When you have eliminated the impossible, whatever remains, however improbable, must be the truth, Sherlock Holmes

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.

The line integral of the vector field $\vec{F}$ along the curve C is defined as: $\int_{C} \vec{F}d\vec{r}$ where $d\vec{r}$ is an infinitesimal vector tangent to the curve, given by: $d\vec{r} = ⟨dx, dy, dz⟩$.

$\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$

Let our vector field $\vec{F} = P\hat{\mathbf{i}}+ Q\hat{\mathbf{j}}+R\hat{\mathbf{k}}$, we define the curl of $\vec{F}$ as $curl(\vec{F})=(R_y-Q_z)\hat{\mathbf{i}} + (P_z-R_x)\hat{\mathbf{j}}+(Q_x-P_y)\hat{\mathbf{k}}$

$curl(\vec{F}) = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ P & Q & R\end{smallmatrix}| = (\frac{∂R}{∂y}-\frac{∂Q}{∂z})\hat{\mathbf{i}}-(\frac{∂R}{∂x}-\frac{∂P}{∂z})\hat{\mathbf{j}} + (\frac{∂Q}{∂x}-\frac{∂P}{∂i})\hat{\mathbf{k}}$

Theorem. The curl of a conservative field is $\vec{0}$. Conversely, if $curl(\vec{F})=\vec{0}$, the components **P, Q, and R have continuous first-order partial derivatives**, and its domain is open and simply-connected, the vector field is conservative.

Given a vector field $\vec{F}$, a smooth surface S, and a simple, closed curve C that forms the boundary of the surface S, Stokes’ Theorem asserts that:

$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS$ where:

- $\oint_C \vec{F} \cdot d\vec{r}$ is the line integral of the vector field $\vec{F}$ around the closed curve C.
- $\int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS$ is the surface integral of the curl of $\vec{F}$ over the surface S.
- ${\hat\mathbf{n}}$ is the unit normal vector to the surface S, indicating the direction perpendicular to the surface.
- dS is the infinitesimal area element of the surface.

- Compute the work (line integral) of the vector field $\vec{F} = z\hat{\mathbf{i}}+x\hat{\mathbf{j}}+y\hat{\mathbf{k}}$ around the unit circle in the xy plane, oriented counterclockwise (Figure B).

The curve C, the unit circle in the xy-plane, is described by the equation: x^{2}+y^{2} = 1.

This calculation can be approached in two different ways:

- Direct computation of the line integral, by parameterizing the curve.
- Using Stokes’ Theorem, applying the vector calculus identity to simplify the work computation over a surface.

The line integral we want to compute is: $\oint_C \vec{F} \cdot d\vec{r} = \oint_C zdx + xdy +ydz$

Parametrize the curve. The unit circle can be parametrized in terms of t, where t runs from 0 to 2π, as follows: $\vec{r}(t)=⟨cos(t), sin(t), 0⟩, 0 ≤ t ≤ 2π$.

Calculating the differential displacement. $d\vec{r}$, $d\vec{r} = \frac{d\vec{r}}{dt}dt = ⟨-sin(t), cos(t), 0⟩dt$

Evaluating $\vec{F}$ along the curve. $\vec{F}(\vec{r}(t))= ⟨0, cos(t), sin(t)⟩$

Thus, the line integral becomes: $\int_{0}^{2π} ⟨0, cos(t), sin(t)⟩·⟨-sin(t), cos(t), 0⟩dt= \int_{0}^{2π} cos^2(t)dt$

**Evaluating the integral**. Recall the trigonometric identity: $cos^2(t)=\frac{1+cos(2t)}{2}⇒ \int_{0}^{2π} cos^2(t)dt = \int_{0}^{2π} \frac{1+cos(2t)}{2}dt = \frac{1}{2}\int_{0}^{2π}dt +\frac{1}{2}\int_{0}^{2π}cos(2t)dt$ =[$\int cos(ax) = \frac{1}{a}sin(ax)$]

$\frac{1}{2}t +\frac{1}{4}sin(2t)\bigg|_{0}^{2π} = \frac{1}{2}2π + \frac{1}{4}sin(4π) -\frac{1}{2}·0 -\frac{1}{4}sin(0) = π + 0 -0 -0 = π.$

We could have simplify the calculations since we already know that the integral of cos(2t) over a full period [0, 2π] is zero.

Let’s use Stoke’s theorem and we are going to use the surface of the paraboloid z = 1 -x^{2} -y^{2} as our chosen surface (Figure C).

According to Stoke’s theorem, the circulation around the curve C is equal to the flux of the curl of F through the surface S,

$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS$

We begin by calculating the curl of $\vec{F}$. $(∇x\vec{F}) = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ z & x & y\end{smallmatrix}| = (1-0)\hat{\mathbf{i}}-(0-1)\hat{\mathbf{j}}+(1-0)\hat{\mathbf{k}} = ⟨1, 1, 1⟩$

Given that z = g(x, y) = 1 -x^{2} -y^{2} ↭ f(x, y) = z -g(x, y) = z -1 + x^{2} + y^{2}, we can use the previously studied formula $\hat{\mathbf{n}}dS = ⟨-g_x, -g_y, 1⟩dxdy = ⟨2x, 2y, 1⟩dxdy$

Recall that the gradient ∇f is like finding the slope of the surface in the x, y, and z directions: ∇f = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z}⟩=⟨2x, 2y, 1⟩$. This gradient vector ⟨2x, 2y, 1⟩ points perpendicular to the surface. The normal vector gives us the direction perpendicular to the surface. We normally use the form $\hat{\mathbf{n}}dS$ which incorporate both the normal vector direction and the differential area, $\hat{\mathbf{n}}dS = ⟨2x, 2y, 1⟩dxdy$.

Simplifying we arrive at: $\int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS = \int \int_{S} ⟨1, 1, 1⟩·⟨2x, 2y, 1⟩dxdy = \int \int_{S} (2x+2y+1)dxdy$

This integral corresponds to the shadow of the surface, namely the unit disk (Figure D), so it is easy to switch to polar coordinates.

Due to symmetry, we can observe that the area corresponding to 2x is zero because it is the area of two parts, it is as negative on the left as it is positive on the right, so both integrals cancelled each other. Similarly, the area corresponding to 2y, so the only part that remains is the constant 1, that is the area of the unit disk, namely π.

Another way of seeing it is switching to polar coordinates (x = rcos(θ), y = rsin(θ), dxdy = rdrdθ): $\int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS = \int \int_{S} ⟨1, 1, 1⟩·⟨2x, 2y, 1⟩dxdy = \int \int_{S} (2x+2y+1)dxdy = \int_{0}^{2π}\int_{0}^{1} (2rcos(θ)+2rsin(θ)+1)dxdy =$[Due to symmetry, the integrals involving cos(θ) and sin(θ) will cancel out over a full period 0 to 2π] $\int_{0}^{2π}\int_{0}^{1} rdrdθ = \int_{0}^{2π}\frac{r^2}{2}\bigg|_{0}^{1} =$

= $\int_{0}^{2π} \frac{1}{2}dθ = \frac{1}{2}θ\bigg|_{0}^{2π}$ = π.

- Evaluate the surface integral $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS$ where the vector field $\vec{F} = z^2\hat{\mathbf{i}}-3xy\hat{\mathbf{j}}+x^3y^3\hat{\mathbf{k}}$, and the surface S is the upper part of the paraboloid z = 5-x
^{2}-y^{2}above the plane z = 1. The surface S is oriented upward (Figure 2)

Given a vector field $\vec{F}$ and a surface S with boundary curve C, Stokes’ Theorem states: $\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS$

**Determining the Boundary Curve C**

The boundary curve C is where the paraboloid intersects the plane z = 1. Setting z = 1 in the equation of the surface z = 5 -x^{2} -y^{2}, we get: 1 = 5 -x^{2}-y^{2}⇒ x^{2} + y^{2} = 4, z = 1. This is the equation of a circle with radius 2 at z = 1.

**Parametrizing the Boundary Curve C**

The parametrization of the curve C, x^{2} + y^{2} = 4, is $\vec{r}(t) = ⟨2cos(t), 2sin(t), 1⟩$ where 0 ≤ t ≤ 2π and t is the parameter.

Vector Field Along the Curve: $\vec{F}(\vec{r}(t)) = ⟨1, -3·2cos(t)2sin(t), 8cos^3(t)8sin^3(t)⟩ = ⟨1, -12cos(t)sin(t), 64cos^3(t)sin^3(t)⟩$

Calculating the differential displacement: $d\vec{r} = \frac{d\vec{r}}{dt}dt = ⟨-2sin(t), 2cos(t), 0⟩dt$

$\oint_C \vec{F} \cdot d\vec{r} = \int_{0}^{2π} ⟨1, -12cos(t)sin(t), 64cos^3(t)sin^3(t)⟩·⟨-2sin(t), 2cos(t), 0⟩dt = \int_{0}^{2π} (-2sin(t) -24sin(t)cos^2(t))dt$

The first integral evaluates to 0 because sin(t) is an odd function over a full period [0,2π]: $ -2\int_{0}^{2π} sin(t)dt = 0$

Similarly, for the second integral, we use the fact that sin(t)cos^2(t)dt is also odd over [0, 2π]: $ -24\int_{0}^{2π} sin(t)cos^2(t)dt = 0$

Thus, the line integral is: $\oint_C \vec{F} \cdot d\vec{r} = 0$. By Stokes’ Theorem, the circulation $\oint_C \vec{F} \cdot d\vec{r}$ is equal to the surface integral of the curl of $\vec{F}$ over the surface S, hence: $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS = 0.$

- Evaluate $\oint_C \vec{F} \cdot d\vec{r}$ where $\vec{F} = z^2\hat{\mathbf{i}}+y^2\hat{\mathbf{j}}+x\hat{\mathbf{k}}$ and C is the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) with counter-clockwise rotation (Figure A).

According to Stoke’s theorem, the circulation around the curve C is equal to the flux of the curl of F through the surface S, where S is any surface with boundary curve C.

$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS$

Now, all we have is the boundary curve for the surface that we’ll need to use in the surface integral. In this case, the simplest choice is the plane defined by the three points A(1, 0, 0), B(0, 1, 0), and C (0, 0, 1).

$\vec{AB} = B -A = (-1, 1, 0), \vec{AC} = C -A = (-1, 0, 1), \hat{\mathbf{n}} = \vec{AB} x \vec{AC} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ -1 & 1 & 0\\ -1 & 0 & 1\end{smallmatrix}| = ⟨1,-(-1), 1⟩ = ⟨1, 1, 1⟩$.

The general equation of a plane is given by: ax +by +cz = d where (a, b, c) is the normal vector. Using A = (1, 0, 0) to find d, 1·1+1·0+1·0=d ⇒ d = 1. Therefore, the equation of the plane is x +y +z = 1. The normal vector is ⟨1, 1, 1⟩, since the plane is orientated upwards the surface that lies on this plane and is bounded by C is a compatible surface because induces a positive (counter-clockwise) direction on C.

The surface is the plane x +y +z = 1 ↭ z = g(x, y) = 1 -x-y. The normal vector can be found from the coefficients of x, y, and z, giving $\hat{\mathbf{n}}=⟨1, 1, 1⟩$ Another way of seeing this: the gradient of f(x, y, z) = z -g(x, y) = z - (1-x-y) = z -1+x+y, ∇f = $⟨\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z}⟩ = ⟨1, 1, 1⟩$.

**Parametrize the surface S.** Using x and y as parameters: $\vec{r}(x, y)=⟨x, y, 1-x-y⟩$

$curl(\vec{F}) = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ z^2 & y^2 & x\end{smallmatrix}| = (\frac{∂x}{∂y}-\frac{∂y^2}{∂z})\hat{\mathbf{i}}-(\frac{∂x}{∂x}-\frac{∂z^2}{∂z})\hat{\mathbf{j}} + (\frac{∂y^2}{∂x}-\frac{∂z^2}{∂y})\hat{\mathbf{k}} = ⟨0-0, -(1-2z), 0-0⟩ = ⟨0, 2z-1, 0⟩$

**Surface Integral**:
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS = \int \int_{S} ⟨0, 2z-1, 0⟩·⟨1, 1, 1⟩dS = \int \int_{S} (2z-1)dS = $

We are going to convert the surface integral to double integral over the xy-region (Figure B). The projection of the surface onto the xy-plane is the triangular region bounded by the lines x = 0, y = 0, and x + y = 1.

Thus, the double integral becomes: $\int_{0}^{1}\int_{0}^{-x+1} 2(1-x-y)-1 dydx = \int_{0}^{1}\int_{0}^{-x+1} (1-2x-2y)dydx$

$\int_{0}^{-x+1} (1-2x-2y)dydx = (y-2xy-y^2)\bigg|_{0}^{-x+1}$

Simplifying the expression: $-x+1-2x(-x+1)-(x^2-2x+1) = -x+1+2x^2-2x-x^2+2x-1 = x^2-x$

Now, integrate with respect to x: $\int_{0}^{-x+1} (x^2-x)dx = \frac{1}{3}x^3-\frac{1}{2}x^2\bigg|_{0}^{1}$

$\frac{1}{3}-\frac{1}{2}=\frac{2}{6}-\frac{3}{6}=\frac{-1}{6}$

Thus, the value of the line integral $\oint_C \vec{F} \cdot d\vec{r} =\frac{-1}{6}$

As we have previously stated, Stokes’ Theorem connects surface integrals of vector fields with line integrals over the boundary of those surfaces. It’s a powerful tool in vector calculus that links two different types of integrals.

In certain cases, if the curl of a vector field is zero, Stokes’ Theorem helps us establish that the field is path-independent. This path-independence implies that the value of the line integral depends only on the endpoints of the path, not the specific path taken between them.

Definition. A region D is called simply-connected if, for any closed loop inside the region D (a path that starts and ends at the same point), then the interior of this loop also lies entirely in D and it can be continuously deformed and contracted into a single point within D without leaving the region.

Deforming a loop refers to gradually changing its shape while keeping it inside the region (or without taking it outside the boundaries of the region). Contracting a loop means shrinking the loop smaller and smaller until it becomes just a single point, all while staying inside the region.

If a region is simply-connected, you can always shrink any loop to a point without crossing any boundaries or encountering any holes.

- ℝ
^{3}. Imagine the entire space around you. Any loop you draw in this space can be contracted to a point because there are no holes. - ℝ
^{3}-z axis. In this region, if you draw a loop around the z-axis, you cannot shrink it to a point without crossing the z-axis, which is not part of the region. This makes the region not simply-connected. - A solid torus are not (doughnut-shaped object). If you draw a loop around the hole of the doughnut (the central void), you can’t shrink that loop to a point without going through the solid part of the doughnut, making the region non-simply-connected (Figure A).
- 💣
**ℝ**. You might think removing a point creates a hole, but it doesn’t create a significant obstacle that prevents shrinking loops. The reason is that a loop can still be shrunk to a point, even if you remove a single point, because you can always push or wiggle the loop around the missing point, so there are no other holes or barriers in ℝ^{3}−{0} is simply connected^{3}-{0}. - The 3D volume of a sphere, meaning the space inside the sphere as well as on the surface, is simply connected. In the 3D sphere, any loop drawn in space can be contracted to a point because you are allowed to use the entire volume of the sphere, including its interior.

Theorem: Path-Independence of Gradient Fields in Simply-Connected Regions

If a vector field $\vec{F}$ is defined in a simply-connected region and curl $\vec{F}$ = 0, then $\vec{F}$ is a gradient field, and consequently, the line integral $\oint_C \vec{F} \cdot d\vec{r}$ is path-independent. This means that the value of the line integral only depends on the endpoints of the path, not on the specific path taken between them.

Proof.

Assume that $curl \vec{F}= ∇×\vec{F} = 0$ in a simply-connected region.

We wat to show the line integral of $\vec{F}$ is path-independent. This means that for any two curves C_{1} and C_{2} that start at the same point P_{0} and end at the same point P_{1} (Figure B), the line integrals over these two paths are the equal: $\int_{C_1} \vec{F}d\vec{r} = \int_{C_2} \vec{F}d\vec{r}.$

**Construct a closed loop.** We can construct a closed loop C, C = C_{1}-C_{2}, by following C_{1} from P_{0} to P_{1}, then following C_{2} in the reverse direction from P_{1} back to P_{0}. This forms a closed loop C, hence we can use Stoke’s Theorem.

This is key: Since the region is simply-connected, any closed loop C lies in the region, and there exists a surface S with C as its boundaries.

$\int_{C_1} \vec{F}d\vec{r}-\int_{C_2} \vec{F}d\vec{r} = \oint_C \vec{F} \cdot d\vec{r}$ =[Stoke’s Theorem] $\int \int_{S} (curl \vec{F})·\hat{\mathbf{n}}dS = [\text{By assumption}]\int \int_{S} 0·\hat{\mathbf{n}}dS = 0⇒ \int_{C_1} \vec{F}d\vec{r}-\int_{C_2} \vec{F}d\vec{r} = 0 ⇒ \int_{C_1} \vec{F}d\vec{r} = \int_{C_2} \vec{F}d\vec{r}$∎

Thus, the line integral of $\vec{F}$ is path-independent. The value of the line integral only depends on the endpoints of the path, not on the specific path chosen between the points. This proves that $\vec{F}$ is a conservative vector field, meaning it is the gradient of some scalar potential function f.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts Mathematics, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn, and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
- MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
- Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.