Vector Calculus | Mathematics - Wyatt's Notes

3.1 Vector Fields

A vector field on $\mathbb{R}^n$ is a function $\mathbf{F} : D \subseteq \mathbb{R}^n \to \mathbb{R}^n$ .

A vector field $\mathbf{F} = (P, Q, R)$ on $\mathbb{R}^3$ is conservative if there exists a scalar Potential $\phi$ such that $\mathbf{F} = \nabla \phi$ .

Theorem 3.1. $\mathbf{F}$ is conservative (on a connected domain) if and only if $\nabla \times \mathbf{F} = \mathbf{0}$ .

Proof. ( $\Rightarrow$ ) If $\mathbf{F} = \nabla \phi$ with $\phi \in C^2$ Then by Clairaut”s theorem $f_{xy} = f_{yx}$ Etc., which directly gives $\nabla \times (\nabla \phi) = \mathbf{0}$ .

( $\Leftarrow$ ) If $\nabla \times \mathbf{F} = \mathbf{0}$ on a connected domain $D$ Then for any Closed curve $C$ in $D$ Stokes’ theorem gives $\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = 0$ . This means line integrals are path-independent, so we can define $\phi(\mathbf{x}) = \int_{\mathbf{x}_0}^{\mathbf{x}} \mathbf{F} \cdot d\mathbf{r}$ (independent of path), And one verifies that $\nabla \phi = \mathbf{F}$ . $\blacksquare$

3.2 Line Integrals

Definition. The line integral of a vector field $\mathbf{F}$ along a curve $C$ parameterised by $\mathbf{r}(t)$ for $a \leq t \leq b$ is

$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt$

Theorem 3.2 (Fundamental Theorem for Line Integrals). If $\mathbf{F} = \nabla \phi$ and $C$ is a Piecewise smooth curve from $A$ to $B$ Then

$\int_C \mathbf{F} \cdot d\mathbf{r} = \phi(B) - \phi(A)$

Proof. Parameterise $C$ by $\mathbf{r}(t)$ for $t \in [a,b]$ with $\mathbf{r}(a) = A$ $\mathbf{r}(b) = B$ .

$\int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \nabla \phi(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt = \int_a^b \frac{d}{dt}\left[\phi(\mathbf{r}(t))\right]\, dt = \phi(\mathbf{r}(b)) - \phi(\mathbf{r}(a)) = \phi(B) - \phi(A)$

By the chain rule. $\blacksquare$

Corollary 3.3. The line integral of a conservative field around any closed curve is zero.

Problem. Evaluate $\int_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = (y,\, x + e^y,\, z + 1)$ and $C$ is the Curve $\mathbf{r}(t) = (t,\, t^2,\, t^3)$ for $0 \leq t \leq 1$ .

Solution

First check if $\mathbf{F}$ is conservative. Compute the curl:

$(\nabla \times \mathbf{F})_x = \frac{\partial (z+1)}{\partial y} - \frac{\partial (x + e^y)}{\partial z} = 0 - 0 = 0$

$(\nabla \times \mathbf{F})_y = \frac{\partial y}{\partial z} - \frac{\partial (z+1)}{\partial x} = 0 - 0 = 0$

$(\nabla \times \mathbf{F})_z = \frac{\partial (x + e^y)}{\partial x} - \frac{\partial y}{\partial y} = 1 - 1 = 0$

Since $\nabla \times \mathbf{F} = \mathbf{0}$ , $\mathbf{F}$ is conservative. Find $\phi$ :

$\frac{\partial \phi}{\partial x} = y \implies \phi = xy + g(y,z)$

$\frac{\partial \phi}{\partial y} = x + g_y = x + e^y \implies g_y = e^y \implies g = e^y + h(z)$

$\frac{\partial \phi}{\partial z} = h'(z) = z + 1 \implies h(z) = \frac{z^2}{2} + z + C$

$\phi(x,y,z) = xy + e^y + \frac{z^2}{2} + z$

Now apply the fundamental theorem:

$\int_C \mathbf{F} \cdot d\mathbf{r} = \phi(1, 1, 1) - \phi(0, 0, 0) = \left(1 + e + \frac{1}{2} + 1\right) - (1 + 1) = e + \frac{1}{2}$

$\blacksquare$

3.3 Green’s Theorem

Theorem 3.4 (Green’s Theorem). Let $C$ be a positively oriented, piecewise smooth, simple closed Curve bounding a region $D$ . If $P$ and $Q$ have continuous partial derivatives on an open set Containing $D$ Then

$\oint_C P\, dx + Q\, dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA$

Proof (for a Type I region). Assume $D$ is a Type I region: $D = \\{(x,y) : a \leq x \leq b,\, g_1(x) \leq y \leq g_2(x)\\}$ . The boundary $C$ consists of Four pieces: bottom $C_1$ Right $C_2$ Top $C_3$ And left $C_4$ .

We first prove $\oint_C P\, dx = -\iint_D \frac{\partial P}{\partial y}\, dA$ .

On $C_1$ : $y = g_1(x)$ , $x$ goes from $a$ to $b$ So $\int_{C_1} P\, dx = \int_a^b P(x, g_1(x))\, dx$ .

On $C_3$ : $y = g_2(x)$ , $x$ goes from $b$ to $a$ So $\int_{C_3} P\, dx = \int_b^a P(x, g_2(x))\, dx = -\int_a^b P(x, g_2(x))\, dx$ .

On $C_2$ and $C_4$ : $x$ is constant, so $dx = 0$ Hence $\int_{C_2} P\, dx = \int_{C_4} P\, dx = 0$ .

Therefore:

$\oint_C P\, dx = \int_a^b P(x, g_1(x))\, dx - \int_a^b P(x, g_2(x))\, dx$

Meanwhile:

$-\iint_D \frac{\partial P}{\partial y}\, dA = -\int_a^b \int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y}\, dy\, dx = -\int_a^b \left[P(x, g_2(x)) - P(x, g_1(x))\right]\, dx$

$= \int_a^b P(x, g_1(x))\, dx - \int_a^b P(x, g_2(x))\, dx = \oint_C P\, dx$

An identical argument (using Type II regions) proves $\oint_C Q\, dy = \iint_D \frac{\partial Q}{\partial x}\, dA$ . Adding the two equalities gives the result. For general regions, decompose $D$ into finitely many Type I and Type II regions and note that the line integrals along shared boundaries cancel. $\blacksquare$

Worked Example. Evaluate $\oint_C (x^2 - y)\, dx + (y^2 + x)\, dy$ where $C$ is the unit circle Traversed counterclockwise.

Solution. By Green’s theorem with $P = x^2 - y$ and $Q = y^2 + x$ :

$\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = -1$

$\oint_C P\, dx + Q\, dy = \iint_D (1 - (-1))\, dA = 2 \iint_D dA = 2 \cdot \pi \cdot 1^2 = 2\pi$

$\blacksquare$

3.4 Curl and Divergence

Definition. Let $\mathbf{F} = (P, Q, R)$ be a $C^1$ vector field on $\mathbb{R}^3$ .

The curl of $\mathbf{F}$ is

$\nabla \times \mathbf{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z},\, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x},\, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$

The divergence of $\mathbf{F}$ is

$\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$

Physical interpretation. If $\mathbf{F}$ represents the velocity field of a fluid:

Curl $\nabla \times \mathbf{F}$ measures the local rotational tendency (vorticity) of the fluid. At a point $\mathbf{p}$ The component $(\nabla \times \mathbf{F}) \cdot \mathbf{n}$ gives twice the angular velocity of a small paddle wheel placed at $\mathbf{p}$ with axis along $\mathbf{n}$ .
Divergence $\nabla \cdot \mathbf{F}$ measures the net rate of outward flux per unit volume at a point. If $\nabla \cdot \mathbf{F} \gt 0$ at $\mathbf{p}$ There is a net source at $\mathbf{p}$ ; if $\nabla \cdot \mathbf{F} \lt 0$ There is a net sink.

Proposition 3.5. For any $C^2$ vector field $\mathbf{F}$ :

$\nabla \cdot (\nabla \times \mathbf{F}) = 0 \quad \mathrm{(div\ of\ curl\ is\ zero)}$

$\nabla \times (\nabla \phi) = \mathbf{0} \quad \mathrm{(curl\ of\ gradient\ is\ zero)}$

Proof. Both follow from Clairaut’s theorem on equality of mixed partials. For the first:

$\nabla \cdot (\nabla \times \mathbf{F}) = \frac{\partial}{\partial x}\left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) + \frac{\partial}{\partial y}\left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) + \frac{\partial}{\partial z}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$

Each pair cancels by Clairaut: $\frac{\partial^2 R}{\partial x\,\partial y} = \frac{\partial^2 R}{\partial y\,\partial x}$ Etc. $\blacksquare$

3.5 Stokes’ Theorem

Theorem 3.6 (Stokes’ Theorem). Let $S$ be an oriented surface with piecewise smooth boundary curve $C$ (positively oriented). If $\mathbf{F}$ has continuous partial derivatives on an open set containing $S$ Then

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$

Where $d\mathbf{S} = \mathbf{n}\, dS$ is the vector surface element with unit normal $\mathbf{n}$ .

Proof (sketch). Parametrise $S$ by $\mathbf{r}(u,v)$ over a region $D$ in the $uv$ -plane. The boundary $C$ of $S$ corresponds to the boundary $\partial D$ of $D$ . The left-hand side becomes:

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \oint_{\partial D} \mathbf{F}(\mathbf{r}(u,v)) \cdot \left(\frac{\partial \mathbf{r}}{\partial u}\, du + \frac{\partial \mathbf{r}}{\partial v}\, dv\right)$

Define $\tilde{P}(u,v) = \mathbf{F}(\mathbf{r}(u,v)) \cdot \mathbf{r}_u$ and $\tilde{Q}(u,v) = \mathbf{F}(\mathbf{r}(u,v)) \cdot \mathbf{r}_v$ . Applying Green’s theorem in the $uv$ -plane:

$\oint_{\partial D} \tilde{P}\, du + \tilde{Q}\, dv = \iint_D \left(\frac{\partial \tilde{Q}}{\partial u} - \frac{\partial \tilde{P}}{\partial v}\right) du\, dv$

Expanding the partial derivatives and using the identity $\mathbf{r}_u \times \mathbf{r}_v = \mathbf{n}\, \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert$ One Verifies that the integrand equals $(\nabla \times \mathbf{F}) \cdot \mathbf{n}\, \lVert \mathbf{r}_u \times \mathbf{r}_v \rVert$ Which gives $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$ . $\blacksquare$

Remark. Green’s theorem is the special case of Stokes’ theorem where $S$ is a planar region in $\mathbb{R}^2$ .

Problem. Use Stokes’ theorem to evaluate $\oint_C \mathbf{F} \cdot d\mathbf{r}$ where $\mathbf{F} = (y^2,\, xz,\, x^2)$ and $C$ is the triangle with vertices $(1,0,0)$ , $(0,1,0)$ $(0,0,1)$ traversed counterclockwise when viewed from above.

Solution

The triangle lies in the plane $x + y + z = 1$ . Compute $\nabla \times \mathbf{F}$ :

$(\nabla \times \mathbf{F})_x = \frac{\partial (x^2)}{\partial y} - \frac{\partial (xz)}{\partial z} = 0 - x = -x$

$(\nabla \times \mathbf{F})_y = \frac{\partial (y^2)}{\partial z} - \frac{\partial (x^2)}{\partial x} = 0 - 2x = -2x$

$(\nabla \times \mathbf{F})_z = \frac{\partial (xz)}{\partial x} - \frac{\partial (y^2)}{\partial y} = z - 2y$

So $\nabla \times \mathbf{F} = (-x,\, -2x,\, z - 2y)$ .

Parametrise the triangle in the $xy$ -plane: $0 \leq x \leq 1$ , $0 \leq y \leq 1 - x$ . On the plane $z = 1 - x - y$ The surface element $dS = \sqrt{3}\, dx\, dy$ .

$\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \frac{1}{\sqrt{3}} \iint_S (-x - 2x + z - 2y)\, dS$

On the plane: $-3x - 2y + z = -3x - 2y + 1 - x - y = -4x - 3y + 1$ .

$= \frac{1}{\sqrt{3}} \int_0^1 \int_0^{1-x} (-4x - 3y + 1)\, \sqrt{3}\, dy\, dx = \int_0^1 \int_0^{1-x} (-4x - 3y + 1)\, dy\, dx$

$= \int_0^1 \left[(-4x + 1)y - \frac{3y^2}{2}\right]_0^{1-x}\, dx = \int_0^1 (-4x + 1)(1 - x) - \frac{3(1-x)^2}{2}\, dx$

$= \int_0^1 \left[4x^2 - 5x + 1 - \frac{3}{2} + 3x - \frac{3x^2}{2}\right]\, dx = \int_0^1 \left[\frac{5x^2}{2} - 2x - \frac{1}{2}\right]\, dx$

$= \left[\frac{5x^3}{6} - x^2 - \frac{x}{2}\right]_0^1 = \frac{5}{6} - 1 - \frac{1}{2} = -\frac{2}{3}$

$\blacksquare$

3.6 Divergence Theorem

Theorem 3.7 (Divergence Theorem / Gauss’s Theorem). Let $E$ be a solid region bounded by a closed Surface $S$ with outward normal $\mathbf{n}$ . If $\mathbf{F}$ has continuous partial derivatives on an Open set containing $E$ Then

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E \nabla \cdot \mathbf{F}\, dV$

Where $\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$ Is the divergence of $\mathbf{F}$ .

Proof (sketch for a Type I region). Assume $E$ is a Type I region: $E = \\{(x,y,z) : (x,y) \in D,\, g_1(x,y) \leq z \leq g_2(x,y)\\}$ . The boundary consists of Bottom surface $S_1$ (normal pointing downward), top surface $S_2$ (normal pointing upward), And the lateral surface $S_3$ (where the normal is horizontal).

We prove the result for the $R$ -component, i.e., $\iint_S R\, \mathbf{k} \cdot d\mathbf{S} = \iiint_E \frac{\partial R}{\partial z}\, dV$ .

The right-hand side:

$\iiint_E \frac{\partial R}{\partial z}\, dV = \iint_D \int_{g_1(x,y)}^{g_2(x,y)} \frac{\partial R}{\partial z}\, dz\, dA = \iint_D \left[R(x,y,g_2) - R(x,y,g_1)\right]\, dA$

On $S_2$ (top): $d\mathbf{S} = (-g_{2x}, -g_{2y}, 1)\, dA$ (upward), so $R\, \mathbf{k} \cdot d\mathbf{S} = R(x,y,g_2)\, dA$ .

On $S_1$ (bottom): $d\mathbf{S} = (g_{1x}, g_{1y}, -1)\, dA$ (downward), so $R\, \mathbf{k} \cdot d\mathbf{S} = -R(x,y,g_1)\, dA$ .

On $S_3$ : $\mathbf{k} \cdot \mathbf{n} = 0$ (the normal is horizontal), so $R\, \mathbf{k} \cdot d\mathbf{S} = 0$ .

Therefore $\iint_S R\, \mathbf{k} \cdot d\mathbf{S} = \iint_D [R(x,y,g_2) - R(x,y,g_1)]\, dA$ Matching the Volume integral. The $P$ and $Q$ components follow by an identical argument for Type II and Type III Regions. For general regions, decompose into finitely many regions of each type. $\blacksquare$

Worked Example. Compute the flux of $\mathbf{F} = (x^3, y^3, z^3)$ through the unit sphere $S$ .

Solution. By the divergence theorem:

$\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2) = 3\rho^2$

Using spherical coordinates:

$\iiint_E 3\rho^2 \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta = 3 \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^4 \sin\phi\, d\rho\, d\phi\, d\theta$

$= 3 \cdot 2\pi \cdot 2 \cdot \frac{1}{5} = \frac{12\pi}{5}$

$\blacksquare$

Problem. Compute the flux of $\mathbf{F} = (x^2,\, y^2,\, z^2)$ outward through the surface of the Cylinder $x^2 + y^2 \leq 1$ , $0 \leq z \leq 2$ .

Solution

By the divergence theorem:

$\nabla \cdot \mathbf{F} = 2x + 2y + 2z$

Use cylindrical coordinates. The region $E'$ is $0 \leq r \leq 1$ , $0 \leq \theta \leq 2\pi$ $0 \leq z \leq 2$ .

$\iiint_E (2x + 2y + 2z)\, dV = \iiint_E 2z\, dV$

Since $\iint_E x\, dV = \iint_E y\, dV = 0$ by symmetry (odd functions over a symmetric domain).

$= 2 \int_0^{2\pi} \int_0^1 \int_0^2 z \cdot r\, dz\, dr\, d\theta = 2 \int_0^{2\pi} \int_0^1 r\left[\frac{z^2}{2}\right]_0^2\, dr\, d\theta$

$= 2 \int_0^{2\pi} \int_0^1 2r\, dr\, d\theta = 2 \int_0^{2\pi} 1\, d\theta = 2 \cdot 2\pi = 4\pi$

$\blacksquare$

3.7 Conservative Fields and Potential Functions

Definition. A vector field $\mathbf{F}$ on a domain $D \subseteq \mathbb{R}^n$ is conservative if There exists a scalar function $\phi : D \to \mathbb{R}$ (called a potential function) such that $\mathbf{F} = \nabla \phi$ .

Proposition 3.8 (Equivalent conditions for conservative fields). Let $\mathbf{F} = (P, Q)$ be a $C^1$ vector field on a connected domain $D \subseteq \mathbb{R}^2$ . The following are equivalent:

$\mathbf{F}$ is conservative: $\mathbf{F} = \nabla \phi$ for some $\phi$ .
$\oint_C \mathbf{F} \cdot d\mathbf{r} = 0$ for every closed curve $C$ in $D$ .
$\int_C \mathbf{F} \cdot d\mathbf{r}$ is path-independent in $D$ .
$\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ everywhere in $D$ .

Procedure for finding a potential function. Given $\mathbf{F} = (P, Q, R)$ with $\nabla \times \mathbf{F} = \mathbf{0}$ :

Integrate $P$ with respect to $x$ : $\phi = \int P\, dx + g(y, z)$ .
Differentiate with respect to $y$ and set equal to $Q$ to determine $g_y$ .
Integrate $g_y$ with respect to $y$ : $g = \int g_y\, dy + h(z)$ .
Differentiate with respect to $z$ and set equal to $R$ to determine $h'(z)$ .
Integrate to find $h(z)$ and assemble $\phi$ .

Problem. Determine whether $\mathbf{F} = (2xy + z^2,\, x^2 + 2yz,\, 2xz + y^2)$ is conservative, And if so, find a potential function.

Solution

Check the curl:

$(\nabla \times \mathbf{F})_x = \frac{\partial}{\partial y}(2xz + y^2) - \frac{\partial}{\partial z}(x^2 + 2yz) = 2y - 2y = 0$

$(\nabla \times \mathbf{F})_y = \frac{\partial}{\partial z}(2xy + z^2) - \frac{\partial}{\partial x}(2xz + y^2) = 2z - 2z = 0$

$(\nabla \times \mathbf{F})_z = \frac{\partial}{\partial x}(x^2 + 2yz) - \frac{\partial}{\partial y}(2xy + z^2) = 2x - 2x = 0$

Since $\nabla \times \mathbf{F} = \mathbf{0}$ , $\mathbf{F}$ is conservative. Find $\phi$ :

$\frac{\partial \phi}{\partial x} = 2xy + z^2 \implies \phi = x^2 y + xz^2 + g(y,z)$

$\frac{\partial \phi}{\partial y} = x^2 + g_y(y,z) = x^2 + 2yz \implies g_y(y,z) = 2yz \implies g(y,z) = y^2 z + h(z)$

$\frac{\partial \phi}{\partial z} = 2xz + y^2 + h'(z)$

This must equal $2xz + y^2$ So $h'(z) = 0$ Giving $h(z) = C$ .

Therefore $\phi(x,y,z) = x^2 y + xz^2 + y^2 z + C$ . $\blacksquare$

3.8 Common Pitfalls

Singularities. When applying Green’s, Stokes’, or the Divergence theorem, verify that the field has continuous partial derivatives on the region (including interior). If there are singularities inside the region, the theorems do not apply directly; the singularity must be handled separately.
** connected domains.** The condition $\nabla \times \mathbf{F} = \mathbf{0}$ guarantees that $\mathbf{F}$ is conservative only on a connected domain. For example, $\mathbf{F} = \frac{(-y, x)}{x^2 + y^2}$ has zero curl on $\mathbb{R}^2 \setminus \\{(0,0)\\}$ but is not conservative there (the domain is not connected).
Orientation. Green’s and Stokes’ theorems require positive orientation (counterclockwise for planar curves, right-hand rule for surfaces). The divergence theorem requires the outward normal. Reversing orientation changes the sign of the result.

flowchart TD
    A[3_Vector Calculus] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Cross-References

Partial Derivatives: The gradient, curl, and divergence are differential operators defined using partial derivatives.
Multiple Integrals: Green’s, Stokes’, and the divergence theorems relate line and surface integrals to double and triple integrals.
Curves and Surfaces: Line integrals and surface integrals require parametrisation of curves and surfaces using the tools from this chapter.
Optimization: Conservative vector fields and potential functions connect to the theory of exact differentials and constrained optimisation.
Quantum Mechanics
Graph Theory
Classical Mechanics
Electromagnetism

3.9 Intuition: What Do Curl, Divergence, and the Integral Theorems Mean?

Divergence measures how much a vector field “flows out” of a point. If the divergence is positive, the point is a source; if negative, it is a sink. The divergence theorem says that the total flow out of a volume equals the total flux through its boundary, which is a conservation law: what flows in must flow out.

Curl measures the local rotation or “swirl” of a vector field. If the curl is non-zero, the field has a tendency to circulate around that point. Stokes’ theorem says that the circulation of a field around a closed curve equals the total curl through any surface bounded by that curve, which generalises the fundamental theorem of calculus to surfaces.

The general pattern connecting all these theorems is: the integral of a derivative over a region equals the integral of the original function over the boundary. This is the multivariable fundamental theorem of calculus. Green’s theorem (planar circulation), Stokes’ theorem (surface circulation), and the divergence theorem (volume flux) are all instances of the same abstract principle applied to different dimensions. A conservative field, one that is the gradient of a potential, has zero curl and zero circulation around closed loops, which is why line integrals become path-independent.

3.10 Relationships Among the Fundamental Theorems

The three major integral theorems of vector calculus are deeply connected:

Remark. Green’s theorem is the planar special case of Stokes’ theorem. Stokes’ theorem relates the Circulation around a curve to the curl through the surface it bounds. The divergence theorem relates The flux through a closed surface to the divergence inside the volume it encloses. Together, these Form the higher-dimensional analogues of the Fundamental Theorem of Calculus:

$\int_a^b f'(x)\, dx = f(b) - f(a) \quad \mathrm{(FTC)}$

$\int_C \nabla \phi \cdot d\mathbf{r} = \phi(B) - \phi(A) \quad \mathrm{(FTLI)}$

$\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \quad \mathrm{(Stokes)}$

$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_E (\nabla \cdot \mathbf{F})\, dV \quad \mathrm{(Divergence)}$

In each case, the integral of a “derivative” over a region equals the integral of the original function Over the boundary of that region. This is the generalised Stokes’ theorem:

$\int_{\partial \Omega} \omega = \int_{\Omega} d\omega$

Where $\Omega$ is a $k$ -dimensional manifold with boundary $\partial \Omega$ , $\omega$ is a $(k-1)$ -form, And $d\omega$ is its exterior derivative.

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