Multivariable Calculus Quiz: Divergence
14 questions · exam conditions
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DivergenceQuestion 1 of 14

A physical system is described by a scalar potential ϕ(x,y,z)=ln⁡(x2+y2+z2)\phi(x,y,z) = \ln(x^2+y^2+z^2). An associated vector field E\mathbf{E} is given by E=−∇ϕ\mathbf{E} = -\nabla \phi. Compute the divergence of E\mathbf{E}.

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2x2+y2+z2\frac{2}{x^2+y^2+z^2}
−2x2+y2+z2-\frac{2}{x^2+y^2+z^2}
−2(x2+y2+z2)2-\frac{2}{(x^2+y^2+z^2)^2}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Divergence

Practice Divergence 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 Divergence, 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.

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Question 1

A physical system is described by a scalar potential ϕ(x,y,z)=ln⁡(x2+y2+z2)\phi(x,y,z) = \ln(x^2+y^2+z^2). An associated vector field E\mathbf{E} is given by E=−∇ϕ\mathbf{E} = -\nabla \phi. Compute the divergence of E\mathbf{E}.

  1. 00
  2. 2x2+y2+z2\frac{2}{x^2+y^2+z^2}
  3. −2x2+y2+z2-\frac{2}{x^2+y^2+z^2} (correct answer)
  4. −2(x2+y2+z2)2-\frac{2}{(x^2+y^2+z^2)^2}
Explanation: We need to compute div(E)=div(−∇ϕ)=−div(∇ϕ)\mathrm{div}(\mathbf{E}) = \mathrm{div}(-\nabla \phi) = -\mathrm{div}(\nabla \phi). The expression div(∇ϕ)\mathrm{div}(\nabla \phi) is the Laplacian of ϕ\phi, denoted ∇2ϕ\nabla^2 \phi. Let r2=x2+y2+z2r^2 = x^2+y^2+z^2. So ϕ=ln⁡(r2)=2ln⁡(r)\phi = \ln(r^2) = 2\ln(r). First, we find the gradient of ϕ\phi: ∇ϕ=⟨2xx2+y2+z2,2yx2+y2+z2,2zx2+y2+z2⟩\nabla \phi = \langle \frac{2x}{x^2+y^2+z^2}, \frac{2y}{x^2+y^2+z^2}, \frac{2z}{x^2+y^2+z^2} \rangle. So, E=⟨−2xr2,−2yr2,−2zr2⟩\mathbf{E} = \langle -\frac{2x}{r^2}, -\frac{2y}{r^2}, -\frac{2z}{r^2} \rangle. Now, we compute the divergence of E\mathbf{E}. We need to find ∂∂x(−2xr2)+∂∂y(−2yr2)+∂∂z(−2zr2)\frac{\partial}{\partial x}(-\frac{2x}{r^2}) + \frac{\partial}{\partial y}(-\frac{2y}{r^2}) + \frac{\partial}{\partial z}(-\frac{2z}{r^2}). Let's compute the first term using the quotient rule: ∂∂x(−2xx2+y2+z2)=−2(x2+y2+z2)−2x(2x)(x2+y2+z2)2=−2x2+2y2+2z2−4x2(r2)2=−−2x2+2y2+2z2r4\frac{\partial}{\partial x}(-\frac{2x}{x^2+y^2+z^2}) = -\frac{2(x^2+y^2+z^2) - 2x(2x)}{(x^2+y^2+z^2)^2} = -\frac{2x^2+2y^2+2z^2-4x^2}{(r^2)^2} = -\frac{-2x^2+2y^2+2z^2}{r^4}. By symmetry, the other two terms are ∂Ey∂y=−2x2−2y2+2z2r4\frac{\partial E_y}{\partial y} = -\frac{2x^2-2y^2+2z^2}{r^4} and ∂Ez∂z=−2x2+2y2−2z2r4\frac{\partial E_z}{\partial z} = -\frac{2x^2+2y^2-2z^2}{r^4}. Summing these three terms: div(E)=−1r4[(−2x2+2y2+2z2)+(2x2−2y2+2z2)+(2x2+2y2−2z2)]=−1r4[2x2+2y2+2z2]=−2r2r4=−2r2=−2x2+y2+z2\mathrm{div}(\mathbf{E}) = -\frac{1}{r^4}[(-2x^2+2y^2+2z^2) + (2x^2-2y^2+2z^2) + (2x^2+2y^2-2z^2)] = -\frac{1}{r^4}[2x^2+2y^2+2z^2] = -\frac{2r^2}{r^4} = -\frac{2}{r^2} = -\frac{2}{x^2+y^2+z^2}.

Question 2

Let r=⟨x,y,z⟩\mathbf{r} = \langle x, y, z \rangle. The gravitational field of a point mass at the origin is proportional to the vector field F=−r∥r∥3\mathbf{F} = -\frac{\mathbf{r}}{\|\mathbf{r}\|^3}. Let G=⟨x2,y2,z2⟩\mathbf{G} = \langle x^2, y^2, z^2 \rangle. What is the divergence of the field H=F+G\mathbf{H} = \mathbf{F} + \mathbf{G} at the point (1,2,−1)(1, 2, -1)?

  1. 0
  2. 4 (correct answer)
  3. 6
  4. 7
Explanation: We use the linearity property of the divergence operator: div(H)=div(F+G)=div(F)+div(G)\mathrm{div}(\mathbf{H}) = \mathrm{div}(\mathbf{F} + \mathbf{G}) = \mathrm{div}(\mathbf{F}) + \mathrm{div}(\mathbf{G}). First, consider the field F=−r∥r∥3\mathbf{F} = -\frac{\mathbf{r}}{\|\mathbf{r}\|^3}. This is a standard field in physics, representing the inverse-square law for gravity or electrostatics. A key property of this field is that its divergence is zero everywhere except at the origin. So, for any point (x,y,z)≠(0,0,0)(x,y,z) \neq (0,0,0), div(F)=0\mathrm{div}(\mathbf{F}) = 0. Since we are evaluating at (1,2,−1)(1, 2, -1), which is not the origin, div(F)=0\mathrm{div}(\mathbf{F})=0. Next, we compute the divergence of G=⟨x2,y2,z2⟩\mathbf{G} = \langle x^2, y^2, z^2 \rangle. div(G)=∂∂x(x2)+∂∂y(y2)+∂∂z(z2)=2x+2y+2z\mathrm{div}(\mathbf{G}) = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(z^2) = 2x + 2y + 2z. Now, we can find the divergence of H\mathbf{H}: div(H)=0+(2x+2y+2z)=2x+2y+2z\mathrm{div}(\mathbf{H}) = 0 + (2x + 2y + 2z) = 2x + 2y + 2z. Finally, we evaluate this expression at the point (1,2,−1)(1, 2, -1): div(H)(1,2,−1)=2(1)+2(2)+2(−1)=2+4−2=4\mathrm{div}(\mathbf{H})(1,2,-1) = 2(1) + 2(2) + 2(-1) = 2 + 4 - 2 = 4.

Question 3

The velocity of a fluid at position (x,y,z)(x,y,z) is described by the vector field v(x,y,z)=⟨x3,y3,−6xyz⟩\mathbf{v}(x,y,z) = \langle x^3, y^3, -6xyz \rangle. The divergence of the velocity field at a point gives the rate of expansion (source) or compression (sink) of the fluid per unit volume at that point.

At the point P=(3,1,1)P=(3, 1, 1), which of the following best describes the fluid flow?

  1. The point is a sink, with a fluid compression rate of 12.
  2. The point is a source, with a fluid expansion rate of 4.
  3. The flow is incompressible at this point.
  4. The point is a source, with a fluid expansion rate of 12. (correct answer)
Explanation: First, we compute the divergence of the velocity field v\mathbf{v}. Let P=x3,Q=y3,R=−6xyzP=x^3, Q=y^3, R=-6xyz. Then div(v)=∂P∂x+∂Q∂y+∂R∂z=3x2+3y2−6xy\mathrm{div}(\mathbf{v}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 3x^2 + 3y^2 - 6xy. This can be factored as 3(x2−2xy+y2)=3(x−y)23(x^2 - 2xy + y^2) = 3(x-y)^2. Next, we evaluate the divergence at the point P=(3,1,1)P=(3, 1, 1): div(v)(3,1,1)=3(3−1)2=3(22)=12\mathrm{div}(\mathbf{v})(3,1,1) = 3(3-1)^2 = 3(2^2) = 12. Since the divergence is positive (12 > 0), the point is a source, meaning there is a net outflow of fluid. The value of the divergence, 12, is the rate of expansion. Thus, the point is a source with a fluid expansion rate of 12.

Question 4

Let F(x,y,z)=⟨x3+sin⁡(z),y3+cos⁡(z),R(x,y,z)⟩\mathbf{F}(x,y,z) = \langle x^3 + \sin(z), y^3 + \cos(z), R(x,y,z) \rangle. If the vector field F\mathbf{F} is incompressible (i.e., has zero divergence everywhere), which of the following could be the component R(x,y,z)R(x,y,z)?

  1. −3z(x2+y2)+xy-3z(x^2+y^2) + xy (correct answer)
  2. −3(x2+y2)-3(x^2+y^2)
  3. z2−3(x2+y2)zz^2 - 3(x^2+y^2)z
  4. cos⁡(z)−sin⁡(z)\cos(z) - \sin(z)
Explanation: For a vector field F=⟨P,Q,R⟩\mathbf{F} = \langle P, Q, R \rangle to be incompressible, its divergence must be zero. div(F)=∂P∂x+∂Q∂y+∂R∂z=0\mathrm{div}(\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = 0. Given P=x3+sin⁡(z)P = x^3 + \sin(z) and Q=y3+cos⁡(z)Q = y^3 + \cos(z), we compute their partial derivatives: ∂P∂x=3x2\frac{\partial P}{\partial x} = 3x^2 ∂Q∂y=3y2\frac{\partial Q}{\partial y} = 3y^2 The divergence equation becomes 3x2+3y2+∂R∂z=03x^2 + 3y^2 + \frac{\partial R}{\partial z} = 0. This implies that the partial derivative of RR with respect to zz must be ∂R∂z=−3x2−3y2=−3(x2+y2)\frac{\partial R}{\partial z} = -3x^2 - 3y^2 = -3(x^2+y^2). To find R(x,y,z)R(x,y,z), we integrate this expression with respect to zz: R(x,y,z)=∫−3(x2+y2) dz=−3(x2+y2)z+g(x,y)R(x,y,z) = \int -3(x^2+y^2) \, dz = -3(x^2+y^2)z + g(x,y), where g(x,y)g(x,y) is an arbitrary function of xx and yy (the 'constant' of integration with respect to zz). We now check the answer choices to see which one has this form. A) R(x,y,z)=−3z(x2+y2)+xyR(x,y,z) = -3z(x^2+y^2) + xy. Here, g(x,y)=xyg(x,y) = xy. This matches the required form. B) For R=−3(x2+y2)R = -3(x^2+y^2), ∂R∂z=0\frac{\partial R}{\partial z} = 0, which is incorrect. C) For R=z2−3(x2+y2)zR = z^2 - 3(x^2+y^2)z, ∂R∂z=2z−3(x2+y2)\frac{\partial R}{\partial z} = 2z - 3(x^2+y^2), which is incorrect. D) For R=cos⁡(z)−sin⁡(z)R = \cos(z) - \sin(z), ∂R∂z=−sin⁡(z)−cos⁡(z)\frac{\partial R}{\partial z} = -\sin(z) - \cos(z), which is incorrect.

Question 5

A long, thin wire runs along the zz-axis. In a steady state, the wire is heated, causing heat to flow radially outward from the wire into the surrounding medium. Let H(x,y,z)\mathbf{H}(x,y,z) be the heat flux vector field, representing the rate and direction of heat flow per unit area.

Consider a point P=(a,b,c)P=(a,b,c) where a2+b2>0a^2+b^2 > 0. Which statement best describes the divergence of the heat flux field, div(H)\mathrm{div}(\mathbf{H}), at point PP?

  1. div(H)>0\mathrm{div}(\mathbf{H}) > 0 because heat is flowing away from the source.
  2. div(H)<0\mathrm{div}(\mathbf{H}) < 0 because heat is being absorbed by the medium.
  3. div(H)=0\mathrm{div}(\mathbf{H}) = 0 because there are no heat sources or sinks at point PP. (correct answer)
  4. div(H)\mathrm{div}(\mathbf{H}) is inversely proportional to the distance from the zz-axis.
Explanation: The divergence of a vector field measures the net outflow or 'source strength' at a point. In this physical scenario, the heat is generated only along the zz-axis (the wire). The point PP is specified to be not on the zz-axis (since a2+b2>0a^2+b^2 > 0). In a steady state, for any small volume around a point PP that does not contain a source or sink of heat, the amount of heat flowing into the volume must exactly equal the amount of heat flowing out. This is a statement of conservation of energy. Therefore, the net outflow is zero, which means the divergence of the heat flux field H\mathbf{H} is zero at any point PP not on the source line. The other options are incorrect: A is wrong because while heat flows away from the zz-axis, PP itself is not a source. B is physically incorrect. D describes the magnitude of the flux field itself (∣H∣|\mathbf{H}|), not its divergence.

Question 6

Consider the vector field G(x,y,z)=(xyz,x2z+y,xz2+y2)\mathbf{G}(x,y,z) = (xyz, x^2z + y, xz^2 + y^2). If SS is the sphere x2+y2+z2=4x^2 + y^2 + z^2 = 4 and we want to find where div(G)\text{div}(\mathbf{G}) achieves its maximum value on SS, which condition must be satisfied at such a point?

  1. yz+1+x=λ(2x)yz + 1 + x = \lambda(2x) and xz+1+2y=λ(2y)xz + 1 + 2y = \lambda(2y) and xy+x2+2z=λ(2z)xy + x^2 + 2z = \lambda(2z)
  2. yz+2xz+x2=4yz + 2xz + x^2 = 4 and x2+y2+z2=4x^2 + y^2 + z^2 = 4
  3. ∇(div(G))=λ∇(x2+y2+z2−4)\nabla(\text{div}(\mathbf{G})) = \lambda \nabla(x^2 + y^2 + z^2 - 4) for some scalar λ\lambda (correct answer)
  4. div(G)=curl(G)⋅n\text{div}(\mathbf{G}) = \text{curl}(\mathbf{G}) \cdot \mathbf{n} where n\mathbf{n} is the unit normal to SS
Explanation: To find the maximum of div(G)\text{div}(\mathbf{G}) on the constraint surface SS, we use Lagrange multipliers. First, div(G)=yz+1+x2\text{div}(\mathbf{G}) = yz + 1 + x^2. To maximize this subject to x2+y2+z2=4x^2 + y^2 + z^2 = 4, we need ∇(yz+1+x2)=λ∇(x2+y2+z2−4)\nabla(yz + 1 + x^2) = \lambda \nabla(x^2 + y^2 + z^2 - 4), which is equivalent to ∇(div(G))=λ∇(x2+y2+z2−4)\nabla(\text{div}(\mathbf{G})) = \lambda \nabla(x^2 + y^2 + z^2 - 4). Choice A incorrectly applies Lagrange multipliers to the vector field components rather than to the divergence. Choice B gives constraint equations but not optimization conditions. Choice D incorrectly relates divergence to curl and normal vector.

Question 7

The vector field H(x,y,z)=(xln⁡(y2+1),yexz,zsin⁡(xy))\mathbf{H}(x,y,z) = (x\ln(y^2 + 1), ye^{xz}, z\sin(xy)) represents a three-dimensional flow. At the point (0,1,π)(0, 1, \pi), what is the instantaneous rate of volume change per unit volume?

  1. ln⁡(2)+e0+sin⁡(0)\ln(2) + e^0 + \sin(0)
  2. ln⁡(2)+1⋅e0+0\ln(2) + 1 \cdot e^0 + 0
  3. ln⁡(2)+π⋅1+sin⁡(0)\ln(2) + \pi \cdot 1 + \sin(0)
  4. ln⁡(2)+e0⋅π+sin⁡(0⋅1)\ln(2) + e^{0 \cdot \pi} + \sin(0 \cdot 1) (correct answer)
Explanation: The instantaneous rate of volume change per unit volume is the divergence. Computing: div(H)=∂∂x(xln⁡(y2+1))+∂∂y(yexz)+∂∂z(zsin⁡(xy))\text{div}(\mathbf{H}) = \frac{\partial}{\partial x}(x\ln(y^2 + 1)) + \frac{\partial}{\partial y}(ye^{xz}) + \frac{\partial}{\partial z}(z\sin(xy)). The partial derivatives are: ∂∂x(xln⁡(y2+1))=ln⁡(y2+1)\frac{\partial}{\partial x}(x\ln(y^2 + 1)) = \ln(y^2 + 1), ∂∂y(yexz)=exz\frac{\partial}{\partial y}(ye^{xz}) = e^{xz}, and ∂∂z(zsin⁡(xy))=sin⁡(xy)\frac{\partial}{\partial z}(z\sin(xy)) = \sin(xy). At (0,1,π)(0,1,\pi): ln⁡(12+1)+e0⋅π+sin⁡(0⋅1)=ln⁡(2)+e0+sin⁡(0)=ln⁡(2)+1+0\ln(1^2 + 1) + e^{0 \cdot \pi} + \sin(0 \cdot 1) = \ln(2) + e^0 + \sin(0) = \ln(2) + 1 + 0. Choice D correctly shows this as ln⁡(2)+e0⋅π+sin⁡(0⋅1)\ln(2) + e^{0 \cdot \pi} + \sin(0 \cdot 1). The other choices have computational errors in evaluating the partial derivatives at the given point.

Question 8

Let F(x,y,z)=(f(x,y,z),g(x,y,z),h(x,y,z))\mathbf{F}(x,y,z) = (f(x,y,z), g(x,y,z), h(x,y,z)) where ff, gg, and hh are twice continuously differentiable. If div(F)=x2+y2+z2\text{div}(\mathbf{F}) = x^2 + y^2 + z^2 and div(curl(F))=6\text{div}(\text{curl}(\mathbf{F})) = 6, what can be concluded?

  1. The given information is consistent and F\mathbf{F} exists as described
  2. The given information is inconsistent because div(curl(F))≠0\text{div}(\text{curl}(\mathbf{F})) \neq 0 (correct answer)
  3. The given information is inconsistent because div(F)\text{div}(\mathbf{F}) cannot equal x2+y2+z2x^2 + y^2 + z^2
  4. More information is needed to determine consistency
Explanation: A fundamental theorem in vector calculus states that the divergence of any curl is always zero: div(curl(F))=0\text{div}(\text{curl}(\mathbf{F})) = 0 for any twice continuously differentiable vector field F\mathbf{F}. This follows from the fact that ∇⋅(∇×F)=0\nabla \cdot (\nabla \times \mathbf{F}) = 0 due to the symmetry of mixed partial derivatives. Since we're told div(curl(F))=6≠0\text{div}(\text{curl}(\mathbf{F})) = 6 \neq 0, this contradicts this fundamental theorem, making the given information mathematically inconsistent. The value of div(F)=x2+y2+z2\text{div}(\mathbf{F}) = x^2 + y^2 + z^2 is perfectly achievable (for instance, by F=(x3/3,y3/3,z3/3)+curl-free terms\mathbf{F} = (x^3/3, y^3/3, z^3/3) + \text{curl-free terms}), so the inconsistency lies solely in the non-zero divergence of curl.

Question 9

A fluid velocity field is given by v(x,y,z)=(x2−y2,2xy+z,x−z2)\mathbf{v}(x,y,z) = (x^2 - y^2, 2xy + z, x - z^2). If the divergence represents the rate of volume expansion per unit volume, at which point does the fluid exhibit the greatest compression (most negative divergence)?

  1. (0,0,1)(0, 0, 1) (correct answer)
  2. (1,0,1)(1, 0, 1)
  3. (0,1,1)(0, 1, 1)
  4. (1,1,0)(1, 1, 0)
Explanation: First, compute the divergence: div(v)=∂∂x(x2−y2)+∂∂y(2xy+z)+∂∂z(x−z2)=2x+2x+(−2z)=4x−2z\text{div}(\mathbf{v}) = \frac{\partial}{\partial x}(x^2 - y^2) + \frac{\partial}{\partial y}(2xy + z) + \frac{\partial}{\partial z}(x - z^2) = 2x + 2x + (-2z) = 4x - 2z. Evaluating at each point: At (0,0,1)(0,0,1): 4(0)−2(1)=−24(0) - 2(1) = -2. At (1,0,1)(1,0,1): 4(1)−2(1)=24(1) - 2(1) = 2. At (0,1,1)(0,1,1): 4(0)−2(1)=−24(0) - 2(1) = -2. At (1,1,0)(1,1,0): 4(1)−2(0)=44(1) - 2(0) = 4. The most negative divergence is −2-2, which occurs at both (0,0,1)(0,0,1) and (0,1,1)(0,1,1). Since only (0,0,1)(0,0,1) is listed as an option, it's the answer.

Question 10

Let f(x,y,z)=ex+yf(x,y,z) = e^{x+y} be a scalar function and F(x,y,z)=⟨y,−x,z⟩\mathbf{F}(x,y,z) = \langle y, -x, z \rangle be a vector field. What is the divergence of the vector field fFf\mathbf{F}?

  1. ex+y(1+y−x)e^{x+y}(1+y-x) (correct answer)
  2. ex+y(y−x)e^{x+y}(y-x)
  3. ex+ye^{x+y}
  4. 00
Explanation: We use the product rule for the divergence of a scalar function ff times a vector field F\mathbf{F}: div(fF)=(∇f)⋅F+fdiv(F)\mathrm{div}(f\mathbf{F}) = (\nabla f) \cdot \mathbf{F} + f \mathrm{div}(\mathbf{F}). First, we compute the required components:
  1. The gradient of ff: ∇f=⟨∂f∂x,∂f∂y,∂f∂z⟩=⟨ex+y,ex+y,0⟩\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle = \langle e^{x+y}, e^{x+y}, 0 \rangle.
  2. The divergence of F\mathbf{F}: div(F)=∂∂x(y)+∂∂y(−x)+∂∂z(z)=0+0+1=1\mathrm{div}(\mathbf{F}) = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(-x) + \frac{\partial}{\partial z}(z) = 0 + 0 + 1 = 1. Now, we apply the product rule formula: (∇f)⋅F=⟨ex+y,ex+y,0⟩⋅⟨y,−x,z⟩=yex+y−xex+y=(y−x)ex+y(\nabla f) \cdot \mathbf{F} = \langle e^{x+y}, e^{x+y}, 0 \rangle \cdot \langle y, -x, z \rangle = y e^{x+y} - x e^{x+y} = (y-x)e^{x+y}. fdiv(F)=(ex+y)(1)=ex+yf \mathrm{div}(\mathbf{F}) = (e^{x+y})(1) = e^{x+y}. Adding these two parts gives: div(fF)=(y−x)ex+y+ex+y=(y−x+1)ex+y\mathrm{div}(f\mathbf{F}) = (y-x)e^{x+y} + e^{x+y} = (y-x+1)e^{x+y}. Distractor B represents only the (∇f)⋅F(\nabla f) \cdot \mathbf{F} term. Distractor C represents only the fdiv(F)f \mathrm{div}(\mathbf{F}) term. Distractor D might result from calculating div(F)=0\mathrm{div}(\mathbf{F})=0 incorrectly.

Question 11

Let F\mathbf{F} be a smooth vector field on R3\mathbb{R}^3. If div(F)=0\mathrm{div}(\mathbf{F})=0 everywhere, which of the following statements must be true?

  1. F\mathbf{F} is a constant vector field.
  2. The curl of F\mathbf{F} must be the zero vector everywhere.
  3. F\mathbf{F} must be the gradient of some scalar function ϕ\phi.
  4. The vector field xFx\mathbf{F} is not necessarily divergence-free. (correct answer)
Explanation: Let's analyze each statement. A) False. div(F)=0\mathrm{div}(\mathbf{F})=0 does not imply F\mathbf{F} is constant. For example, F=⟨y,−x,0⟩\mathbf{F} = \langle y, -x, 0 \rangle has div(F)=0+0+0=0\mathrm{div}(\mathbf{F}) = 0+0+0=0, but it is not constant. B) False. Divergence and curl are independent conditions. The same example F=⟨y,−x,0⟩\mathbf{F} = \langle y, -x, 0 \rangle has div(F)=0\mathrm{div}(\mathbf{F})=0 but curl(F)=⟨0,0,−2⟩≠0\mathrm{curl}(\mathbf{F}) = \langle 0,0,-2 \rangle \neq \mathbf{0}. C) False. A vector field that is the gradient of a scalar function is called conservative and satisfies curl(F)=0\mathrm{curl}(\mathbf{F})=0, not div(F)=0\mathrm{div}(\mathbf{F})=0. D) True. Using the product rule: div(xF)=(∇x)⋅F+xdiv(F)=⟨1,0,0⟩⋅F+x(0)=F1\mathrm{div}(x\mathbf{F}) = (\nabla x) \cdot \mathbf{F} + x \mathrm{div}(\mathbf{F}) = \langle 1,0,0 \rangle \cdot \mathbf{F} + x(0) = F_1. For xFx\mathbf{F} to be divergence-free, we would need F1=0F_1=0 everywhere, which is not required by div(F)=0\mathrm{div}(\mathbf{F})=0.

Question 12

Let F(x,y,z)=⟨x2y,y2z,−xyz⟩\mathbf{F}(x,y,z) = \langle x^2y, y^2z, -xyz \rangle be a vector field. The set of all points (x,y,z)(x,y,z) where the field is incompressible (i.e., divergence-free) consists of:

  1. the xyxy-plane and the yzyz-plane.
  2. the xzxz-plane and the plane defined by x+2z=0x+2z=0. (correct answer)
  3. only the origin (0,0,0)(0,0,0).
  4. the line defined by the intersection of y=0y=0 and x+2z=0x+2z=0.
Explanation: The divergence of a vector field F=⟨P,Q,R⟩\mathbf{F} = \langle P, Q, R \rangle is given by div(F)=∂P∂x+∂Q∂y+∂R∂z\mathrm{div}(\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}. For the given field F(x,y,z)=⟨x2y,y2z,−xyz⟩\mathbf{F}(x,y,z) = \langle x^2y, y^2z, -xyz \rangle, we have P=x2yP=x^2y, Q=y2zQ=y^2z, and R=−xyzR=-xyz. The partial derivatives are ∂P∂x=2xy\frac{\partial P}{\partial x} = 2xy, ∂Q∂y=2yz\frac{\partial Q}{\partial y} = 2yz, and ∂R∂z=−xy\frac{\partial R}{\partial z} = -xy. The divergence is div(F)=2xy+2yz−xy=xy+2yz\mathrm{div}(\mathbf{F}) = 2xy + 2yz - xy = xy + 2yz. The field is incompressible where its divergence is zero. So we set div(F)=0\mathrm{div}(\mathbf{F}) = 0: xy+2yz=0xy + 2yz = 0. Factoring out yy, we get y(x+2z)=0y(x+2z)=0. This equation holds if y=0y=0 or if x+2z=0x+2z=0. The equation y=0y=0 describes the xzxz-plane. The equation x+2z=0x+2z=0 describes another plane. The set of points is the union of these two planes.

Question 13

For the two-dimensional vector field F(x,y)=⟨xy2,−x2y⟩\mathbf{F}(x,y) = \langle xy^2, -x^2y \rangle, identify the region in the xyxy-plane where the flow is expanding.

  1. The region where ∣x∣>∣y∣|x| > |y|.
  2. The region where ∣y∣>∣x∣|y| > |x|. (correct answer)
  3. The first and third quadrants.
  4. The entire xyxy-plane except for the axes.
Explanation: A flow is expanding where its divergence is positive. For a 2D vector field F(x,y)=⟨P(x,y),Q(x,y)⟩\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle, the divergence is div(F)=∂P∂x+∂Q∂y\mathrm{div}(\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}. Here, P(x,y)=xy2P(x,y) = xy^2 and Q(x,y)=−x2yQ(x,y) = -x^2y. Their partial derivatives are ∂P∂x=y2\frac{\partial P}{\partial x} = y^2 and ∂Q∂y=−x2\frac{\partial Q}{\partial y} = -x^2. So, div(F)=y2−x2\mathrm{div}(\mathbf{F}) = y^2 - x^2. The flow is expanding where div(F)>0\mathrm{div}(\mathbf{F}) > 0. This gives the inequality y2−x2>0y^2 - x^2 > 0, which simplifies to y2>x2y^2 > x^2. Taking the square root of both sides gives y2>x2\sqrt{y^2} > \sqrt{x^2}, which is equivalent to ∣y∣>∣x∣|y| > |x|. This inequality describes the regions between the lines y=xy=x and y=−xy=-x that contain the yy-axis.

Question 14

Consider the vector field G(x,y,z)=⟨xexy−xexz,yeyz−yexy,zexz−zeyz⟩\mathbf{G}(x,y,z) = \langle xe^{xy} - xe^{xz}, ye^{yz} - ye^{xy}, ze^{xz} - ze^{yz} \rangle. What is the divergence of G\mathbf{G}?

  1. exy+eyz+exze^{xy} + e^{yz} + e^{xz}
  2. exy(1+xy)−exz(1+xz)e^{xy}(1+xy) - e^{xz}(1+xz)
  3. 0 (correct answer)
  4. 3−(exy+eyz+exz)3 - (e^{xy} + e^{yz} + e^{xz})
Explanation: This question can be solved by recognizing a key vector identity or by direct computation. The identity is that for any twice continuously differentiable vector field F\mathbf{F}, the divergence of its curl is always zero: div(curl(F))=0\mathrm{div}(\mathrm{curl}(\mathbf{F})) = 0. Let's see if G\mathbf{G} can be expressed as the curl of some field F\mathbf{F}. Consider F=⟨eyz,exz,exy⟩\mathbf{F} = \langle e^{yz}, e^{xz}, e^{xy} \rangle. Its curl is curl(F)=⟨∂∂y(exy)−∂∂z(exz),∂∂z(eyz)−∂∂x(exy),∂∂x(exz)−∂∂y(eyz)⟩=⟨xexy−xexz,yeyz−yexy,zexz−zeyz⟩\mathrm{curl}(\mathbf{F}) = \langle \frac{\partial}{\partial y}(e^{xy}) - \frac{\partial}{\partial z}(e^{xz}), \frac{\partial}{\partial z}(e^{yz}) - \frac{\partial}{\partial x}(e^{xy}), \frac{\partial}{\partial x}(e^{xz}) - \frac{\partial}{\partial y}(e^{yz}) \rangle = \langle xe^{xy} - xe^{xz}, ye^{yz} - ye^{xy}, ze^{xz} - ze^{yz} \rangle. This is exactly the given field G\mathbf{G}. Since G=curl(F)\mathbf{G} = \mathrm{curl}(\mathbf{F}), its divergence must be 0. Direct computation is also possible but lengthy and prone to error: ∂∂x(xexy−xexz)=(exy+xyexy)−(exz+xzexz)\frac{\partial}{\partial x}(xe^{xy} - xe^{xz}) = (e^{xy}+xye^{xy}) - (e^{xz}+xze^{xz}), ∂∂y(yeyz−yexy)=(eyz+yzeyz)−(exy+xyexy)\frac{\partial}{\partial y}(ye^{yz} - ye^{xy}) = (e^{yz}+yze^{yz}) - (e^{xy}+xye^{xy}), ∂∂z(zexz−zeyz)=(exz+xzexz)−(eyz+yzeyz)\frac{\partial}{\partial z}(ze^{xz} - ze^{yz}) = (e^{xz}+xze^{xz}) - (e^{yz}+yze^{yz}). Summing these three expressions, all terms cancel out, yielding 0.