Vector Calculus

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 $\nabla \phi = \mathbf{F}$. $\blacksquare$

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:

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

$\blacksquare$

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 $Q = y^2 + x$:

$$\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = -1$$ $$\oint_C P\, dx + Q\, dy = \iint_D (1 - (-1))\, dA = 2 \iint_D dA = 2 \cdot \pi \cdot 1^2 = 2\pi$$

$\blacksquare$

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. $\blacksquare$

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

$$= \frac{1}{\sqrt{3}} \int_0^1 \int_0^{1-x} (-4x - 3y + 1)\, \sqrt{3}\, dy\, dx = \int_0^1 \int_0^{1-x} (-4x - 3y + 1)\, dy\, dx$$ $$= \int_0^1 \left[(-4x + 1)y - \frac{3y^2}{2}\right]_0^{1-x}\, dx = \int_0^1 (-4x + 1)(1 - x) - \frac{3(1-x)^2}{2}\, dx$$ $$= \int_0^1 \left[4x^2 - 5x + 1 - \frac{3}{2} + 3x - \frac{3x^2}{2}\right]\, dx = \int_0^1 \left[\frac{5x^2}{2} - 2x - \frac{1}{2}\right]\, dx$$ $$= \left[\frac{5x^3}{6} - x^2 - \frac{x}{2}\right]_0^1 = \frac{5}{6} - 1 - \frac{1}{2} = -\frac{2}{3}$$

$\blacksquare$

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).

$$= 2 \int_0^{2\pi} \int_0^1 \int_0^2 z \cdot r\, dz\, dr\, d\theta = 2 \int_0^{2\pi} \int_0^1 r\left[\frac{z^2}{2}\right]_0^2\, dr\, d\theta$$ $$= 2 \int_0^{2\pi} \int_0^1 2r\, dr\, d\theta = 2 \int_0^{2\pi} 1\, d\theta = 2 \cdot 2\pi = 4\pi$$

$\blacksquare$

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.

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

• Curves and Surfaces
• Multiple Integrals
• Optimization
• Partial Derivatives