Multivariable Calculus

Vector

Arclength of a curve: $L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^2} dx$

$\vec{c}(t) = \vec{p_1} + t(\vec{p_2} - \vec{p_1})$

Dot Product

$\vec{u} \cdot \vec{v}= |\vec{u}||\vec{v}|\cos\theta$
Parallel: $\vec{u} \cdot \vec{v}= |\vec{u}||\vec{v}|$.
Perpendicular: $\vec{u} \cdot \vec{v}= 0$.

Cross Product

$\vec{u} = \langle x_1, y_1, z_1 \rangle \\ \vec{v} = \langle x_2, y_2, z_2 \rangle \\ \vec{u} \times \vec{v} = \langle y_1z_2-z_1y_2, z_1x_2-x_1z_2, x_1y_2-y_1x_2 \rangle \\ |\vec{u} \times \vec{v}|= |\vec{u}||\vec{v}|\sin\theta$ The magnitude of the cross product is the area of a parallelogram formed by these two vectors.

Plane Equation

$ax + by + cz = d$
$\vec{p} = \vec{w} + s\vec{u} + t\vec{v}$

Parallel with: $ax + by + cz = 0$
$\vec{p} = s\vec{u} + t\vec{v}$

Move by $\langle 2, 3, 4 \rangle$
$\vec{p} = \langle 2, 3, 4 \rangle + \vec{w} + s\vec{u} + t\vec{v}$
$a(x - 2) + b(y - 3) + c(z - 4) = d$

Perpendicular: $\vec{n} = \langle a, b, c \rangle$.
$\vec{n} = \vec{u} \times \vec{v}$.
$\vec{n} = \nabla f = f_r \times f_\theta$.

Two planes are parallel when their normal vectors are parallel.

Derivatives

$\vec{v}(t)' = \langle \vec{x}(t)', \vec{y}(t)', \vec{z}(t)' \rangle$.
$\int\vec{v}(t) = \langle \int\vec{x}(t), \int\vec{y}(t), \int\vec{z}(t) \rangle$.
$(\vec{u}\cdot \vec{v})' = \vec{u}'\cdot \vec{v} + \vec{u}\cdot \vec{v}'$.

Curvature: $\kappa(t) = \frac{|\vec{v}(t)' \times \vec{v}(t)''|}{|\vec{v}(t)'|^3}$

Arc-length parameterization $s(t) = \int_{t_0}^t |\vec{r}(u)'|du \ \\ t = t(s)$.
Replace $t$ with $t(s)$, $\vec{r}(t)$ -> $\vec{r}(t(s)) = \vec{r}(s)$; $\vec{r}(s)' = 1$.
Curvature is the same as acceleration.

Partial Derivatives

Clairaut's Theorem: The mixed partials with different orders are equal: $f_{xy} = f_{yx}$

$(fg)'''' = f''''g + 4f'''g' + 6f''g'' + 4f'g''' + fg''''$

$\frac{df}{dt} = \frac{df}{dx}\cdot \frac{dx}{dt} + \frac{df}{dy}\cdot \frac{dy}{dt} + \frac{df}{dz}\cdot \frac{dz}{dt}$

$F(x, y, z) = 0 \\ \frac{dF}{dy}\cdot \frac{dy}{dz} + \frac{dF}{dz} = 0 \\ \frac{dy}{dz} = -\frac{F_z}{F_y} \\$

$f(x, y), x(u, v), y(u, v)$ $\begin{bmatrix} f_u & f_v \end{bmatrix} = \begin{bmatrix} f_x & f_y \end{bmatrix} \begin{bmatrix} x_u & x_v \\ y_u & y_v \\ \end{bmatrix}$

Optimization

Gradient vector: $\nabla f(x, y) = \langle f_x, f_y \rangle$.
Movement vector: $\vec{v} = \langle dx, dy \rangle$.

Tangent Plane

$dz = f_xdx + f_ydy = \nabla f\cdot \vec{v}$.
$\Delta z \approx f_x\Delta x + f_y\Delta y$.
$z = z_0 + f_x(x - x_0) + f_y(y - y_0)$.

Directional Derivative

$Slope = \frac{rise}{run} = \frac{\nabla f\cdot \vec{v}}{|\vec{v}|} = \nabla f \cdot \vec{u} = |\nabla f||\vec{u}| \cos\theta$

Movement in the direction of the gradient has a maximum slope equal to the magnitude of the gradient. Movement perpendicular to the gradient has a slope of 0.

Critical Points

$f(x, y)$, $\nabla f = 0$

Second Derivative Test $D(x_0, y_0) = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \\ \end{vmatrix} = f_{xx}f_{yy} - {f_{xy}}^2$ $D > 0, f_{xx} > 0$, relative minimum; $D > 0, f_{xx} < 0$, relative maximum; $D < 0$, saddle point; $D = 0$, don't know.

Lagrange Multipliers

Constraint function: $g(x, y) = 0$ $\nabla f = \lambda \nabla g$ $f_x = \lambda g_x \\ f_y = \lambda g_y \\ g(x, y) = 0$

Double Integrals

Over rectangular region $R = [a, b] \times [c, d]$, a Cartesian product of intervals.

Definition: $\int\int_R f(x, y) dA, \int\int_R f(x, y) dx\ dy$

Cavalieri's Principle (the slice method): $\text{Volume} = \int_a^b A(x) dx$, where $A(x)$ is the area of the cross section at $x$.
Therefore, we have,

$$\begin{aligned} \int\int_R f(x, y) dA &= \int_a^b \left( \int_c^d f(x, y) dy \right) dx \\ &= \int_c^d \left( \int_a^b f(x, y) dx \right) dy \\ \end{aligned}$$

Fubini's Theorem: If $f$ is continuous on a rectangular region $R$, then $\int\int_R f(x, y) dA = \int_a^b \int_c^d f(x, y) dy\ dx = \int_c^d \int_a^b f(x, y) dx\ dy$

If $f(x, y) = g(x)h(y)$, then $\int\int_R f(x, y) dA = \left( \int_a^b g(x) dx \right) \left( \int_c^d h(y) dy \right)$

Properties:

1. Any continuous function defined on a closed rectangle $R$ is integrable.
2. If the set of discontinuities lies on a finite union of graphs of continuous functions, then $f$ is integrable on $R$.

Mean Value Theorem for Double Integrals: If $f$ is continuous on a closed, bounded region $R$ with area $A$, then there exists a point $(x_0, y_0)$ in $R$ such that $\int\int_R f(x, y) dA = f(x_0, y_0) \cdot A(R)$

Theorems

$\int_a^b d - c\ dx = \int_a^b \int_c^d 1 dydx$, then $\int_a^b \int_c^d f(x, y)\ dydx$.
Estimating: If $g \le f \le h$, then $\int\int_R g dA \le \int\int_R f dA \le \int\int_R h dA$

Change of Coordinates

Double Integral

$(x, y) \to (u, v)$ where $x = f_x(u, v)$ and $y = f_y(u, v)$

Jacobian Determinant:

$$J = \frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \end{vmatrix} = \frac{\partial x}{\partial u} \cdot \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \cdot \frac{\partial y}{\partial u}$$

The change of variables formula is: $\int\int_R f(x, y)\ dx dy = \int\int_S f(f_x(u, v), f_y(u, v)) \left| \frac{\partial(x, y)}{\partial(u, v)} \right| du dv$

Triple Integral

$(x, y, z) \to (u, v, w)$ where $x = f_x(u, v, w)$, $y = f_y(u, v, w)$, and $z = f_z(u, v, w)$

Jacobian Determinant:

$$J = \frac{\partial(x, y, z)}{\partial(u, v, w)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \\ \end{vmatrix}$$

The change of variables formula is: $\int\int\int_R f(x, y, z)\ dx dy dz = \int\int\int_S f(f_x, f_y, f_z) \left| \frac{\partial(x, y, z)}{\partial(u, v, w)} \right| du dv dw$

The Gaussian integral: $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$

Typical Coordinate Systems

Polar Coordinate $(r, \theta)$

$(x, y) = (r\cos\theta, r\sin\theta)$, $\frac{ \partial(x, y) }{ \partial(r, \theta) } = r$

$\int_0^{2\pi}\int_0^R f(r, \theta) r\ dr d\theta$

Ellipse

$(\frac{x}{a})^2 + (\frac{y}{b})^2 = R^2$

$(x, y) = (ar\cos\theta, br\sin\theta)$, $\frac{ \partial(x, y) }{ \partial(r, \theta) } = abr$

$\int_0^{2\pi}\int_0^R f(r, \theta) abr\ dr d\theta$

Cylindrical Coordinate $(r, \theta, z)$

$(x, y, z) = (r\cos\theta, r\sin\theta, z)$, $\frac{ \partial(x, y, z) }{ \partial(r, \theta, z) } = r$

$\int_{z_0}^{z_1}\int_0^{2\pi}\int_0^R f(r, \theta, z) r \ dr d\theta dz$

Spherical Coordinate $(r, \theta, \phi)$

$(x, y, z) = (r\sin\phi\cos\theta, r\sin\phi\sin\theta, r\cos\phi)$, $\frac{ \partial(x, y, z) }{ \partial(r, \theta, \phi) } = r^2\sin\phi$

$\int_0^{\pi}\int_0^{2\pi}\int_0^R f(r, \theta, \phi) r^2\sin\phi\ dr d\theta d\phi$

Line Integral (Path Integral)

Definition: Integrals over curves.

$\vec{r}(t) = \langle x(t), y(t) \rangle, a \le t \le b \\ C$ is the curve parameterized by $\vec{r}(t)$

$ds = |\vec{r}(t)'| dt \\ d\vec{r} = \vec{r}(t)' dt$ $\int_C f(x, y) ds = \int_a^b f(\vec{r}(t)) |\vec{r}(t)'| dt$ $\int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}(t)' dt$

$\vec{r}(t) = \langle x(t), y(t) \rangle, f(x, y)$.
$\int_a^b f(\vec{r}(t))|\vec{r}(t)'|dt$

$\vec{F}(x, y) = \langle P(x, y), Q(x, y) \rangle$.
$\int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}(t)'dt$

$T$ is the unit tangent vector to $C$ at the point $\vec{r}(t)$. $\vec{T} = \frac{\vec{r}(t)'}{|\vec{r}(t)'|}$

Reparameterization

If $C$ is parameterized by $\vec{r}(t), a \le t \le b$ and also by $\vec{p}(s), c \le s \le d$, where $\vec{r}(t) = \vec{p}(g(t))$ for some differentiable function $g(t)$ with $g(a) = c$ and $g(b) = d$, then $\int_C f(x, y) ds = \int_a^b f(\vec{r}(t)) |\vec{r}(t)'| dt = \int_c^d f(\vec{p}(s)) |\vec{p}(s)'| ds$ $\int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}(t)' dt = \int_c^d \vec{F}(\vec{p}(s)) \cdot \vec{p}(s)' ds$

Line Integrals of Gradient Fields

If $\vec{F} = \nabla f$, then $\int_C \vec{F} \cdot d\vec{r} = f(\vec{r}(b)) - f(\vec{r}(a))$

Images

Let A be a matrix with det and let be the linear mapping of to given by (matrix multiplication). Then transforms parallelograms into parallelograms and vertices into vertices. Moreover, if is a parallelogram, must be a parallelogram.

A linear transformation matrix $A$ is one-to-one and onto if and only if $\det(A) \neq 0$.

$T$ maps $[0, 1] \times [0, 1]$ to the parallelogram spanned by the vectors $\langle a, c \rangle$ and $\langle b, d \rangle$. $T(u, v) = \langle au + bv, cu + dv \rangle$. The area of the parallelogram is $|\det(A)| = |ad - bc|$.

Applications

Average Value

The average value of $f$ over a region $R$ is given by $f_{avg} = \frac{\int\int_R f(x, y) dx dy}{\int\int_R dx dy}$

Center of Mass

The total mass $M$ of a lamina with density function $\delta(x, y)$ over a region $R$ is given by $M = \int\int_R \delta(x, y) dA$

The center of mass $(\bar{x}, \bar{y})$ of a lamina with density function $\delta(x, y)$ over a region $R$ is given by $\bar{x} = \frac{\int\int_R x \delta(x, y) dA}{M}, \quad \bar{y} = \frac{\int\int_R y \delta(x, y) dA}{M}$

Moment of Inertia

Moments of inertia about the coordinate axes with uniform density $\delta$ are given by $I_x = \delta \int\int\int_R (y^2 + z^2) dV, \quad I_y = \delta \int\int\int_R (x^2 + z^2) dV, \quad I_z = \delta \int\int\int_R (x^2 + y^2) dV$

Gravitational Force

The gravitational force exerted on a point mass $m$ located at the point $(x_0, y_0, z_0)$ by a solid region $R$ with density function $\delta(x, y, z)$ is given by $\vec{F} = Gm \int\int\int_R \frac{\delta(x, y, z)}{r^2} \hat{r} dV$ $|F| = Gm \int\int\int_R \frac{\delta(x, y, z)}{r} dV$ where $G$ is the gravitational constant, $r$ is the distance from the point mass to the point $(x, y, z)$, and $\hat{r}$ is the unit vector pointing from $(x, y, z)$ to $(x_0, y_0, z_0)$.

Isaac Newton's Breathtaking Proposition 71

A uniform spherical shell of mass $M$ exerts a gravitational force on a point mass $m$ located outside the shell as if all the shell's mass were concentrated at its center. If the point mass is located inside the shell, the net gravitational force exerted on it by the shell is zero.

Surface Integral

A parameterization of a surface $S$ is a vector function $\vec{r}(u, v) = \langle x(u, v), y(u, v), z(u, v) \rangle$ defined on a region $R$ in the $uv$-plane such that $\vec{r}$ is one-to-one and $\vec{r}(R) = S$.

$T_u = \vec{r}_u = \frac{\partial \vec{r}}{\partial u}$ and $T_v = \vec{r}_v = \frac{\partial \vec{r}}{\partial v}$ are tangent vectors to the surface, and $\vec{n} = \frac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|}$ is the unit normal vector to the surface. The magnitude of the cross product $|\vec{r}_u \times \vec{r}_v|$ gives the area of the parallelogram spanned by the tangent vectors, which is used to compute the surface area element $dS$.

An equation of the tangent plane to the surface at a point $\vec{r}(u_0, v_0)$ is given by $\vec{n} \cdot (\vec{r} - \vec{r}(u_0, v_0)) = 0$ $n_1(x - x_0) + n_2(y - y_0) + n_3(z - z_0) = 0$

The surface $S$ is regular if $T_u \times T_v \neq \vec{0}$ for all $(u, v)$ in $R$. This condition ensures that the surface does not have any singularities or self-intersections, allowing for a well-defined normal vector and surface area element.

Surface area of $S$ is given by

$$A(S) = \iint_R |\vec{r}_u \times \vec{r}_v| du dv \\ = \iint_R \sqrt{\left( \frac{\partial(y, z)}{\partial(u, v)} \right)^2 + \left( \frac{\partial(z, x)}{\partial(u, v)} \right)^2 + \left( \frac{\partial(x, y)}{\partial(u, v)} \right)^2} \, du dv$$

Parametic equation of a surface: $ax + by + cz = d$ $\vec{r}(u, v) = \vec{w} + s\vec{u} + t\vec{v}$ Parametic surface integral: $\iint_S f(x, y, z) dS = \iint_R f(\vec{r}(u, v)) |\vec{r}_u \times \vec{r}_v| du dv$ Vector field surface integral: $\iint_S \vec{F} \cdot d\vec{S} = \iint_R \vec{F}(\vec{r}(u, v)) \cdot (\vec{r}_u \times \vec{r}_v) du dv$

$d\vec{S} = (\vec{r}_u \times \vec{r}_v) du dv$

How to find the parameterization of a surface defined by an implicit equation $f(x, y, z) = 0$?

1. Find the normal vector $\vec{n} = \nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle$.
2. Find two tangent vectors $\vec{u}$ and $\vec{v}$ such that $\vec{n} \cdot \vec{u} = 0$ and $\vec{n} \cdot \vec{v} = 0$.
3. Find a point $\vec{w}$ on the surface.
4. The parameterization is given by $\vec{r}(s, t) = \vec{w} + s\vec{u} + t\vec{v}$.

Area of a surface defined by $f(x, y, z) = 0$ is given by $A(S) = \iint_R \frac{|\nabla f|}{|\nabla f \cdot \vec{k}|} du dv$ where $\vec{k}$ is the unit vector in the $z$-direction. This formula arises from the fact that the area element $dS$ can be expressed in terms of the gradient of $f$ and the projection of the surface onto the $xy$-plane. The term $|\nabla f|$ accounts for the rate of change of the surface, while $|\nabla f \cdot \vec{k}|$ accounts for the angle between the surface and the $xy$-plane, ensuring that the area is correctly scaled based on the orientation of the surface. $A = \iint_R \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} dx dy$

For rotation about the $z$-axis, the surface area is given by $A = \iint_R 2\pi \sqrt{x^2 + y^2} \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} dx dy$

For rotation about the $y$-axis, the surface area is given by $A = \iint_R 2\pi x \sqrt{1 + \left( \frac{\partial y}{\partial x} \right)^2} dx$

Graph

If $z = f(x, y)$, then $|r_u \times r_v| = \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2}$

Sphere

$dS = r^2\sin\phi\ d\phi d\theta$

Green's Theorem

A simple closed curve $C$ has two orientations:

• Counterclockwise orientation (positive orientation), $C^+$
• Clockwise orientation (negative orientation), $C^-$

$\int_C \vec{F} \cdot d\vec{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$

Lemma

Lemma 1. Let $D$ be a $y$-simple region and let $C$ be its boundary. Suppose $P: D \to \mathbb{R}$ is of class $C^1$. Then

$\int_{C^+} P \, dx = - \iint_D \frac{\partial P}{\partial y} \, dx \, dy$

(The left-hand side denotes the line integral $\int_{C^+} P \, dx + Q \, dy$, where $Q = 0$.)

Lemma 2. Let $D$ be an $x$-simple region and let $C$ be its boundary. Suppose $Q: D \to \mathbb{R}$ is of class $C^1$. Then

$\int_{C^+} Q \, dy = \iint_D \frac{\partial Q}{\partial x} \, dx \, dy$

(The left-hand side denotes the line integral $\int_{C^+} P \, dx + Q \, dy$, where $P = 0$.)

Theorems

Theorem 1: Suppose $D$ is a simple region and $C$ is its boundary, and $P: D \to \mathbb{R}$ and $Q: D \to \mathbb{R}$ are of class $C^1$. Then $\int_{C^+} P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy$ $\int_{C^-} P \, dx + Q \, dy = - \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy$

Theorem 2: Area via Green's Theorem. If $C$ is a simple closed curve that bounds a region to which Green's theorem applies, then the area of the region $D$ bounded by $C = \partial D$ is $A = \frac{1}{2} \oint_{\partial D} \left( x \, dy - y \, dx \right)$

Theorem 3: Vector Form of Green's Theorem. Let $D \subset \mathbb{R}^2$ be a region to which Green's theorem applies, let $\partial D$ be its (positively oriented) boundary, and let $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ be a $C^1$ vector field on $D$. Then $\oint_{\partial D} \mathbf{F} \cdot d\mathbf{s} = \iint_D (\operatorname{curl}\mathbf{F}) \cdot \mathbf{k} \, dA = \iint_D (\nabla \times \mathbf{F}) \cdot \mathbf{k} \, dA$

Theorem 4: Divergence Theorem in the Plane. Let $D \subset \mathbb{R}^2$ be a region to which Green's theorem applies and let $\partial D$ be its boundary. Let $\mathbf{n}$ denote the outward unit normal to $\partial D$. If $\mathbf{c} : [a, b] \to \mathbb{R}^2$, $t \mapsto \mathbf{c}(t) = (x(t), y(t))$ is a positively oriented parametrization of $\partial D$, then $\mathbf{n}$ is given by $\mathbf{n} = \frac{(y'(t), -x'(t))}{\sqrt{[x'(t)]^2 + [y'(t)]^2}}$ (see Figure 18.9). Let $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ be a $C^1$ vector field on $D$. Then $\oint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, ds = \iint_D \operatorname{div} \mathbf{F} \, dA$

Stokes' Theorem

div: $\text{div } \vec{F} = \nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$.
curl: $\text{curl } \vec{F} = \nabla \times \vec{F} = \langle \frac{\partial }{\partial y}F_z - \frac{\partial }{\partial z}F_y, \frac{\partial }{\partial z}F_x - \frac{\partial }{\partial x}F_z, \frac{\partial }{\partial x}F_y - \frac{\partial }{\partial y}F_x \rangle$.
flux: $\iint_S \vec{F} \cdot d\vec{S} = \iint_S \vec{F} \cdot \vec{n} dS$.
flux of curl: $\iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$.

$\vec{n} = \frac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|}$ is the unit normal vector to the surface.

Stokes' Theorem for Graph

$\int_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_S \text{curl } (\vec{F}) \cdot \vec{n} dA$

Stokes' Theorem for Parametrized Surface

$\iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \int_{\partial S} \vec{F} \cdot d\vec{r}$

Conservative Vector Fields

If $\vec{F} = \nabla f$, then $\int_{S} \vec{F} \cdot d\vec{r} = f(\vec{r}(b)) - f(\vec{r}(a))$.

Theorem 1: Conservative Fields. Let $\vec{F}$ be a $C^1$ vector field defined on $\mathbb{R}^3$, except possibly for a finite number of points. The following conditions on $\vec{F}$ are all equivalent:

1. For any oriented simple closed curve $C$, $\displaystyle \int_C \vec{F} \cdot d\vec{s} = 0$.
2. For any two oriented simple curves $C_1$ and $C_2$ that have the same endpoints, $\displaystyle \int_{C_1} \vec{F} \cdot d\vec{s} = \int_{C_2} \vec{F} \cdot d\vec{s}$.
3. $\vec{F}$ is the gradient of some function $f$; that is, $\vec{F} = \nabla f$ (and if $\vec{F}$ has one or more exceptional points where it fails to be defined, $f$ is also undefined there).
4. $\nabla \times \vec{F} = \vec{0}$.

A vector field satisfying one (and, hence, all) of the conditions 1–4 is called a conservative vector field.

Corollary 1: If $\vec{F}$ is a $C^1$ vector field on $\mathbb{R}^2$ of the form $P\vec{i} + Q\vec{j}$ that satisfies $\displaystyle \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, then $\vec{F} = \nabla f$ for some $f$ on $\mathbb{R}^2$.

Theorem 2: If $\vec{F}$ is a $C^1$ vector field on all of $\mathbb{R}^3$ with $\nabla \cdot \vec{F} = 0$, then there exists a $C^1$ vector field $\vec{G}$ with $\vec{F} = \nabla \times \vec{G}$.

If $F$ has a potential function $f$, then it satisfies the cross-partial property: $\frac{\partial F_x}{\partial y} = \frac{\partial F_y}{\partial x}$.

Gauss' Theorem

Gauss' Divergence Theorem

Let $W$ be a symmetric elementary region in space. $\iiint_W (\nabla \cdot \vec{F}) dV = \iint_{\partial W} \vec{F} \cdot d\vec{S}$ $\iiint_W (\text{div } \vec{F}) dV = \iint_{\partial W} (\vec{F} \cdot \vec{n}) dS$

Gauss' Law

Let $M$ be a symmetric elementary region in $\mathbb{R}^3$. If $(0, 0, 0) \notin \partial M$, then

$$\iint_{\partial M} \frac{\vec{r} \cdot \vec{n}}{r^3} \, dS = \begin{cases} 4\pi, & \text{if } (0, 0, 0) \in M, \\ 0, & \text{if } (0, 0, 0) \notin M. \end{cases}$$

where

$$\vec{r}(x, y, z) = x\vec{i} + y\vec{j} + z\vec{k}, \\ r(x, y, z) = \|\vec{r}(x, y, z)\| = \sqrt{x^2 + y^2 + z^2}.$$