Multivariable Calculus Quiz: Divergence Theorem
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Divergence TheoremQuestion 1 of 14

Let SS be the portion of the paraboloid z=4−x2−y2z = 4 - x^2 - y^2 for z≥0z \ge 0, with upward orientation. Evaluate the flux integral ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S} for the vector field F(x,y,z)=⟨xz,yz,x2+y2⟩\mathbf{F}(x,y,z) = \langle xz, yz, x^2 + y^2 \rangle.

88π/388\pi/3
64π/364\pi/3
40π/340\pi/3
−8π-8\pi
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Divergence Theorem

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

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

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Question 1

Let SS be the portion of the paraboloid z=4−x2−y2z = 4 - x^2 - y^2 for z≥0z \ge 0, with upward orientation. Evaluate the flux integral ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S} for the vector field F(x,y,z)=⟨xz,yz,x2+y2⟩\mathbf{F}(x,y,z) = \langle xz, yz, x^2 + y^2 \rangle.

  1. 88π/388\pi/3 (correct answer)
  2. 64π/364\pi/3
  3. 40π/340\pi/3
  4. −8π-8\pi
Explanation: The surface SS is open. To use the Divergence Theorem, we close the surface with the disk ScapS_{cap} defined by x2+y2≤4x^2+y^2 \le 4 at z=0z=0, oriented downward to ensure the total orientation is outward from the enclosed solid EE. The total flux through the closed surface S∪ScapS \cup S_{cap} is given by ∭Ediv(F) dV\iiint_E \text{div}(\mathbf{F}) \, dV. The divergence of F\mathbf{F} is div(F)=∂∂x(xz)+∂∂y(yz)+∂∂z(x2+y2)=z+z+0=2z\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(xz) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(x^2+y^2) = z + z + 0 = 2z. Using cylindrical coordinates, the volume integral is ∫02π∫02∫04−r2(2z)r dz dr dθ=2π∫02r[z2]04−r2 dr=2π∫02r(4−r2)2 dr=64π/3\int_0^{2\pi} \int_0^2 \int_0^{4-r^2} (2z) r \, dz \, dr \, d\theta = 2\pi \int_0^2 r[z^2]_0^{4-r^2} \, dr = 2\pi \int_0^2 r(4-r^2)^2 \, dr = 64\pi/3. This is the total flux. The flux through the cap ScapS_{cap} (where z=0z=0 and the normal is n=⟨0,0,−1⟩\mathbf{n}=\langle 0,0,-1 \rangle) is ∬Scap⟨0,0,x2+y2⟩⋅⟨0,0,−1⟩ dA=∬Scap−(x2+y2) dA=−∫02π∫02r2⋅r dr dθ=−8π\iint_{S_{cap}} \langle 0, 0, x^2+y^2 \rangle \cdot \langle 0,0,-1 \rangle \, dA = \iint_{S_{cap}} -(x^2+y^2) \, dA = -\int_0^{2\pi}\int_0^2 r^2 \cdot r \, dr \, d\theta = -8\pi. The flux through SS is the total flux minus the flux through the cap: 64π/3−(−8π)=88π/364\pi/3 - (-8\pi) = 88\pi/3.

Question 2

Let F\mathbf{F} be a continuously differentiable vector field in R3\mathbb{R}^3. The flux of F\mathbf{F} out of any sphere centered at the origin is found to be proportional to the cube of the sphere's radius RR, i.e., Φ(R)=CR3\Phi(R) = CR^3 for some constant CC. What is the divergence of F\mathbf{F} at the origin, div(F)(0,0,0)\text{div}(\mathbf{F})(0,0,0)?

  1. 00
  2. CC
  3. 3C/(4π)3C/(4\pi) (correct answer)
  4. It cannot be determined from the given information.
Explanation: By the Divergence Theorem, the flux Φ(R)\Phi(R) through a sphere of radius RR is equal to the triple integral of its divergence over the enclosed ball BRB_R. So, ∭BRdiv(F) dV=CR3\iiint_{B_R} \text{div}(\mathbf{F}) \, dV = CR^3. By the Mean Value Theorem for integrals, there exists a point PP in BRB_R such that div(F)(P)⋅Volume(BR)=CR3\text{div}(\mathbf{F})(P) \cdot \text{Volume}(B_R) = CR^3. The volume of the ball is 43πR3\frac{4}{3}\pi R^3. Thus, div(F)(P)⋅43πR3=CR3\text{div}(\mathbf{F})(P) \cdot \frac{4}{3}\pi R^3 = CR^3, which simplifies to div(F)(P)=3C4π\text{div}(\mathbf{F})(P) = \frac{3C}{4\pi}. As we take the limit R→0R \to 0, the point PP must approach the origin. Since F\mathbf{F} is continuously differentiable, its divergence is continuous, so div(F)(0,0,0)=lim⁡R→0div(F)(P)=3C4π\text{div}(\mathbf{F})(0,0,0) = \lim_{R\to 0} \text{div}(\mathbf{F})(P) = \frac{3C}{4\pi}.

Question 3

Let F=⟨ey2sin⁡(z)+3x,x5cos⁡(z)−2y,ln⁡(x2+y2+1)+5z⟩\mathbf{F} = \langle e^{y^2}\sin(z) + 3x, x^5\cos(z) - 2y, \ln(x^2+y^2+1) + 5z \rangle. What is the outward flux of F\mathbf{F} through the surface of the cube with vertices at (±1,±1,±1)(\pm 1, \pm 1, \pm 1)?

  1. 00
  2. 66
  3. 4848 (correct answer)
  4. 144144
Explanation: Directly calculating the flux over the six faces of the cube would be extremely complicated. The Divergence Theorem is the appropriate tool. We first compute the divergence of F\mathbf{F}: div(F)=∂∂x(ey2sin⁡(z)+3x)+∂∂y(x5cos⁡(z)−2y)+∂∂z(ln⁡(x2+y2+1)+5z)=3−2+5=6\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(e^{y^2}\sin(z) + 3x) + \frac{\partial}{\partial y}(x^5\cos(z) - 2y) + \frac{\partial}{\partial z}(\ln(x^2+y^2+1) + 5z) = 3 - 2 + 5 = 6. Since the divergence is a constant, the flux is ∭E6 dV=6⋅Volume(E)\iiint_E 6 \, dV = 6 \cdot \text{Volume}(E). The cube has vertices at (±1,±1,±1)(\pm 1, \pm 1, \pm 1), so its side length is 2. The volume is 23=82^3 = 8. Therefore, the flux is 6×8=486 \times 8 = 48.

Question 4

Let EE be the region between the concentric spheres x2+y2+z2=1x^2+y^2+z^2=1 and x2+y2+z2=4x^2+y^2+z^2=4. Find the total outward flux of the vector field F=r∣r∣3\mathbf{F} = \frac{\mathbf{r}}{|\mathbf{r}|^3} through the boundary of EE, where r=⟨x,y,z⟩\mathbf{r} = \langle x,y,z \rangle.

  1. 00 (correct answer)
  2. 4π4\pi
  3. 8π8\pi
  4. 12π12\pi
Explanation: The vector field is F=⟨x,y,z⟩(x2+y2+z2)3/2\mathbf{F} = \frac{\langle x, y, z \rangle}{(x^2+y^2+z^2)^{3/2}}. The divergence of this field is div(F)=0\text{div}(\mathbf{F}) = 0 for all r≠0\mathbf{r} \neq \mathbf{0}. The region EE is defined by 1≤x2+y2+z2≤41 \le x^2+y^2+z^2 \le 4, which does not contain the origin. Since F\mathbf{F} is continuously differentiable on EE and its boundary, we can apply the Divergence Theorem. The total outward flux is ∭Ediv(F) dV=∭E0 dV=0\iiint_E \text{div}(\mathbf{F}) \, dV = \iiint_E 0 \, dV = 0. Alternatively, one can calculate the flux through the outer sphere (radius 2, outward normal) and the inner sphere (radius 1, inward normal relative to origin). The flux through a sphere of radius RR is 4π4\pi. For the outer sphere, the flux is 4π4\pi. For the inner sphere, the outward normal from the region points towards the origin, yielding a flux of −4π-4\pi. The sum is 4π−4π=04\pi - 4\pi = 0.

Question 5

Let EE be the region defined by 1≤x2+y2+z2≤41 \le x^2+y^2+z^2 \le 4. The outward flux of a radial vector field F=f(ρ)eρ\mathbf{F} = f(\rho)\mathbf{e}_\rho through the boundary of EE is 14π14\pi. Here, ρ=x2+y2+z2\rho = \sqrt{x^2+y^2+z^2} and eρ\mathbf{e}_\rho is the radial unit vector. Which of the following could be the function f(ρ)f(\rho)?

  1. f(ρ)=ρf(\rho) = \rho
  2. f(ρ)=ρ/2f(\rho) = \rho/2 (correct answer)
  3. f(ρ)=1/ρf(\rho) = 1/\rho
  4. f(ρ)=1/ρ3f(\rho) = 1/\rho^3
Explanation: The flux can be found using the Divergence Theorem or by direct integration over the two spherical surfaces. Using direct integration, the flux through a sphere of radius ρ\rho is ∬SF⋅dS=∬Sf(ρ)eρ⋅eρ dA=f(ρ)⋅(Surface Area)=f(ρ)⋅4πρ2\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S f(\rho)\mathbf{e}_\rho \cdot \mathbf{e}_\rho \, dA = f(\rho) \cdot (\text{Surface Area}) = f(\rho) \cdot 4\pi\rho^2. The total outward flux from the region EE is the flux through the outer sphere (radius 2) minus the flux through the inner sphere (radius 1). Flux = f(2)⋅4π(22)−f(1)⋅4π(12)=16πf(2)−4πf(1)=4π(4f(2)−f(1))f(2) \cdot 4\pi(2^2) - f(1) \cdot 4\pi(1^2) = 16\pi f(2) - 4\pi f(1) = 4\pi(4f(2) - f(1)). We are given that the flux is 14π14\pi. So, 4π(4f(2)−f(1))=14π4\pi(4f(2) - f(1)) = 14\pi, which simplifies to 4f(2)−f(1)=14/4=7/24f(2) - f(1) = 14/4 = 7/2. We test the given options: For f(ρ)=ρ/2f(\rho)=\rho/2, we have 4f(2)−f(1)=4(2/2)−(1/2)=4(1)−1/2=7/24f(2) - f(1) = 4(2/2) - (1/2) = 4(1) - 1/2 = 7/2. This matches the condition.

Question 6

Let EE be the solid tetrahedron with vertices at (0,0,0)(0,0,0), (1,0,0)(1,0,0), (0,2,0)(0,2,0), and (0,0,3)(0,0,3). Evaluate the outward flux of the vector field F=⟨x2,2y,z⟩\mathbf{F} = \langle x^2, 2y, z \rangle through the boundary of EE.

  1. 33
  2. 7/27/2 (correct answer)
  3. 11/311/3
  4. 55
Explanation: Using the Divergence Theorem, the flux is ∭Ediv(F) dV\iiint_E \text{div}(\mathbf{F}) \, dV. First, div(F)=∂∂x(x2)+∂∂y(2y)+∂∂z(z)=2x+2+1=2x+3\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial y}(2y) + \frac{\partial}{\partial z}(z) = 2x + 2 + 1 = 2x+3. The flux integral is ∭E(2x+3) dV=∭E2x dV+∭E3 dV\iiint_E (2x+3) \, dV = \iiint_E 2x \, dV + \iiint_E 3 \, dV. The volume of a tetrahedron with a vertex at the origin and other vertices on the axes at (a,0,0),(0,b,0),(0,0,c)(a,0,0), (0,b,0), (0,0,c) is V=abc/6V = abc/6. Here, V=(1)(2)(3)/6=1V = (1)(2)(3)/6 = 1. So, ∭E3 dV=3V=3\iiint_E 3 \, dV = 3V = 3. The integral ∭E2x dV\iiint_E 2x \, dV is 2Vxˉ2V\bar{x}, where xˉ\bar{x} is the x-coordinate of the centroid of the tetrahedron. The centroid is the average of the vertices: xˉ=(0+1+0+0)/4=1/4\bar{x} = (0+1+0+0)/4 = 1/4. Thus, ∭E2x dV=2(1)(1/4)=1/2\iiint_E 2x \, dV = 2(1)(1/4) = 1/2. The total flux is 3+1/2=7/23 + 1/2 = 7/2.

Question 7

Let S1S_1 be the lateral surface of the cone z=x2+y2z = \sqrt{x^2+y^2} for 0≤z≤20 \le z \le 2, oriented upward. Let S2S_2 be the disk x2+y2≤4x^2+y^2 \le 4 in the plane z=2z=2, also oriented upward. For the vector field F=⟨3x−y,3y+z,3z−x⟩\mathbf{F} = \langle 3x-y, 3y+z, 3z-x \rangle, what is the flux ∬S1F⋅dS\iint_{S_1} \mathbf{F} \cdot d\mathbf{S}?

  1. 00 (correct answer)
  2. 8π8\pi
  3. 24π24\pi
  4. −24π-24\pi
Explanation: Let EE be the solid cone bounded by S1S_1 and S2S_2. The boundary of EE with outward orientation is Sout=S2,up∪S1,downS_{out} = S_{2,up} \cup S_{1,down}. By the Divergence Theorem, the total outward flux is ∭Ediv(F) dV\iiint_E \text{div}(\mathbf{F}) \, dV. Here, div(F)=3+3+3=9\text{div}(\mathbf{F}) = 3+3+3=9. The volume of the cone is V=13πr2h=13π(22)(2)=8π/3V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (2^2)(2) = 8\pi/3. The total flux is 9V=9(8π/3)=24π9V = 9(8\pi/3) = 24\pi. So, ∬S2,upF⋅dS+∬S1,downF⋅dS=24π\iint_{S_{2,up}} \mathbf{F} \cdot d\mathbf{S} + \iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 24\pi. Let's calculate the flux through the top disk S2S_2. The normal is n=⟨0,0,1⟩\mathbf{n}=\langle 0,0,1 \rangle. On S2S_2, z=2z=2, so F=⟨3x−y,3y+2,6−x⟩\mathbf{F} = \langle 3x-y, 3y+2, 6-x \rangle. The flux is ∬S2(6−x) dA=6⋅Area(S2)−∬S2x dA=6(π22)−0=24π\iint_{S_2} (6-x) \, dA = 6 \cdot \text{Area}(S_2) - \iint_{S_2} x \, dA = 6(\pi 2^2) - 0 = 24\pi. Substituting this into the Divergence Theorem equation: 24π+∬S1,downF⋅dS=24π24\pi + \iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 24\pi, which implies ∬S1,downF⋅dS=0\iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = 0. The question asks for the flux through S1S_1 oriented upward, which is the opposite orientation: ∬S1,upF⋅dS=−∬S1,downF⋅dS=−0=0\iint_{S_{1,up}} \mathbf{F} \cdot d\mathbf{S} = -\iint_{S_{1,down}} \mathbf{F} \cdot d\mathbf{S} = -0 = 0.

Question 8

Let F\mathbf{F} be a vector field with constant divergence div(F)=k>0\text{div}(\mathbf{F}) = k > 0. Let S1S_1 be the surface of a sphere of radius RR and S2S_2 be the surface of a cube of side length 2R2R, both centered at the origin. Let Φ1\Phi_1 and Φ2\Phi_2 be the outward fluxes through S1S_1 and S2S_2 respectively. Which of the following statements is true?

  1. Φ1>Φ2\Phi_1 > \Phi_2
  2. Φ1<Φ2\Phi_1 < \Phi_2 (correct answer)
  3. Φ1=Φ2\Phi_1 = \Phi_2
  4. The relationship cannot be determined without knowing F\mathbf{F}.
Explanation: According to the Divergence Theorem, the flux of a vector field through a closed surface is the triple integral of its divergence over the enclosed volume. Since the divergence is a constant kk, the flux is Φ=∭Ek dV=k⋅Volume(E)\Phi = \iiint_E k \, dV = k \cdot \text{Volume}(E). For the sphere S1S_1, the enclosed volume is V1=43πR3V_1 = \frac{4}{3}\pi R^3. So, Φ1=k⋅43πR3\Phi_1 = k \cdot \frac{4}{3}\pi R^3. For the cube S2S_2 with side length 2R2R, the enclosed volume is V2=(2R)3=8R3V_2 = (2R)^3 = 8R^3. So, Φ2=k⋅8R3\Phi_2 = k \cdot 8R^3. To compare Φ1\Phi_1 and Φ2\Phi_2, we compare their volumes. We need to compare 43π\frac{4}{3}\pi with 8. Since π≈3.14159\pi \approx 3.14159, 43π≈4.189\frac{4}{3}\pi \approx 4.189. As 4.189<84.189 < 8, we have V1<V2V_1 < V_2. Since k>0k>0, it follows that Φ1<Φ2\Phi_1 < \Phi_2.

Question 9

Let EE be the solid region in the first octant bounded by the coordinate planes, the parabolic cylinder z=1−x2z = 1 - x^2, and the plane y=2y=2. Calculate the outward flux of F=⟨xy,y2,xz⟩\mathbf{F} = \langle xy, y^2, xz \rangle through the boundary of EE.

  1. 11/611/6
  2. 44
  3. 9/29/2 (correct answer)
  4. 11/211/2
Explanation: Using the Divergence Theorem, the flux is the volume integral of the divergence. div(F)=∂∂x(xy)+∂∂y(y2)+∂∂z(xz)=y+2y+x=3y+x\text{div}(\mathbf{F}) = \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(y^2) + \frac{\partial}{\partial z}(xz) = y + 2y + x = 3y+x. The region EE is described by the inequalities 0≤x≤10 \le x \le 1 (since z=1−x2z=1-x^2 must be non-negative), 0≤y≤20 \le y \le 2, and 0≤z≤1−x20 \le z \le 1-x^2. The flux is ∭E(3y+x) dV=∫01∫02∫01−x2(3y+x) dz dy dx\iiint_E (3y+x) \, dV = \int_0^1 \int_0^2 \int_0^{1-x^2} (3y+x) \, dz \, dy \, dx. The innermost integral (with respect to zz) is (3y+x)(1−x2)(3y+x)(1-x^2). The middle integral (with respect to yy) is ∫02(3y+x)(1−x2) dy=(1−x2)[32y2+xy]02=(1−x2)(6+2x)\int_0^2 (3y+x)(1-x^2) \, dy = (1-x^2) [\frac{3}{2}y^2 + xy]_0^2 = (1-x^2)(6+2x). The outermost integral is ∫01(1−x2)(6+2x) dx=∫01(6+2x−6x2−2x3) dx=[6x+x2−2x3−12x4]01=6+1−2−1/2=9/2\int_0^1 (1-x^2)(6+2x) \, dx = \int_0^1 (6+2x-6x^2-2x^3) \, dx = [6x+x^2-2x^3-\frac{1}{2}x^4]_0^1 = 6+1-2-1/2 = 9/2.

Question 10

Let EE be the region bounded by z=x2+y2z = \sqrt{x^2 + y^2} and z=2z = 2, and let F=(yz,xz,xy)\mathbf{F} = (yz, xz, xy). When applying the divergence theorem to find ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S} where SS is the boundary of EE oriented outward, what is the correct setup?

  1. ∭E0 dV=0\iiint_E 0 \, dV = 0 since ∇⋅F=z+z+0=2z≠0\nabla \cdot \mathbf{F} = z + z + 0 = 2z \neq 0
  2. ∭E0 dV=0\iiint_E 0 \, dV = 0 since ∇⋅F=0+0+0=0\nabla \cdot \mathbf{F} = 0 + 0 + 0 = 0 (correct answer)
  3. ∫02π∫02∫r22z⋅r dz dr dθ\int_0^{2\pi}\int_0^2\int_r^2 2z \cdot r \, dz \, dr \, d\theta since ∇⋅F=2z\nabla \cdot \mathbf{F} = 2z
  4. ∫02π∫02∫020⋅r dz dr dθ\int_0^{2\pi}\int_0^2\int_0^2 0 \cdot r \, dz \, dr \, d\theta since ∇⋅F=0\nabla \cdot \mathbf{F} = 0
Explanation: First, compute ∇⋅F=∂∂x(yz)+∂∂y(xz)+∂∂z(xy)=0+0+0=0\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(yz) + \frac{\partial}{\partial y}(xz) + \frac{\partial}{\partial z}(xy) = 0 + 0 + 0 = 0. Since the divergence is zero everywhere, the divergence theorem gives ∬SF⋅dS=∭E0 dV=0\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E 0 \, dV = 0. Choice A incorrectly computes the divergence. Choice C has the wrong divergence and wrong integration limits (should be from x2+y2\sqrt{x^2+y^2} to 22, or rr to 22 in cylindrical). Choice D has the correct divergence but wrong limits.

Question 11

Consider the vector field F(x,y,z)=(x3,y3,z3)\mathbf{F}(x,y,z) = (x^3, y^3, z^3) and the surface SS consisting of the part of the sphere x2+y2+z2=1x^2 + y^2 + z^2 = 1 where z≥0z \geq 0, together with the disk x2+y2≤1,z=0x^2 + y^2 \leq 1, z = 0. If both surfaces are oriented so their normal vectors point away from the origin, what is ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S}?

  1. 3π2\frac{3\pi}{2} because we integrate 3(x2+y2+z2)3(x^2+y^2+z^2) over the hemisphere
  2. 3π5\frac{3\pi}{5} because we integrate 3(x2+y2+z2)3(x^2+y^2+z^2) over the hemisphere using spherical coordinates
  3. 00 because the flux through the hemisphere cancels the flux through the disk
  4. 6π5\frac{6\pi}{5} because we apply divergence theorem to the closed hemisphere (correct answer)
Explanation: The surface SS is closed (hemisphere + disk), so we can apply the divergence theorem. We have ∇⋅F=3x2+3y2+3z2=3(x2+y2+z2)\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2). The region EE is the solid hemisphere x2+y2+z2≤1,z≥0x^2 + y^2 + z^2 \leq 1, z \geq 0. Using spherical coordinates: ∭E3ρ2⋅ρ2sin⁡ϕ dρ dϕ dθ=3∫02π∫0π/2∫01ρ4sin⁡ϕ dρ dϕ dθ=3⋅2π⋅1⋅15=6π5\iiint_E 3\rho^2 \cdot \rho^2\sin\phi \, d\rho \, d\phi \, d\theta = 3\int_0^{2\pi}\int_0^{\pi/2}\int_0^1 \rho^4\sin\phi \, d\rho \, d\phi \, d\theta = 3 \cdot 2\pi \cdot 1 \cdot \frac{1}{5} = \frac{6\pi}{5}. Choice A has wrong computation. Choice B uses the wrong final calculation. Choice C incorrectly assumes cancellation.

Question 12

The divergence theorem is used to convert the surface integral ∬S(x2i+y2j+z2k)⋅dS\iint_S (x^2\mathbf{i} + y^2\mathbf{j} + z^2\mathbf{k}) \cdot d\mathbf{S} over the closed surface SS bounding region EE into ∭E2(x+y+z) dV\iiint_E 2(x+y+z) \, dV. If this conversion is correct and EE is the unit cube [0,1]3[0,1]^3, what constraint must SS satisfy?

  1. SS must be the boundary of [0,1]3[0,1]^3 since that's the only region where ∇⋅(x2,y2,z2)=2(x+y+z)\nabla \cdot (x^2, y^2, z^2) = 2(x+y+z)
  2. SS can be any closed surface bounding a region EE since ∇⋅(x2,y2,z2)=2x+2y+2z=2(x+y+z)\nabla \cdot (x^2, y^2, z^2) = 2x + 2y + 2z = 2(x+y+z) everywhere (correct answer)
  3. SS must enclose the origin since the divergence formula changes sign there
  4. SS must be smooth and orientable, but the region EE can be any region where the divergence equals 2(x+y+z)2(x+y+z)
Explanation: Let's verify the divergence: ∇⋅(x2,y2,z2)=∂x2∂x+∂y2∂y+∂z2∂z=2x+2y+2z=2(x+y+z)\nabla \cdot (x^2, y^2, z^2) = \frac{\partial x^2}{\partial x} + \frac{\partial y^2}{\partial y} + \frac{\partial z^2}{\partial z} = 2x + 2y + 2z = 2(x+y+z). This computation is valid everywhere in R3\mathbb{R}^3, so the divergence theorem applies to any closed, smooth, orientable surface SS bounding any region EE. The specific choice of E=[0,1]3E = [0,1]^3 is just for evaluation purposes. Choice A incorrectly restricts the domain. Choice C incorrectly suggests the divergence depends on the origin. Choice D is partially correct but misses that the divergence is 2(x+y+z)2(x+y+z) everywhere, not just in special regions.

Question 13

A vector field F\mathbf{F} satisfies ∇⋅F=xyz\nabla \cdot \mathbf{F} = xyz in a region EE. If EE is the ellipsoid x24+y29+z216≤1\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} \leq 1, then by the divergence theorem, ∬SF⋅dS=∭Exyz dV\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E xyz \, dV where SS is the boundary of EE. What is this triple integral?

  1. 00 because xyzxyz is an odd function in each variable over a symmetric region (correct answer)
  2. 64π9\frac{64\pi}{9} because the ellipsoid has volume 4π3⋅2⋅3⋅4=32π\frac{4\pi}{3} \cdot 2 \cdot 3 \cdot 4 = 32\pi and average value 23\frac{2}{3}
  3. 32π32\pi because we integrate xyzxyz over the entire ellipsoid volume
  4. 32π3\frac{32\pi}{3} because the integral equals the ellipsoid volume times the average divergence
Explanation: The ellipsoid x24+y29+z216≤1\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} \leq 1 is symmetric about each coordinate plane (the yzyz-plane, xzxz-plane, and xyxy-plane). The function f(x,y,z)=xyzf(x,y,z) = xyz is odd in xx: f(−x,y,z)=−xyz=−f(x,y,z)f(-x,y,z) = -xyz = -f(x,y,z). Similarly, it's odd in yy and zz. When integrating an odd function over a region symmetric about the origin, the integral equals zero because the contributions from opposite sides cancel. Choice B incorrectly assumes a non-zero average value. Choices C and D ignore the symmetry property.

Question 14

Evaluate the flux of the vector field F=⟨x3,y3,z3⟩\mathbf{F} = \langle x^3, y^3, z^3 \rangle through the surface of the unit sphere x2+y2+z2=1x^2+y^2+z^2=1.

  1. 00
  2. 4π/34\pi/3
  3. 4π4\pi
  4. 12π/512\pi/5 (correct answer)
Explanation: When you encounter flux problems through closed surfaces, immediately think about the Divergence Theorem, which converts difficult surface integrals into often-simpler volume integrals. The theorem states that ∬SF⋅n dS=∭E∇⋅F dV\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV where EE is the region enclosed by surface SS. For F=⟨x3,y3,z3⟩\mathbf{F} = \langle x^3, y^3, z^3 \rangle, we need the divergence: ∇⋅F=∂∂x(x3)+∂∂y(y3)+∂∂z(z3)=3x2+3y2+3z2=3(x2+y2+z2)\nabla \cdot \mathbf{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 = 3(x^2 + y^2 + z^2) The flux becomes: ∭E3(x2+y2+z2) dV\iiint_E 3(x^2 + y^2 + z^2) \, dV Converting to spherical coordinates where x2+y2+z2=ρ2x^2 + y^2 + z^2 = \rho^2 and dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta: ∫02π∫0π∫013ρ2⋅ρ2sin⁡ϕ dρ dϕ dθ=3∫02πdθ∫0πsin⁡ϕ dϕ∫01ρ4 dρ\int_0^{2\pi} \int_0^{\pi} \int_0^1 3\rho^2 \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = 3 \int_0^{2\pi} d\theta \int_0^{\pi} \sin\phi \, d\phi \int_0^1 \rho^4 \, d\rho =3⋅2π⋅2⋅15=12π5= 3 \cdot 2\pi \cdot 2 \cdot \frac{1}{5} = \frac{12\pi}{5} Choice A (00) would result from incorrectly assuming the field has zero divergence. Choice B (4π/34\pi/3) comes from forgetting the factor of 3 in the divergence. Choice C (4π4\pi) results from miscalculating the ρ4\rho^4 integral as ρ3\rho^3. Remember: for flux through closed surfaces, always check if the Divergence Theorem simplifies your work—it usually does when the divergence has a nice form.