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. 0 is not a number, 1 plus one equals zero in 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.
A vector field is an assignment of a vector $\vec{F}$ to each point (x, y) in a space, i.e., $\vec{F} = M\vec{i}+N\vec{j}$ where M and N are functions of x and y.
A vector field on a plane can be visualized as a collection of arrows, each attached to a point on the plane. These arrows represent vectors with specific magnitudes and directions.
Work is defined as the energy transferred when a force acts on an object and displaces it along a path. In the context of vector fields, we calculate the work done by a force field along a curve or trajectory C using a line integral. The work done by a force field $\vec{F}$ along a curve C is: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy = \int_{C} \vec{F}·\hat{\mathbf{T}}ds$, where $\hat{\mathbf{T}}$ is the unit tangent vector.
A vector field is conservative if there exist a scalar function such that $\vec{F}$ = ∇f (the vector field is its gradient). This scalar function is known or referred to as the potential function associated with the vector field.
Theorem. Fundamental theorem of calculus for line integral. If $\vec{F}$ is a conservative vector field in a simply connected region of space (i.e., a region with no holes), and if f is a scalar potential function for $\vec{F}$ in that region, then $\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(P_1)-f(P_0)$ where P_{0} and P_{1} are the initial and final points of the curve C, respectively.
Conservative force. A force $\vec{F}$ is considered conservative if the work done by the force around any closed curve C is zero. Mathematically, this is expressed as $\int_{C} \vec{F}·d\vec{r} = 0$.
Path independence. A force field is path-independent, meaning the work done by the force in moving an object from one point to another is the same, regardless of the path taken between the two points.
Gradient Field. A vector field $\vec{F}$ is a gradient field if it can be expressed as the gradient of a scalar potential function f. In mathematical terms, this means $\vec{F} = ∇f$, where f is a scalar function and the vector field $\vec{F}$ has components $\vec{F} = ⟨M, N⟩ = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}⟩ = ⟨f_x, f_y⟩$. Here, f is the potential function associated with the vector field $\vec{F}$. The lineal integral of $\vec{F}$ along a path C measures the work done by the vector field in moving an object along the path C. If $\vec{F}$ is a gradient field, then: $\int_{C} \vec{F}·d\vec{r} = f(P_1)-f(P_0).$
Exact differential. In the context of differential forms, a differential expression Mdx+Ndy is called an exact differential if there exist a scalar function f(x, y) such that df = $\frac{∂f}{∂x}dx+\frac{∂f}{∂y}dy$ = Mdx + Ndy. This implies that the vector field $\vec{F} = ⟨M, N⟩$ is conservative, and there exists a potential function f such that $\vec{F} = ∇f$.
If a differential is exact, then the line integral of $\vec{F}$ over any path C can be evaluated by simply finding the difference in the potential function values at the endpoints of the path.
These four properties —conservative force, path independence, gradient field, and exact differential— are different perspectives of the same fundamental concept: the conservativeness of a vector field.
The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If $\frac{∂M}{∂y} = \frac{∂N}{∂x}$, and if the functions M and N have continuous first partial derivatives across the entire domain D, and the domain D is open and simply connected, then $\vec{F}$ is a gradient field, meaning it is conservative.
Green’s Theorem. If C is a positively oriented (counterclockwise) simple closed curve enclosing a region R, and $\vec{F} = ⟨M(x, y), N(x,y)$ is a vector field that is defined and has continuous partial derivatives on an open region containing R, then Green's Theorem states that: $\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R curl(\vec{F}) dA ↭ \oint_C Mdx + N dy = \iint_R (N_x-M_y)dA = \iint_R (\frac{∂N}{∂x}-\frac{∂M}{∂y})dA$ where
Green’s Theorem for Flux. If C is a positively oriented (counterclockwise), simple closed curve that encloses a region R and if $\vec{F} = ⟨P, Q⟩$ is a continuously differentiable vector field defined on an open region that contains R, then the flux of $\vec{F}$ across C is equal to the double integral of the divergence of $\vec{F}$ over R. Mathematically, this is expressed as: $\oint_C \vec{F}·\vec{n}d\vec{s} = \int \int_{R} div \vec{F}dA$ where the divergence of $\vec{F}$ is given by $div \vec{F} = P_x + Q_y = \frac{∂P}{∂x}+\frac{∂Q}{∂y}$.
Triple integrals are powerful tools in calculus, particularly useful for calculating various physical quantities over a region in three-dimensional space. These integrals allow us to compute volume, mass, and other physical quantities over a solid region bounded by surfaces in 3d space.
A triple integral extends the concept of integration to three dimensions. When you have a function f(x, y, z) that varies over a 3D region R, you can use a triple integral to sum up the values of f(x, y, z) over that region.
The triple integral of a function f(x, y, z) over a region R in three-dimensional space is expressed as $\int \int \int_{R} f(x, y, z)dV$ where
To evaluate a triple integral, you need to set up the bounds of integration, which describe the region R over which you’re integrating. The process involves: (1) Identifying the region. (2) Choosing the order of integration: In Cartesian coordinates, you typically integrate with respect to x, y, and z. However, the order can be rearranged. (3) Setting up the limits: These limits are the ranges for x, y, and z.
Triple integrals can be interpreted or computed in various coordinate systems depending on the symmetry of the region R. Choosing the appropriate coordinate system is crucial as it can either simplify or complicate the calculation by aligning the integration limits with the geometry of the region.
Conversion to Cartesian Coordinates: z = z, x = rcos(θ), y = rsin(θ), and the differential volume element is: dV = rdrdθdz.
They are ideal for regions with spherical symmetry such as spheres, hemispheres, or cones. In this system, the coordinates or parameters are:
The differential volume element dV = ρ^{2}sin(ϕ)dρdθdϕ. Conversion to Cartesian Coordinates: x = rcos(θ) = ρsin(φ)cos(θ), y = rsin(θ) = ρsin(φ)sin(θ), and z = ρcos(φ).
The divergence theorem, also known as Gauss’s theorem or the Gauss-Green theorem, is a fundamental result in Calculus. It provides a powerful tool for converting a flux of a vector field $\vec{F}$ through a closed surface S enclosing a region in space D into the divergence of the vector field within the region D. (Figure B).
Divergence Theorem. Let D be a bounded region in three-dimensional space and let S be the closed surface that forms the boundary of D. The surface S is oriented with and 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: $\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} div \vec{F} dV =$[Notation] $ \int \int \int_{D} ∇·\vec{F} dV$
where:
Let’s apply the divergence theorem to a simple example. Consider the vector field $\vec{F}= z\hat{\mathbf{k}}$ where ${\mathbf{k}}$ is the unit vector in the z-direction. Suppose we want to calculate the flux of $\vec{F}$ through a sphere of radius a.
Using the Divergence Theorem: $\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} ∇·\vec{F} dV = \int \int \int_{D} ∇·(z\hat{\mathbf{k}}) dV$
Compute the Divergence of $\vec{F}$. The divergence of $\vec{F}$ is $∇·(z\hat{\mathbf{k}}) = \frac{∂z}{∂z} = 1$. The divergence of $\vec{F}$ is 1, which is constant throughout the region D.
Apply the Divergence Theorem. Thus, the flux through the sphere is: $\int \int \int_{D} 1dV = vol(D) = \frac{4}{3}πa^3.$
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. It quantifies the sources and sinks within the field.
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 $∇⋅\vec{F}>0$ 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 $∇⋅\vec{F}<0$ suggests a net inflow of fluid towards the point implying a sink.
The Divergence Theorem states that $\oint_S \vec{F} \cdot d\vec{S}$ 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} ∇·\vec{F} dV$, that is, the total amount of sources and sinks within the region D.
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 (∇). The “Nabla” or “del” operator, denoted by ∇, is a vector differential operator used to express various mathematical operations involving differentiation and gradients in vector calculus, $∇ = ⟨\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}⟩, ∇f = ⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z}⟩$, and more importantly, $∇·\vec{F} =⟨\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}⟩·⟨P, Q, R⟩ = \frac{∂P}{∂x} + \frac{∂Q}{∂y} + \frac{∂R}{∂z} = div(\vec{F})$.
The divergence $∇·\vec{F}$ 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.
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 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 $\vec{F} = ⟨0, 0, R⟩$, 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 $\oint_S ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS = \int \int \int_{D} \frac{∂R}{∂z}V = \int \int \int_{D} R_zdV$ 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 and bounded below and above by the surfaces z = z_{1}(x, y), z = 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:
$\int \int \int_{D} R_zdV = \int \int_{U} (\int_{z_1(x, y)}^{z_2(x, y)} \frac{∂R}{∂z}dz)dxdy$ =[Applying the Fundamental Theorem of Calculus to the inner integral gives:] $\int \int_{U} (R(x, y, z_2(x, y)-R(x, y, z_1(x, y))))dxdy$ 🚀
Next, consider the surface integral on the left-hand side of the equation:
$\oint_S ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS$ =[The surface S can be broken down into the contributions from the bottom surface S_{bottom}, the top surface S_{top}, and the sides S_{sides}:] $\oint_{SBottom} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS + \oint_{Stop} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS + 2\oint_{SSides} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS$ [Recall that we have seen that we can compute $\hat{\mathbf{n}}·dS$ when it is in a graph z = z_{2}(x, y), $\hat{\mathbf{n}}·dS = ⟨-\frac{∂z_2}{∂x}, -\frac{∂z_2}{∂y}, 1⟩dxdy$]
$\oint_{Stop} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS = \oint_{Stop} ⟨0, 0, R⟩·⟨-\frac{∂z_2}{∂x}, -\frac{∂z_2}{∂y}, 1⟩dxdy = \oint_{Stop} Rdxdy = \int \int_{U} R(x, y, z_2(x, y))dxdy$
Mutates Mutandis, when we want to compute the button surface, $\hat{\mathbf{n}}·dS = -⟨-\frac{∂z_1}{∂x}, -\frac{∂z_1}{∂y}, 1⟩dxdy = ⟨+\frac{∂z_1}{∂x}, +\frac{∂z_1}{∂y}, -1⟩dxdy$ and this is so because we have established that $\hat{\mathbf{n}}$ is the downwards-pointing normal vector at the bottom surface.
$\oint_{SBottom} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS = \oint_{Stop} ⟨0, 0, R⟩·⟨\frac{∂z_2}{∂x}, \frac{∂z_2}{∂y}, -1⟩dxdy = \oint_{Stop} -Rdxdy = \int \int_{U} -R(x, y, z_1(x, y))dxdy$
$\oint_{SSides} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS = 0$ because on the side surfaces, the vector ⟨0, 0, R⟩ is parallel to the z-axis, and the normal vector $\hat{\mathbf{n}}$ lies in the xy-plane, making it perpendicular to the sides. Therefore, their dot product is zero, contributing nothing to the flux, so flux through the sides is zero.
Combining results. Summing the contributions from all parts, the surface integral becomes:
$\oint_S ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS = \oint_{SBottom} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS + \oint_{Stop} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS + 2\oint_{SSides} ⟨0, 0, R⟩·\hat{\mathbf{n}}·dS = \int \int_{U} (R(x, y, z_2(x, y)-R(x, y, z_1(x, y))))dxdy$ 🚀
This matches the volume integral computed earlier for the right-hand side of the equation, recall $\int \int \int_{D} R_zdV = \int \int_{U} (\int_{z_1(x, y)}^{z_2(x, y)} \frac{∂R}{∂z}dz)dxdy$ =[Applying the Fundamental Theorem of Calculus:] $\int \int_{U} (R(x, y, z_2(x, y)-R(x, y, z_1(x, y))))dxdy$. Thus, the Divergence Theorem is proven for the specific case where $\vec{F} = ⟨0, 0, R⟩$.
Step 2. General Case. For a general vector field $\vec{F}= ⟨P, Q, R⟩$, 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:
$\oint_S \vec{F} \cdot d\vec{S} = \oint_S ⟨P, Q, R⟩·\hat{\mathbf{n}}dS = \int \int \int_{D} div \vec{F} dV = \int \int \int_{D} (P_x+Q_y+R_z)dV = \int \int \int_{D} ∇·\vec{F}dV$
Step 3. If the region D is not vertically simple, 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.
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 (also known as Gauss’s Theorem) is a fundamental result in calculus that relates the flux of a vector field through a closed surface to the divergence of the field within the region bounded by that surface. Let D be the solid cone enclosed by S. We need to verify that:
$\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} div \vec{F} dV =$[Notation] $ \int \int \int_{D} ∇·\vec{F} dV$
by calculating each integral separately.
Calculate the Divergence of $\vec{F}$: $∇⋅\vec{F} = P_x +Q_y + R_z =[\text{Given: P = x−y, Q = x+z, R = z−y}] = \frac{∂}{∂x}(x-y)+\frac{∂}{∂y}(x+z)+\frac{∂}{∂z}(z-y) = 1 + 0 + 1 = 2$
Compute the volume integral of the Divergence over the region D:
$\int \int \int_{D} div \vec{F} dV = \int \int \int_{D} P_x +Q_y + R_z = \int \int \int_{D} 2 = 2·Vol(D)$
The formula for the volume of a right circular cone is Vol(D) = $\frac{π·r^2·h}{3}$=[In our case, height (h) = Radius at z = 1 (r = 1) = 1] $\frac{π}{3}$
Therefore, $\int \int \int_{D} div \vec{F} dV = 2·vol(D) = \frac{2π}{3}$
The surface integral $\oint_S \vec{F} \cdot d\vec{S}$ 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: $\vec{r}(u, v) = ⟨ucos(v), usin(v), 1⟩$ where 0 ≤ u ≤ 1 (radius), 0 ≤ v ≤ 2π (angle).
Plug these values in the vector field. $\vec{F(\vec{r}(u, v))} = ⟨x−y, x+z, z−y⟩ = ⟨ucos(v)-usin(v), ucos(v)+1, 1-usin(v)⟩$
Compute the Partial Derivatives. We need to find the partial derivatives of $\vec{r}$ with respect to u and v:
$\frac{∂\vec{r}}{∂u} = ⟨cos(v), sin(v), 0⟩, \frac{∂\vec{r}}{∂v} = ⟨-usin(v), ucos(v), 0⟩$
Compute the Cross Product
$\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v} = |\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ cos(v) & sin(v) & 0\\ -usin(v) & ucos(v) & 0\end{smallmatrix}| = ⟨0, 0, u⟩$. 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: Flux_{S₁} $\int \int_{S_1} \vec{F}·d\vec{S_1} = \int \int_{S_1} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})dudv = \int_{0}^{2π}\int_{0}^{1} ⟨ucos(v)-usin(v), ucos(v)+1, 1-usin(v)⟩·⟨0, 0, u⟩dudv = \int_{0}^{2π}\int_{0}^{1} (u-u^2sin(v))dudv$
Evaluate the inner integral: $\int_{0}^{1} (u-u^2sin(v))du = \frac{u^2}{2}-\frac{u^3sin(v)}{3}\bigg|_{0}^{1} = \frac{1}{2}-\frac{1}{3}sin(v)$
Evaluate the outer integral: $\int_{0}^{2π} \frac{1}{2}-\frac{1}{3}sin(v)dv = \frac{1}{2}v\bigg|_{0}^{2π} = π$.
Notice the second part of the integral $\int_{0}^{2π} -\frac{1}{3}sin(v)dv = 0$ since sin(v) integrates to 0 over a full period. Flux_{S₁} = π.
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 $\vec{r}(u, v) = ⟨ucos(v), usin(v), u⟩$, where 0 ≤ u ≤ 1 (radius), 0 ≤ v ≤ 2π (angle)
$x^2+y^2=z^2⇒u^2cos^2(v)+u^2sin^2(v)=z^2⇒z = \sqrt{u^2cos^2(v)+u^2sin^2(v)} = \sqrt{u^2} = u$
Plug these values in the vector field. $\vec{F}(\vec{r}(u, v)) = = ⟨x−y, x+z, z−y⟩ = ⟨ucos(v)-usin(v), ucos(v)+u, u-usin(v)⟩$
Compute the Partial Derivatives. We need to find the partial derivatives of $\vec{r}(u, v)$ with respect to u and v:
$\frac{∂\vec{r}}{∂u} = ⟨cos(v), sin(v), 1⟩, \frac{∂\vec{r}}{∂v} = ⟨-usin(v), ucos(v), 0⟩$
Compute the Cross Product
$\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v} = |\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ cos(v) & sin(v) & 1\\ -usin(v) & ucos(v) & 0\end{smallmatrix}| = ⟨-ucos(v), -usin(v), u⟩$ Notice that the normal vector points inward, so for the outward orientation, we take the negative ⟨ucos(v), usin(v), -u⟩.
Calculate the flux
Flux_{S₂} = $\int \int_{D} \vec{F}·d\vec{S_2} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int \int_{D} \vec{F}·(\frac{∂\vec{r}}{∂u}x\frac{∂\vec{r}}{∂v})dudv = \int_{0}^{2π}\int_{0}^{1} ⟨ucos(v)-usin(v), ucos(v)+u, u-usin(v)⟩·⟨ucos(v), usin(v), -u⟩dudv = \int_{0}^{2π}\int_{0}^{1} (u^2cos^2(v)-u^2cos(v)sin(v)+u^2cos(v)sin(v)+u^2sin(v)-u^2+u^2sin(v))dudv = \int_{0}^{2π}\int_{0}^{1} (u^2cos^2(v)+2u^2sin(v)-u^2)$
Let’s evaluate each of these integrals separately.
$\int_{0}^{2π}\int_{0}^{1} u^2cos^2(v)dudv = \int_{0}^{2π}cos^2(v)(\frac{u^3}{3}\bigg|_{0}^{1})$
$= \frac{1}{3}\int_{0}^{2π} cos^2(v)dv = \frac{1}{3}\int_{0}^{2π} \frac{1+cos(2v)}{2}dv = \frac{1}{6}\int_{0}^{2π} (1+cos(2v))dv = \frac{1}{6}\int_{0}^{2π} dv = \frac{1}{6}v\bigg|_{0}^{2π} = \frac{π}{3}$. 💣 Notice: The integral of cos(2v) over one period 2π is zero
$\int_{0}^{2π}\int_{0}^{1} 2u^2sin(v)dudv = 2\int_{0}^{2π}sin(v)\int_{0}^{1} u^2dudv = 2\int_{0}^{2π}sin(v) (\frac{u^3}{3})\bigg|_{0}^{1}$
$=\frac{2}{3}·\int_{0}^{2π}sin(v)dv = 0. $
$\int_{0}^{2π}\int_{0}^{1}-u^2dudv = \int_{0}^{2π}-\frac{u^3}{3}\bigg|_{0}^{1}$
$= -\frac{1}{3}·\int_{0}^{2π}dv = \frac{1}{3}v\bigg|_{0}^{2π} = -\frac{2π}{3}$
Thus, the flux through S_{2} is Flux_{S₂} = $\frac{π}{3}-\frac{2π}{3} = \frac{-π}{3}$
Total Flux Across S
The total flux across S is $\int \int_{D} \vec{F}·d\vec{S} = Flux_{S₁} + Flux_{S₂} = π +\frac{-π}{3} = \frac{2π}{3}$. We have computed both the volume integral and the surface integral. Both integrals yield the same value, $\frac{2π}{3}$ so we have verified the divergence theorem.
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: $\oint_S \vec{F} \cdot d\vec{S}$ 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:
$\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} div \vec{F} dV =$[Notation] $ \int \int \int_{D} ∇·\vec{F} dV$
where D is the volume enclosed by the surface S, and $∇·\vec{F}$ is the divergence of the vector field.
Calculate the Divergence of $\vec{F}$: $∇⋅\vec{F} = P_x +Q_y + R_z =[\text{Given: P = zarctan(y²), Q = z³ln(x²+1), R = 1}] = \frac{∂}{∂x}(zarctan(y^2))+\frac{∂}{∂y}(z^3ln(x^2+1))+\frac{∂}{∂z}(z) =$[Since zarctan(y²) and z³ln(x²+1) do not depend on x and y respectively, the first two partial derivatives are zero] 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 ⇒ x^{2} + y^{2} = 1. This is a circle of radius 1 in the xy-plane.
We need to calculate $\int \int \int_{D} ∇·\vec{F} dV$, the total flux over the solid of the paraboloid between z = 1 and z = 2, then subtract flux on the base z = 2.
Given the symmetry about the z-axis, it’s convenient to switch to cylindrical coordinates. In cylindrical coordinates: x = rcos(θ), y = rsin(θ), z = z, and dV = rdrdθdz.
The limits of integration are:
Compute the Volume Integral:
$ \int \int \int_{D} ∇·\vec{F} dV = \int_{0}^{2π} \int_{0}^{1} \int_{1}^{2-r^2} 1·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 θ: $\int_{0}^{2π} \frac{1}{4}dθ = \frac{1}{4}θ\bigg|_{0}^{2π} = 2π·\frac{1}{4} = \frac{π}{2}$. So, the total flux through the paraboloid is $\frac{π}{2}$.
Calculate the flux on the “base” z = 1.
The “base” of the surface is the disk at z = 1, with normal vector $\hat{\mathbf{n}} = ⟨0, 0, -1⟩$. It can also be calculated from the equation of the plane z = 1, $⟨\frac{∂z}{∂x}, \frac{∂z}{∂y}, -1⟩$ and these partial derivatives are obviously zero.
Flux_on_base = $ \int \int_{S} \vec{F}·\hat{\mathbf{n}} dS = \int \int_{S} ⟨arctan(y^2), ln(x^2+1), 1⟩·(0, 0, -1) dS = \int \int_{D} -1dA =$ -Area of the disk = -π
The flux through the paraboloid (not including the base) is given by the total flux minus the flux on the base:
Flux through paraboloid = total_flux - flux_on_base = $\frac{π}{2}-(-π) = \frac{π}{2}+\frac{2π}{2} = \frac{3π}{2}$