Surface Integrals of Vector Fields. Flux.

One who asks a question is a fool for a minute; one who does not remains a fool forever, Chinese proverb

Vector Fields, Work, and Green’s Theorem

A vector field on a plane is an assignment of a vector $\vec{F}$ to every point $(x, y)$ in a given space. Mathematically, it is expressed as: $\vec{F}(x, y) = \langle M(x, y), N(x, y) \rangle = M\hat{i} + N\hat{j}$ where $M$ and $N$ are scalar 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. In physics, vector fields often represent forces (like gravity or electromagnetism). If we want to calculate the work done by this force field in moving an object along a specific trajectory $C$, we use a line integral.

The work $W$ done by a force field $\vec{F}$ along a curve $C$ is the integral of the tangential component of the force along the path: $W = \int_C \vec{F} \cdot d\vec{r} = \int_C (M \, dx + N \, dy)$

The Four Faces of Conservative Fields

A vector field is considered conservative if it is the gradient of a scalar potential function $f(x,y)$, meaning $\boxed{\vec{F} = \nabla f}$.

If a vector field is continuous on an open, simply-connected region (a region with no “holes”), the following four properties are mathematically equivalent. If one is true, they are all true:

1. Gradient Field: There exists a scalar potential function $f$ such that $\vec{F} = \nabla f = \langle f_x, f_y \rangle$.
2. Path Independence: The work done by the force in moving an object from point $A$ to point $B$ is identical regardless of the path taken.
3. Conservative Force: The work done by the force around any closed loop $C$ is exactly zero ($\oint_C \vec{F} \cdot d\vec{r} = 0$).
4. Exact Differential: The differential expression $M \, dx + N \, dy$ is exact, meaning it is equal to the total differential $df = f_x \, dx + f_y \, dy$.

For these conservative fields, we can bypass complex line integrals using the Fundamental Theorem of Line Integrals. The work is simply the difference in potential energy between the endpoints: $\boxed{\int_C \nabla f \cdot d\vec{r} = f(B) - f(A)}$

Criteria for a Conservative Vector Field

Before diving into flux, it is crucial to understand conservative vector fields. Let $\vec{F}(x, y) = \langle P(x, y), Q(x, y) \rangle$ be a 2D vector field.

The criterion for checking whether a vector field $\vec{F}$ is conservative can be summarized as follows: If the cross-partials are equal, meaning $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, and if the scalar functions $M$ and $N$ have continuous first partial derivatives across an open, simply connected domain $D$ (a region with no

This conservative property is deeply linked to Green’s Theorem for Circulation (Work).

Green’s Theorem. If $C$ is a positively oriented (counterclockwise) simple closed curve enclosing a region $R$, and $\vec{F} = \langle P(x, y), Q(x,y) \rangle$ is a vector field that is defined and has continuous partial derivatives on an open region containing R, then Green's Theorem states that: $\oint_C \vec{F} \cdot{} d\vec{r} = \iint_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$ where

• The line integral $\oint_C \vec{F} \cdot{} d\vec{r}$ represents the work (or line integral) done by the vector field $\vec{F}$ as you move around the curve $C$.
• $\iint_R (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$ represents the double integral over the region $R$, with the integrand being the difference between the partial derivatives of $N$ and $M$. It computes the microscopic curl (rotation) at every point inside the region $R$.

If the field is conservative, $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$, the double integral becomes zero. Consequently, the work done around any closed loop in a conservative field is always zero.

Flux

Flux is an important concept in multivariable calculus that allows us to quantify how much of a vector field passes through a given curve or surface. 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 ($\int_{C} \vec{F}·\vec{dr}$). While work line integrals measure how much a vector field pushes you along a path, flux measures how much of the field crosses perpendicularly through the path.

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: $\int_{C} \vec{F} \cdot \vec{dr}$ where $\vec{dr}$ is a small displacement vector along the curve $C$, and $\vec{F} \cdot \vec{dr}$ represents the dot product of the vector field $\vec{F}$ with the displacement vector $\vec{dr}$.

The Physical Intuition

Imagine a river where the flowing water is represented by a vector field $\vec{F}$.

• If you swim along a specific path $C$, the extent to which the current helps or hinders your forward progress is the Work.
• If you stretch a porous fishing net across the river along path $C$, the total volume of water passing through the net per second is the Flux.

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, $\boxed{Flux = \oint_{C} (\vec{F}·\hat{\mathbf{n}})ds}$ where:

• $\vec{F}$ is the vector field.
• $\hat{\mathbf{n}}$ is the unit normal vector to the curve $C$, meaning it is perpendicular to the curve and has a length of 1. It is the outward normal of a counterclockwise curve. By convention for closed curves, $\hat{\mathbf{n}}$ always points outward, away from the enclosed region.
• $ds$ represents an infinitesimal area element along the curve $C$ (Figure 1).

To approximate this, imagine breaking the curve $C$ into tiny segments of length $\Delta s_i$. The flux through one tiny segment is roughly $\vec{F}_i \cdot \hat{\mathbf{n}}_i \Delta s_i$.

Taking the limit as the segment sizes shrink to zero yields the exact integral: $\text{Flux} = \lim_{n \to \infty} \sum_{i=1}^{n} (\vec{F}_i \cdot \hat{\mathbf{n}}_i) \Delta s_i$. In other words, the total flux is then the sum of these contributions as the segment size $\Delta s \to 0$ approaches zero:

Recall that work is computed as: $\oint_{C} \vec{F} \cdot d\vec{r} = \oint_{C} (\vec{F} \cdot \hat{\mathbf{T}})ds$, 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.

Imagine the curve C as a thin fence in a flowing fluid, 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 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 $\Delta S$) 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 $\Delta S$. Summing these areas over the entire curve gives the total flux.

Solved examples

• The Radial Field (Maximum Flux). Let $C$ be a circle of radius $a$ centered at the origin, oriented counterclockwise. The vector field is $\vec{F} = \langle x, y \rangle$, which points radially outward, directly away from the origin (Figure 4).

Solution.

1. Vector Field and Normal Vector Relationship. The vector field $\vec{F}$ is radially outward, meaning at every point (x, y) the vector $\vec{F}$ points directly away from the origin, hence it is aligned with the normal vector $\hat{\mathbf{n}}$ of the circle. Therefore, $\vec{F}||\vec{n}$. Because $\vec{F}$ and $\vec{n}$ are parallel, the dot product $\vec{F}·\vec{n}$ simplifies significantly: $\vec{F}·\vec{n} = |\vec{F}|·|\vec{n}|·cos(0) = |\vec{F}|·1·1 = |\vec{F}|$. The magnitude of $\vec{F}$ at any point on the circle of radius “a” is simply the distance from the origin, which is simply the radius “a” of the circle.
2. Setting up the integral for Flux. We now compute the flux of $\vec{F}$ across the curve C. Flux is given by the line integral of the dot product of the vector field $\vec{F}$ and the unit normal vector $\vec{n}$ over the curve C: Flux = $\int_{C} \vec{F}·\vec{n}·ds = \int_{C} a·ds = a·\int_{C} ds = a·length(C)$ = [The arc length of the circle C with radius a is the circumference of the circle.] $a \cdot 2\pi a = 2\pi a^2$.

   The result $2\pi a^2$ 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. If $\vec{F}$ represents a flow of some quantity (like fluid or heat), then $2\pi a^2$ is the total amount of that quantity passing through the boundary of the circle per unit time.

• The Rotational Field (Zero Flux). Let $C$ be the same circle of radius a centered at the origin, oriented counterclockwise. The vector field is $\vec{F} = \langle -y, x \rangle = -y\vec{i}+x\vec{j}$ is defined to rotate counterclockwise around the origin (Figure A). We are asked to compute the flux of this vector field across the circle $C$.

1. Geometric Intuition. 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 ($\vec{-i}$).

2. 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.

3. Dot product calculation: Since $\vec{F} \perp \vec{n} \implies \vec{F} \cdot \vec{n} = \vert \vec{F} \vert \times \vert \vec{n} \vert \times \cos(90^\circ) = \vert \vec{F} \vert \cdot 1 \cdot 0 = 0$.

4. 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 water is swirling in a perfect whirlpool inside the circle. None of the water is escaping the boundary, hence the net outflow (flux) is absolutely zero.

• Consider the vector field $\vec{F} = \langle y-x, x \rangle$ and the curve $C$ defined by $\vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle$ for $0 \le t \le 2\pi$. Compute the flux of $\vec{F}$ across C (Refer to Figure iv for a visual representation and aid in understanding it).

1. Understand the problem. C: Given the parametrization $\vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle$ where t ranges from $0$ to $2\pi$. This describes a circle of radius 2 centered at the origin, traversed counterclockwise.

2. Determine the Normal Vector. First, compute the derivative of $\vec{r}(t)$ with respect to t, which gives the tangent vector, $\vec{r'}(t) = \frac{d}{dt}(2\cos(t), 2\sin(t)) = \langle -2\sin(t), 2\cos(t) \rangle$.
   For flux calculations, we need the outward-pointing normal vector. To find this, we rotate the tangent vector by $90 \deg$ clockwise, $\vec{n} = \langle 2\cos(t), 2\sin(t) \rangle$. This has length 2. The unit outward normal is $\vec{n}_{unit}(t) = \langle \cos(t), \sin(t) \rangle$, that is, the position vector direction.

3. Evaluate the vector field along C: $\vec{F}(\vec{r}(t)) = \vec{F}(2\cos(t), 2\sin(t)) = \langle 2\sin(t)-2\cos(t), 2\cos(t) \rangle$

4. Set up the flux integral: $Flux = \int_{C} (\vec{F} \cdot \vec{n})ds = \int_{C} (\vec{F} \cdot \vec{n}_{unit}(t)) \vert \vec{r'}(t) \vert dt$.
   Compute the dot product: $\vec{F}(\vec{r}(t)) \cdot \vec{n}_{unit}(t) = \langle 2\sin(t)-2\cos(t), 2\cos(t) \rangle \cdot \langle \cos(t), \sin(t) \rangle =[\text{Simplify}] (2\sin(t)-2\cos(t))(\cos(t)) +(2\cos(t))(\sin(t))dt = 2\sin(t)\cos(t)-2\cos^2(t)+2\cos(t)\sin(t) = 4\sin(t)\cos(t) -2\cos^2(t)$ Now multiply by $\vert \vec{r'}(t) \vert = \sqrt{(-2\sin(t))^2+ (2\cos(t))^2} = \sqrt{4\sin^2(t) + 4\cos^2(t)} = \sqrt{4} = 2$:
   So $ds = \vert \vec{r'}(t) \vert dt = 2dt$.
   $Flux = \int_{0}^{2\pi} (4\sin(t)\cos(t) -2\cos^2(t))\times 2 dt = \int_{0}^{2\pi} -4\cos^2(t)dt + 8\cos(t)\sin(t)$

5. Evaluate the integral.
   Use the identity $\sin(t)\cos(t) = \frac{1}{2}\sin(2t)$. Then, $\int_{0}^{2\pi} 8\cos(t)\sin(t) = 8 \times \frac{1}{2} \int_{0}^{2\pi} \sin(2t) dt = 4\int_{0}^{2\pi} \sin(2t) dt = -2\cos(2t)|_0^2\pi = -2(cos(4\pi) -\cos(0)) = -2(1 -1) = 0$. So this term contributes zero.
   Using the identity $cos^2(t) = \frac{1+cos(2t)}{2}$, $-4\int_{0}^{2π} cos^2(t) = -2\int_{0}^{2π} (1+cos(2t))dt = -2[t+\frac{sin(2t)}{2}]\bigg|_{0}^{2π} = (-2)2π = -4π$

6. Check via Green’s theorem. For a closed curve $C$ bounding a region $D$, the flux of $\vec{F} = \langle P, Q \rangle$ across $C$ with outward normal is: $\text{Flux} = \oint_C \vec{F}\cdot \vec{n}\, ds = \iint_D \nabla\cdot \vec{F}\, dA.$
   Here: $P(x,y) = y - x, \quad Q(x,y) = x.$
   So: $\nabla\cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} = \frac{\partial (y - x)}{\partial x} + \frac{\partial x}{\partial y} = -1 + 0 = -1$.
   The region $D$ is the disk of radius $2$, so its area is: $\text{Area}(D) = \pi(2^2) = 4\pi.$
   Thus: $ \text{Flux} = \iint_D (-1)\, dA = -1 \cdot \text{Area}(D) = -4\pi.$
   Same result —this confirms the computation and ties it directly to divergence.

7. Physical meaning of the sign. The flux is negative: $\text{Flux} = -4\pi.$ With the normal pointing outward, a negative flux means:

• The net flow is inward across the curve.
• More of the field is crossing the circle into the disk than out of it.
• So in the “fluid” picture, $\vec{F}$ is pushing fluid into the circle, not out of it.

Calculating Flux using components

Flux is a measure of how much a vector field “flows” through a curve. To easily compute the flux of a vector field $\vec{F}$ across a curve $C$ in the $xy$-plane, we can express it using the components of the vector field and the curve.

1. A curve $C$ in the plane is traced out by a position vector $\vec r(s)=\langle x(s),\,y(s)\rangle$ where $s$ is arc length.
   Arc length is the “natural” parameter: moving one unit in $s$ means you physically moved one unit along the curve.
   Because of that, the derivative $\frac{d\vec r}{ds}$ is automatically a unit vector. It tells you the direction in which the curve is heading at each point.
   This is the unit tangent vector: $\hat{\mathbf T}=\frac{d\vec r}{ds}=\left\langle \frac{dx}{ds},\,\frac{dy}{ds}\right\rangle$.

2. Flux requires a normal direction — the direction perpendicular to the curve.
   In 2D, “perpendicular” is extremely simple: you rotate the tangent vector by 90°.
   But you must rotate in the correct direction. For a counterclockwise closed curve:
   $\star$ the tangent vector $\hat{\mathbf T}$ points forward along the curve.
   $\star$ the outward normal must point away from the region enclosed by the curve.
   To get that outward normal, the convention is to rotate the tangent vector 90° clockwise.
   Clockwise rotation of a vector $\langle x, y\rangle$ is $\langle y,\,-x\rangle.$
   Next, we apply this to the tangent vector: $\hat{\mathbf n} = \left\langle \frac{dy}{ds},\, -\frac{dx}{ds} \right\rangle.$
   This vector is: (i) perpendicular to $\hat{\mathbf T}$, (ii) unit length (because $(\hat{\mathbf T}$ is unit), and (iii) pointing outward for counterclockwise orientation.

3. Why this rotation works (quick check)
   Take the dot product: $\hat{\mathbf T}\cdot \hat{\mathbf n} = \left\langle \frac{dx}{ds},\frac{dy}{ds}\right\rangle \cdot \left\langle \frac{dy}{ds}, -\frac{dx}{ds}\right\rangle = \frac{dx}{ds}\frac{dy}{ds}- \frac{dy}{ds}\frac{dx}{ds} =0.$
   So they are orthogonal, as required.
   And the length: $\left\|\hat{\mathbf n}\right\| = \sqrt{\left(\frac{dy}{ds}\right)^2+\left(-\frac{dx}{ds}\right)^2} = \sqrt{\left(\frac{dx}{ds}\right)^2+\left(\frac{dy}{ds}\right)^2} =1$, because $\hat{\mathbf T}$ is a unit vector.

4. Imagine walking along the curve: (i) Your body faces the tangent direction ($\hat{\mathbf T}$). (ii) Your right arm points outward from the region (for counterclockwise orientation).
   Rotating your body’s forward direction 90° clockwise gives exactly the direction your right arm points. That’s the outward normal.

5. When we multiply this unit normal vector by the differential arc length $ds$, the $ds$ denominators cancel out perfectly: $\hat{\mathbf{n}} ds = \langle dy, -dx \rangle$.

6. Therefore, if our vector field is defined as $\vec{F} = \langle P, Q \rangle$, the flux of $\vec{F}$ across $C$ can be calculated via a straightforward dot product: $\int_{C} (\vec{F} \cdot \hat{\mathbf{n}}) ds = \int_{C} \langle P, Q \rangle \cdot \langle dy, -dx \rangle = \int_{C} -Q dx + P dy$

The dot product of $\vec{F}$ and $\hat{\mathbf{n}} ds$ mathematically filters out any movement along the curve, isolating only the component of $\vec{F}$ that is “pushing” perpendicularly across $C$.

Green’s Theorem for Flux

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 2D region it encloses. Specifically, Green’s Theorem for Flux provides a powerful way to relate the macroscopic flux of a vector field across a closed boundary to the microscopic 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 $\vec{F} = \langle P, Q \rangle$ is a continuously differentiable vector field defined on an open region that contains $R$, then the outward 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: $\boxed{\oint_C (\vec{F} \cdot \vec{n})ds = \int \int_{R} div \vec{F}dA}$ where:

• The divergence of $\vec{F}$ is given by $div \vec{F} = \nabla \cdot \vec{F} = P_x + Q_y = \frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}$.
• $\vec{F}$ is the vector field.
• $\hat{\mathbf{n}}$ is the outward-pointing unit normal vector to $C$.
• $ds$ is the differential of the arc length along $C$.
• $dA$ is the differential area element ($dx \, dy$) in $R$.

The visual explanation (Figure C) helps to understand that the flux integral (Figure C) measures the total net flow of the vector field $\vec{F} = \langle P, Q \rangle$ leaving a region through its boundary $C$ is exactly equal to the sum of all the tiny sources and sinks generated inside the region ($R$).

Note: $\nabla \cdot \vec{F} = P_x + Q_y$, $P_x$ is the horizontal component of the flow. It measures how much the horizontal flow increases as you move right. If $P_x \gt 0$, then fluid is spreading outward horizontally - like stretching rubber. If $P_x \lt 0$, fluid is squeezing inward horizontally. Similarly, $Q_y$ is the vertical expansion. Add them: $P_x + Q_y$ = total expansion or contraction at the point (locally).

Proof.

1. We want to prove that $\oint_C (\vec{F} \cdot \hat{\mathbf{n}}) ds = \iint_{R} \text{div} \vec{F} \, dA$.
2. In component form, this is equivalent $\oint_C -Qdx + Pdy = \int \int_{R} (P_x + Q_y)dA$.
3. We can prove this by directly invoking the standard Circulation/Work form of Green’s Theorem: $\oint_C M dx + N dy = \iint_R (N_x - M_y) dA$.
4. Let us define $M = -Q$ and $N = P$. Substituting these into the circulation theorem yields: $\oint_C -Q dx + P dy = \iint_R \left( \frac{\partial (P)}{\partial x} - \frac{\partial (-Q)}{\partial y} \right) dA$
   $\oint_C -Q dx + P dy = \iint_R \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} \right) dA \quad \blacksquare$

Exercises

• The Radial Field (Uniform Expansion).

Consider the vector field $\vec{F} = x\vec{i}+y\vec{j}$, which points radially away from the origin. We want to find the flux of this vector field across a circle $C$ of radius $a$, centered at the origin, and oriented counterclockwise (Figure 4).

1. Compute the Divergence. First, we calculate the divergence of the vector field: $\vec{F} = \langle P, Q \rangle, div \vec{F} = P_x + Q_y = \frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y) = 1 + 1 = 2.$
2. Apply Green’s Theorem: Green’s Theorem relates the flux of a vector field $\vec{F}$ across a closed curve $C$ to the double integral of the divergence of the vector field $\vec{F}$ over the region $R$ enclosed by $C$. Mathematically, this is expressed as: $\int_{C} (\vec{F} \cdot \hat{\mathbf{n}}) ds = \int \int_{R} div \vec{F}dA$
3. Set up and Evaluate the double integral. $\int \int_{R} div \vec{F}dA = \int \int_{R} 2dA = 2\int \int_{R} dA = 2area(R)$ =[The area of the region $R$ (which is a circle of radius a) is $\pi a^2$] $2\pi a^2$.
4. Conclusion. The flux of the vector field $\vec{F}$ across the circle $C$ is: $\int_{C} (\vec{F} \cdot \hat{\mathbf{n}}) ds =2\pi a^2$.
5. Intuition. A constant divergence of 2 means the field is uniformly generating outward flow everywhere in space. The larger the circle (area), the more “source material” is trapped inside, resulting in higher outward flux.

• Let’s consider the vector field $\vec{F} = 2x\vec{i}+y\vec{j}$. We want to determine the flux of this vector field over a triangular region with vertices $(0, 0)$, $(2, 0)$, and $(2, 2)$ (see Figure i).

1. Compute the Divergence of the vector field: The given vector field is $\vec{F} = 2x\vec{i}+y\vec{j} = \langle P, Q \rangle$ where $P = 2x$ and $Q = y$. The divergence of a vector field $\vec{F}$ is defined as: $div \vec{F} = P_x + Q_y = \frac{\partial}{\partial x}(2x)+\frac{\partial}{\partial y}(y) = 2 + 1 = 3.$
2. Apply Green’s Theorem: Green’s Theorem relates the flux of a vector field $\vec{F}$ across a closed curve $C$ to the double integral of the divergence of the vector field $\vec{F}$ over the region $R$ enclosed by $C$. Mathematically, this is expressed as: $\int_{C} (\vec{F} \cdot \hat{\mathbf{n}}) ds = \int \int_{R} div \vec{F}dA$
3. Set up and Evaluate the double integral. Substitute, $div \vec{F} = 3. \int \int_{R} div \vec{F}dA = \int \int_{R} 3dA$ =[The region $R$ is defined by the triangular area with vertices at (0, 0), (2, 0), and (2, 2). To set up the double integral, we observe that for a given $x, y$ ranges from $0$ to $x$, and $x$ ranges from $0$ to $2$.] $\int_{0}^{2}\int_{0}^{x} 3 \cdot dydx$
4. Compute the inner integral: $\int_{0}^{x} 3 \cdot dy = 3y\bigg|_{0}^{x} = 3x -0 = 3x$.
5. Compute the outer integral: $\int_{0}^{2} 3xdx = \frac{3x^2}{2}\bigg|_{0}^{2} = \frac{3*2^2}{2} -0 = 6.$
6. Conclusion. The flux of $\vec{F}$ across C is: $\oint_C \vec{F}·\vec{n}ds =6$.

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.

• Let’s consider the vector field $\vec{F} = -3y\vec{i}+3x\vec{j}$. We want to determine the flux of this vector field across the unit circle $C$ oriented counterclockwise.

1. Compute the Divergence of the vector field: $\vec{F} = \langle P, Q \rangle, div \vec{F} = P_x + Q_y = \frac{\partial}{\partial x}(-3y)+\frac{\partial}{\partial y}(3x) = 0 + 0 = 0.$
2. Apply Green’s Theorem: Green’s Theorem relates the flux of a vector field $\vec{F}$ across a closed curve C to the double integral of the divergence of the vector field $\vec{F}$ over the region R enclosed by C. Mathematically, this is expressed as: $\int_{C} (\vec{F} \cdot \hat{\mathbf{n}}) ds = \int \int_{R} div \vec{F}dA$
3. Set up and Evaluate the double integral. Substitute, $div \vec{F} = 0. \int \int_{R} div \vec{F}dA = \int \int_{R} 0dA = \int_{0}^{2π}\int_{0}^{1} 0 \cdot rdrdθ = 0$. This indicates that the total flux across the unit circle is zero.
4. Conclusion. The flux of $\vec{F}$ across C is: $\oint_C \vec{F} \cdot \vec{n}ds =0$. This result occurs because the given vector field is a rotational field, meaning that nothing is flowing in or out of the unit circle (no net flow goes through the boundary of the region), instead everything is just flowing around the circle (parallel to the boundary edge forever). Thus, the net flux through the circle is zero.

A Deeper Explanation of Divergence

The divergence of a vector field measures how much the flow is expanding or contracting at a point within the field. It is fundamentally a microscopic property.

Imagine a fluid flowing through a region of space.

1. Positive Divergence (Source). If more fluid is flowing out of a point than flowing into it, the divergence at that point is positive, indicating a source. It acts like a geyser or a star emitting radiation.
2. Negative Divergence (Sink). Conversely, if more fluid is converging into a point than leaving, the divergence is negative, indicating a sink. It acts like a drain or a black hole.
3. Zero Divergence: If exactly as much fluid enters a point as leaves it, the divergence is zero.

Mathematically, the divergence at a point $(x_0, y_0)$ is formally defined as the limit of the net flux per unit area as the area shrinks to that point: $\text{div} \vec{F}(x_0, y_0) = \lim_{\Delta A \to 0} \frac{1}{\Delta A} \oint_{\partial A} (\vec{F} \cdot \hat{\mathbf{n}}) ds$.

Green’s Theorem is simply the macroscopic summation of this limit over a large region.

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.

Limitations of Green’s Theorem

Green’s Theorem is incredibly powerful, but it comes with a strict topological requirement: The vector field $\vec{F}$ must be continuously differentiable everywhere inside the region $R$. If there is even a single point where the field is undefined (a singularity or a “hole”), blindly applying Green’s Theorem will yield incorrect, physically impossible results.

Consider the radial point-source vector field: $\vec{F} = \frac{x}{x^2+y^2}\hat{\mathbf{\imath}} + \frac{y}{x^2+y^2}\hat{\mathbf{\jmath}}$

This field represents a perfect point source at the origin, blasting material outward in all directions. Notice that $\vec{F}$ is undefined at $(0,0)$.

Let’s compute the divergence for any point $(x,y) \neq (0,0)$:
$\frac{\partial P}{\partial x} = \frac{1(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$
$\frac{\partial Q}{\partial y} = \frac{1(x^2+y^2) - y(2y)}{(x^2+y^2)^2} = \frac{x^2-y^2}{(x^2+y^2)^2}$
$\text{div} \vec{F} = \frac{y^2-x^2}{(x^2+y^2)^2} + \frac{x^2-y^2}{(x^2+y^2)^2} = 0$

If we naively apply Green’s Theorem to find the flux through the unit circle $C$, we would integrate the divergence: $\iint_R \text{div} \vec{F} \, dA = \iint_R 0 \, dA = 0$

The math claims the flux is zero. But wait! If we calculate the line integral directly along the unit circle (where $x^2+y^2 = 1$, so $\vec{F} = \langle x, y \rangle$ and $\hat{\mathbf{n}} = \langle x, y \rangle$): $\oint_C (\vec{F} \cdot \hat{\mathbf{n}}) ds = \oint_C (x^2 + y^2) ds = \oint_C 1 \, ds = 2\pi$

Direct calculation yields $2\pi$, but Green’s Theorem yields $0$.

Why? Because the field has a singularity at $(0,0)$. The region $R$ is not a valid, simply connected domain for this field.

The Solution: Punctured Regions

To rescue Green’s Theorem, we must isolate the singularity. We do this by creating a “hole” in our region.

Imagine an outer counterclockwise curve $C_1$ (e.g., our unit circle) and a tiny clockwise inner curve $C_2$ tightly wrapping the origin. If we connect them with a slit, we create a region $R'$ that looks like a washer. Because the origin $(0,0)$ is completely outside $R'$, $\vec{F}$ is perfectly continuous everywhere inside $R'$.

Applying Green’s Theorem to this new region: $\iint_{R'} \text{div} \vec{F} \, dA = \oint_{C_1} (\vec{F} \cdot \hat{\mathbf{n}}) ds + \oint_{C_2} (\vec{F} \cdot \hat{\mathbf{n}}) ds$

Since $\text{div} \vec{F} = 0$ everywhere inside $R'$, the double integral is $0$. Therefore, $\oint_{C_1} (\vec{F} \cdot \hat{\mathbf{n}}) ds = -\oint_{C_2} (\vec{F} \cdot \hat{\mathbf{n}}) ds$

Because $C_2$ is oriented clockwise (negative), reversing its direction to counterclockwise yields: $\oint_{C_1} (\vec{F} \cdot \hat{\mathbf{n}}) ds = \oint_{\text{counterclockwise } C_2} (\vec{F} \cdot \hat{\mathbf{n}}) ds$.

This profound result proves that any simple closed curve enclosing the origin will have a flux of $\oint_{C_1} (\vec{F} \cdot \hat{\mathbf{n}}) ds = 2\pi$ (it was previously calculated). By surgically cutting out the singularity, we turn Green’s Theorem’s limitation into a powerful tool for analyzing complex topologies.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
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3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
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5. Michael Penn, and MathMajor.
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