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Sixth rule A, Almost all properties apply to the empty set, most of them to finite sets, and dividing by zero is a terrible idea. zero is not a number, $1 + 1 = 0$ in $\mathbb{Z}_2$, and $\sqrt{2}$ and $-\sqrt{2}$ are algebraically the same. Galois Theory and Partial differential equations are not for the faint of heart. Absolute numbers have no meaning, Apocalypse, Anawim, #justtothepoint.
The flux of a vector field $\vec{F}$ across a surface $S$ is defined as the dot product of the vector field and the unit normal vector $\hat{\mathbf{n}}$ to the curve. This is expressed mathematically as $\int_{C} \vec{F} \cdot \hat{\mathbf{n}}ds$.
In vector calculus, calculating the flux of a vector field across a complex 3D surface can be a computationally grueling task involving tedious parameterizations. The Divergence Theorem (also known as Gauss’s Theorem or the Gauss-Green Theorem) provides a brilliant workaround.
It provides a powerful tool for converting the surface integral of a vector field's flux through a closed surface $S$ into a significantly easier triple integral over the solid volume enclosed by that surface (Figure B).
Divergence Theorem. Let $D$ be a bounded, solid region in three-dimensional space, and let $S$ be the closed surface that forms the boundary of $D$. The surface $S$ is oriented with an outward-pointing unit normal vector $\hat{\mathbf{n}}.$ If $\vec{F}$ is a continuously differentiable vector field defined on an open region containing $D$, then the outward flux of the vector field $\vec{F}$ across the closed surface $S$ is given by the triple integral of the divergence of the vector field $\vec{F}$ over the region $D$: $\oiint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} \operatorname{div}\vec{F} dV \underbrace{=}_\text{Notation} \int \int \int_{D} \nabla \cdot\vec{F} dV$ where:
Imagine a two-dimensional vector field $\vec{F} = \langle P(x,y), Q(x,y) \rangle$ representing the velocity of flowing water. We want to know if fluid is being spontaneously created (a source) or drained away (a sink) at a specific coordinate $(x,y)$.
To measure this, we place a microscopic square control volume—a tiny box—around the point $(x,y)$. Let the side length of this square be an infinitesimally small number $h$.
By calculating the total net outflow of fluid through the four walls of this box, we can determine the flux density at that exact point. To find the outward flow across any boundary, we take the dot product of the velocity field $\vec{F}$ and the outward unit normal vector $\hat{\mathbf{n}}$.
🧩 1. Horizontal Flow Imbalance (The x-direction). Let’s analyze the left and right walls of our tiny box.
To find the net horizontal outflow, we add the contributions from both walls: $\text{Horizontal Outflow} \approx P(x+h,y)h - P(x,y)h = [P(x+h,y) - P(x,y)]h$
Here, we apply the foundational definition of the derivative. For a tiny step $h$, the change in the function $P$ is equal to its partial derivative $P_x$ multiplied by the step size: $P(x+h, y) - P(x,y) \approx \frac{\partial P}{\partial x}h$. Substituting this into our equation: $\boxed{\text{Horizontal Outflow} \approx \left( \frac{\partial P}{\partial x}h \right)h = \frac{\partial P}{\partial x}h^2}$
🧩 2. Vertical Flow Imbalance (The y-direction). Now, we apply the exact same logic to the top and bottom walls.
The net vertical outflow is: $\text{Vertical Outflow} \approx [Q(x, y+h) - Q(x,y)]h$ Using the partial derivative definition for $Q_y$: $\text{Vertical Outflow} \approx \left( \frac{\partial Q}{\partial y}h \right)h = \frac{\partial Q}{\partial y}h^2$
🌋 3. Normalizing the Flux. To find the total net outflow from the entire box, we add the horizontal and vertical contributions together: $\text{Total Net Outflow} \approx \frac{\partial P}{\partial x}h^2 + \frac{\partial Q}{\partial y}h^2 = \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right)h^2$
This formula tells us how much fluid is leaving a box of area $h^2$. However, we don’t want the outflow of a box; we want the outflow property of a single point. To get the normalized flux density, we divide the total net outflow by the area of the box ($h^2$) and take the limit as the box shrinks to zero: $\lim_{h \to 0} \frac{\text{Total Net Outflow}}{\text{Area}} = \frac{\left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right)h^2}{h^2} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} = \nabla \cdot \vec{F}$
This elegantly proves that divergence is literally the normalized flux imbalance at a single microscopic point. Divergence measures sources and sinks within the field. It is literally the normalized flux imbalance at a point
⭐ The divergence of a vector field $\vec{F}$ provides a measure of how much “stuff” is expanding or contracting at a point in space. This “stuff” could represent various physical quantities, such as fluid, heat, or electromagnetic fields, depending on the context. Specifically, the divergence quantifies the net rate at which a quantity is flowing out of or into a small volume surrounding a point.
Imagine an incompressible fluid (meaning that the mass of the fluid occupies a fixed volume) moving through space, with its velocity at any point represented by the vector field $\vec{F}$. The fluid could be water, air, or any other substance that flows. The divergence of $\vec{F}$ tells us whether the fluid is “expanding” out of a point (like a source) or “contracting” into a point (like a sink).
A positive divergence $\nabla \cdot\vec{F} \gt 0$ indicates that there is a net outflow of fluid from the point, implying the presence of a source at that point. Conversely, a negative divergence $\nabla \cdot\vec{F} \lt 0$ suggests a net inflow of fluid towards the point implying a sink.
The Divergence Theorem states that $\oiint_S \vec{F} \cdot d\vec{S}$ the total flux of the vector field $\vec{F}$ through the boundary surface S, representing the net outflow of the fluid through S, is equal to the integral of the divergence of the vector field $\vec{F}$ over the region D, $ \int \int \int_{D} \nabla \cdot\vec{F} dV$, that is, the total amount of sources and sinks within the region D (the sum of all the sources minus the sinks).
This relationship is immensely useful because, in many practical situations, calculating the flux over a surface can be difficult and quite challenging, but integrating over a region (especially if the divergence is simple or uniform) might be easier.
Notation: The Nabla operator ($\nabla$). The “Nabla” or “del” operator, denoted by $\nabla$ , is a vector differential operator used to express various mathematical operations involving differentiation and gradients in vector calculus, $\nabla = \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z} \rangle , \nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle $, and more importantly, $\nabla \cdot\vec{F} = \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z} \rangle \cdot \langle P, Q, R \rangle = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = \operatorname{div}(\vec{F})$.
The divergence $\nabla \cdot\vec{F}$ is a scalar quantity that describes the net rate of flow out of a point. It’s like taking the dot product of the “del” operator with the vector field.
Let’s apply the divergence theorem to a simple example. Let the vector field be the radial field $\vec{F} = \langle x, y, z \rangle$. Let the solid region $D$ be a sphere of radius $a$ centered at the origin, meaning the boundary surface $S$ is given by $x^2 + y^2 + z^2 = a^2$.
On the surface of a sphere of radius $a$, the outward unit normal vector points directly away from the origin: $\hat{\mathbf{n}} = \frac{\langle x, y, z \rangle}{a}$.
The dot product of the vector field and the normal vector is: $\vec{F} \cdot \hat{\mathbf{n}} = \langle x, y, z \rangle \cdot \frac{\langle x, y, z \rangle}{a} = \frac{x^2 + y^2 + z^2}{a}$
Because we are strictly evaluating this on the surface of the sphere, we know $x^2 + y^2 + z^2 = a^2$. Therefore, $\vec{F} \cdot \hat{\mathbf{n}} = \frac{a^2}{a} = a$
The flux integral becomes: $\oiint_S (\vec{F} \cdot \hat{\mathbf{n}}) \, dS = \oiint_S a \, dS = a \oiint_S dS$
The integral of $dS$ over a sphere is just its surface area ($4\pi a^2$). $\text{Flux} = a(4\pi a^2) = 4\pi a^3$
Let’s bypass the surface entirely and compute the divergence of $\vec{F}$: $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z) = 1 + 1 + 1 = 3$
Now, integrate this scalar over the solid spherical volume $D$: $\iiint_{D} 3 \, dV = 3 \iiint_{D} dV$
The integral of $dV$ is simply the total volume of the sphere ($\frac{4}{3}\pi a^3$). Then, $\text{Flux} = 3 \left( \frac{4}{3}\pi a^3 \right) = 4\pi a^3$. Conclusion: Both methods yield $4\pi a^3$, but the Divergence Theorem reduced a multivariable calculus parameterization problem into basic geometry!
The Divergence Theorem is a fundamental result in vector calculus that connects the flux of a vector field across a closed surface to the divergence of the field inside the region enclosed by that surface. Here’s a step-by-step and rigorous proof of the Divergence Theorem.
Step 1. Simplifying the problem
We begin by proving the Divergence Theorem for a specific case where the vector field is as follows $\vec{F} = \langle 0, 0, R \rangle$, meaning that the field has only a z-component. We will later generalize this proof to any vector field by considering each component of the vector field separately.
We aim to prove that $\oiint_S \langle 0, 0, R\rangle \cdot\hat{\mathbf{n}}dS = \int \int \int_{D} \frac{\partial R}{\partial z}dV = \int \int \int_{D} R_zdV$ where D is the region enclosed by the surface S, and $\hat{\mathbf{n}}$ is the outward-pointing unit normal vector on S.
Assume the region D is vertically simple, meaning it is bounded on the left and right by vertical lines (it has a clear, flat-ish top and bottom; more formally, every line parallel to the z-axis that meets D, does so in a single segment) and bounded below and above by the surfaces $z = z_1(x, y), z_2(x, y)$ respectively, and the projection of $D$ onto the $xy$-plane is a region $U$ (see Figure C)
First, let’s consider the right-hand side of the equation and compute the volume integral: $\int \int \int_{D} R_zdV = \int \int_{U} (\int_{z_1(x, y)}^{z_2(x, y)} \frac{\partial R}{\partial z}dz)dxdy$
By the Fundamental Theorem of Calculus, the inner integral evaluates exactly to the function $R$ evaluated at the upper and lower boundaries: $\iint_U \big[ R(x,y,z_2(x,y)) - R(x,y,z_1(x,y)) \big]\,dx\,dy$. Therefore, $\iiint_{D} \frac{\partial R}{\partial z} \, dV = \iint_{U} \Big[ R(x,y,z_2(x,y)) - R(x,y,z_1(x,y)) \Big] dx\,dy \quad \mathbf{[Equation \; 1]}$
Now let’s compute the left side. The total closed boundary $S$ consists of three parts: the top, the bottom, and the sides. $\iint_{S_{\text{bottom}}} \langle 0, 0, R\rangle \cdot\hat{\mathbf{n}}dS + \iint_{S_{\text{top}}} \langle 0, 0, R\rangle \cdot\hat{\mathbf{n}}dS + \iint_{S_{\text{sides}}} \langle 0, 0, R\rangle \cdot\hat{\mathbf{n}}dS$
Adding the contributions of the top, bottom, and sides together gives the total surface integral: $\oiint_S \langle 0, 0, R \rangle \cdot \hat{\mathbf{n}} \, dS = \iint_{U} \Big[ R(x,y,z_2(x,y)) - R(x,y,z_1(x,y)) \Big] dx\,dy \quad \mathbf{[Equation \; 2]}$
Because [Equation 1] equals [Equation 2], the theorem is proven for the z-component!
Step 2. General Case. For a general vector field $\vec{F}= \langle P, Q, R \rangle$, the proof follows a similar approach. The idea is to treat each component $P$, $Q$, and $R$ separately and apply the same reasoning to each component of $\vec{F}$ as in the simplified case. The total flux through the surface S is: $\oiint_S \vec{F} \cdot d\vec{S} = \oiint_S \langle P, Q, R\rangle \cdot\hat{\mathbf{n}}dS = \int \int \int_{D} div \vec{F} dV = \int \int \int_{D} (P_x+Q_y+R_z)dV = \int \int \int_{D} \nabla \cdot\vec{F}dV$
Step 3. If the region D is not vertically simple (like a twisted, entangled, doughnut -torus-), we can divide or decompose it into simple (smaller) pieces or subregions that are vertically simple (Figure D). By applying the Divergence Theorem to each subregion and summing the results, we account for the entire volume D. The contributions from the shared boundaries between subregions cancel out because the normal vectors on these boundaries point in opposite directions (the outward normal for one sub-region is the inward normal for its neighbor), leaving only the flux across the true outer boundary $S$, meaning $\vec{F}\cdot\hat{\mathbf{n}}\,dS + \vec{F}\cdot(-\hat{\mathbf{n}})\,dS = 0$ (outward normal of one subregion = inward normal of its neighbor).
Conclusion: The proof shows that the flux of a vector field $\vec{F}$ through a closed surface $S$ is equal to the triple integral of the divergence of $\vec{F}$ over the region $D$. This result is foundational in Calculus, and has broad applications in physics, engineering, and mathematics, linking the behavior of a vector field over a surface to its behavior within a volume.
The Divergence Theorem (Gauss’s Theorem) is one of the most powerful tools in vector calculus. It states that the outward flux of a vector field $\vec{F}$ across a closed surface $S$ is perfectly equal to the volume integral of the field’s divergence over the enclosed solid region $D$: $\boxed{\oiint_S \vec{F} \cdot d\vec{S} = \iiint_{D} (\nabla \cdot \vec{F}) \, dV}$
Let the vector field be $\vec{F} = \langle x - y, x + z, z - y \rangle$. Let the closed surface $S$ consist of the conical surface $x^2 + y^2 = z^2$ (for $0 \le z \le 1$) and the circular flat top at $z = 1$ (see the figure A). Assume this surface $S$ is positively oriented (outward, i.e., for closed surfaces enclosing a volume $D$, the normal vector points out from the enclosed volume). Verify the Divergence Theorem by computing both the volume integral and the surface integral.
🧩 Step 1. Compute the Divergence and Volume Integral
Calculate the Divergence of $\vec{F}$: $\nabla \cdot\vec{F} = P_x +Q_y + R_z =[\text{Given: P = x−y, Q = x+z, R = z−y}] = \frac{\partial }{\partial x}(x-y)+\frac{\partial }{\partial y}(x+z)+\frac{\partial }{\partial z}(z-y) = 1 + 0 + 1 = 2$
Compute the volume integral of the Divergence over the region D. Because the divergence is a constant $2$, the volume integral becomes incredibly simple: $\int \int \int_{D} div \vec{F} dV = \int \int \int_{D} (P_x +Q_y + R_z)\, dV = \int \int \int_{D} 2 = 2\cdot Vol(D)$
The formula for the volume of a right circular cone is $\operatorname{Vol}(D) = \frac{\pi r^2 h}{3} = \frac{\pi}{3}$=[In our case, height (h) = Radius at z = 1 (r = 1) = 1] $\frac{\pi}{3}$
Therefore, the volume integral is $\int \int \int_{D} div \vec{F} dV = 2\cdot vol(D) = \frac{2\pi}{3}$
🧩 Step 2. Compute the Surface Integral.
The surface integral $\oiint_S \vec{F} \cdot d\vec{S}$ is the total flux of $\vec{F}$ through S. To compute this, we are going to divide it into its two constituent parts $S_1$ and $S_2$:
Surface Integral over $S_1$
Parametrization of the Surface $S_1$: The circular top at z = 1.
To parametrize $S_1$, we can use polar coordinates in the plane z = 1: $\vec{r}(u, v) = \langle u\cos(v), u\sin(v), 1 \rangle $ where $0 \le u \le 1$ (radius), $0 \le v \le 2\pi$ (angle).
Plug these values in the vector field. $\vec{F}(\vec{r}(u, v)) = \langle x−y, x+z, z−y \rangle = \langle u\cos(v)-u\sin(v), u\cos(v)+1, 1-u\sin(v) \rangle $
Compute the Partial Derivatives. We need to find the partial derivatives of $\vec{r}$ with respect to $u$ and $v$: $\frac{\partial \vec{r}}{\partial u} = \langle \cos(v), \sin(v), 0 \rangle , \frac{\partial \vec{r}}{\partial v} = \langle -u\sin(v), u\cos(v), 0 \rangle $
Compute the Cross Product
$\frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k}\\\ \cos(v) & \sin(v) & 0\\\ -u\sin(v) & u\cos(v) & 0\end{vmatrix} = \langle 0, 0, u \rangle $. This vector is the differential surface vector $d\vec{S}$ for $S_1$. Since $S_1$ is the top of the cone, the normal vector points upwards along the positive z-axis.
Calculate the flux through $S_1$
The flux through $S_1$ is: $\operatorname{Flux}_{S_1} = \int \int_{S_1} \vec{F}\cdot d\vec{S_1} = \int \int_{S_1} \vec{F}\cdot \hat{\mathbf{n}}dS = \int \int_{D} \vec{F}\cdot (\frac{\partial \vec{r}}{\partial u}x\frac{\partial \vec{r}}{\partial v})dudv$
$= \int_{0}^{2\pi}\int_{0}^{1} \langle u\cos(v)-u\sin(v), u\cos(v)+1, 1-u\sin(v)\rangle \cdot\langle 0, 0, u \rangle dudv = \int_{0}^{2\pi}\int_{0}^{1} (u-u^ 2\sin(v))dudv$
Evaluate the inner integral: $\int_{0}^{1} (u-u^ 2\sin(v))du = \frac{u^2}{2}-\frac{u^3sin(v)}{3}\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{3}sin(v)$
Evaluate the outer integral: $\int_{0}^{2\pi} \frac{1}{2}-\frac{1}{3}sin(v)dv = \frac{1}{2}v\bigg|_{0}^{2\pi} = \pi$.
Notice the second part of the integral $\int_{0}^{2\pi} -\frac{1}{3}sin(v)dv = 0$ since $\sin(v)$ integrates to 0 over a full period. $\operatorname{Flux}_{S_1} = \pi$.
Surface Integral over $S_2$ (Conical Surface)
Parametrization of the Surface $S_2$
The conical surface $S_2$ ($x^2+y^2=z^2$) can be parametrized by $\vec{r}(u, v) = \langle u\cos(v), u\sin(v), u \rangle $, where $0 \le u \le 1$ (radius), $0 \le v \le 2\pi$ (angle)
$x^2+y^2=z^2 \implies u^2\cos^2(v) + u^2\sin^2(v) = z^2 \implies z = \sqrt{u^2\cos^2(v)+u^2\sin^2(v)} = \sqrt{u^2} = u$
Plug these values in the vector field. $\vec{F}(\vec{r}(u, v)) = \langle x−y, x+z, z−y \rangle = \langle u\cos(v)-u\sin(v), u\cos(v)+u, u-u\sin(v) \rangle $
Compute the Partial Derivatives. We need to find the partial derivatives of $\vec{r}(u, v)$ with respect to u and v:
$\frac{\partial \vec{r}}{\partial u} = \langle cos(v), sin(v), 1 \rangle , \frac{\partial \vec{r}}{\partial v} = \langle -u\sin(v), u\cos(v), 0 \rangle $
Compute the Cross Product
$\frac{\partial \vec{r}}{\partial u}x\frac{\partial \vec{r}}{\partial v} = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k}\\\ cos(v) & sin(v) & 1\\\ -u\sin(v) & u\cos(v) & 0\end{vmatrix} = \langle -u\cos(v), -u\sin(v), u \rangle $
Notice that the $z$-component is positive ($u$). For a cone opening upward, a positive $z$ points inward toward the z-axis. To get the outward normal, we must multiply by $-1$: $\langle u\cos(v), u\sin(v), -u \rangle, dS_2 = \langle u\cos(v), u\sin(v), -u \rangle\,du \, dv$.
Calculate the flux
$\operatorname{Flux}_{S_2} = \int \int_{D} \vec{F}\cdot d\vec{S_2} = \int \int_{D} \vec{F}\cdot \hat{\mathbf{n}}dS = \int \int_{D} \vec{F}\cdot (\frac{\partial \vec{r}}{\partial u}x\frac{\partial \vec{r}}{\partial v})dudv = \int_{0}^{2\pi}\int_{0}^{1} \langle u\cos(v)-u\sin(v), u\cos(v)+u, u-u\sin(v)\rangle \cdot\langle u\cos(v), u\sin(v), -u \rangle dudv$
Take the dot product $\vec{F} \cdot d\vec{S}_2$: $= \int_{0}^{2\pi}\int_{0}^{1} (u^2\cos^2(v)-u^2\cos(v)\sin(v)+u^2\cos(v)\sin(v)+u^ 2\sin(v)-u^2+u^ 2\sin(v))dudv = \int_{0}^{2\pi}\int_{0}^{1} (u^2\cos^2(v)+2u^ 2\sin(v)-u^2)dudv$
Let’s evaluate each of these integrals separately.
$\int_{0}^{2\pi}\int_{0}^{1} u^2\cos^2(v)dudv = \int_{0}^{2\pi}\cos^2(v)\frac{u^3}{3}\bigg|_{0}^{1}$
$= \frac{1}{3}\int_{0}^{2\pi} cos^2(v)dv = \frac{1}{3}\int_{0}^{2\pi} \frac{1+cos(2v)}{2}dv = \frac{1}{6}\int_{0}^{2\pi} (1+cos(2v))dv = \frac{1}{6}\int_{0}^{2\pi} dv = \frac{1}{6}v\bigg|_{0}^{2\pi} = \frac{π}{3}$. 💣 Notice: The integral of $\cos(2v)$ over one period $2\pi$ is zero.
$\int_{0}^{2\pi}\int_{0}^{1} 2u^ 2\sin(v)dudv = 2\int_0^{2\pi} \sin v \left(\int_0^1 u^2\,du\right) dv = 2\int_{0}^{2\pi}sin(v) (\frac{u^3}{3})\bigg|_{0}^{1}$
$=\frac{2}{3}\cdot \int_{0}^{2\pi}sin(v)dv = 0. $
$\int_{0}^{2\pi}\int_{0}^{1}-u^2dudv = \int_{0}^{2\pi}-\frac{u^3}{3}\bigg|_{0}^{1}$
$= -\frac{1}{3}\cdot \int_{0}^{2\pi}dv = \frac{1}{3}v\bigg|_{0}^{2\pi} = -\frac{2\pi}{3}$
Summing these parts: $\text{Flux}_{S_2} = \frac{1}{3}(\pi + 0 - 2\pi) = -\frac{\pi}{3}$
Total Flux Across S
Conclusion of Verification. The total surface flux is the sum of the two boundaries: $\oiint_S \vec{F} \cdot d\vec{S} = \text{Flux}_{S_1} + \text{Flux}_{S_2} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$. We have computed both the volume integral and the surface integral. Both integrals yield the same value, $\frac{2\pi}{3}$, so we have verified the divergence theorem.
💥 Notice a critical detail: The paraboloid is an open surface (it has a gaping hole at the bottom where $z=1$). The Divergence Theorem only works for closed surfaces. To use the theorem as a shortcut, we artificially “cap off” the surface by adding a flat base disk at $z=1$. This creates a closed solid region $D$. By the Divergence Theorem: $\text{Flux}_{Total} = \text{Flux}_{Paraboloid} + \text{Flux}_{Base} = \iiint_{D} (\nabla \cdot \vec{F}) \, dV$.
Therefore, to find the flux across just the paraboloid, we can compute the volume integral, and simply subtract the flux of the artificial base!
🧩 Step 1: Compute the Volume Integral ($\text{Flux}_{Total}$)
The paraboloid $x^2 + y^2 + z = 2$ is a downward-opening surface with its vertex at $(0, 0, 2)$.
At $z = 2$, the radius is $0$ (point). At $z = 1$, the radius is $1$. At $z = 0$, the radius is $\sqrt{2}$. As $z$ continues to decrease, the circles grow larger, reflecting the behavior of the paraboloid.
The flux of a vector field through a surface is a measure of how much of the field “flows” through the surface. Mathematically, it is represented as: $\oiint_S \vec{F} \cdot d\vec{S}$ where $\vec{F}$ is the vector field, $S$ is the surface, and $d\vec{S}$ is a vector representing an infinitesimal element of the surface, pointing outward (for a closed surface).
In this problem, instead of directly calculating the flux through the surface, we can use the Divergence Theorem, which relates the flux through a closed surface to the divergence of the vector field over the volume enclosed by the surface. The Divergence Theorem states: $\oiint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} div \vec{F} dV =$[Notation] $ \int \int \int_{D} \nabla \cdot\vec{F} dV$ where $D$ is the volume enclosed by the surface $S$, and $\nabla \cdot\vec{F}$ is the divergence of the vector field.
Calculate the Divergence of $\vec{F}$: $\nabla \cdot\vec{F} = P_x +Q_y + R_z$
Given $P = z\arctan(y^2), Q = z^3\ln(x^2+1), R = z$: $= \frac{\partial}{\partial x}(z\arctan(y^2))+\frac{\partial }{\partial y}(z^3ln(x^2+1))+\frac{\partial }{\partial z}(z) =$
Because the first two components do not contain $x$ and $y$ respectively, their partial derivatives are exactly $0$: $\nabla \cdot\vec{F} = 0 + 0 + 1 = 1$.
Determine the Region of Integration. The intersection of the paraboloid $x^2 + y^2 + z = 2$ with the plane $z = 1$ gives the boundary of the region: $x^2 + y^2 + 1 = 2 \implies x^2 + y^2 = 1$. This is a circle of radius $1$ in the $xy$-plane.
We need to calculate $\int \int \int_{D} \nabla \cdot\vec{F} dV$, the total flux over the solid of the paraboloid between $z = 1$ and $z = 2$, then subtract flux on the base $z = 1$.
Given the symmetry about the $z$-axis, it’s convenient to switch to cylindrical coordinates. In cylindrical coordinates: $x = r\cos(\theta), y = r\sin(\theta), z = z$, and $dV = r\, dr\, d\theta\, dz$.
The limits of integration are:
Compute the Volume Integral:
$ \int \int \int_{D} \nabla \cdot\vec{F} dV = \int_{0}^{2\pi} \int_{0}^{1} \int_{1}^{2-r^2} 1\cdot rdzdrdθ$
Inner integration with respect to z: $\int_{1}^{2-r^2} rdz = r\int_{1}^{2-r^2} dz = rz\bigg|_{1}^{2-r^2}= r[(2-r^2)-1] = r(1-r^2)$
Next, integrate with respect to r: $\int_{0}^{1} r(1-r^2)dr = \int_{0}^{1} (r-r^3)dr = (\frac{r^2}{2}-\frac{r^4}{4})\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{4} = \frac{2}{4}-\frac{1}{4} = \frac{1}{4}$
Finally, integrate with respect to $\theta$: $\int_{0}^{2\pi} \frac{1}{4}d\theta = \frac{1}{4}θ\bigg|_{0}^{2\pi} = 2\pi\cdot \frac{1}{4} = \frac{\pi}{2}$. So, the total flux out of the closed solid is $\frac{\pi}{2}$.
🧩 Step 2. Calculate the flux on the “base” $z = 1$.
The “base” of the surface is the disk at $z = 1$. To ensure the normal vector points outward from our newly formed solid, it must point straight down: $\hat{\mathbf{n}} = \langle 0, 0, -1 \rangle$.
Evaluate the flux integral over the base: $\iint_{S_{base}} \vec{F} \cdot \hat{\mathbf{n}} \, dS = \iint_{S_{base}} \langle \dots, \dots, z \rangle \cdot \langle 0, 0, -1 \rangle \, dS$
Because the dot product multiplies the horrific, demoralizing $x$ and $y$ components by $0$, they vanish entirely. We are left with: $\iint_{S_{base}} -z \, dS$
Since we are on the plane $z=1$, this simplifies to: $\iint_{S_{base}} -1 \, dA = -1 \times \text{Area of the Disk}$
The disk has a radius of $1$, so its area is $\pi(1)^2 = \pi$. Therefore, $\text{Flux}_{Base} = -\pi$
🧩 Step 3: Solve for the Paraboloid. We know that $\text{Flux}_{Total} = \text{Flux}_{Paraboloid} + \text{Flux}_{Base}$.Substitute our calculated values: $\frac{\pi}{2} = \text{Flux}_{Paraboloid} + (-\pi)$
$\text{Flux}_{Paraboloid} = \frac{\pi}{2} + \pi = \frac{3\pi}{2}$
By strategically closing the surface, we bypassed an agonizing surface integral and found the solution using basic volume integration and a trivial flat base.
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