All those who seem stupid, they are, and also so are half of those who do not, Quevedo.
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).
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$
The flux of a vector field through a surface is a measure of how much of the field “flows” through the surface. In this problem, instead of directly calculating the flux through each face of the tetrahedron, 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.
$\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} div \vec{F} dV =$[Notation] $ \int \int \int_{D} ∇·\vec{F} dV$
Calculate the Divergence of $\vec{F}$: $∇⋅\vec{F} = P_x +Q_y + R_z =[\text{Given: P = 3xy, Q = y², R = -x²y⁴}] = \frac{∂}{∂x}(3xy)+\frac{∂}{∂y}(y^2)+\frac{∂}{∂z}(-x^2y^4) = 3y + 2y + 0 = 5y.$
Set Up the Volume Integral
Next, we need to integrate the divergence 5y over the volume V of the tetrahedron. We’ll use Cartesian coordinates for this.
The tetrahedron is bounded by the coordinate planes x = 0, y = 0, z = 0, and the plane x + y + z = 1. The limits of integration are:
Thus, the volume element dV is dxdydz and the volume integral becomes:
The volume integral of the divergence over the region D is given by: $\int \int \int_{D} (∇·\vec{F})dV = \int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-x-y} 5ydzdydz = \frac{5}{24}$
Evaluate the integral
The solid is enclosed by the two hemispheres and the disk in the xy-plane, forming a shell-like region.
The flux of a vector field through a surface is a measure of how much of the field “flows” through the surface. In this problem, instead of directly calculating the flux through each component of the surface, we can simplify by using 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.
$\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} div \vec{F} dV =$[Notation] $ \int \int \int_{D} ∇·\vec{F} dV$
Calculate the Divergence of $\vec{F}$: $∇⋅\vec{F} = P_x +Q_y + R_z =[\text{Given: P = x³+ysin(z), Q = y³+zsin(x), R = 3z}] = \frac{∂}{∂x}(x^3+ysin(z))+\frac{∂}{∂y}(y^3+zsin(x))+\frac{∂}{∂z}(3z) = 3x^2 + 3y^2 + 3 = 3(x^2+y^2+1)$
Set Up the Volume Integral
The volume integral of the divergence over the region D is given by: $\int \int \int_{D} (∇·\vec{F})dV = \int \int \int_{D} 3(x^2+y^2+1)dV$
We will convert this integral into spherical coordinates for easier evaluation. In spherical coordinates, the relationships between Cartesian coordinates and spherical coordinates are:
In spherical coordinates, the integrand $3(x^2+y^2+1) = 3(r^2+1) = 3(ρ^2sin^2(φ)+1)$ and dV = ρ^{2}sin(ϕ)dρdθdϕ.
The limits of integration are:
The integral becomes:
$\int \int \int_{D} (∇·\vec{F})dV = \int \int \int_{D} 3(x^2+y^2+1)dV = 3\int_{0}^{2π} \int_{0}^{\frac{π}{2}} \int_{1}^{2} (ρ^2sin^2(φ)+1)ρ^2sin(ϕ)dρdθdϕ = 6π \int_{0}^{\frac{π}{2}} \int_{1}^{2} (ρ^4sin^3(φ)+ ρ^2sin(φ)) dρdϕ$
Integrating with respect to ρ: $\int_{1}^{2} (ρ^4sin^3(φ)+ ρ^2sin(φ)) dρ = sin^3(φ)\frac{ρ^5}{5} + sin(φ)\frac{ρ^3}{3}\bigg|_{1}^{2} = \frac{32}{5}sin^3(φ) + \frac{8}{3}sin(φ) -(\frac{1}{5}sin^3(φ) +\frac{1}{3}sin(φ)) = \frac{31}{5}sin^3(φ) +\frac{7}{3}sin(φ)$
$\int_{0}^{\frac{π}{2}} \frac{31}{5}sin^3(φ) +\frac{7}{3}sin(φ) dφ$
$\frac{31}{5}\int_{0}^{\frac{π}{2}} sin^3(φ)dφ =[(i)] \frac{31}{5} · \frac{2}{3} = \frac{62}{15}$ (iii)
$\int_{0}^{\frac{π}{2}} sin^3(φ)dφ = \int_{0}^{\frac{π}{2}} sin(φ)(1-cos^2(φ)) =[(ii)] 1 - \frac{1}{3} = \frac{2}{3}$ (i)
$\int_{0}^{\frac{π}{2}} -sin(φ)cos^2(φ)dφ$ =[u = cos(φ), du = -sin(φ)dφ] $\int_{0}^{\frac{π}{2}} u^2du = \frac{u^3}{3} = \frac{cos(φ)^3}{3}\bigg|_{0}^{\frac{π}{2}} = -\frac{1}{3}$ (ii)
$\frac{7}{3}\int_{0}^{\frac{π}{2}} sin(φ) dφ = \frac{7}{3}cos(φ)\bigg|_{0}^{\frac{π}{2}} = \frac{7}{3}$ (iv).
Adding these $\int_{0}^{\frac{π}{2}} \frac{31}{5}sin^3(φ) +\frac{7}{3}sin(φ) dφ =[(iv), (iii)] \frac{7}{3} + \frac{62}{15} = \frac{35}{15}+\frac{62}{15} = \frac{97}{15}$. Finally, multiplying by 6π = $6π·\frac{97}{15} = \frac{582π}{15} = \frac{194π}{5}$
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 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 $\vec{F}$: $∇⋅\vec{F} = P_x +Q_y + R_z =[\text{Given: P = xz, Q = yz, R = 3z²}] = \frac{∂}{∂x}(xz)+\frac{∂}{∂y}(yz)+\frac{∂}{∂z}(3z^2) = z + z + 6z = 8z.$
The volume integral of the divergence over the region D is given by: $\int \int \int_{D} (∇·\vec{F})dV = \int \int \int_{D} 8zdV$
To evaluate this, we switch to cylindrical coordinates because the region is symmetric about the z-axis. In cylindrical coordinates: x = rcos(θ), y = rsin(θ), z = z, and dV = rdrdθdz.
The region is bound by the paraboloid z = r^{2} (where r = $\sqrt{x^2+y^2}$) and the plane z = 1. Thus, the limits of integration are: 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1 (the unit circle in the xy-plane; z = 1 corresponds to r^{2} = 1, so r = 1), and z from the paraboloid (z = x^{2} + y^{2} = r^{2}) to 1 (the solid is bounded by the plane z = 1).
Now, the volume integral becomes:
$\int \int \int_{D} div \vec{F} dV = \int_{0}^{2π} \int_{0}^{1} \int_{r^2}^{1} 8z·rdzdrdθ.$
We first integrate with respect to z: $\int_{r^2}^{1} 8zrdz = 8r\frac{z^2}{2}\bigg|_{r^2}^{1} = 8r(\frac{1}{2}-\frac{r^4}{2}) = 4r(1-r^4)$.
Next, we integrate with respect to r: $\int_{0}^{1} 4r(1-r^4)dr = 4(\frac{r^2}{2}-\frac{r^6}{6})\bigg|_{0}^{1} = 4(\frac{1}{2}-\frac{1}{6}) = 4(\frac{3}{6}-\frac{1}{6}) = 4·\frac{2}{6} = \frac{8}{6} = \frac{4}{3}$
Finally, integrate with respect to θ: $\int_{0}^{2π} \frac{4}{3}dθ = \frac{4}{3}·2π = \frac{8π}{3}$. Thus, the volume integral is $\int \int \int_{D} div \vec{F} dV = \frac{8π}{3}$
$\oint_S \vec{F} \cdot d\vec{S} = \oint_S \vec{F} \cdot \hat{\mathbf{n}}dS$
The surface S is composed of two parts:
Surface Integral over S_{1} (Top Disk)
For the disk at z = 1, $\vec{F} = ⟨xz, yz, 3z^2⟩ =[\text{Substitute z = 1}] = ⟨x, y, 3⟩$
The normal vector $\hat{\mathbf{n}}$ for this surface (z = 1) points directly upwards in the z-direction, so: $\hat{\mathbf{n}} = ⟨-z_x, -z_y, 1⟩$ =[Since the surface is a flat plane at z = 1, the components of z_{x} and z_{y} are both zero] = ⟨0, 0, 1⟩
The surface integral over S_{1} is: $\oint_{S_1} \vec{F} \cdot \hat{\mathbf{n}}dS = \oint_{D} ⟨x, y, 3⟩·⟨0, 0, 1⟩dA = 3dA$ = 3(area of D) =[D is the unit circle in the xy-plane] 3π.
Surface Integral over S_{2} (Paraboloid)
For the paraboloid z = x^{2} +y^{2}, the vector field is: $\vec{F} = ⟨xz, yz, 3z^2⟩ = ⟨x(x^2+y^2), y(x^2+y^2), 3(x^2+y^2)^2⟩$
The normal vector $\hat{\mathbf{n}}$ for the paraboloid can be computed using the gradient of the function defining the surface z = x^{2}+y^{2}: $\hat{\mathbf{n}} = ⟨z_x, z_y, -1⟩ = ⟨2x, 2y, -1⟩$. Since the surface is oriented outward, the z-component of the normal vector should be negative to indicate outward orientation.
Compute the dot product: $\vec{F}·\hat{\mathbf{n}} = ⟨x(x^2+y^2), y(x^2+y^2), 3(x^2+y^2)^2⟩·⟨2x, 2y, -1⟩ = 2x^2(x^2+y^2)+2y^2(x^2+y^2)- 3(x^2+y^2)^2 = (x^2+y^2)[2x^2+2y^2-3(x^2+y^2)] = (x^2+y^2)·-(x^2+y^2)=-(x^2+y^2)^2$
The surface integral over S_{2} is:
$\oint_{S_2} \vec{F} \cdot \hat{\mathbf{n}}dS = \oint_{D} -(x^2+y^2)^2dA$, where $\hat{\mathbf{n}}$ is the outward-pointing normal vector on the surface S_{2} and dA is the differential area element on the surface.
To simplify the computation, we switch to polar coordinates, where the Cartesian coordinates (x, y) are replaced by polar coordinates (r, θ), with r = $\sqrt{x^2+y^2}$ and θ as the angle in the xy-plane. In polar coordinates:
Substituting these into the integral, we have:
$\oint_{D} -(x^2+y^2)^2dA = \int_{0}^{2π}\int_{0}^{1} -(r^2)^2·rdrdθ = \int_{0}^{2π}\int_{0}^{1} -r^5drdθ$
Integrate with respect to r: $\int_{0}^{1} -r^5dr = -\frac{r^6}{6}\bigg|_{0}^{1} = \frac{-1}{6}$.
Then, integrate with respect to θ: $\oint_{S_2} \vec{F} \cdot \hat{\mathbf{n}}dS = \int_{0}^{2π} \frac{-1}{6}dθ = \frac{-1}{6}·2π = -\frac{π}{3}$
The total flux across the surface S is: $\oint_S \vec{F} \cdot \hat{\mathbf{n}}dS = \oint_{S_1} \vec{F} \cdot \hat{\mathbf{n}}dS + \oint_{S_2} \vec{F} \cdot \hat{\mathbf{n}}dS = 3π -\frac{π}{3} = \frac{9π}{3}-\frac{π}{3} = \frac{8π}{3}$.
Since both integral (volume and surface integral) are equal, the Divergence Theorem is verified for the given vector field and region.
The diffusion equation models the process by which quantities such as heat, mass, or charge spread out over time in a medium. This spreading process is due to diffusion, where particles naturally move from regions of high concentration to regions of low concentration. Examples of diffusion include the spread of smoke in air, the mixing of dye in water, or the distribution of heat in a solid material.
What is diffusion? It is the spontaneous movement of particles or substances as they spread out from regions of high concentration to regions of low concentration, driven by the tendency of systems to move towards equilibrium. Mathematically, the diffusion process can be described using a partial differential equation.
More specifically, if u(x, y, z, t) represents the concentration of a substance or temperature at a point P(x, y, z) at time t, the diffusion equation is: $\frac{∂u}{∂t} = k·∇^2u$ where:
$\frac{∂u}{∂t}$= $k∇^2u = k∇·∇u$ = $k(\frac{∂^2u}{∂x^2}+\frac{∂^2u}{∂y^2}+\frac{∂^2u}{∂z^2})$. This equation is also know as the heat equation when applied to the diffusion of temperature in a solid.
Diffusion describes how particles or heat move from areas of high concentration to areas of low concentration. For instance, if you drop a drop of ink in water, the ink particles will spread out until they are evenly distributed throughout the water.
Let’s derive the diffusion equation starting with the concept of flux. Suppose $\vec{F}$ represents the flow of a substance, such as smoke or heat. The vector $\vec{F}$ indicates both the direction and the magnitude of the flow. Observation tells us that the substance flows from regions of high concentration to regions of low concentration, which can be mathematically expressed as: $\vec{F} = -k∇u$ (🚀).
Here, $\vec{F}$ is proportional to $-∇u$ where $∇u$ is the gradient of u, indicating the direction of the greatest rate of increase of u, and the negative sign indicates that the flux moves from high to low concentration.
To find the change in concentration u over time, consider a region D of the medium, bounded by a surface S. The flux of the substance out of this region through the surface can be calculated using the surface integral: (Figure A) $\oint_S \vec{F} \cdot d\vec{S}$
By the divergence theorem, this surface integral can be converted into a volume integral over D:
$\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} ∇⋅\vec{F}dV$. This expression represents the total flux of the substance through the surface S.
$\oint_S \vec{F} \cdot d\vec{S}$ is the amount of the substance (e.g., smoke) within region S per unit time. According to the principle of conservation of mass (or energy), the flux out of the region must equal the negative rate of change of the total amount of the substance within D, $\oint_S \vec{F} \cdot d\vec{S} = -\frac{d}{dt}(\int \int \int_{D} udV)$ where
By the Divergence Theorem we know that
$\oint_S \vec{F} \cdot d\vec{S} = \oint_S \vec{F} ·\hat{\mathbf{n}}·dS = \int \int \int_{D} div(\vec{F})dV = -\frac{d}{dt}(\int \int \int_{D} udV) = -\int \int \int_{D} \frac{∂u}{∂t}dV$ Basically, the idea behind the last step is that $\frac{d}{dt} (\sum_{i} u(x_i,y_i,z_i,t)ΔV_i)$ -this is the total amount of substance and how it changes over time-= $\sum_{i} \frac{∂u}{∂t}(x_i,y_i,z_i,t)ΔV_i$ -the previous term is equal to the derivatives of the amount of substance inside each little box (small region)-.
Notice that $\int \int \int_{D} div(\vec{F})dV = -\int \int \int_{D} \frac{∂u}{∂t}dV ⇒ \frac{1}{vol(D)}\int \int \int_{D} div(\vec{F})dV = \frac{1}{vol(D)}\int \int \int_{D} -\frac{∂u}{∂t}dV$. In words, the average of $div(\vec{F})$ in D is equal to the average of $-\frac{∂u}{∂t}$ in any region D of V ⇒[For this equality to hold for any arbitrary region D, the integrands must be equal] $div(\vec{F}) = ∇⋅\vec{F} = -\frac{∂u}{∂t}$
Substituting the earlier expression, $\vec{F}=-k∇u$, into the above equation: $∇⋅-k∇u = -\frac{∂u}{∂t}$. Simplifying we get, $-k∇^2u=-\frac{∂u}{∂t}$.
Removing the negative signs and solving for $\frac{∂u}{∂t}$, we arrive at the diffusion equation $\frac{∂u}{∂t}=k∇^2u$, which describes how the concentration u changes over time due to diffusion and has a wide range of applications: heat transfer (describe how heat spread through a solid material), chemical diffusion (modelling the spread of pollutants in air or water), biological processes (it helps to understand how substances like oxygen diffuse through tissues) or finance (it models the evolution of prices in markets).