Vector Calculus

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 $0 \leq t \leq 1$.

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.

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 $x^2 + y^2 \leq 1$, $0 \leq z \leq 2$.

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:

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

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

1. Integrate $P$ with respect to $x$: $\phi = \int P\, dx + g(y, z)$.
2. Differentiate with respect to $y$ and set equal to $Q$ to determine $g_y$.
3. Integrate $g_y$ with respect to $y$: $g = \int g_y\, dy + h(z)$.
4. Differentiate with respect to $z$ and set equal to $R$ to determine $h'(z)$.
5. 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.

3.8 Common Pitfalls

:::caution 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.

3.9 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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