One who asks a question is a fool for a minute; one who does not remains a fool forever, Chinese proverb
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
Flux is an important concept in multivariable calculus that allows us to quantify how much of a vector field passes through a given curve. This concept is particularly useful in physics and engineering, where it helps describe phenomena such as fluid flow, electromagnetic fields, and heat transfer.
Before diving into flux, it’s essential to understand the concept of a line integral. A line integral extends the idea of integrating a function f(x) of one variable x over an interval [a, b] ($\int_{a}^{b} f(x)dx$) to integrating a function or vector field $\vec{F}$ over a curve C in two or three dimensions ($\int_{C} \vec{F}·\vec{dr}$).
A line integral gives us the ability to integrate multivariable functions and vector fields over arbitrary curves in a plane or in space. In general, for a vector field $\vec{F}$, the line integral along a curve C is given by: $\int_{C} \vec{F}·\vec{dr}$ where $\vec{dr}$ is a small displacement vector along the curve C, and $\vec{F}·\vec{dr}$ represents the dot product of the vector field $\vec{F}$ with the displacement vector $\vec{dr}$.
Flux is a specific type of line integral. Rather than integrating the vector field along the curve, as we do when calculating work, flux measures how much of the vector field passes through the curve C. More precisely: the flux of a vector field $\vec{F}$ across a curve C is defined as the line integral of the dot product of the vector field $\vec{F}$ and the unit normal vector $\hat{\mathbf{n}}$ to the curve C, that is, $\oint_{C} \vec{F}·\hat{\mathbf{n}}·ds$. In this equation:
If we break the curve C into small pieces or segments Δs. For each segment, the flux is given by: $ΔFlux = \vec{F}·\vec{n}·ΔS$
The total flux is then the sum of these contributions as the segment size Δs approaches zero:
Flux = $\lim_{\Delta s \to 0} (\sum \vec{F}·\vec{n}·ΔS)$.
Recall that work is computed as: $\oint_{C} \vec{F}·d\vec{r} = \oint_{C} \vec{F}·\hat{\mathbf{T}}·ds$, i.e., the dot product of the vector field with the unit tangential vector (in the direction of motion) to the curve. I am loosely speaking summing the tangential component of $\vec{F}$ along the curve C, that is, how much the vector field $\vec{F}$ aligns with the direction of the curve C. It measures when I move along the curve how much I am going with or against $\vec{F}$, i.e., the work done by the field when moving along the curve.
On the other hand, flux measures when I move along the curve how much the field is crossing the curve. I am loosely speaking summing the (perpendicular) normal component of $\vec{F}$ along the curve C.
It represents the amount of a vector field (e.g., a fluid flow) passing through a curve. If the field is a flow of water, for instance, the flux measures how much fluid passes through the curve C per unit time or represents the volume of water flowing through the surface per unit time. It quantifies to the amount of fluid, energy, or other quantity that flows through a curve in a given direction per unit time.
To be more precise, what flows across C from left-to-right (relative to the orientation of $\vec{n}$) is counted positively, and right-to-left is counted negatively, so in fact, it is the net flow per unit time.
Observe Figure 2 and 3 (2 rotated), we are zooming over a little piece of my curve C (length ΔS) and a fluid flows to the right. How much fluid, energy or whatever passes through this piece of my curve over time? It is a parallelogram with Area = base · height = $ΔS· (\vec{F}·\vec{n})$. This area represents the amount of fluid (or any other quantity described by $\vec{F}$) passing through the segment Δs. Summing these areas over the entire curve gives the total flux.
Solution.
The result 2πa^{2} represents the total amount of the vector field that “crosses” the curve C. Since $\vec{F}$ is radially outward and C is a circle centered at the origin, the entire field contributes to the flux uniformly across the circle. It $\vec{F}$ represents a flow of some quantity (like fluid or heat), then 2πa^{2} is the total amount of that quantity passing through the boundary of the circle per unit time.
Understanding the Vector Field. The vector field $\vec{F} = -y\vec{i}+x\vec{j}$ describes a vector that rotates counterclockwise around the origin. Specifically, at any point (x, y) in the plane, $\vec{F}$ has a component of −y in the x-direction and a component of x in the y-direction. This vector field can be visualized as always being tangent to circles centered at the origin. For example, at the point (x, y) = (1, 0), $\vec{F}$ points straight up ($\vec{j}$), and at the point (0, 1), $\vec{F}$ points left ($\vec{-i}$).
Vector Field and Normal Vector Relationship. Along the curve C, the unit normal vector $\hat{\mathbf{n}}$ is radially outward from the origin, while the tangent vector $\hat{\mathbf{T}}$ is tangent to the circle, perpendicular to $\hat{\mathbf{n}}$. Because $\vec{F}$ is tangent to the circle (it rotates counterclockwise around the origin) and $\hat{\mathbf{n}}$ is perpendicular to $\hat{\mathbf{T}}$, the vector field $\vec{F}$ aligns with the tangent vector $\hat{\mathbf{T}}$ and is perpendicular to the normal vector $\hat{\mathbf{n}}$ at every point on the circle.
Dot product calculation: Since $\vec{F} ⊥ \vec{n} ⇒ \vec{F}·\vec{n} = |\vec{F}|·|\vec{n}|·cos(0) = |\vec{F}|·1·0 = 0$.
Setting up the integral: $\int_{C} \vec{F}·\vec{n}·ds = \int_{C} 0ds = 0$
The result indicates that there is no flux of the vector field $\vec{F}$ across the curve C. This makes sense because $\vec{F}$ represents a rotational field that is always tangent to the curve C and does not have any component that would “cross” the boundary of C. Physical Interpretation: If $\vec{F}$ were a velocity field representing fluid flow, the zero flux would mean that no fluid is passing through the curve C.
The second part of the integral is zero because cos(t)sin(t) is odd over a period.
Using the identity $cos^2(t) = \frac{1+cos(2t)}{2}$, $-4\int_{0}^{2π} cos^2(t) = -2\int_{0}^{2π} (1+cos(2t))dt = -2[t+\frac{sin(2t)}{2}]\bigg|_{0}^{2π} = (-2)2π = -4π$
Flux is a measure of how much a vector field “flows” through a curve. To compute the flux of a vector field $\vec{F}$ across a curve C in the plane, we can express it using the components of the vector field and the curve.
Remember that $d\vec{r} = \hat{\mathbf{T}}·ds = ⟨dx, dy⟩$ where $\hat{\mathbf{T}} = ⟨\frac{dx}{ds}, \frac{dy}{ds}⟩$ is the unit tangent vector to the curve C and ds is the differential arc length.
The unit normal vector $\vec{n}$ is obtained by rotating $\hat{\mathbf{T}}$ 90° clockwise. Thus, $\vec{n}ds = ⟨dy, -dx⟩$ (Figure B). The differential vector $\vec{n}ds$ represent the normal component of the curve’s differential element.
Therefore, if $\vec{F} = ⟨P, Q⟩$, then the flux of $\vec{F}$ across C can be calculated as follows $\int_{C} \vec{F}·\vec{n}ds =$ $\int_{C} ⟨P, Q⟩·⟨dy, -dx⟩ =$$\int_{C} -Qdx + Pdy.$
The dot product of $\vec{F}$ and $\hat{\mathbf{n}}ds$ give us the component of $\vec{F}$ that is “pushing” across the curve C.
Green’s Theorem is a fundamental result in vector calculus that connects the behavior of a vector field along a closed curve to its behavior over the region it encloses. Specifically, Green’s Theorem for Flux provides a powerful way to relate the flux of a vector field across a closed curve to the divergence of the vector field within the enclosed region.
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 visual explanation (Figure C) helps to understand that the flux integral $\oint_C \vec{F}·\vec{n}·d\vec{s}$ (Figure C) measures the total flow of the vector field $\vec{F} = ⟨P, Q⟩$ across the boundary C, how much of the vector field leaves or enters the region R through the boundary curve C.
Proof.
We want to prove $\oint_C \vec{F}·\vec{n}d\vec{s} = \int \int_{R} div \vec{F}dA$ ↭[In component form, this is equivalent] $\oint_C -Qdx + Pdy = \int \int_{R} (P_x + Q_y)dA$.
To use Green’s Theorem, we rename -Q = M, P = N ⇒ $\oint_C \vec{F}·\vec{n}·d\vec{s} = \oint_C Mdx + Ndy$ =[By Green’s Theorem, which relates a line integral around a closed curve to a double integral over the region it encloses] $\int \int_{R} (N_x-M_y)dA$ =[Substituting back N = P and M = −Q] $\int \int_{R} (P_x+Q_y)dA$ ∎
This result intuitively corresponds to the fact that the vector field is spreading outward uniformly from the origin, with the flux depending directly on the area of the circle.
This example illustrates how to use Green’s Theorem to calculate the flux of a vector field across a closed curve. The vector field has a constant divergence of 3. By applying Green’s Theorem, we computed the flux across a triangular region, resulting in a value of 6. This value represents the total “flow” of the vector field through the boundary of the triangle.
The divergence of a vector field measures how much the flow is expanding or contracting at a point within the field. Imagine a fluid flowing through a region of space. If more fluid is flowing out of a point than flowing into it, the divergence at that point is positive, indicating a source. Conversely, if more fluid is converging into a point than leaving, the divergence is negative, indicating a sink.
Mathematically, divergence represents the source rate or the amount at which something (like fluid, heat, or electrical charge) is added to or removed from the system per unit time and per unit area. It acts as a measure of the presence of sources (positive divergence) or sinks (negative divergence) within the vector field.
In the previous example, the divergence of $\vec{F}$ is zero, indicating that there are no sources (places where the vector field spreads out) or sinks (places where the vector field converges) within the unit circle. Thus, the net flux across the boundary of the unit circle is zero because the vector field is purely rotational —meaning it just spins around the origin without expanding or contracting.
Green’s Theorem is a powerful tool in vector calculus, but it has its limitations.
Both forms or versions of Green theorems relate a line integral around a closed curve C to a double integral over the region R enclosed by C, $\oint_{C} \vec{F}·\vec{n}·ds = \int \int_R div\vec{F}dA$ and $\oint_C \vec{F}·\hat{\mathbf{T}}·ds = \int \int_R curl\vec{F}dA$, but only work when the vector field ($\vec{F}$) and its partial derivative are defined and continuous throughout the region R and on its boundary C.
Counterexample. Consider the vector field $\vec{F} = \frac{-y\vec{i}+x\vec{j}}{x^2+y^2}$. This vector field $\vec{F}$ is not defined at the origin, because the denominator x^{2}+y^{2} becomes zero at (0, 0). However, everywhere else, the curl of $\vec{F}$ is zero: $curl(\vec{F}) = \frac{∂}{∂x}(\frac{x}{x^2+y^2})-\frac{∂}{∂y}(\frac{-y}{x^2+y^2}) = 0$
In this case, the region R_{1} (which does not include the origin) allows the application of the Green’s Theorem: $\oint_{C} \vec{F}d\vec{r} = \int \int_{R_1} curl\vec{F}dA = 0$.
There are situations as illustrated in Figure D where we can not use Green theorems directly. If we consider a curve C′ that encloses the origin (where $\vec{F}$ is not defined), Green’s Theorem cannot be directly applied. This is because the field and its derivatives are not defined at the origin, violating the conditions required by Green’s Theorem.
Green’s Theorem can still be applied in scenarios where the vector field is not defined at certain points within a region. This involves splitting the region into parts that avoid these singularities or using punctured regions, meaning a small region around the singularity that can be excluded, a region where the theorem can be applied to the remaining parts. The difference between the integrals over the outer boundary and the inner boundary (around the excluded region) can be considered.