Mistakes are a great educator when one is honest enough to admit them and willing to learn from them, Aleksandr Solzhenitsynv.
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).
In mathematics and physics, scalar fields and vector fields are two fundamental concepts used to describe varying quantities in space.
Definition. A vector field in three-dimensional space assigns a vector to each point in that space. Mathematically, a vector field in 3D can be represented as $\vec{F} = ⟨P, Q, R⟩$ where P, Q, and R are functions that depend on the spatial coordinates x, y, and z.
In Physics, a force field is a map that shows the distribution of forces over a particular area of space. Examples include magnetic fields (represent the force exerted by a magnet or current on moving charges), electric fields (represent the force experienced by a charged particle in the presence of other charges), velocity fields, gradient fields, and gravitational fields.
The gravitational force is a classic example of a vector field. The gravitational attraction is the force that objects with mass exert on each other. Newton’s law of universal gravitation states that every particle (a small object that can be described by several properties, such as volume, density, or mass) in the universe attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
This is expressed mathematically as: $\vec{F} = -\frac{G·M·m}{r^2}\hat{\mathbf{r}}$ where:
The magnitude of the gravitational force can also be expressed as: $|\vec{F}| ∝ \frac{c}{r^2}$ where $|\vec{F}|$ is the magnitude of the gravitational force, c is a constant representing the strength of the gravitational force (it depends on the masses involved) and r is the distance between the two objects.
Consider a solid mass M located at the origin (0, 0, 0) in three dimensional space. The gravitational force exerted by this mass M on a mass m located at (x, y, z) results in a force $\vec{F}$ directed towards the origin, with magnitude $|\vec{F}| ∝ \frac{c}{ρ^2}$ where $ρ = \sqrt{x^2+y^2+z^2}$ (Figure A)
More specifically, this force can be written as: $\vec{F} = \frac{-G·M·m⟨x, y, z⟩}{ρ^3} = \frac{-c⟨x, y, z⟩}{ρ^3}$ where we divide by ρ^{3} because we are taking into consideration the unit vector $\hat{\mathbf{r}} = \frac{⟨x, y, z⟩}{ρ}$ and c = G·M·m is a constant.
In mathematics and physics, flux is a concept that quantifies how much of a vector field “flows” through a surface or a curve. It is an essential tool in fields such as fluid dynamics, electromagnetism, and thermodynamics, where understanding how quantities like fluid, electricity, or heat move through a given area is crucial.
In two dimensions, the flux is another line integral. Specifically, the flux of a vector field $\vec{F}$ across a curve C 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 $\int_{C} \vec{F}·\hat{\mathbf{n}}·ds$ where:
In this context, if $\vec{F}$ represent a velocity field (like the velocity of a fluid), flux measures how much fluid is passing through the curve per unit time (the rate at which fluid passes through C). Imagine a stream of water flowing across a fence; the flux would tell you how much water crosses that fence per unit time.
When moving from 2D to 3D, the concept of flux naturally extends from curves to surfaces. For a surface S in 3D space, we want to determine how much fluid passes through S per unit time (Figure 5).
Consider a surface S in 3D space, and let $\vec{F}$ be a vector field (such as the velocity of fluid flow or an electric field). Consider a small area element dS on the surface S. The flux through dS is given by the velocity of the fluid at dS multiplied by the area of dS. The component of the vector field $\vec{F}$ that is perpendicular to the surface is given by $\vec{F}·\hat{\mathbf{n}}$. Multiplying this by the area of dS give us the volume of fluid passing through dS per unit time.
In the context of vector fields in space, flux is represented by a surface integral, not a line integral. It measures how much of the vector field crosses a given surface.
Definition. If we have a vector field $\vec{F}$ and a surface S in space, with $\hat{\mathbf{n}}$ being the unit normal vector, the flux through S is given by: Flux = $\int \int_{S} \vec{F}·\hat{\mathbf{n}}·dS$ (Figure C) where:
This integral sums up the contributions of the vector field across every infinitesimal piece of the surface, giving the total flux through the entire surface.
The ease case is where the vector field is tangent to the surface. In this scenario, the flux is zero because there is no component of the vector field passing through the surface. This is because the dot product $\vec{F}·\hat{\mathbf{n}}$ will be zero when $\vec{F}$ is tangent to the surface.
Flux is a measure of how much of a vector field passes through a given surface. The vector field $\vec{F}$ represent a radial field, meaning that at each point in space, the vector points directly away from the origin.
The flux of a vector field $\vec{F}$ through a surface S is given by the surface integral: $\int \int_{S} \vec{F}·\hat{\mathbf{n}}·dS = \int \int_{S} \vec{F}· \vec{dS}$ where $\vec{dS} = \hat{\mathbf{n}}·dS$ is the vector area element and $\hat{\mathbf{n}}$ is the unit normal vector to the surface S at each point.
Determining the Unit Normal vector. On the surface of the sphere, the position vector $\vec{r} = ⟨x, y, z⟩$ points radially outward from the origin and is normal to the surface. Therefore, the unit normal vector $\hat{\mathbf{n}}$ (Figure D) at any point on the sphere is given by $\hat{\mathbf{n}} = \frac{⟨x, y, z⟩}{|⟨x, y, z⟩|}$ =[Since the radius of the sphere is a, we have a = $\sqrt{x^2+y^2+z^2} = |⟨x, y, z⟩|$] $\frac{1}{a}·⟨x, y, z⟩$
Calculating the Dot product $\vec{F}·\hat{\mathbf{n}}$. The normal vector and our vector field are parallel to each other. $\vec{F}·\hat{\mathbf{n}} = |\vec{F}|·|\hat{\mathbf{n}}|·cos(0) = ⟨x, y, z⟩·\frac{⟨x, y, z⟩}{a}·1 = \frac{x^2+y^2+z^2}{a}·1 =[\text{ Since we are on the surface of the sphere where } x^2+y^2+z^2 = a^2] \frac{a^2}{a} = a.$
Therefore, the flux through the surface S of the sphere is then given by the integral: $\int \int_{S} \vec{F}·\hat{\mathbf{n}}·dS =[\text{Substituting the result from the dot product:}] \int \int_{S} adS$ =[Since a is constant over the surface of the sphere, it can be factored out of the integral] $a\int \int_{S} dS = a ·Area(S)$ =[The surface area S of a sphere with radius a is 4πa^{2}] $a·4πa^2 = 4πa^3$.
This result is intuitive because the vector field is radial and its magnitude increases linearly as we move away from the origin, making the flux proportional to the cube of the radius of the sphere.
The flux of a vector field through a surface S is a measure of how much of the field passes through that surface. The vector fields points in the z-direction, through the surface of a sphere with radius a centered at the origin.
The flux through a surface S is given by the surface integral: $\int \int_{S} \vec{F}·\hat{\mathbf{n}}·dS = \int \int_{S} \vec{F}· \vec{dS}$ where $\vec{dS} = \hat{\mathbf{n}}·dS$ is the vector area element and $\hat{\mathbf{n}}$ is the unit normal vector to the surface.
Normal vector. On the surface of the sphere, the position vector $\vec{r} = ⟨x, y, z⟩$ points radially outward from the origin and is normal to the surface. Therefore, the unit normal vector $\hat{\mathbf{n}}$ (Figure D) at any point on the sphere is given by $\hat{\mathbf{n}} = \frac{⟨x, y, z⟩}{|⟨x, y, z⟩|}$ =[Since the radius of the sphere is a, we have a = $\sqrt{x^2+y^2+z^2} = |⟨x, y, z⟩|$] $\frac{1}{a}·⟨x, y, z⟩$
Calculating the dot product $\vec{F}·\hat{\mathbf{n}}$. Given that $\vec{F} = z\vec{k}$, $\vec{F}·\hat{\mathbf{n}} = ⟨0, 0, z⟩·\frac{⟨x, y, z⟩}{a} = \frac{z^2}{a}$.
Therefore, the flux through the surface S of the sphere is then: $\int \int_{S} \vec{F}·\hat{\mathbf{n}}·dS = \int \int_{S} \frac{z^2}{a}·dS$
To calculate the flux, we need to integrate $\frac{z^2}{a}$ over the surface of the sphere. To do this, we use spherical coordinates (r, θ, φ).
Surface Integral in Spherical Coordinates: In spherical coordinates, the surface element dS on a sphere of radius a (we are not counting the volume element, which would be dρ) is given by: (Figure E), dS = (adφ)·(a·sin(φ)·dθ) = a^{2}sin(φ)dφdθ and we also know that z = acos(φ). Substituting this into the expression for the flux, we have:
So, the integral becomes:
$\int \int_{S} \frac{z^2}{a}·dS = \int_{0}^{2π} \int_{0}^{π} \frac{a^2cos^2(φ)}{a}a^2sin(φ)dφdθ = a^3\int_{0}^{2π} \int_{0}^{π} cos^2(φ)sin(φ)dφdθ =$
$\int_{0}^{π} cos^2(φ)sin(φ)dφ = -\frac{1}{3}cos^3(φ)\bigg|_{0}^{π} =\frac{2}{3}$
Another way of seeing this integral is as follows:
$\int_{0}^{π} cos^2(φ)sin(φ)dφ$ =[To solve this, let u = cos(φ), hence du = -sin(φ)dφ. When φ = 0, u = 1, and when φ = π, u = -1] $\int_{1}^{-1} u^2du = \frac{u^3}{3}\bigg|_{1}^{-1}=\frac{1}{3}-\frac{-1}{3}= \frac{2}{3}$
Finally, the integral simplifies to:
$\int \int_{S} \frac{z^2}{a}·dS = a^3\int_{0}^{2π} \frac{2}{3}dθ = \frac{a^3·2}{3}·θ\bigg|_{0}^{2π} = \frac{4}{3}πa^3.$
In multivariable calculus, surface integrals are used to calculate various quantities over surfaces, such as flux. To work effectively with surface integrals, it’s important to understand the different surface elements dS and their associated unit normal vectors $\hat{\mathbf{n}}$ for different types of surfaces. Below, we outline some common surfaces, detailing their respective surface elements and normal vectors:
The unit normal vector $\hat{\mathbf{n}}$ for this plane is perpendicular to the surface. Since the plane is horizontal, the normal vector points either straight up or straight down, depending on the orientation. These directions correspond to: $\hat{\mathbf{n}} = ± \vec{k}$ where $\vec{k}$ is the unit vector in the z-direction. The positive sign indicates the vector pointing upward, and the negative sign indicates the vector pointing downward.
The unit normal vector $\hat{\mathbf{n}}$ for this plane is perpendicular to the surface and can point either in the positive or negative x-direction: $\hat{\mathbf{n}} = ±\vec{i}$ where $\vec{i}$ is the unit vector in the x direction. The choice of sign depends on whether the vector points towards the positive or negative x-axis.
The unit normal vector $\hat{\mathbf{n}}$ at any point on the surface of the sphere is radial, meaning that it points radially outward from the center or inward towards the center, depending on the orientation, and can be expressed as: $\hat{\mathbf{n}} = ± \frac{⟨x, y, z⟩}{a}$.
The unit normal vector $\hat{\mathbf{n}}$ at any point on the curved surface of the cylinder is $\hat{\mathbf{n}} = ± \frac{1}{a}⟨x, y, 0⟩$. This indicates that the normal vector points radially outward or inward from the z-axis, depending on the orientation.
When working with surface integrals, it is often necessary to compute the surface area element dS and the corresponding unit normal vector $\hat{\mathbf{n}}$ for a given surface. Let’s consider the surface area defined by the function z = f(x, y), which represent a graph over a region in the xy-plane (Figure 4). We aim to determine the surface element dS and the corresponding unit normal vector $\hat{\mathbf{n}}$.
For the surface defined by z = f(x, y), the surface element vector is given by: $\hat{\mathbf{n}}dS = ±⟨-f_x, -f_y, 1⟩dxdy$ (🚀) where:
The last equation is a bit trick (Figure 0, 01, 02) where R is the shadow of our surface S in the xy-plane.
Consider a small region (rectangle) R in the xy-plane, with dimensions (sides) Δx and Δy. The corresponding patch on the surface S is above this rectangle. For a sufficiently small Δx and Δy, the patch on the surface can be approximated as a parallelogram.
What is its area and its normal vector? The surface area of the parallelogram can be found using the cross product of two vectors, $\vec{u}$ and $\vec{v}$, that define its two sides. The magnitude of this cross product gives us the area $ΔS = |\vec{u} x \vec{v}|$, while the direction of the cross product give us the orientation of the normal vector $±\vec{u} x \vec{v} = ΔS·\hat{\mathbf{n}}$.
💣The magnitude of the cross product of two vectors is the area of the parallelogram they enclose, and its direction is perpendicular to the plane containing those vectors, $\vec{u}x\vec{v} = ±\hat{\mathbf{n}}dS.$
$\vec{u}$ is a vector from (x, y, f(x, y)) to (x + Δx, y, f(x+Δx, y)). Using a linear approximation for f(x + Δx, y): f(x + Δx, y) ≈ f(x, y) + Δx·f_{x}. Thus, $\vec{u}≈ ⟨Δx, 0, f(x+Δx, y)-f(x, y)⟩ = ⟨Δx, 0, f_xΔx⟩ = Δx⟨1, 0, f_x⟩$
Analogously, $\vec{v}$ is a vector from (x, y, f(x, y)) to (x, y + Δy, f(x, y +Δy)). Similarly, using a linear approximation for f(x, y + Δy) ≈ f(x, y) + Δy·f_{y}. Thus, $\vec{v}≈⟨0, Δy, f(x, y +Δy)-f(x,y)⟩ = ⟨0, Δy, f_yΔy⟩ = Δy⟨0, 1, f_y⟩$
The cross product $\vec{u}x\vec{v}$ gives both the area of the parallelogram and the direction of the normal vector:
$±\hat{\mathbf{n}}dS = \vec{u}x\vec{v} = (\begin{smallmatrix}\vec{i} & \vec{j} & \vec{k}\\ 1 & 0 & f_x\\ 0 & 1 & f_y\end{smallmatrix})ΔxΔy = ⟨-f_x, -f_y, 1⟩ΔxΔy$.
In the infinitesimal limit, where Δx and Δy become dx and dy we got the previous formula: $\hat{\mathbf{n}}dS = ±⟨-f_x, -f_y, 1⟩dxdy$. This expression correctly captures both the surface area element dS and the direction of the normal vector. This formula is crucial for setting up surface integrals for surfaces defined as graph of functions.
Remember, the ± sign allows us to choose the orientation of the normal vector, which can point either upward or downward, depending on the problem’s requirements.
The flux of a vector field through a surface represents the amount of the field passing through the surface. Since the surface is oriented downward, we need to take this orientation into account when setting up the surface integral. The surface z = x^{2}+y^{2}-9 is a paraboloid that opens upwards, has its vertex at (0, 0, -9), and has circular cross-sections perpendicular to the z-axis.
Vector Field Simplification. Given the vector field $\vec{F} = ⟨x^2y, -3xz, 4y^3⟩$ we will express the components of $\vec{F}$ in terms of z using the surface equation z = x^{2}+y^{2}-9 [This simplification allows us to work with $\vec{F}$ directly on the giving surface, facilitating the calculation of the flux]: $\vec{F} = ⟨x^2y, -3x(x^2+y^2-9), 4y^3⟩ =[\text{This simplifies to:}] ⟨x^2y, -3x^3-3xy^2+27x, 4y^3⟩$
Surface Normal Vector
As we have previously stated, for a surface defined by z = f(x, y), the surface normal vector $\hat{\mathbf{n}}$ and the surface element dS are given by: $\hat{\mathbf{n}}dS = ±⟨-f_x, -f_y, 1⟩dxdy$ where f(x, y) = x^{2}+y^{2} -9, $f_x=\frac{∂f}{∂y} = 2x, f_y=\frac{∂f}{∂y} = 2y$ are the partial derivatives with respect to x and y, respectively.
Since the surface is orientated downwards, we use the negative sign: $\hat{\mathbf{n}}dS = -⟨-f_x, -f_y, 1⟩dxdy = -⟨-2x, -2y, 1⟩dxdy = ⟨2x, 2y, -1⟩dxdy$
Calculate the flux
We calculate the flux through the surface over the region D = [0, 2] × [0, 1] using the formula:
Flux = $\int \int_{D} \vec{F}·d\vec{S} = \int \int_{D} \vec{F}·\hat{\mathbf{n}}dS = \int_{0}^{2} \int_{0}^{1} ⟨x^2y, -3x^3-3xy^2+27x, 4y^3⟩·⟨2x, 2y, -1⟩dxdy = \int_{0}^{2} \int_{0}^{1} (2x^3y-6x^3y-6xy^3+54xy-4y^3)dydx = \int_{0}^{2} \int_{0}^{1} (-4x^3y-6xy^3+54xy-4y^3)dydx$
Inner integral:
$\int_{0}^{1} (-4x^3y-6xy^3+54xy-4y^3)dy = -4x^3\frac{y^2}{2}-6x\frac{y^4}{4}+54x\frac{y^2}{2}-4\frac{y^4}{4}\bigg|_{0}^{1} = -4x^3\frac{1}{2}-6x\frac{1}{4}+54x\frac{1}{2}-4\frac{1}{4} = -2x^3 -\frac{3x}{2}+27x-1$
Outer integral:
$\int_{0}^{2} -2x^3 -\frac{3x}{2}+27x-1 dx = -2\frac{x^4}{4}-\frac{3}{2}\frac{x^2}{2}+27\frac{x^2}{2}-x\bigg|_{0}^{2} = -2·\frac{16}{4}-\frac{3}{2}2+27·2-2 = -8-3 +54-2 = 41.$