Units, Orientation & Signs - Multivariable Calculus

1. 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?
   Answer: $W_1 \neq W_2$ and both are positive. 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.
2. 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 $S$.
   Answer: The flux is negative. 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.
3. 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?
   Answer: Upward-pointing. 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.
4. 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?
   Answer: Mass flow rate through the surface; units: kg/s 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.
5. 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 $I$?
   Answer: Electric potential difference, in Volts ($V$). 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.
6. 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?
   Answer: The orientation must be checked since this gives the inward normal for a bowl-shaped surface 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.
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.
   Answer: The total flux is positive. 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.
8. 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?
   Answer: Units should be m³/s; 8.5 m³/s converts to 8500 L/s correctly 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.
9. 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?
   Answer: Both students made errors since the work should be path-independent and equal to the potential difference 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 $(0,0,0)$ to $(1,1,1)$. 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.
10. 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 $\Phi = \oiint_S \vec{F} \cdot d\vec{S}$?
    Answer: The flux is positive. 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.
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 $C$?
    Answer: Counter-clockwise as viewed from above. 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.
12. 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?
    Answer: Units: newton-meters; Sign: the field does negative work, removing energy from the object 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.
13. 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?
    Answer: Heat flows into the region; check that the normal orientation matches the intended physical direction 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.
14. 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?
    Answer: Positive (net flow is outward). 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.
15. 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?
    Answer: The force field, on average, acted against the direction of motion. 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.
16. 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?
    Answer: Circulation around the boundary of the surface; units: m²/s 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.
17. 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?
    Answer: Confused flux magnitude with field strength; $\Phi = 0.05$ Wb indicates moderate flux through the surface 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.
18. 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?
    Answer: $W > 0$ 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.
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 $\Phi$?
    Answer: Watts ($W$), representing the rate of energy transfer. 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.

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