Flux Integrals - Calculus 3

1. Consider the vector field $\mathbf{F}(x,y,z) = \langle xy, z^2, x+y \rangle$ and the surface $S$ defined by $z = 4 - x^2 - y^2$ for $z \geq 0$, oriented with upward-pointing normal vectors. If the flux integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$ is evaluated using the parameterization $\mathbf{r}(u,v) = \langle u, v, 4-u^2-v^2 \rangle$, what is the integrand after computing $\mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v)$?
   Answer: $2u^2v + 2v(4-u^2-v^2)^2 + (u+v)$ First, compute the partial derivatives: $\mathbf{r}_u = \langle 1, 0, -2u \rangle$ and $\mathbf{r}_v = \langle 0, 1, -2v \rangle$. The cross product is $\mathbf{r}_u \times \mathbf{r}_v = \langle 2u, 2v, 1 \rangle$ (upward orientation). Next, evaluate the vector field: $\mathbf{F}(\mathbf{r}(u,v)) = \langle uv, (4-u^2-v^2)^2, u+v \rangle$. The dot product gives: $uv \cdot 2u + (4-u^2-v^2)^2 \cdot 2v + (u+v) \cdot 1 = 2u^2v + 2v(4-u^2-v^2)^2 + (u+v)$. Choice A shows the computation before simplification. Choice C has an incorrect sign on the last term. Choice D uses the wrong orientation (downward normal).
2. A surface $S$ is defined implicitly by $F(x,y,z) = x^2 + 2y^2 + z^2 - 6 = 0$ and oriented so that the normal vector points toward the origin. The flux of $\mathbf{G}(x,y,z) = \langle x^3, y^3, z^3 \rangle$ through this surface can be computed using the formula $\iint_S \mathbf{G} \cdot \mathbf{n} \, dS = \iint_D \mathbf{G} \cdot \frac{\nabla F}{|\nabla F|} \frac{|\nabla F|}{|F_z|} \, dx \, dy$ when projecting onto the xy-plane. However, this approach has a critical limitation. What is it?
   Answer: The ellipsoid $x^2 + 2y^2 + z^2 = 6$ cannot be written as a single-valued function $z = g(x,y)$ over its entire projection onto the xy-plane When computing surface integrals using the formula given, you're essentially projecting the surface onto a coordinate plane and integrating over that projection. This approach requires the surface to be expressible as a single-valued function over the projection region. The surface $x^2 + 2y^2 + z^2 = 6$ is an ellipsoid. To use the xy-projection formula, you'd need to solve for $z$ as a function of $x$ and $y$: $z = \pm\sqrt{6 - x^2 - 2y^2}$. The $\pm$ reveals the fundamental problem—for most points $(x,y)$ in the projection, there are two corresponding $z$-values (upper and lower halves of the ellipsoid). You cannot write the entire surface as $z = g(x,y)$ for a single function $g$. This makes option D correct. Option A is wrong because orientation issues, while requiring sign corrections, don't prevent the method from working—they're manageable with careful attention to direction. Option B misidentifies where $\nabla F = 0$. Since $\nabla F = \langle 2x, 4y, 2z \rangle$, it's zero only at the origin $(0,0,0)$, but the origin lies inside the ellipsoid (since $0 + 0 + 0 < 6$), not on the surface itself. Option C incorrectly suggests convergence problems with the divergence, but the surface is bounded and closed, so the integral will converge. Study tip: When using coordinate projections for surface integrals, always check if the surface can be written as a single-valued function over the projection. Closed surfaces like spheres and ellipsoids typically require splitting into multiple pieces or using alternative methods like the divergence theorem.
3. Consider the vector field $\mathbf{F}(x,y,z) = \langle yz, xz + z^2, xy + 2yz \rangle$ and the surface $S$ defined by $z = \sqrt{x^2 + y^2}$ for $0 \leq z \leq 2$, oriented with upward-pointing normals. When computing the flux using cylindrical coordinates $x = r\cos\theta, y = r\sin\theta, z = r$, what is the correct setup for the flux integral?
   Answer: $\int_0^{2\pi} \int_0^2 \mathbf{F}(r\cos\theta, r\sin\theta, r) \cdot \langle -\cos\theta, -\sin\theta, 1 \rangle r\sqrt{2} \, dr \, d\theta$ Using parameterization $\mathbf{r}(r,\theta) = \langle r\cos\theta, r\sin\theta, r \rangle$, we get $\mathbf{r}_r = \langle \cos\theta, \sin\theta, 1 \rangle$ and $\mathbf{r}_\theta = \langle -r\sin\theta, r\cos\theta, 0 \rangle$. The cross product $\mathbf{r}_r \times \mathbf{r}_\theta = \langle -r\cos\theta, -r\sin\theta, r \rangle = r\langle -\cos\theta, -\sin\theta, 1 \rangle$ points upward (positive z-component). The magnitude is $r\sqrt{2}$, so the unit normal is $\frac{1}{\sqrt{2}}\langle -\cos\theta, -\sin\theta, 1 \rangle$. The surface element is $d\mathbf{S} = \langle -\cos\theta, -\sin\theta, 1 \rangle r\sqrt{2} \, dr \, d\theta$. Choice A omits the $\sqrt{2}$ factor. Choice B has the wrong normal direction. Choice D has an incorrect normal vector.
4. The flux of vector field $\mathbf{F} = \langle P, Q, R \rangle$ through a surface $S$ can sometimes be computed more easily by recognizing special properties. Consider $\mathbf{F}(x,y,z) = \langle 3x^2y, x^3 + z^3, 3z^2y \rangle$ and the closed surface consisting of the cylinder $x^2 + y^2 = 1$ for $0 \leq z \leq 2$, including its top and bottom circular faces. Before computing any integrals, what can we determine about the flux through this surface?
   Answer: The flux is zero because $\nabla \cdot \mathbf{F} = 6xy + 3z^2 + 6zy$, which integrates to zero over the cylindrical region due to symmetry Computing the divergence: $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(3x^2y) + \frac{\partial}{\partial y}(x^3 + z^3) + \frac{\partial}{\partial z}(3z^2y) = 6xy + 0 + 6zy = 6y(x+z)$. By the divergence theorem, the flux equals $\iiint 6y(x+z) \, dV$. In cylindrical coordinates with $x = r\cos\theta, y = r\sin\theta$: $\iiint 6r\sin\theta(r\cos\theta + z) r \, dr \, d\theta \, dz = \int_0^{2\pi} \int_0^1 \int_0^2 6r^3\sin\theta\cos\theta + 6r^2z\sin\theta \, dz \, dr \, d\theta$. Since $\int_0^{2\pi} \sin\theta \, d\theta = 0$ and $\int_0^{2\pi} \sin\theta\cos\theta \, d\theta = 0$, the entire integral is zero. Choice B is wrong because the field is not conservative ($\nabla \times \mathbf{F} \neq 0$). Choice C has an incorrect calculation. Choice D misses the symmetry argument.
5. A fluid flows with velocity field $\mathbf{v}(x,y,z) = \langle z, x+y, y^2 \rangle$. The flux through the triangular surface with vertices at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ represents the net volume flow rate through this surface. If the surface is parameterized as $\mathbf{r}(s,t) = (1-s-t,s,t)$ where $s,t \geq 0$ and $s+t \leq 1$, and the normal vector points away from the origin, what is the value of the flux integral?
   Answer: $\frac{1}{3}$ First, find the partial derivatives: $\mathbf{r}_s = \langle -1, 1, 0 \rangle$ and $\mathbf{r}_t = \langle -1, 0, 1 \rangle$. The cross product is $\mathbf{r}_s \times \mathbf{r}_t = \langle 1, 1, 1 \rangle$, which points away from the origin. Next, substitute the parameterization into the velocity field: $\mathbf{v}(\mathbf{r}(s,t)) = \langle t, (1-s-t)+s, s^2 \rangle = \langle t, 1-t, s^2 \rangle$. The flux integral becomes $\iint \mathbf{v} \cdot (\mathbf{r}_s \times \mathbf{r}_t) \, ds \, dt = \iint (t + (1-t) + s^2) \, ds \, dt = \iint (1 + s^2) \, ds \, dt$. Evaluating: $\int_0^1 \int_0^{1-s} (1 + s^2) \, dt \, ds = \int_0^1 (1 + s^2)(1-s) \, ds = \int_0^1 (1 - s + s^2 - s^3) \, ds = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{1}{3}$.
6. The flux of the vector field $\mathbf{F} = \langle 2x, -y, 3z \rangle$ through the closed surface of a cube with vertices at $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$, $(2,2,0)$, $(2,0,2)$, $(0,2,2)$, and $(2,2,2)$ can be computed using the divergence theorem. However, if we wanted to verify this by computing the flux through each face separately, what would be the contribution from the face where $x = 2$?
   Answer: $\int_0^2 \int_0^2 4 \, dy \, dz = 16$ On the face where $x = 2$, the outward normal vector is $\mathbf{n} = \langle 1, 0, 0 \rangle$. The vector field on this face becomes $\mathbf{F}(2,y,z) = \langle 4, -y, 3z \rangle$. The flux through this face is $\iint \mathbf{F} \cdot \mathbf{n} \, dS = \iint 4 \, dS = \int_0^2 \int_0^2 4 \, dy \, dz = 16$. Choice B uses the wrong coefficient (2 instead of 4). Choice C omits the x-component and uses the wrong normal. Choice D incorrectly adds all components together.
7. The electric field $\mathbf{E}(x,y,z) = \frac{1}{(x^2+y^2+z^2)^{3/2}} \langle x, y, z \rangle$ represents a point charge at the origin. Consider the flux through the portion of the ellipsoid $\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$ where $z \geq 0$. Due to the symmetry and properties of this field, which approach would be most efficient for computing this flux?
   Answer: Use the divergence theorem on the region bounded by the upper ellipsoid and the elliptical disk in the xy-plane The vector field $\mathbf{E} = \frac{\mathbf{r}}{|\mathbf{r}|^3}$ has divergence zero everywhere except at the origin (where it's undefined). Using the divergence theorem on the region bounded by the upper ellipsoid and the elliptical disk $\frac{x^2}{4} + \frac{y^2}{9} \leq 1, z = 0$ gives us that the flux through the ellipsoid equals the flux through the disk (with opposite orientation). The flux through the disk is easier to compute. Choice A would work but is unnecessarily complicated. Choice B is incorrect because Gauss's law applies to closed surfaces, not open ones. Choice D is misleading because the scaling doesn't preserve the vector field's form.

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