Multivariable Calculus Quiz: Flux Across Surfaces
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Flux Across SurfacesQuestion 1 of 8

Consider a vector field F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))\mathbf{F}(x,y,z) = (P(x,y,z), Q(x,y,z), R(x,y,z)) where PP, QQ, and RR are continuously differentiable functions. If the flux of F\mathbf{F} through every closed surface in a region is zero, and you need to compute the flux through an open surface SS with boundary curve CC, which approach is most appropriate?

Apply Stokes' theorem directly to convert the surface integral to a line integral around the boundary CC
Parametrize the surface SS directly and compute ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S} since no simplification is possible
Use the divergence theorem by extending SS to a closed surface and noting that the total flux is zero
Since ∇⋅F=0\nabla \cdot \mathbf{F} = 0 in the region, the flux through SS equals the flux through any other surface with the same boundary CC
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Flux Across Surfaces

Practice Flux Across Surfaces 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 Flux Across Surfaces, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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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

Consider a vector field F(x,y,z)=(P(x,y,z),Q(x,y,z),R(x,y,z))\mathbf{F}(x,y,z) = (P(x,y,z), Q(x,y,z), R(x,y,z)) where PP, QQ, and RR are continuously differentiable functions. If the flux of F\mathbf{F} through every closed surface in a region is zero, and you need to compute the flux through an open surface SS with boundary curve CC, which approach is most appropriate?

  1. Apply Stokes' theorem directly to convert the surface integral to a line integral around the boundary CC
  2. Parametrize the surface SS directly and compute ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S} since no simplification is possible
  3. Use the divergence theorem by extending SS to a closed surface and noting that the total flux is zero
  4. Since ∇⋅F=0\nabla \cdot \mathbf{F} = 0 in the region, the flux through SS equals the flux through any other surface with the same boundary CC (correct answer)
Explanation: When you encounter a vector field where the flux through every closed surface is zero, you're dealing with a solenoidal (divergence-free) field. This condition tells you that ∇⋅F=0\nabla \cdot \mathbf{F} = 0 throughout the region, which has profound implications for flux calculations. The key insight is that when ∇⋅F=0\nabla \cdot \mathbf{F} = 0, the flux through any surface depends only on its boundary, not the specific surface itself. If two surfaces S1S_1 and S2S_2 share the same boundary curve CC, then by the divergence theorem applied to the closed surface formed by S1S_1 and −S2-S_2, we get ∬S1F⋅dS−∬S2F⋅dS=∭V∇⋅F dV=0\iint_{S_1} \mathbf{F} \cdot d\mathbf{S} - \iint_{S_2} \mathbf{F} \cdot d\mathbf{S} = \iiint_V \nabla \cdot \mathbf{F} \, dV = 0. Therefore, both surfaces have equal flux. Answer D captures this principle perfectly. A is incorrect because Stokes' theorem relates circulation (line integrals) to curl, not flux to divergence. B misses the simplification opportunity entirely—direct parametrization would be unnecessarily complex when we can exploit the divergence-free property. C has the right idea about using closed surfaces but describes an overly complicated approach when the simpler surface-substitution method works directly. Study tip: When you see "flux through every closed surface is zero," immediately think "divergence-free field" and "surface independence." This means you can always replace a complicated surface with a simpler one having the same boundary—often a flat disk or plane region.

Question 2

Consider the vector field F(x,y,z)=(2x,−y,z2)\mathbf{F}(x,y,z) = (2x, -y, z^2) and the surface SS consisting of the portion of the paraboloid z=x2+y2z = x^2 + y^2 for 0≤z≤40 \leq z \leq 4, oriented with upward-pointing normal vectors. If the flux of F\mathbf{F} across SS is computed using the surface integral ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S}, which expression correctly represents this flux?

  1. ∬D(2x+y+(x2+y2)2)1+4x2+4y2 dx dy\iint_D (2x + y + (x^2 + y^2)^2) \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy where D:x2+y2≤4D: x^2 + y^2 \leq 4
  2. ∬D(2x−y+(x2+y2)2)1+4x2+4y2 dx dy\iint_D (2x - y + (x^2 + y^2)^2) \sqrt{1 + 4x^2 + 4y^2} \, dx \, dy where D:x2+y2≤4D: x^2 + y^2 \leq 4
  3. ∬D(−4x2+2xy+(x2+y2)2) dx dy\iint_D (-4x^2 + 2xy + (x^2 + y^2)^2) \, dx \, dy where D:x2+y2≤4D: x^2 + y^2 \leq 4 (correct answer)
  4. ∬D(−4x2−2xy+(x2+y2)2) dx dy\iint_D (-4x^2 - 2xy + (x^2 + y^2)^2) \, dx \, dy where D:x2+y2≤4D: x^2 + y^2 \leq 4
Explanation: For the surface z=g(x,y)=x2+y2z = g(x,y) = x^2 + y^2, the upward normal vector is n=(−gx,−gy,1)1+gx2+gy2=(−2x,−2y,1)1+4x2+4y2\mathbf{n} = \frac{(-g_x, -g_y, 1)}{\sqrt{1 + g_x^2 + g_y^2}} = \frac{(-2x, -2y, 1)}{\sqrt{1 + 4x^2 + 4y^2}}. The flux integral becomes ∬DF⋅n dS=∬D(2x,−y,(x2+y2)2)⋅(−2x,−2y,1) dx dy=∬D(−4x2+2xy+(x2+y2)2) dx dy\iint_D \mathbf{F} \cdot \mathbf{n} \, dS = \iint_D (2x, -y, (x^2+y^2)^2) \cdot (-2x, -2y, 1) \, dx \, dy = \iint_D (-4x^2 + 2xy + (x^2+y^2)^2) \, dx \, dy. Choice A has the wrong sign for the y-component. Choice B has both wrong signs and includes the unnecessary 1+4x2+4y2\sqrt{1+4x^2+4y^2} factor. Choice D has the wrong sign for the xy term.

Question 3

A surface SS is defined parametrically by r(u,v)=(ucos⁡v,usin⁡v,u2)\mathbf{r}(u,v) = (u\cos v, u\sin v, u^2) for 0≤u≤20 \leq u \leq 2 and 0≤v≤π0 \leq v \leq \pi. If the vector field F(x,y,z)=(0,0,xyz)\mathbf{F}(x,y,z) = (0, 0, xyz) flows through this surface, and the surface is oriented using the normal vector ru×rv\mathbf{r}_u \times \mathbf{r}_v, what is the flux of F\mathbf{F} through SS?

  1. ∫0π∫02u4cos⁡vsin⁡v du dv\int_0^\pi \int_0^2 u^4 \cos v \sin v \, du \, dv
  2. ∫0π∫02u5cos⁡vsin⁡v du dv\int_0^\pi \int_0^2 u^5 \cos v \sin v \, du \, dv (correct answer)
  3. ∫0π∫02u3cos⁡vsin⁡v du dv\int_0^\pi \int_0^2 u^3 \cos v \sin v \, du \, dv
  4. ∫0π∫02u6cos⁡vsin⁡v du dv\int_0^\pi \int_0^2 u^6 \cos v \sin v \, du \, dv
Explanation: First, we find ru=(cos⁡v,sin⁡v,2u)\mathbf{r}_u = (\cos v, \sin v, 2u) and rv=(−usin⁡v,ucos⁡v,0)\mathbf{r}_v = (-u\sin v, u\cos v, 0). Then ru×rv=(−2u2cos⁡v,−2u2sin⁡v,u)\mathbf{r}_u \times \mathbf{r}_v = (-2u^2\cos v, -2u^2\sin v, u). On the surface, F=(0,0,ucos⁡v⋅usin⁡v⋅u2)=(0,0,u4cos⁡vsin⁡v)\mathbf{F} = (0, 0, u\cos v \cdot u\sin v \cdot u^2) = (0, 0, u^4\cos v \sin v). The flux integral is ∬SF⋅dS=∫0π∫02(0,0,u4cos⁡vsin⁡v)⋅(−2u2cos⁡v,−2u2sin⁡v,u) du dv=∫0π∫02u5cos⁡vsin⁡v du dv\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^\pi \int_0^2 (0, 0, u^4\cos v \sin v) \cdot (-2u^2\cos v, -2u^2\sin v, u) \, du \, dv = \int_0^\pi \int_0^2 u^5\cos v \sin v \, du \, dv. Choice A forgets one factor of uu. Choice C is missing u2u^2. Choice D has an extra factor of uu.

Question 4

The flux of F(x,y,z)=(y2,xz,z2−x)\mathbf{F}(x,y,z) = (y^2, xz, z^2-x) through the triangular surface with vertices at (1,0,0)(1,0,0), (0,1,0)(0,1,0), and (0,0,1)(0,0,1), oriented so that the normal vector points away from the origin, is being computed. Which statement about this computation is correct?

  1. The surface can be parametrized as r(u,v)=(1−u−v,u,v)\mathbf{r}(u,v) = (1-u-v, u, v) with u,v≥0u,v \geq 0 and u+v≤1u+v \leq 1, and the normal vector ru×rv=(1,1,1)\mathbf{r}_u \times \mathbf{r}_v = (1,1,1) points toward the origin
  2. The flux integral requires computing ∬D(u2+(1−u−v)v+v2−(1−u−v))3 du dv\iint_D (u^2 + (1-u-v)v + v^2 - (1-u-v)) \sqrt{3} \, du \, dv over the triangular region DD
  3. The surface lies on the plane x+y+z=1x + y + z = 1, and using the upward normal (1,1,1)/3(1,1,1)/\sqrt{3} gives the correct orientation away from the origin
  4. The surface can be parametrized as r(u,v)=(1−u−v,u,v)\mathbf{r}(u,v) = (1-u-v, u, v) with u,v≥0u,v \geq 0 and u+v≤1u+v \leq 1, and the normal vector ru×rv=(1,1,1)\mathbf{r}_u \times \mathbf{r}_v = (1,1,1) points away from the origin (correct answer)
Explanation: When computing flux through a surface, you need three key components: a parametrization of the surface, the correct normal vector with proper orientation, and the flux integral setup. The triangular surface has vertices at (1,0,0)(1,0,0), (0,1,0)(0,1,0), and (0,0,1)(0,0,1). These points all satisfy x+y+z=1x + y + z = 1, confirming the surface lies on this plane. The parametrization r(u,v)=(1−u−v,u,v)\mathbf{r}(u,v) = (1-u-v, u, v) correctly represents this triangle: when u=v=0u = v = 0, you get (1,0,0)(1,0,0); when u=1,v=0u = 1, v = 0, you get (0,1,0)(0,1,0); and when u=0,v=1u = 0, v = 1, you get (0,0,1)(0,0,1). To find the normal vector, compute ru=(−1,1,0)\mathbf{r}_u = (-1,1,0) and rv=(−1,0,1)\mathbf{r}_v = (-1,0,1), giving ru×rv=(1,1,1)\mathbf{r}_u \times \mathbf{r}_v = (1,1,1). To check orientation, pick a point on the surface like (13,13,13)(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) and verify that this point plus the normal vector (1,1,1)(1,1,1) moves away from the origin. Indeed, (13,13,13)+(1,1,1)=(43,43,43)(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) + (1,1,1) = (\frac{4}{3}, \frac{4}{3}, \frac{4}{3}) is farther from the origin. Choice A incorrectly claims (1,1,1)(1,1,1) points toward the origin. Choice B shows an incorrect flux integral computation. Choice C incorrectly suggests using the unit normal (1,1,1)3\frac{(1,1,1)}{\sqrt{3}} instead of (1,1,1)(1,1,1) itself. Choice D correctly identifies both the parametrization and proper orientation. Study tip: Always verify normal vector orientation by checking whether adding the normal to a surface point moves you toward or away from the reference point.

Question 5

Consider the surface SS defined by x2+y2+z2=1x^2 + y^2 + z^2 = 1 with z≥0z \geq 0 (upper hemisphere), and let F(x,y,z)=(z,0,x)\mathbf{F}(x,y,z) = (z, 0, x). To find the flux of F\mathbf{F} through SS oriented with outward normal vectors, a student computes ∇⋅F=0\nabla \cdot \mathbf{F} = 0 and concludes the flux is zero by the divergence theorem. What is wrong with this reasoning?

  1. The divergence theorem applies to closed surfaces, but SS is not closed; the circular base at z=0z = 0 must be included (correct answer)
  2. The divergence theorem cannot be applied because F\mathbf{F} is not continuously differentiable on the hemisphere
  3. The divergence calculation is incorrect; ∇⋅F=1\nabla \cdot \mathbf{F} = 1, so the flux should be 2π3\frac{2\pi}{3}
  4. The orientation specified (outward normal) is inconsistent with the surface parametrization typically used for hemispheres
Explanation: When you encounter flux problems involving the divergence theorem, always check whether the surface satisfies the theorem's requirements. The divergence theorem states that ∬SF⋅n dS=∭E∇⋅F dV\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV, but it only applies to closed surfaces that completely enclose a region. The student's error lies in applying the divergence theorem to an incomplete surface. While the divergence calculation ∇⋅F=∂z∂x+∂0∂y+∂x∂z=0\nabla \cdot \mathbf{F} = \frac{\partial z}{\partial x} + \frac{\partial 0}{\partial y} + \frac{\partial x}{\partial z} = 0 is correct, the upper hemisphere alone doesn't form a closed surface—it has an open circular boundary at z=0z = 0. To use the divergence theorem, you must include the disk x2+y2≤1,z=0x^2 + y^2 \leq 1, z = 0 to create a complete closed surface enclosing the solid hemisphere. Looking at the wrong answers: (B) is incorrect because F=(z,0,x)\mathbf{F} = (z, 0, x) is continuously differentiable everywhere, including on the hemisphere. (C) contains a calculation error—the divergence is indeed zero, not 1, since ∂x∂z=0\frac{\partial x}{\partial z} = 0. (D) is wrong because outward normal orientation is perfectly valid and standard for flux calculations through hemispheres. Answer: (A) Study tip: Before applying the divergence theorem, always verify the surface is closed. If you have a partial surface like a hemisphere, cap, or portion of a cylinder, you'll need to add the missing pieces to create a closed boundary, or use direct surface integration instead.

Question 6

A surface SS is given by the graph of z=f(x,y)z = f(x,y) over a region DD in the xy-plane, where ff has continuous partial derivatives. The flux of F(x,y,z)=(P,Q,R)\mathbf{F}(x,y,z) = (P, Q, R) through SS with upward orientation is computed as ∬D(−Pfx−Qfy+R) dx dy\iint_D (-P f_x - Q f_y + R) \, dx \, dy. If instead we want the flux through the same surface but with downward orientation, the integral becomes:

  1. ∬D(Pfx+Qfy−R) dx dy\iint_D (P f_x + Q f_y - R) \, dx \, dy (correct answer)
  2. ∬D(−Pfx−Qfy+R)1+fx2+fy2 dx dy\iint_D (-P f_x - Q f_y + R) \sqrt{1 + f_x^2 + f_y^2} \, dx \, dy
  3. ∬DPfx+Qfy−R1+fx2+fy2 dx dy\iint_D \frac{P f_x + Q f_y - R}{\sqrt{1 + f_x^2 + f_y^2}} \, dx \, dy
  4. ∬D(Pfx+Qfy+R) dx dy\iint_D (P f_x + Q f_y + R) \, dx \, dy
Explanation: For a surface z=f(x,y)z = f(x,y) with upward orientation, the unit normal vector is n=(−fx,−fy,1)1+fx2+fy2\mathbf{n} = \frac{(-f_x, -f_y, 1)}{\sqrt{1 + f_x^2 + f_y^2}} and dS=n dS=(−fx,−fy,1) dx dyd\mathbf{S} = \mathbf{n} \, dS = (-f_x, -f_y, 1) \, dx \, dy. The flux integral becomes ∬DF⋅dS=∬D(P,Q,R)⋅(−fx,−fy,1) dx dy=∬D(−Pfx−Qfy+R) dx dy\iint_D \mathbf{F} \cdot d\mathbf{S} = \iint_D (P, Q, R) \cdot (-f_x, -f_y, 1) \, dx \, dy = \iint_D (-P f_x - Q f_y + R) \, dx \, dy. For downward orientation, we use −n-\mathbf{n}, so dS=(fx,fy,−1) dx dyd\mathbf{S} = (f_x, f_y, -1) \, dx \, dy, giving ∬D(P,Q,R)⋅(fx,fy,−1) dx dy=∬D(Pfx+Qfy−R) dx dy\iint_D (P, Q, R) \cdot (f_x, f_y, -1) \, dx \, dy = \iint_D (P f_x + Q f_y - R) \, dx \, dy. Choice B incorrectly includes the magnitude factor 1+fx2+fy2\sqrt{1 + f_x^2 + f_y^2}. Choice C incorrectly divides by this factor. Choice D has the wrong sign on the RR term.

Question 7

Consider the hemisphere S:x2+y2+z2=9S: x^2 + y^2 + z^2 = 9 with z≥0z \geq 0, oriented with outward normal vectors. The flux of the vector field F(x,y,z)=(x3,y3,z3)\mathbf{F}(x,y,z) = (x^3, y^3, z^3) through this hemisphere (not including the circular base) can be computed most efficiently using which approach?

  1. Direct parametrization using spherical coordinates and computing ∬SF⋅dS\iint_S \mathbf{F} \cdot d\mathbf{S} over the curved surface only
  2. Applying divergence theorem to the entire hemisphere volume, then subtracting the flux through the circular base (correct answer)
  3. Converting to cylindrical coordinates and integrating over the projected circular region in the xy-plane
  4. Using Green's theorem after projecting the vector field onto the xy-plane and integrating around the boundary
Explanation: Since ∇⋅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), which is constant (=27= 27) on the sphere, the divergence theorem is most efficient. We compute the flux through the closed hemisphere (including the base), then subtract the flux through the base. The total flux through the closed surface is ∭V27 dV=27⋅2π3⋅33=486π\iiint_V 27 \, dV = 27 \cdot \frac{2\pi}{3} \cdot 3^3 = 486\pi. The flux through the base is ∬x2+y2≤9z3 dx dy=0\iint_{x^2+y^2 \leq 9} z^3 \, dx \, dy = 0 (since z=0z = 0 on the base). Choice A would work but is computationally intensive. Choice C is not a standard approach for surface integrals. Choice D incorrectly applies Green's theorem, which is for line integrals.

Question 8

Two surfaces S1S_1 and S2S_2 have the same boundary curve CC but different orientations. If S1S_1 is oriented with normal vectors n1\mathbf{n}_1 and S2S_2 is oriented with normal vectors n2=−n1\mathbf{n}_2 = -\mathbf{n}_1, and if Φ1\Phi_1 and Φ2\Phi_2 represent the flux of a vector field F\mathbf{F} through S1S_1 and S2S_2 respectively, which relationship must hold?

  1. Φ1+Φ2=0\Phi_1 + \Phi_2 = 0 regardless of the vector field F\mathbf{F}, provided ∇⋅F=0\nabla \cdot \mathbf{F} = 0
  2. Φ1=Φ2\Phi_1 = \Phi_2 if and only if the surfaces S1S_1 and S2S_2 are geometrically identical
  3. Φ1+Φ2=0\Phi_1 + \Phi_2 = 0 if the surfaces S1S_1 and S2S_2 are geometrically identical, regardless of the vector field (correct answer)
  4. Φ1=Φ2\Phi_1 = \Phi_2 regardless of the surfaces, provided they have the same boundary and ∇⋅F=0\nabla \cdot \mathbf{F} = 0
Explanation: If S1S_1 and S2S_2 are geometrically identical surfaces with opposite orientations (n2=−n1\mathbf{n}_2 = -\mathbf{n}_1), then Φ2=∬S2F⋅n2 dS=∬S1F⋅(−n1) dS=−∬S1F⋅n1 dS=−Φ1\Phi_2 = \iint_{S_2} \mathbf{F} \cdot \mathbf{n}_2 \, dS = \iint_{S_1} \mathbf{F} \cdot (-\mathbf{n}_1) \, dS = -\iint_{S_1} \mathbf{F} \cdot \mathbf{n}_1 \, dS = -\Phi_1. Therefore Φ1+Φ2=0\Phi_1 + \Phi_2 = 0 regardless of the vector field, as long as the surfaces are geometrically identical. Choice A is incorrect because it doesn't require the surfaces to be identical. Choice B is wrong because identical surfaces with opposite orientations give opposite flux values, not equal ones. Choice D is incorrect because even with ∇⋅F=0\nabla \cdot \mathbf{F} = 0, different surfaces can have different flux values; the condition only guarantees that flux depends on boundary, not that it's equal for opposite orientations.