Divergence Theorem & Flux - Calculus 3

1. Let $S$ be the closed surface consisting of the hemisphere $x^2 + y^2 + z^2 = 4$ with $z \geq 0$ together with its circular base in the $xy$-plane. Consider the vector field $\mathbf{F} = \langle x^3 + yz, y^3 - xz, z^3 + xy \rangle$. If the flux through just the hemispherical portion of $S$ is $24\pi$, what is the flux through the circular base?
   Answer: $-8\pi$ Using the Divergence Theorem, the total flux through the closed surface equals the triple integral of $\nabla \cdot \mathbf{F}$ over the solid hemisphere. We have $\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2)$. The integral becomes $\iiint_V 3(x^2 + y^2 + z^2) \, dV = 3\iiint_V \rho^2 \, dV$ in spherical coordinates. This evaluates to $3 \cdot \frac{4}{5} \cdot \frac{4\pi}{3} \cdot 2^5 = 32\pi$. Since total flux = flux through hemisphere + flux through base, we have $32\pi = 24\pi + \text{flux through base}$, so the flux through the base is $8\pi$. However, the outward normal to the base points downward ($-\mathbf{k}$), so the flux is $-8\pi$.
2. Let $\mathbf{F} = \langle f(x,y,z), g(x,y,z), h(x,y,z) \rangle$ be a $C^2$ vector field defined in a region containing the solid ellipsoid $\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} \leq 1$. If $\nabla \cdot \mathbf{F} = xyz$ throughout this region, what is the flux of $\mathbf{F}$ through the ellipsoid surface (with outward orientation)?
   Answer: $0$ By the Divergence Theorem, the flux equals $\iiint_V xyz \, dV$ where $V$ is the solid ellipsoid. To evaluate this integral, we use the transformation $u = \frac{x}{2}$, $v = \frac{y}{3}$, $w = \frac{z}{4}$, which maps the ellipsoid to the unit ball $u^2 + v^2 + w^2 \leq 1$. The Jacobian is $\frac{\partial(x,y,z)}{\partial(u,v,w)} = 2 \cdot 3 \cdot 4 = 24$, and $xyz = (2u)(3v)(4w) = 24uvw$. The integral becomes: $\iiint_V xyz \, dV = \int\int\int 24uvw \cdot 24 \, du \, dv \, dw = 576 \iiint_{u^2+v^2+w^2 \leq 1} uvw \, du \, dv \, dw$. In the unit ball, by symmetry, $\iiint uvw \, du \, dv \, dw = 0$ because the integrand is an odd function in each variable and the domain is symmetric about each coordinate plane. Therefore, the flux is $0$. The other choices would result from errors like forgetting the symmetry argument, miscalculating the Jacobian, or confusing this with integration of $|xyz|$.
3. A vector field $\mathbf{G}$ has the property that for any closed surface $S$ enclosing volume $V$, the outward flux satisfies $\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = 3 \cdot \text{Volume}(V)$. Consider the region $R$ between two concentric spheres of radii $r = 1$ and $r = 2$. If the outward flux of $\mathbf{G}$ through the outer sphere is $\Phi_{\text{out}}$ and the flux through the inner sphere (with normal pointing toward the center) is $\Phi_{\text{in}}$, what is $\Phi_{\text{out}} + \Phi_{\text{in}}$?
   Answer: $28\pi$ From the given property, $\nabla \cdot \mathbf{G} = 3$ throughout space (by the Divergence Theorem). The volume of the region $R$ between the spheres is $\frac{4\pi}{3}(2^3 - 1^3) = \frac{4\pi}{3}(8 - 1) = \frac{28\pi}{3}$. Applying the Divergence Theorem to region $R$ with the boundary consisting of the outer sphere (outward normal) and inner sphere (inward normal): $\iint_{\text{outer}} \mathbf{G} \cdot \mathbf{n} \, dS + \iint_{\text{inner}} \mathbf{G} \cdot (-\mathbf{n}) \, dS = \iiint_R 3 \, dV = 3 \cdot \frac{28\pi}{3} = 28\pi$. The flux through the inner sphere with inward normal is $\iint_{\text{inner}} \mathbf{G} \cdot (-\mathbf{n}) \, dS = -\iint_{\text{inner}} \mathbf{G} \cdot \mathbf{n} \, dS = -\Phi_{\text{in}}$ (since $\Phi_{\text{in}}$ is defined with normal pointing toward center, which is $-\mathbf{n}$). Therefore: $\Phi_{\text{out}} - \Phi_{\text{in}} = 28\pi$. But wait, we want $\Phi_{\text{out}} + \Phi_{\text{in}}$. Let me reconsider the problem setup. If $\Phi_{\text{in}}$ is the flux through the inner sphere with normal pointing toward the center, then in the Divergence Theorem application, this contributes as $+\Phi_{\text{in}}$. So we have $\Phi_{\text{out}} + \Phi_{\text{in}} = 28\pi$.
4. Let $S$ be the surface of the solid region bounded by $z = x^2 + y^2$ and $z = 4$, oriented with outward normal. If $\mathbf{F}$ is a vector field such that $\nabla \cdot \mathbf{F} = 3$ throughout the region, what additional information is needed to determine the flux of $\mathbf{F}$ through just the paraboloid portion $z = x^2 + y^2$ (not including the disk at $z = 4$)?
   Answer: The flux through the disk at $z = 4$ is needed, since the flux through the paraboloid equals the total flux minus the flux through the disk When you encounter flux problems involving closed surfaces and the Divergence Theorem, think strategically about how to break down the surface into manageable pieces. The Divergence Theorem tells us that $\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V \nabla \cdot \mathbf{F} \, dV$ for the entire closed surface $S$. Since $\nabla \cdot \mathbf{F} = 3$ is constant, we can compute the total flux through the closed surface as $3 \times \text{Volume}$. The volume of the region between the paraboloid $z = x^2 + y^2$ and the plane $z = 4$ is $8\pi$ (found using cylindrical coordinates), so the total flux is $24\pi$. However, the closed surface $S$ consists of two parts: the paraboloid portion and the disk at $z = 4$. Since flux through the entire surface equals flux through the paraboloid plus flux through the disk, we need the flux through the disk to find the flux through just the paraboloid portion. Choice A is incorrect because knowing only the total flux still leaves us with two unknowns (paraboloid flux and disk flux). Choice B misses that while we can compute total flux, we can't directly compute flux through just part of the surface without knowing the vector field explicitly. Choice C incorrectly suggests we need both pieces of information when just the disk flux suffices (we can already calculate the volume and total flux). Study tip: In Divergence Theorem problems, remember that the theorem gives you the total flux through a closed surface. To find flux through one piece, you need information about the flux through all other pieces.
5. Consider a closed surface $S$ enclosing volume $V$, and let $\mathbf{F} = \langle P, Q, R \rangle$ where $P$, $Q$, and $R$ are twice continuously differentiable. If $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = 0$ and $\iiint_V |\nabla \cdot \mathbf{F}|^2 \, dV = 5$, what can be concluded about $\nabla \cdot \mathbf{F}$ in the region $V$?
   Answer: $\nabla \cdot \mathbf{F}$ changes sign within $V$ such that its integral over $V$ is zero When you encounter problems involving surface integrals and divergence, think about the relationship between local and global properties of vector fields. The key insight here is connecting the Divergence Theorem with properties of integrals. By the Divergence Theorem, $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V \nabla \cdot \mathbf{F} \, dV$. Since the surface integral equals zero, we know that $\iiint_V \nabla \cdot \mathbf{F} \, dV = 0$. This means the total "sources" and "sinks" within $V$ balance out perfectly. However, we're also given that $\iiint_V |\nabla \cdot \mathbf{F}|^2 \, dV = 5 > 0$. Since $|\nabla \cdot \mathbf{F}|^2 \geq 0$ everywhere, this integral can only be positive if $\nabla \cdot \mathbf{F} \neq 0$ somewhere in $V$. The correct answer is C: $\nabla \cdot \mathbf{F}$ changes sign within $V$ such that its integral is zero. Option A is wrong because if $\nabla \cdot \mathbf{F} = 0$ everywhere, then $\iiint_V |\nabla \cdot \mathbf{F}|^2 \, dV$ would equal zero, not 5. Option B incorrectly assumes $\nabla \cdot \mathbf{F}$ is constant—if it were constant and nonzero, the flux couldn't be zero. Option D is too weak; we know more than just that $\nabla \cdot \mathbf{F} = 0$ at some point—we know it must have both positive and negative values to satisfy both given conditions simultaneously. Remember: when flux is zero but the divergence is clearly nonzero somewhere, look for regions where positive and negative divergence cancel out globally.

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