumed zero flux incorrectly; any positive value indicates net outward magnetic field
Misread the units; the flux should be measured in tesla-meters rather than webers
Confused flux magnitude with field strength; $\Phi = 0.05$ Wb indicates moderate flux through the surface (correct answer)
Calculated total field instead of flux; magnetic flux requires integration over the surface area

Explanation: When you encounter magnetic flux problems, remember that flux measures the net flow of magnetic field lines through a surface, not the absence or presence of the field itself. The key insight is understanding what different flux values actually mean. The student's reasoning contains a fundamental misconception about interpreting flux values. A flux of

\Phi = 0.05

webers (Wb) indicates there is a net magnetic field passing through the surface. Flux only equals zero when the magnetic field lines entering and exiting the surface exactly cancel out, or when there's no magnetic field present at all. Since

\Phi = 0.05

Wb is positive, this tells us there's more magnetic field flowing outward through the surface than inward, representing a moderate amount of net flux. Looking at the wrong answers: (A) incorrectly suggests the student assumed zero flux, but the problem states

\Phi = 0.05

, which is clearly non-zero. (B) misses the point entirely—the units are correct (tesla × m² = webers), and this isn't a units problem. (D) doesn't apply since the student isn't confusing integration with direct measurement; they're misinterpreting what the flux value means. The correct answer is (C) because the student confused the magnitude of flux with field strength, failing to recognize that

\Phi = 0.05

Wb represents meaningful magnetic flux through the surface. Study tip: Remember that magnetic flux is like water flow through a net—zero means equal in and out, while any non-zero value indicates net flow in one direction.

Question 14

The work integral $W = \int_C \vec{F} \cdot d\vec{r}$ is computed for $\vec{F} = (3x^2y, x^3 + 2y, 5z^4)$ along curve $C$ from $(0,0,0)$ to $(1,1,1)$ . Two students get different answers: Student A gets $W = 3.25$ and Student B gets $W = 4.75$ . Given that $\vec{F}$ is conservative, what can be concluded?

The students used different paths, but both answers are valid since work is path-independent for conservative fields
Student A is correct since conservative fields always produce positive work along upward paths
Both students made errors since the work should be path-independent and equal to the potential difference (correct answer)
Student B is correct since the field components suggest the work should exceed 4 units

Explanation: When you encounter work integrals with conservative vector fields, the fundamental principle is that work depends only on the starting and ending points, not the path taken. This means all correct calculations between the same endpoints must yield identical results. For a conservative field, work equals the potential difference:

W = \phi(1,1,1) - \phi(0,0,0)

, where

\phi

is the potential function. Since

\vec{F} = (3x^2y, x^3 + 2y, 5z^4)

is given as conservative, there's exactly one correct answer for any path from

(0,0,0)

(1,1,1)

. When two students get different values (3.25 vs 4.75), at least one must be wrong, and possibly both are wrong. Option A incorrectly suggests both answers could be valid. This contradicts the path-independence property of conservative fields—there should be only one correct value regardless of the chosen path. Option B makes an unfounded claim about conservative fields always producing positive work on "upward paths" and arbitrarily declares Student A correct without justification. Option D arbitrarily favors Student B based on a vague intuition about field components, which isn't a valid mathematical argument. The correct conclusion is C: both students likely made computational errors since conservative fields guarantee path-independence, meaning there's a unique correct answer. Study tip: Always verify that a field is truly conservative by checking

\nabla \times \vec{F} = 0

, then use the potential function method rather than line integration to avoid computational errors.

Question 15

Consider the surface integral $\iint_S (\nabla \times \vec{F}) \cdot \vec{n} \, dS$ where $\vec{F}$ represents fluid velocity in m/s, $S$ is a surface in m², and $\vec{n}$ is the unit normal. If this integral equals $-2.4$ , what does this represent and what are the units?

Net vorticity flux through the surface; units: rad/s
Total rotational kinetic energy flux; units: m²/s²
Net angular velocity through the surface; units: s⁻¹
Circulation around the boundary of the surface; units: m²/s (correct answer)

Explanation: When you encounter a surface integral involving the curl of a vector field, you're looking at a fundamental relationship in vector calculus. This particular integral

\iint_S (\nabla \times \vec{F}) \cdot \vec{n} \, dS

is directly connected to Stokes' theorem, which states that the flux of curl through a surface equals the circulation of the original vector field around the surface's boundary. Since

\vec{F}

represents fluid velocity (m/s), this integral measures the circulation around the boundary curve of surface

S

. The curl

\nabla \times \vec{F}

captures the local rotation of the fluid, and integrating its normal component over the surface gives you the total circulation around the perimeter. The units work out to m²/s: velocity (m/s) integrated over area (m²). The answer is D. Here's why the other options miss the mark: A confuses the physical interpretation - while curl relates to vorticity, this integral specifically measures circulation, not vorticity flux, and the units would be s⁻¹, not rad/s. B incorrectly suggests this measures energy flux; kinetic energy involves velocity squared and mass, giving different units entirely. C misunderstands both the concept and units - this isn't measuring angular velocity (which would have units s⁻¹), but rather circulation with units m²/s. Remember this key connection: whenever you see the surface integral of curl dotted with the normal vector, think Stokes' theorem and circulation around the boundary. This relationship appears frequently in fluid dynamics and electromagnetic theory.

Question 16

A velocity field $\vec{v} = (\sin y, x\cos y, e^z)$ in m/s has flux computed through a surface as $\iint_S \vec{v} \cdot \vec{n} \, dS = 8.5$ . A student reports this as "volumetric flow rate = 8.5 L/s". What corrections are needed?

Units should be m³/s, and conversion gives 8500 L/s, not 8.5 L/s
Units are correct but the conversion factor is wrong; 8.5 m³/s equals 8.5 × 10³ L/s
The calculation is dimensionally incorrect; velocity flux has units m²/s, not volume flow
Units should be m³/s; 8.5 m³/s converts to 8500 L/s correctly (correct answer)

Explanation: When you encounter flux calculations through surfaces, you're dealing with the fundamental concept of flow rate through a boundary. The flux integral

\iint_S \vec{v} \cdot \vec{n} \, dS

measures how much "stuff" flows through surface S per unit time. Since

\vec{v}

has units of m/s and

dS

has units of m², the flux integral yields units of m³/s, which is indeed a volumetric flow rate. The calculation

\iint_S \vec{v} \cdot \vec{n} \, dS = 8.5

means 8.5 m³/s of fluid flows through the surface. To convert to liters: since 1 m³ = 1000 L, we have 8.5 m³/s × 1000 L/m³ = 8500 L/s. Looking at the wrong answers: Choice A incorrectly states the units should be m³/s (which is actually correct) but gives the right conversion. Choice B claims the units are correct as reported (L/s) when the student failed to convert from the calculated m³/s, and uses an incorrect conversion factor of 10³ instead of 1000. Choice C makes a fundamental error by claiming velocity flux has units m²/s - this confuses flux (m³/s) with circulation or other line integrals. The student's error was reporting 8.5 L/s instead of recognizing that the flux calculation gives 8.5 m³/s, which converts to 8500 L/s. Study tip: Always check units in flux problems. Velocity field flux through a surface always gives volumetric flow rate (volume/time), and remember the factor of 1000 when converting m³ to liters.

Question 17

A vector field $\vec{F}(x,y,z) = 3xy\vec{i} + (x^2-z^2)\vec{j} + 2yz\vec{k}$ represents a force field measured in newtons when coordinates are in meters. The work done by this field along a curve from point A to point B is calculated as $W = \int_C \vec{F} \cdot d\vec{r}$ . If the computed value is $-15$ , what are the correct units for this work, and what does the negative sign indicate?

Units: newton-meters; Sign: the field does negative work, removing energy from the object (correct answer)
Units: newtons per meter; Sign: the field opposes the direction of motion along the entire path
Units: newton-meters; Sign: the path integral was computed in the wrong direction
Units: newtons; Sign: the net force component is opposite to the displacement direction

Explanation: Work has units of force × distance = newton-meters (joules). The negative sign indicates the field does negative work, meaning energy is removed from the object moving along the path. Choice B has wrong units (force/distance rather than force×distance). Choice C incorrectly assumes the sign always indicates computational direction error. Choice D has wrong units and misinterprets the physical meaning.

Question 18

Consider the flux integral $\iint_S \vec{F} \cdot \vec{n} \, dS$ where $\vec{F} = \rho x\vec{i} + \rho y\vec{j} + \rho z\vec{k}$ represents fluid velocity in m/s, $\rho$ is density in kg/m³, and $S$ is a closed surface oriented outward. If $\rho = 2$ kg/m³ and the computed flux is $48$ , what physical quantity does this represent and what are its units?

Mass flow rate through the surface; units: kg·m²/s
Mass flow rate through the surface; units: kg/s (correct answer)
Volume flow rate through the surface; units: m³/s
Momentum flux through the surface; units: kg·m/s²

Explanation: The vector field

\rho\vec{v}

represents momentum density (mass flux density). The flux integral gives mass flow rate with units (kg/m³)(m/s)(m²) = kg/s. Choice A has wrong units (includes extra m²). Choice C ignores the density factor. Choice D confuses momentum flux with mass flow rate and has wrong units.

Question 19

A surface $S$ is parametrized as $\vec{r}(u,v) = (u\cos v, u\sin v, u^2)$ for $0 \leq u \leq 2$ , $0 \leq v \leq \pi$ . When computing flux $\iint_S \vec{F} \cdot \vec{n} \, dS$ where $\vec{F} = z\vec{k}$ , a student calculates $\vec{n} = \vec{r}_u \times \vec{r}_v = (-2u^2\cos v, -2u^2\sin v, u)$ . What issue must be addressed before proceeding with the flux calculation?

The normal vector magnitude must be computed to ensure proper scaling of the area element
The orientation must be checked since this gives the inward normal for a bowl-shaped surface (correct answer)
The parametrization domain must be extended to $0 \leq v \leq 2\pi$ for a complete surface
The cross product order must be reversed because $u$ should increase before $v$ in the parametrization

Explanation: The surface is a paraboloid bowl opening upward. The computed normal

\vec{n}

points inward (negative

z

-component dominates below the surface). For most physical applications, we need to specify whether we want inward or outward orientation. Choice A is incorrect since

\vec{n}

is already the correct vector for flux (includes magnitude). Choice C is wrong as

v \in [0,\pi]

gives a half-bowl, which may be intended. Choice D misunderstands that cross product order affects orientation, not parametrization validity.