All things are difficult before they are easy, Thomas Fuller.

Curiosity is the driving force between knowledge and discovery, Arturo Perez Reverte.

What about the future?" “We’ll talk about the future when it gets here,” Arturo Perez Reverte.

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 $curl(\vec{F})=(R_y-Q_z)\hat{\mathbf{i}} + (P_z-R_x)\hat{\mathbf{j}}+(Q_x-P_y)\hat{\mathbf{k}}$

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

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.

It’s a powerful result in calculus that bridges the gap and establishes a connection between two seemingly different types of integrals: **the line integral of a vector field $\vec{F}$ around a simple, closed curve C and tbe surface integral of the curl of that vector field over a surface S bounded by that curve C**. S is any surface bounded by C.

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.

It states that the total circulation (line integral) of the vector field $\vec{F}$ around the closed curve C (how much the vector field flows along the curve) is equal to the flux (surface integral) of the curl of that vector field through any surface S bounded by the curve (how much the field is spinning or rotating across the surface)..

In simple terms, Stokes’ Theorem equates the **circulation around the boundary of a surface to the total amount of rotation of the field over the surface itself**. Imagine a fluid flowing over a surface S. Stokes’ Theorem tells us that the total circulation of the fluid along the boundary curve
C of the surface (line integral) is the same as the sum of all the tiny local rotations (curl) of the fluid at every point on the surface (surface integral).

An important aspect of applying Stokes’ Theorem is ensuring the correct orientation of the surface S and the curve C. The orientation must be chosen such that the surface and curve are compatible, meaning the surface normal and the direction of the curve are aligned properly.

In particular, the direction of the normal vector $\hat{\mathbf{n}}$ must be chosen so that it is compatible with the orientation of the curve and the surface. **If I am traversing the curve C in a specific direction (the positive direction**, i.e., in the direction that I have chosen to orientate C), **imagine walking along the curve with the surface on your left-hand side, then the normal vector $\hat{\mathbf{n}}$ to the surface must point upwards** (away from the surface) (Figure F).

Orientation and the Right-Hand Rule. A mnemonic, trick, or useful way to remember this is by using the right hand rule: align your thumb along the direction of the curve positively, the index finger tangent to the surface S and towards the interior of the surface, and the middle finger will point in the direction of the normal vector $\hat{\mathbf{n}}$ (Figures i and ii).

- Verify Stokes’ Theorem for the vector field $\vec{F} = ⟨P, Q, R⟩ = ⟨y, z, x⟩$ over a surface S bounded by the curve C, which is a unit circle in the xy-plane.

**Compute the curl of $\vec{F}$**

The curl of a vector field is given by:

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

Stokes’ Theorem states that:

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

Suppose the curve C is in the xy-plane going counter-clockwise. Let’s chose a surface S that is inside (or bounded by) the curve C (Figure iii), i.e., S is the unit disk in the xy-plane, z = 0.

Since the surface lies in the xy-plane, the normal vector is $\hat{\mathbf{n}} = \hat{\mathbf{k}}$, pointing in the positive z-direction. The surface area element in this case is dS = dxdy.

$\int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS = \int \int_{S} ⟨-1, -1, -1⟩·\hat{\mathbf{k}}dxdy = \int \int_{S} -1dxdy =$[The surface S is a unit disk, so the area of the surface is π (the area of a circle with radius 1):] -π.

Line integral Around the Curve C.

- Parametrize the curve C: $\vec{r}(t)=⟨cos(t), sin(t), 0⟩$, 0 ≤ t ≤ 2π.
- Evaluate $\vec{F}$ along the curve C: $\vec{F}(\vec{r}(t)) = ⟨sin(t), 0, cos(t)⟩$
- The differential element $d\vec{r}$ is: $d\vec{r} = ⟨-sin(t), cos(t), 0⟩dt$
- Finally, the line integral is: $\oint_C \vec{F} \cdot d\vec{r} = \int_{0}^{2π} ⟨sin(t), 0, cos(t)⟩·⟨-sin(t), cos(t), 0⟩dt = \int_{0}^{2π} -sin^2(t)$

$\int_{0}^{2π} -sin^2(t)$ =[Using the identity $sin^2(t)=\frac{1-cos(2t)}{2}$] $-\int_{0}^{2π} \frac{1-cos(-2t)}{2}dt = \frac{-1}{2}(\int_{0}^{2π}1·dt - \int_{0}^{2π} cos(2t)dt)$

The integral of cos(2t) over one period [0, 2π] is zero, $\int_{0}^{2π} cos(2t)dt = 0$

$\frac{-1}{2}\int_{0}^{2π}1·dt = \frac{-1}{2}t\bigg|_{0}^{2π} = \frac{-1}{2}2π = -π$∎

Both the surface integral and the line integral yield the same result, confirming Stokes’ Theorem.

- Verify Stoke’s Theorem for the vector field $\vec{F} = ⟨P, Q, R⟩ = ⟨2y, -z, 3⟩$ where S is the positively oriented surface of the portion of z = 4-x
^{2}-y^{2}(a paraboloid) that lies inside the cylinder x^{2}+y^{2}= 1 (Figure 1). The curve C is the boundary of this surface, which is the intersection of the paraboloid with the cylinder.

Stokes’ Theorem states that:

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

**Parametrize the curve C**. The boundary curve C is the intersection of the surface $z = 4 -x^2-y^2$ (paraboloid) and the cylinder (like a soda can) x^{2} + y^{2} = 1. At the boundary (intersection), z = 4 - 1 = 3, so C lies in the plane z = 3. The surface S is the part of the upside-down bowl inside the cylinder (soda can), which is above z = 3, where the bowl and the soda can (cylinder) intersect. Thus, the curve C lies in the plane z = 3 and is a circle of radius 1.

The parametrization of the curve C is $\vec{r}(t)= ⟨cos(t), sin(t), 3⟩$, 0 ≤ t ≤ 2π.

**Evaluate $\vec{F}$ on C**: $\vec{F}(\vec{r}(t)) = ⟨2sin(t), -3, 3⟩$

**Compute $d\vec{r}$**: $d\vec{r} = \frac{d\vec{r}}{dt}dt = ⟨-sin(t), cos(t), 0⟩dt$

**Compute the line integral**: $\oint_C \vec{F} \cdot d\vec{r} = \int_{0}^{2π} ⟨2sin(t), -3, 3⟩·⟨-sin(t), cos(t), 0⟩dtdt = \int_{0}^{2π} (-2sin^2(t)-3cos(t))dt = -2\int_{0}^{2π} sin^2(t)dt -3\int_{0}^{2π} cos(t)dt$

We already know that the integral of cos(2t) over one period [0, 2π] is zero: $\int_{0}^{2π} cos(t)dt = 0$

$-2\int_{0}^{2π} sin^2(t)dt$ =[We use the identity $sin^2(t)=\frac{1-cos(2t)}{2}$] $-2\frac{1}{2} \int_{0}^{2π} (1 -cos(2t))dt = -\int_{0}^{2π} dt = -t\bigg|_{0}^{2π}$ =

= -2π since we know that $\int_{0}^{2π} cos(2t)dt = 0$.

To verify the theorem, we now evaluate $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS$

**Compute the curl of $\vec{F}$**

$curl(\vec{F}) = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ 2y & -z & 3\end{smallmatrix}| = (\frac{∂3}{∂y}-\frac{∂(-z)}{∂z})\hat{\mathbf{i}}-(\frac{∂3}{∂x}-\frac{∂2y}{∂z})\hat{\mathbf{j}} + (\frac{∂(-z)}{∂x}-\frac{∂2y}{∂y})\hat{\mathbf{k}} = ⟨1, 0, -2⟩$

**Normal vector $\hat{\mathbf{n}}$ to the surface**. The surface S is given by z = 4 -x^{2}-y^{2}. The normal vector to the surface can be found by computing the gradient of the scalar function f(x, y, z) (it represents how f changes in each direction) = z -(4 -x^{2}-y^{2}) = z -4 + x^{2} + y^{2}

If we have a surface z=f(x,y), e.g. z = 4 -x^{2} -y^{2}. I can rewrite it as z - f(x,y) = 0. So the graph of the function f(x, y) is a level surface of the new function g(x, y, z) = z - f(x,y). The gradient of any function is always perpendicular to its level sets, so the 3D gradient of this new function g(x, y, z) = z -f(x,y) is perpendicular to the level surface defined by our original graph.

∇f = $⟨\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z}⟩ =$ ⟨2x, 2y, 1⟩. This is the normal vector, but we should normalize it to get the unit normal vector. However, we don’t need to normalize this vector since we’re only interested in the direction for the surface integral.

**Compute the Surface integral**.

$(∇x\vec{F})\hat{\mathbf{n}} = ⟨1, 0, -2⟩⟨2x, 2y, 1⟩ = 2x -2.$

$\int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS = \int \int_{S} (2x-2)dS =$ where we need to integrate over the region x^{2} + y^{2} ≤ 1.

Switching to polar coordinates, where x = rcos(θ) and y = rsin(θ), dA = rdrdθ. The bounds for r are from 0 to 1, and for θ from 0 to 2π.

$=\int_{S} (2x-2)dA = \int_{0}^{2π}\int_{0}^{1} (2rcos(θ)-2)rdrdθ = \int_{0}^{2π}\int_{0}^{1} (2r^2cos(θ)-2r)drdθ$

Evaluate the inner integral with respect to r:

$\int_{0}^{1} 2r^2cos(θ)dr = 2cos(θ)\frac{r^3}{3}\bigg|_{0}^{1} = 2cos(θ)\frac{1}{3} = \frac{2}{3}cos(θ)$

$\int_{0}^{1} -2rdr = -r^2\bigg|_{0}^{1}=-1$

So the surface integral becomes:

$\int_{0}^{2π} (\frac{2}{3}cos(θ)-1)dθ$

Separate and evaluate each term:

$\frac{2}{3}\int_{0}^{2π} cos(θ)dθ = \frac{2}{3}·0 = 0$

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

So the surface integral is: $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS = -2π$

Conclusion: Both the surface integral and the line integral yield the same result, confirming Stokes’ Theorem.

- Compute $\oint_C \vec{F} \cdot d\vec{r}$ 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 (Refer to Figure B for a visual representation and aid in understanding it).

We are going to use Stoke’s Theorem to compute the line integral. Given a vector field $\vec{F}$, a smooth surface S, and a simple, closed curve C that forms the boundary of the surface S, it asserts that:

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

Let’s calculate the curl of $\vec{F}$: $curl(\vec{F}) = ∇ x \vec{F} = (\frac{∂R}{∂y}-\frac{∂Q}{∂z})\hat{\mathbf{i}}-(\frac{∂R}{∂x}-\frac{∂P}{∂z})\hat{\mathbf{j}} + (\frac{∂Q}{∂x}-\frac{∂P}{∂y})\hat{\mathbf{k}} = (0-0)\hat{\mathbf{i}}-(1-2z)\hat{\mathbf{j}}+(0-0)\hat{\mathbf{k}} = (2z -1)\hat{\mathbf{j}}$ where P = z^{2}, Q = y^{2}, and R = x.

The surface S is a triangle with vertices at (1, 0, 0), (0, 1, 0), and (0, 0, 1). The equation of the plane that contains these points is: x + y + z = 1. This can be rearranged as: z = g(x, y) = 1 -x -y.

The general form of a plane in 3D is: Ax + By + Cz = D. Since the point (1, 0, 0) lies on the plane, it must satisfy the equation. Plugging this point into the general form, A = D. Similarly (0, 1, 0), B = D, and (0, 0, 1), C = D. Thus, we have A = B = C = D, then the equation is (dividing by A the general form) x + y + z = 1.

The unit normal vector $\hat{\mathbf{n}}$ to this plane can be found by taking the gradient of the scalar function f(x, y, z) = z -g(x, y) = z -1 +x +y. ∇f = $⟨\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z}⟩ = ⟨1, 1, 1⟩$

This normal vector is constant across the surface and does not require normalization because Stokes’ Theorem only requires the direction of the normal vector.

$\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS$ =[Since the surface S is defined by the triangle in the plane x +y +z = 1, we can project this triangle onto the xy-plane. The projection is a triangular region D (we plug z = 0 into the equation of the plane) x +y ≤ 1, bounded by 0 ≤ x ≤ 1, and given a fixed x, y ranges from 0 to 1 -x] $\int \int_{D} (2z -1)\hat{\mathbf{j}}·⟨1, 1, 1⟩dA = \int_{0}^{1} \int_{0}^{-x+1} (2z -1)dydx =[\text{Substituting z = 1 −x −y into the integral:}] \int_{0}^{1} \int_{0}^{-x+1} [2(1-x-y)-1]dydx =[\text{Simplifying the integrand:}] \int_{0}^{1} \int_{0}^{-x+1} (1-2x-2y)dydx$

**Compute the Double Integral**

Calculate the inner integral with respect to y: $\int_{0}^{-x+1} (1-2x-2y)dy = y -2xy-y^2\bigg|_{0}^{-x+1} = (-x+1)-2(-x+1)x-(-x+1)^2 = -x +1 + 2x^2-2x-(x^2+1-2x) = -x +1 + 2x^2-2x-x^2-1+2x = x^2-x $

Calculate the outer integral with respect to x: $\int_{0}^{1} (x^2-x)dx = \frac{x^3}{3}-\frac{x^2}{2}\bigg|_{0}^{1} = \frac{1}{3}-\frac{1}{2} = -\frac{1}{6}$

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

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

- $\oint_C \vec{F} \cdot{} d\vec{r}$ represents the work or line integral of the vector field $\vec{F}$ around the curve C.
- $\iint_R (\frac{∂Q}{∂x}-\frac{∂P}{∂y})dA$ represents the double integral over the region R, with the integrand being the difference between the partial derivatives of Q and P.

Stokes’ Theorem generalizes Green’s Theorem to three dimensions. 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:

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

To see how Stokes’ Theorem generalizes Green’s Theorem, let’s assume that the curve C lies in the xy-plane and is oriented counter-clockwise. We chose a surface S that is bounded by the curve C (Figure iii). In this particular case, the unit normal vector $\hat{\mathbf{n}}$ to the surface is $\hat{\mathbf{k}}$, which is the unit vector in the z-direction. Additionally, the area element dS is dxdy.

Applying Stokes’ Theorem, we have:

$\oint_C \vec{F} \cdot d\vec{r} =$[Given the vector field $\vec{F}$ = ⟨P, Q, R⟩, and that the surface S lies in the xy-plane, z = 0, dz = 0, hence the line integral satisfies] $\oint_C Pdx + Qdy$ =[By Stoke’s Theorem] $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS = \int \int_{S} (∇x\vec{F})·\hat{\mathbf{k}}dS$

Notice that $(∇x\vec{F})·\hat{\mathbf{k}}$ is the z-component of curl $\vec{F}$ and recall $curl(\vec{F})=(R_y-Q_z)\hat{\mathbf{i}} + (P_z-R_x)\hat{\mathbf{j}}+(Q_x-P_y)\hat{\mathbf{k}}$

Hence, $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{k}}dS = \int \int_{S} (Q_x-P_y)dS = \int \int_{S} (Q_x-P_y)dxdy$

Combining these results, we get:

$\oint_C \vec{F} \cdot d\vec{r} = \oint_C Pdx + Qdy = \int \int_{S} (Q_x-P_y)dxdy = \int \int_{S} (\frac{∂Q}{∂x}-\frac{∂P}{∂y})dA$⇒ **Stokes’ theorem is indeed a generalization of Green’s theorem to higher dimensions**. Green’s Theorem is restricted to two-dimensional regions in the plane (e.g., xy-plane), while Stokes’ Theorem applies to any surface in ℝ^{3} bounded by a closed curve.

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

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

Proof

**Special Case: The XY-Plane**

To start, let’s prove Stokes’ Theorem for the special case where the curve C and surface S lie in the xy-plane. In this case, Stokes’ Theorem reduces to Green’s Theorem. Here’s how:

- Suppose the surface S is in the xy-plane, so the surface lies flat along z = 0.
- The vector field $\vec{F} = ⟨P(x,y),Q(x,y),R(x,y)⟩$ has components P, Q, and R. Since the surface is in the xy-plane, R(x, y) doesn’t contribute to the line integral, and the differential change in z, dz = 0.

Thus, the line integral around C becomes: $\oint_C \vec{F} \cdot d\vec{r} = \oint_C Pdx + Qdy =[\text{By Green’s Theorem}] \int \int_{S} (Q_x-P_y)dxdy = \int \int_{S} (\frac{∂Q}{∂x}-\frac{∂P}{∂y})dA$.

Notice that this is the z-component of the curl of $\vec{F}$, which is:

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

$\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS =$[Since $\hat{\mathbf{k}}$ is the unit vector in the z-direction, and dS = dxdy in the xy-plane] $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{k}}dS = \int \int_{S}(\frac{∂Q}{∂x}-\frac{∂P}{∂y})\hat{\mathbf{k}}dA = \oint_C \vec{F} \cdot d\vec{r}$

**General Surface S**
Now, let’s move to the more general case where S is any arbitrary smooth surface S with a boundary curve C in three-dimensional space, not restricted to the xy-plane.

To prove Stoke’s Theorem for a general surface, the key🔑 idea is to divide or decompose the surface S into many infinitesimal (very small), nearly flat pieces, similar to a mosaic or mesh (Figure A). This technique allows us to approximate each small piece as a flat surface, so Stokes' Theorem can be applied locally to each piece.

**Partition the surface.**Suppose the surface S is divided into many small surface elements (tiny pieces) S_{i}, each with a small boundary curve C_{i}. These infinitesimal elements can be considered nearly flat, making it possible to apply Stokes’ Theorem locally.**Apply Stokes’ Theorem locally**. For each small surface element S_{i}with boundary C_{i}, by Stoke’s Theorem, the line integral around the small boundary C_{i}is equal to the flux of the curl through the corresponding surface element S_{i}: $\oint_{C_i} \vec{F} \cdot d\vec{r} = \int \int_{S_i} (∇x\vec{F})·\hat{\mathbf{n_i}}dS$ where $\hat{\mathbf{n_i}}$ is the unit normal vector to the surface element S_{i}.

Next, sum the contributions from all these small elements or pieces. The total line integral around the boundary of the entire surface S is the sum of the line integrals around each small boundary C_{i}: $\oint_{C} \vec{F} \cdot d\vec{r} = \sum_{i} \oint_{C_i} \vec{F} \cdot d\vec{r}$

Similarly, the total flux of the curl of $\vec{F}$ through the surface S is the sum of the fluxes through each small surface element or piece S_{i}: $\int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS = \sum_{i} \int \int_{S_i} (∇x\vec{F})·\hat{\mathbf{n_i}}dS$.

As **we sum the line integrals over all the small surface elements, we get some extra junk. The contributions along the interior borders between adjacent pieces cancel each other out**. This is because, for any two adjacent pieces, the boundary curve between them is traversed in opposite directions for each piece. Therefore, the line integrals over the interior boundaries cancel each other, leaving only the line integral along the outer boundary C of the entire surface S. Therefore, the sum of these contribution yield the work done solely along the outer boundary C.

Therefore, Stoke’s Theorem holds true as the line integral around the boundary C is equal to the flux of the curl through the entire surface S, regardless of the shape or orientation of S.

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].

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- 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.