Computing Flux via Divergence Theorem - Calculus 3

1. Consider the vector field $\vec{F} = \langle x + y^2z, y + x^2z, z + x^2y^2 \rangle$ and the closed surface $S$ that bounds the region between two concentric spheres: $1 \leq x^2 + y^2 + z^2 \leq 4$. The surface $S$ consists of the outer sphere (radius 2) with outward normal and the inner sphere (radius 1) with inward normal. If we want to apply the Divergence Theorem directly to find the flux through $S$, what is the correct setup?
   Answer: $\iiint_{1 \leq x^2+y^2+z^2 \leq 4} (1 + 1 + 1) \, dV = 3 \cdot \frac{28\pi}{3} = 28\pi$ To apply the Divergence Theorem, we need $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x + y^2z) + \frac{\partial}{\partial y}(y + x^2z) + \frac{\partial}{\partial z}(z + x^2y^2) = 1 + 1 + 1 = 3$. The region is the spherical shell with volume $\frac{4\pi}{3}(2^3 - 1^3) = \frac{4\pi}{3} \cdot 7 = \frac{28\pi}{3}$. So the flux is $3 \cdot \frac{28\pi}{3} = 28\pi$. Choice B incorrectly includes $x^2y^2$ from the original field. Choice C incorrectly doubles the $x^2y^2$ term. Choice D incorrectly computes partial derivatives, getting terms from the original components rather than their derivatives.
2. A vector field $\vec{F}$ has the property that its divergence is $\nabla \cdot \vec{F} = x^2 + y^2 + z^2$. Consider three regions: (I) the unit ball $x^2 + y^2 + z^2 \leq 1$, (II) the cube $[-1,1]^3$, and (III) the cylindrical region $x^2 + y^2 \leq 1, -1 \leq z \leq 1$. If $\Phi_I$, $\Phi_{II}$, and $\Phi_{III}$ represent the outward flux through the boundaries of regions I, II, and III respectively, which ordering is correct?
   Answer: $\Phi_I < \Phi_{III} < \Phi_{II}$ When you encounter a vector field with known divergence and need to compare flux through different surfaces, immediately think of the Divergence Theorem: $\iint_S \vec{F} \cdot d\vec{A} = \iiint_V \nabla \cdot \vec{F} \, dV$. The outward flux through each boundary equals the triple integral of the divergence over the enclosed volume. Since $\nabla \cdot \vec{F} = x^2 + y^2 + z^2$, you need to evaluate $\iiint_V (x^2 + y^2 + z^2) \, dV$ for each region. Notice that this integrand is always non-negative and increases with distance from the origin. For the unit ball (I), using spherical coordinates: $\Phi_I = \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2 \cdot \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta = \frac{4\pi}{5}$. For the cube (II), the integrand reaches its maximum value of 3 at corners like $(1,1,1)$, and the volume is 8, giving $\Phi_{II} = 8$. For the cylinder (III), the integrand reaches maximum value 2 at points like $(1,0,1)$, and the volume is $2\pi$, giving $\Phi_{III} = 2\pi \approx 6.28$. Therefore: $\Phi_I = \frac{4\pi}{5} \approx 2.51 < \Phi_{III} = 2\pi \approx 6.28 < \Phi_{II} = 8$. Choice A incorrectly orders the cylinder and cube. Choice C completely reverses the order. Choice D incorrectly places the cylinder as smallest. Strategy tip: For divergence problems, always use the Divergence Theorem to convert surface integrals to volume integrals. Compare how the divergence function behaves over each region's volume and extent.
3. Let $\vec{F} = \langle P(x,y,z), Q(x,y,z), R(x,y,z) \rangle$ be a vector field where $\nabla \cdot \vec{F} = 6$ throughout a region $E$. The boundary surface $S$ of $E$ consists of two parts: $S_1$ with area 10 and $S_2$ with area 15. If the flux of $\vec{F}$ through $S_1$ is 30 (outward), and the volume of $E$ is 20, what is the flux through $S_2$ (outward)?
   Answer: 90 When you encounter a problem involving flux through boundary surfaces with known divergence, immediately think of the Divergence Theorem, which connects surface integrals to volume integrals. The Divergence Theorem states that $\iint_S \vec{F} \cdot \hat{n} \, dS = \iiint_E \nabla \cdot \vec{F} \, dV$ for a closed surface $S$ bounding region $E$. Since $\nabla \cdot \vec{F} = 6$ is constant throughout $E$ with volume 20, the total outward flux through the entire boundary is $6 \times 20 = 120$. The complete boundary $S$ consists of $S_1$ and $S_2$, so the total flux equals the sum of fluxes through each part: $\text{Flux through } S_1 + \text{Flux through } S_2 = 120$. Given that the flux through $S_1$ is 30 (outward), we have $30 + \text{Flux through } S_2 = 120$, yielding $\text{Flux through } S_2 = 90$. Answer choice A (150) incorrectly adds the divergence to the area of $S_2$: $6 \times 15 + 6 \times 10$. Choice B (120) represents the total flux through the entire boundary, not just $S_2$. Choice C (-30) might come from incorrectly assuming the flux through $S_2$ has opposite sign to $S_1$, but both are outward fluxes. Study tip: When you see divergence problems with multiple boundary surfaces, always apply the Divergence Theorem to find total flux first, then use the constraint that individual surface fluxes must sum to this total.
4. Let $\vec{F} = \langle yz, xz, xy + 2z \rangle$ and let $S$ be the boundary of the region $E = \{(x,y,z) : x^2 + y^2 \leq 1, 0 \leq z \leq x^2 + y^2\}$, oriented with outward normals. The surface $S$ consists of three parts: the bottom disk, the top paraboloid, and a curved connecting surface. What is the flux of $\vec{F}$ through $S$?
   Answer: $\pi$ Using the Divergence Theorem, the flux equals $\iiint_E \nabla \cdot \vec{F} \, dV$. We compute $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(yz) + \frac{\partial}{\partial y}(xz) + \frac{\partial}{\partial z}(xy + 2z) = 0 + 0 + 2 = 2$. The region $E$ has volume $\iiint_E 1 \, dV = \int_0^{2\pi} \int_0^1 \int_0^{r^2} r \, dz \, dr \, d\theta = \int_0^{2\pi} \int_0^1 r \cdot r^2 \, dr \, d\theta = \int_0^{2\pi} \int_0^1 r^3 \, dr \, d\theta = 2\pi \cdot \frac{1}{4} = \frac{\pi}{2}$. Therefore, the flux is $2 \cdot \frac{\pi}{2} = \pi$.
5. Consider the vector field $\vec{F} = \langle x^3, y^3, z^3 \rangle$ and the surface $S$ consisting of the portion of the sphere $x^2 + y^2 + z^2 = 9$ where $z \geq 0$, together with the disk $x^2 + y^2 \leq 9$ in the $xy$-plane. Both surfaces are oriented with outward-pointing normals. Which expression correctly represents the flux of $\vec{F}$ through $S$ using the Divergence Theorem?
   Answer: $\iiint_E (3x^2 + 3y^2 + 3z^2) \, dV$ where $E$ is the upper hemisphere The Divergence Theorem states that flux through a closed surface equals $\iiint_E \nabla \cdot \vec{F} \, dV$. Here, $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(z^3) = 3x^2 + 3y^2 + 3z^2$. The region $E$ is the upper hemisphere (the region enclosed by $S$). Choice A has the wrong divergence expression. Choice B uses the wrong region (full sphere instead of upper hemisphere). Choice C gives the flux integral setup but doesn't use the Divergence Theorem as requested.
6. Let $S$ be the closed surface of the tetrahedron with vertices at $(0,0,0)$, $(2,0,0)$, $(0,3,0)$, and $(0,0,1)$, oriented with outward normals. For the vector field $\vec{F} = \langle x^2, y^2, z^2 \rangle$, a student attempts to use the Divergence Theorem but makes an error in setting up the integral. Which of the following represents the most likely incorrect setup?
   Answer: $\int_0^2 \int_0^3 \int_0^1 (2x + 2y + 2z) \, dz \, dy \, dx$ The correct application of the Divergence Theorem requires $\nabla \cdot \vec{F} = 2x + 2y + 2z$ and integration over the tetrahedron region. The tetrahedron is defined by $x \geq 0, y \geq 0, z \geq 0$ and $\frac{x}{2} + \frac{y}{3} + z \leq 1$. Choice A gives the correct integrand and bounds. Choice B reorders the integration correctly with proper bounds. Choice C is the most likely student error: using a rectangular box $[0,2] \times [0,3] \times [0,1]$ instead of the actual tetrahedron region, which significantly overestimates the integration domain. Choice D uses the wrong integrand (the original field instead of its divergence).
7. Consider the vector field $\vec{F} = \langle e^{x+y}, \sin(xy), z^3 - z \rangle$ and the cube $[0,1] \times [0,1] \times [0,1]$ with outward orientation. A student correctly identifies that $\nabla \cdot \vec{F} = e^{x+y} + x\cos(xy) + 3z^2 - 1$ but then claims the Divergence Theorem cannot be applied because "the divergence is too complicated to integrate." What is the flaw in this reasoning?
   Answer: The Divergence Theorem can still be applied; computational difficulty doesn't invalidate the theorem's applicability The Divergence Theorem applies to any vector field with continuous partial derivatives on a region bounded by a piecewise smooth closed surface, regardless of how complicated the divergence expression is. The computational difficulty of evaluating $\iiint (e^{x+y} + x\cos(xy) + 3z^2 - 1) \, dV$ doesn't affect the theorem's validity. Choice A is incorrect - the divergence calculation is correct ($\frac{\partial}{\partial y}(\sin(xy)) = x\cos(xy)$). Choice B is wrong - the divergence can be any function. Choice D is incorrect - cubes are standard examples of valid regions for the Divergence Theorem.
8. Consider the vector field $\vec{F} = \langle f(x), g(y), h(z) \rangle$ where $f'(x) = x$, $g'(y) = 2y$, and $h'(z) = 3z^2$. Let $S$ be the boundary of the region $E = \{(x,y,z) : 0 \leq x \leq 2, 0 \leq y \leq 3, 0 \leq z \leq 1\}$, oriented outward. What additional information is needed to compute the flux of $\vec{F}$ through $S$ using the Divergence Theorem?
   Answer: No additional information is needed; the flux can be computed as $\int_0^2 \int_0^3 \int_0^1 (x + 2y + 3z^2) \, dz \, dy \, dx$ The Divergence Theorem states that flux = $\iiint_E \nabla \cdot \vec{F} \, dV$. Here, $\nabla \cdot \vec{F} = f'(x) + g'(y) + h'(z) = x + 2y + 3z^2$. This depends only on the derivatives of the component functions, not the functions themselves. The integral $\int_0^2 \int_0^3 \int_0^1 (x + 2y + 3z^2) \, dz \, dy \, dx$ can be evaluated to get the flux. Choice A is incorrect because only the derivatives matter for divergence. Choice C is wrong because the Divergence Theorem handles the surface orientation automatically. Choice D is unnecessary since the derivatives are given as simple polynomials, implying sufficient smoothness.

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