# WHY: From scalar to vector

# WHY: Why we should know about vectors?

• A tool used to express objects in higher dimensions.
• A thing that contains more information: magnitude and direction
• A sequence of numbers (objects) in order used for more generalized analysis.

# WHAT: What is the vector?

• Vector addition / subtraction: $\vec{v} + \vec{u} = \vec{u} + \vec{v} = \sum_D (v_i + u_i)\hat{D_i} = (v_x + u_x)\hat{i} + (v_y + u_y)\hat{j} + \ldots = \langle v_x + u_x, v_y + u_y, \ldots \rangle$ ($-\vec{u} = \langle -u_x, -u_y, \ldots \rangle$); $(\vec{v}_1 + \vec{v}_2) + \vec{v}_3 = \vec{v}_1 + (\vec{v}_2 + \vec{v}_3)$
• Vector multiplication
  • Scalar multiplication: $c\vec{v} = \langle cv_x, cv_y, \ldots \rangle$
  • Dot product: $\vec{v} \cdot \vec{u} = \sum_D v_i u_i = |\vec{v}||\vec{u}| \cos\langle\vec{v}, \vec{u}\rangle \Rightarrow \cos\langle\vec{v}, \vec{u}\rangle = \hat{v} \cdot \hat{u}$

    As for why numerical multiplication can relate to geometry trigonometry interpretation, here is a very good explanation: duality - Dot Product in Linear transformations

    • Component(Scalar): $\text{comp}_{\vec{a}}{\vec{b}} = |\vec{b}|\cos\theta = \vec{b} \cdot \hat{a}$
    • Projection(Vector): $\text{proj}_{\vec{a}}{\vec{b}} = |\vec{b}|\cos\theta = (\vec{b} \cdot \hat{a}) \hat{a}$
  • Cross product(USE LEFT FIST RULE TO FIND THE DIRECTION Left-Fist Rule)
    • $\vec{v} \times \vec{u} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ v_x & v_y & v_z \\ u_x & u_y & u_z \end{vmatrix} = (v_yu_z - v_zu_y)\hat{i} + (v_zu_x - v_xu_z)\hat{j} + (v_xu_y - v_yu_x)\hat{k}$
    • Area: $|\vec{v} \times \vec{u}|$, Volume(scalar triple profuct): $|\vec{v} \cdot (\vec{u} \times \vec{w})| = |(\vec{v} \times \vec{u}) \cdot \vec{w}| = \begin{vmatrix} v_x & v_y & v_z \\ u_x & u_y & u_z \\ w_x & w_y & w_z \end{vmatrix}$
• Vector unitization: $\displaystyle \vec{a} = |\vec{a}|\hat{a} = a\hat{a} = \sqrt{a_x^2 + a_y^2 + \ldots} \cdots \frac{\langle a_x, a_y, \ldots \rangle}{\sqrt{a_x^2 + a_y^2 + \ldots}}$

# WHY: Why should we apply calculus to vectors?

• More applicable in the real world, like the 3D real world, linear algebra, etc.

• Vector calculus in two/three dimensions is largely applied in fluid mechanics, electromagnetism, etc.

# WHAT: What is the scalar field?

$f(x_1, x_2, x_3, \ldots)$

# Example of scalar fields

• The height of a mountain: $h = f(x, y)$
• The temperature inside a room: $T = f(x, y, z)$
• The density of objects/water surface/air: $\rho = f(x_1, x_2, \ldots)$

# HOW: How to express objects (curves, surfaces, etc.) in a scalar field?

$f: \{(x_1, x_2, \ldots) | f_1(x_1, x_2, \ldots) = 0, f_2(x_1, x_2, \ldots) = 0, \ldots\}$

$f_1(x_1, x_2, \ldots) = 0$ has many examples, like $x_2 - b - kx_1 = 0$ (it is the same as popular function $x_2 = b + kx_1$), $\displaystyle\frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} - 1 = 0$, $(x_1 - a)^2 + (x_2 - b)^2 + (x_3 - c)^2 - R^2 = 0$, etc.

# E.g. from X-Y-Z coordinates to vectors

# WHY: Why should we use X-Y-Z coordinate

• It is the generalization of the X-Y coordinate that is easy to learn and understand
• It is more visualizable using some graphic software or even drawn by hand
• The process of the generalization from X-Y to X-Y-Z is the same as the generalization from X-Y-Z to higher dimensions. In the higher dimensions, the meaning of vector is generalized to just a set of points: $\vec{v} = (x_1, x_2, \ldots)$

# WHAT: What are points, lines, and planes in the coordinate

• Points/Vectors(3 variables - 3 constraints = 0 dimensions): $\{(x, y, z) | f_1(x, y, z) = a, f_2(x, y, z) = b, f_3(x, y, z) = c\}$ E.g. $O = (0, 0, 0)$ OR $O: \{(x, y, z) | x = 0, y = 0, z = 0\}$, $\hat{u} = (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$

1. $f_1$, $f_2$, and $f_3$ is independent (independent: you cannot turn $f_1 = a$ to $f_2 = b$ OR there is only one solution to the functions)
2. Vector is just a set of numbers, it could be used for showing direction(vectors in three dimensions), matrix(computer science programming), etc.

• Lines/Curves(3 variables - 2 constraints = 1 dimensions): $\{(x, y, z) | f_1(x, y, z) = a, f_2(x, y, z) = b\}$

• Planes/Surfaces(3 variables - 1 constraints = 2 dimensions): $\{(x, y, z) | f_1(x, y, z) = a\}$

• Cubes/Volumes(3 variables - 0 constraints (only partial constraints) = 3 dimensions): $\{(x, y, z) | f(x, y, z) \lessgtr a\}$

Partial Constraints, not absolutely equal constraints, or constraints with intervals, still gives some freedom to one particular dimensions/some lines to move along, so the points would not be fixed and spawn greater dimensions.

# HOW: How to present in the coordinate - Parametrized Objects

• Points/Vectors: $\vec{p} = \vec{r}_0$

• lines: $\vec{r} = \vec{r}_0 + t\vec{v}$ (from original point to go along the line)

  • $x = x_0 + v_xt, y = y_0 + v_yt, z = z_0 + v_zt$
  • $t = (x - x_0)/v_x = (y - y_0)/v_y = (z - z_0)/v_z$

• planes: $\vec{n} \cdot (\vec{r} - \vec{r}_0)$ (from original direction to go perpendicularly the direction) $\Rightarrow ax + by + cz + d = 0$ or $\vec{r}_0 +u\vec{r}_1 + v\vec{r}_2$

  • cylinder: Circular ~ $x^2 + y^2 = 1$; Parabolic ~ $z = x^2$; Elliptic ~ $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
  • Sphere $(x - a)^2 + (y - b)^2 + (z - c)^2 = R^2$; Ellipsoid $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$; Elliptic paraboloid $\displaystyle \frac{z}{c} = \frac{x^2}{a^2} + \frac{y^2}{b^2}$; Hyperbolic paraboloid(saddle) $\displaystyle \frac{z}{c} = \frac{x^2}{a^2} - \frac{y^2}{b^2}$

# HOW: How to visualize vectors in dimensions?

From $y = f(x_1, x_2, \cdots)$ to $y = f(\vec{x})$ for $\vec{x} = \langle x_1, x_2, \cdots \rangle$; mappings of $\R^n \rightarrow \R$

• Contour map, or Level curves: slice along the $z$ (or $y$ above)

• e.i. $t = f(x, y, z) = x^2 - y^2 - z^2$: hyperboloid of one sheet $\rightarrow$ cone $\rightarrow$ hyperboloid of one sheet (cell division, droplet, …)

# HOW: How to do calculus in a scalar field?

# Single derivatives

$\vec{r}(t) = \langle F(t), G(t), H(t) \rangle = F(t)\hat{i} + G(t)\hat{j} + H(t)\hat{k}$

• Derivative: $\vec{v}(t) = \vec{r}’(t) = \langle F’(t), G’(t), H’(t) \rangle = F’(t)\hat{i} + G’(t)\hat{j} + H’(t)\hat{k}$
  • $\frac{\mathrm{d}}{\mathrm{d}t}(f(t)\vec{u}(t)) = f’(t)\vec{u}(t) + f(t)\vec{u}’(t)$($f$ could be either scalar or vector function, while it is vector, both dot and cross product could be applied)

Why the rule of product could be applied to dot and cross product? As for dot product, we regard it as multiple independent product process, and combine it after the product.

• Integral: $\vec{r}(t) = \int\vec{v}(t)\mathrm{d}t = \langle \int f(t) \,\mathrm{d}t, \int g(t) \,\mathrm{d}t, \int h(t) \,\mathrm{d}t \rangle = \int f(t) \,\mathrm{d}t \hat{i} + \int g(t) \,\mathrm{d}t \hat{j} + \int h(t) \,\mathrm{d}t \hat{k}$
  • Arc Length: $\displaystyle l = \int 1 ds = \int \frac{\mathrm{d}s}{\mathrm{d}t} \,\mathrm{d}t = \int |\vec{r}’(t)| \,\mathrm{d}t = \int \sqrt{\frac{\mathrm{d}F^2}{\mathrm{d}t^2} + \frac{\mathrm{d}G^2}{\mathrm{d}t^2} + \frac{\mathrm{d}H^2}{\mathrm{d}t^2}} \,\mathrm{d}t$
  • Curvature: $\displaystyle \kappa(t) = |\frac{\mathrm{d}\hat{T}}{\mathrm{d}s}| = \frac{\mathrm{d}\hat{T}/\mathrm{d}t}{\mathrm{d}s/\mathrm{d}t} = \frac{|\hat{T}’(t)|}{|\vec{r}’(t)|}, \hat{T} = \frac{\vec{r}’}{|\vec{r}’|} \Rightarrow \kappa(t) = \frac{|\vec{r}’ \times \vec{r}’’|}{|\vec{r}’|^3}$ (relate to determinant)
    • proof of $\displaystyle |\hat{T}’| = |\frac{\vec{r}’’ \times \vec{r}’}{\,\mathrm{d}s^2/\,\mathrm{d}t^2}|: \hat{T}’ = \frac{\vec{r}’’\,\mathrm{d}s/\,\mathrm{d}t - \vec{r}’\,\mathrm{d}^2s/\,\mathrm{d}t^2}{\,\mathrm{d}s^2/\,\mathrm{d}t^2} \Rightarrow |\hat{T}’| = |\hat{T} \times \hat{T}’|$

# HOW: How to apply to space?

• Tangent vector to a space curve: $\hat{T} = \vec{r}’/|\vec{r}’|$
• Normal vector to a tangent vector: $\hat{N} = \hat{T}’/|\hat{T}’| = \hat{T}’/|\kappa \vec{r}’|$
• Binormal vector to the vectors: $\hat{B} = \hat{N} \times \hat{T}$

Build the coordinate with $\hat{T}, \hat{N}, \hat{B}$: $\vec{r} \Rightarrow \hat{T} \Rightarrow \hat{N} \Rightarrow \hat{B}$

$\vec{a} = \vec{v}’ = (v\hat{T}) = v’\hat{T} + v\hat{T}’ = v’\hat{T} + v^2\kappa\hat{N}$

$\Rightarrow \vec{a} = a_T\hat{T} + a_N\hat{N}, a_T = v’ = \frac{|\vec{r}’ \cdot \vec{r}’’|}{|\vec{r}’|}, a_N = \kappa v^2 = \frac{|\vec{r}’ \times \vec{r}’’|}{|\vec{r}’|}$

Note. How to understand $\frac{\mathrm{d}}{\mathrm{d}t} \hat{i} \cdot \hat{i} = 0$: as the magnitude of $\hat{i}$ never change, the only change of $\hat{i}$ is its direction. And the change of its direction only present in the tendency of the orthogonal effect vector.

# Partial Derivatives

# WHY: Why a function could be differentiated? - Differentiability

• Limit & Continuity: $\displaystyle \lim_{(x, y) \rightarrow (a, b)}f(x, y) = L$ (no matter any directions approaching to $(a, b)$, f(x, y) is continuous at $(a, b)$ if and only if those values are the same)

  • The existence of the Limit: common directions - along y-axis ($x = 0$), x-axis, $x = y$, $r^2 = x^2 + y^2, x = r\cos\theta, y = r\sin\theta$, etc. (e.i. using L’Hôpital’s rule after fixing one dimension) $$ \lim_{(x, y) \rightarrow (a, b)}f(x, y) = L $$

  When $f \rightarrow L$ no matter what direction $(x, y) \rightarrow (a, b)$, the limit of $f$ in $(a, b)$ exists.

  It is easy to prove the limit is not exists. If $f(x, y) \rightarrow L_1$ as $(x, y) \rightarrow (a, b)$ along a path $C_1$ and $f(x, y) \rightarrow L_2$ as $(x, y) \rightarrow (a, b)$ along a path $C_2$, where $L_1 \neq L_2$, then $\lim_{(x, y) \rightarrow (a, b)}f(x, y)$ does not exist.

  • Limit along axis: let $y = 0$ or $x = 0$ Limit along the specific line: $y = x$

• Continuity: $\lim_{(x, y) \rightarrow (a, b)}f(x, y) = f(a, b)$

# WHAT: What are Partial Derivatives

• Algebra: Definition $$ \frac{\partial}{\partial x}f = f_x(a, b) = \lim_{h \rightarrow 0}\frac{f(a + h, b) - f(a, b)}{h} = g’(a) \ (g(x) = g(x, b)) = \lim_{h \rightarrow 0}\frac{g(a + h) - g(a)}{h} $$ $$ \frac{\partial}{\partial y}f = f_y(a, b) = \lim_{h \rightarrow 0}\frac{f(a, b + h) - f(a, b)}{h} = g’(b) \ (g(y) = g(a, y)) = \lim_{h \rightarrow 0}\frac{g(b + h) - g(b)}{h} $$

  • Increment: $\Delta y = f(x_0 + \,\mathrm{d}x, y_0 + \,\mathrm{d}x) - f(x_0, y_0) \Rightarrow$ differentiable: $\Delta{z} = f_x(a, b)\Delta{x} + f_y(a, b)\Delta{y} + \epsilon_1\Delta{x} + \epsilon_2\Delta{y}, (\Delta{x}, \Delta{y}) \rightarrow (0, 0) \Rightarrow (\epsilon_1, \epsilon_2) \rightarrow (0, 0)$
  • Differentials: $\displaystyle \,\mathrm{d}z = \frac{\partial z}{\partial x} \,\mathrm{d}x + \frac{\partial z}{\partial y} \,\mathrm{d}y \Leftarrow \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{\partial z}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial z}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}$

• Geometry

  • Geometry interpretation: Monotonicity - $f_x$, $f_y$; Concativity - $f_{xx}, f_{yy}$; “Twistivity” - $f_{xy}, f_{yx}$
  • Tangent Planes & Linear Approximations: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \Rightarrow z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ (normal vector with cross product of the x component and y component)

# HOW: How to do Partial Derivatives

• *Chain Rule $$ (1) z = f(x, y): \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{\partial f}{\partial x} \frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y} \frac{\mathrm{d}y}{\mathrm{d}t} $$ $$ (2) \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}, \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} $$ $$ (3) \text{General}: \frac{\partial u}{\partial t_i} = \frac{\partial u}{\partial x_1} \frac{\partial x_1}{\partial t_i} + \frac{\partial u}{\partial x_2} \frac{\partial x_2}{\partial t_i} + \frac{\partial u}{\partial x_3} \frac{\partial x_3}{\partial t_i} + \ldots $$ $$ *(3): \frac{\partial u}{\partial x_1} = \frac{\partial u}{\partial x_1} \frac{\partial x_1}{\partial x_1} + \frac{\partial u}{\partial x_2} \frac{\partial x_2}{\partial x_1} + \frac{\partial u}{\partial x_3} \frac{\partial x_3}{\partial x_1} + \ldots \\ \text{since $x_1, x_2, x_3, \ldots$ are independent} \Rightarrow \frac{\partial x_n}{\partial x_1} = 0, \frac{\partial x_1}{\partial x_1} = 1 \Rightarrow \frac{\partial u}{\partial x_1} = \frac{\partial u}{\partial x_1} $$

# HOW: How to Apply to Vector

$$ D_u{f(x, y)} = \vec\nabla{f(x, y)} \cdot \hat{u} $$

• $D_u{f(x, y)} = \vec{P’Q’}$ - Gradient Vector: moving in a tangent plant with a specific direction
  • $\mathrm{grad}f(x, y) = \vec\nabla{f(x, y)}$ represents tangent plant: $\vec\nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} + \ldots$
    • $f(x_1, x_2, \ldots) \Rightarrow \mathbb{R}^n \rightarrow \mathbb{R}$
    • $\vec\nabla{f} \Rightarrow \mathbb{R} \rightarrow \mathbb{R}^n$
    • $\text{div } f = \vec\nabla{f} = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k}$
    • $\text{curl } \vec{F} = \vec\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix} = (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z})\hat{i} + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x})\hat{j} + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})\hat{k}$
    • $\mathrm{div}\,\mathrm{curl}\,\textbf{F} = \mathbf{0}$, $\mathrm{curl}\,\mathrm{grad}\,f = 0$
  • $\vec{u}$ represents a direction (in unit), or $\vec{u} = \langle \hat{a}, \hat{b} \rangle$
  • $\vec\nabla{f(x, y)}$ presents the fastest increasing direction/perpendicular to the contour line.

# HOW: How to Apply to Finding Maximum and Minimum

1. $\displaystyle \frac{\partial f}{\partial x} = 0, \frac{\partial f}{\partial y} = 0 \Leftarrow \frac{\partial}{\partial t} = a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}, \frac{\partial f}{\partial \forall t} = 0$

2. Second Derivative Test: The relationship between $\displaystyle D(x_0, y_0) = \begin{vmatrix} \frac{\partial^2}{\partial x^2} & \frac{\partial^2}{\partial x \partial y} \\ \frac{\partial^2}{\partial y \partial x} & \frac{\partial^2}{\partial y^2} \end{vmatrix}f = \mathbf{H}f = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - f^2_{xy}(x_0, y_0)$ and $f_{xx}(x_0, y_0)$ (or second derivatives in any direction)

   • $\mathbf{H}f(x_0, y_0) > 0 \Rightarrow$ concavity find local extremum
     • $f_{xx} > 0$: local minimum
     • $f_{xx} < 0$: local maximum
   • $\mathbf{H}f(x_0, y_0) < 0 \Rightarrow$ concavity find saddle point
   • $\mathbf{H}f(x_0, y_0) = 0 \Rightarrow$ second derivative provides not enough info

• A geometry interpretation of it: Khan Academy: Second partial derivative test

• *Hessian matrix: Wikipedia: Hessian matrix

• Lagrange Multipliers $\lambda$: $\vec\nabla f = \lambda \vec\nabla g$ (Simple idea: the highest point of the path is tangent to the highest contour)

  • Generalization: $\vec\nabla f(x_1, x_2, x_3, \ldots) = \lambda \vec\nabla g(x_1, x_2, x_3, \ldots) + \mu \vec\nabla h(x_1, x_2, x_3, \ldots) + \ldots$
  • Or, finding local max/min of $f$ under $L(x_1, x_2, \ldots) = 0$ is equivalant to finding local max/min of $g(x_1, x_2 \ldots, x_n, \lambda) = f - \lambda L$

# Iterated Integrals

# WHAT: What is the double integral

$\iint\limits_R{f(x, y)}\,\mathrm{d}A = \lim_{max \Delta x_i, \Delta y_i \rightarrow 0}\sum_{i = 1}^{m}\sum_{j = 1}^{n}f(x_{ij}^* , y_{ij}^* ) \Delta A_{i,j}$ (Double Riemann Sum, Midpoint: $x^* , y^ * \rightarrow \bar{x}, \bar{y}$)

# HOW: How to switch iterating order?

$$ \int\int{f(x, y)}\,\mathrm{d}A (A =[a, b] \times [c, d]) = \int_a^b[\int_c^df(x, y)\,\mathrm{d}y]\,\mathrm{d}x = \int_c^d[\int_a^bf(x, y)\,\mathrm{d}x]\,\mathrm{d}y $$

$$ \iiint f(x, y, z) \,\mathrm{d}V (V = [0, 1] \times [\sqrt{x}, 1] \times [0, 1 - y]) \\ =\int_0^1\int_{\sqrt{x}}^1\int_0^{1-y} f(x, y, z) \,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x = \int_0^1\int_0^{y^2}\int_0^{1-y} f(x, y, z) \,\mathrm{d}z\,\mathrm{d}x\,\mathrm{d}y \\ = \int_0^1\int_0^{1 - \sqrt{x}}\int_{\sqrt{x}}^{1 - z} f(x, y, z) \,\mathrm{d}y\,\mathrm{d}z\,\mathrm{d}x = \int_0^1\int_0^{(1 - z)^2}\int_{\sqrt{x}}^{1 - z} f(x, y, z) \,\mathrm{d}y\,\mathrm{d}x\,\mathrm{d}z \\ = \int_0^1\int_0^{1 - z}\int_0^{y^2} f(x, y, z) \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = \int_0^1\int_0^{1 - y}\int_0^{y^2} f(x, y, z) \,\mathrm{d}x\,\mathrm{d}z\,\mathrm{d}y $$

# HOW: How to turn variables in different coordinates?

$$\iint_\mathbb{R} f(x, y) \partial x \partial y = \iint_\mathbb{R} f(u, v) |\frac{\partial(x, y)}{\partial(u, v)}| \partial u \partial v, |\frac{\partial(x, y)}{\partial(u, v)}| = \begin{vmatrix} \partial x / \partial u & \partial x / \partial v \\ \partial y / \partial u & \partial y / \partial v \end{vmatrix}$$

from polar coordinate, to cylindar, then to sphere

$$\iint f(x, y) \,\mathrm{d}x\,\mathrm{d}y = \iint f(r\cos\theta, r\sin\theta) \cdot r \,\mathrm{d} r \,\mathrm{d}\theta$$

$$\iiint f(x, y, z) \,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = \iiint f(r\cos\theta, r\sin\theta, z) \cdot r \,\mathrm{d} r \,\mathrm{d}\theta \,\mathrm{d}z \\ = \iiint f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi) \cdot \rho^2\sin\phi \,\mathrm{d} \rho \,\mathrm{d}\theta \,\mathrm{d}\phi$$

# HOW: How to apply to reality

• Volume Region of $x_1, x_2, y_1(x), y_2(x)$, Assume $x_1 \leq x_2, y_1 \leq y_2$:

  $$ V = \iiint\,\mathrm{d}V = \int_{y_1(x)}^{y_2(x)}\int_{x_1}^{x_2}\int_0^{f(x, y)}\,\mathrm{d}V = \int_{x_1}^{x_2}[\int_{y_1(x)}^{y_2(x)}f(x, y)\,\mathrm{d}y]\,\mathrm{d}x$$

• Density and Mass Mass of a volume in $x_1, x_2, y_1(x), y_2(x), z_1(x,y), z_2(x, y)$:

  $$M = \iiint\int_0^{\rho(x, y, z)} \,\mathrm{d}\rho\,\mathrm{d}V = \iiint\rho(x, y, z)\,\mathrm{d}V$$

# WHAT: What is the vector field?

$\mathbf{F}(x_1, x_2, x_3, \ldots)$ or $\vec{F}(x_1, x_2, x_3, \ldots)$

# Example of vector fields

• The velocity of water/air (fluid): $\vec{v} = \mathbf{F}(x_1, x_2, x_3, \ldots)$
• Gravitational field (force): $\displaystyle \vec{F} = \mathbf{F}(x_1, x_2, x_3) = \frac{GMm}{r^2}\hat{r}$ ($\hat{r}$ is a unit vector)
• Electric field: $\displaystyle \vec{E} = \mathbf{F}(x_1, x_2, x_3) = k\frac{Q}{r^2}\hat{r}$
• Magnitic field: $\displaystyle \vec{B} = \mathbf{F}(x_1, x_2, x_3) = \frac{\mu_0q\vec{v}\times\hat{r}}{4\pi r^2}$

# WHAT: What are vector fields

• Vector Fields: $\vec{F}(x, y) = I(x, y)\hat{i} + J(x, y)\hat{j} = \langle I(x, y), J(x, y) \rangle (in \ \R^2)$

• Gradient Fields: $\vec\nabla{f(x, y)} = f_x(x, y)\hat{i} + f_y(x, y)\hat{j}$

# Conservative vector fields

$$\forall C_1, C_2 = (P_1 \rightarrow P_2) \in D, \int_{C_1} \mathbf{F} \cdot \,\mathrm{d} \mathbf{r} = \int_{C_2} \mathbf{F} \cdot \,\mathrm{d} \mathbf{r} \Leftrightarrow \exists f, \vec\nabla{f} = \mathbf{F}$$

$D$: an open($\forall P \in D,\exists \circ P \in D$) connected($\forall P_1, P_2 \in D, \exists C = (P_1 \rightarrow P_2) \in D$) region; the right side means $\mathbf{F}$ is a conservative vector field; under this situation, we call $\int_{C} \mathbf{F} \cdot \,\mathrm{d} \mathbf{r}$ is independent of path, or line integrals of conservative vector fields are independent of path.

$$\exists f, \vec\nabla{f} = \mathbf{F} \Leftrightarrow \mathbf{F} = P\mathbf{i} + Q\mathbf{j}, \partial P / \partial y = \partial Q / \partial x \,\mathbf{or}\, \vec\nabla \times \mathbf{F} = \mathbf{0}$$

Clairaut’s Theorem: $\partial^2f/\partial x\partial y = \partial^2f/\partial y \partial x$

# HOW: How to do calculus in vector field?

# HOW: How to integrate along the line - Line(Path/Contour) Integrals

$$\displaystyle \int_C \vec{F} \cdot \,\mathrm{d}\vec{r} = \int \vec{F}(\vec{r}(t)) \cdot \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\,\mathrm{d}t$$

$$ \int_{\mathrm{curve}\,C}f(x, y)\,\mathrm{d}s = \lim_{\max \Delta{s}_i \rightarrow 0}\sum_{i = 1}^{n}f(x_i^* , y_i^*)\Delta{s}_i, \mathrm{curve}\, C: s = \int_a^b\sqrt{(\frac{\mathrm{d}x}{\mathrm{d}t})^2 + (\frac{\mathrm{d}y}{\mathrm{d}t})^2}\mathrm{d}t $$ $$ \Rightarrow \int_C f(x, y)\,\mathrm{d}s = \int_a^b f(x(t), y(t))\sqrt{(\frac{\mathrm{d}x}{\mathrm{d}t})^2 + (\frac{\mathrm{d}y}{\mathrm{d}t})^2}\mathrm{d}t $$

• Vector Fields $$ W = \int_C \vec{F} \cdot \,\mathrm{d}\vec{r} = \int_a^b{\vec{F}(\vec{r}(t)) \cdot \vec{r’}(t)\mathrm{d}t} $$

  (Substitution: $\vec{r}(t) = \vec{r} = \int_C \vec{F} \cdot \vec{T} \,\mathrm{d}s$)

• Conservative $$ F(x, y) = P\hat{i} + Q\hat{j}, F \text{ is conservative} \Leftrightarrow \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \text{ thoughout } D $$

  • independent of path: $\int_C \vec{F} \cdot d \vec{r} \text{ is independent of path} \in D \Leftrightarrow \forall C \in D, \int_C \vec{F} \cdot d \vec{r} = 0$
    • $C$: Closed Path - its terminal point coincides with its initial point
  • conservative vector field: $\int_C \vec{F} \cdot d \vec{r} \text{ is independent of path} \in D \Leftrightarrow F \text{ is a conservative vector field}$ ($\vec\nabla{f} = \mathbf{F}$)

• Relationship with double integral: Green’s Theorem $$ \oint_C \vec{F} \cdot \,\mathrm{d}\vec{r} = \oint_C P \,\mathrm{d}x + Q \,\mathrm{d}y = \iint_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})\,\mathrm{d}A = \int_{\partial D}{P \,\mathrm{d}x + Q \,\mathrm{d}y} $$

  • Positive orientation: a simple closed curve $C$ refers to a single counterclockwise traversal of $C$
  • $\oint_C P \,\mathrm{d}x = \oint_{g_2(x)}^{g_1(x)} P \,\mathrm{d}x = - \oint_{g_1(x)}^{g_2(x)} P \,\mathrm{d}x = -\int_{g_1(x)}^{g_2(x)} \int_a^b \frac{\partial P}{\partial y} \,\mathrm{d}y \,\mathrm{d}x = \iint_D(-\frac{\partial P}{\partial y}) \,\mathrm{d}A$

# HOW: How to integrate along the surface - Surface Integrals

# WHAT: What are surfaces and areas

• Parametric Surfaces: $\mathbf{r}(u, v) = x(u, v)\mathbf{i} + y(u, v)\mathbf{j} + z(u, v)\mathbf{k}$
  • Revolution: eg. along x-axis - $\mathbf{r}(u, v) = u\mathbf{i} + f(u)\cos v\mathbf{j} + f(u)\sin v\mathbf{k}$
• Tangent Planes: there are two directions for the tangent planes of parametric surfaces - u-direction and v-direction
  • $\mathbf{r}_u(u, v) = \frac{\partial}{\partial u}(x(u, v)\mathbf{i} + y(u, v)\mathbf{j} + z(u, v)\mathbf{k})$
  • $\mathbf{r}_v(u, v) = \frac{\partial}{\partial v}(x(u, v)\mathbf{i} + y(u, v)\mathbf{j} + z(u, v)\mathbf{k})$
  • $\mathbf{n} = \mathbf{r}_u(u, v) \times \mathbf{r}_v(u, v)$
  • tangent planes: $\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0)$
• Surface Area (Integral of Tangent Planes)
  • Recall arc length: $l = \int_C |(x’(t), y’(t))| \,\mathrm{d}t = \int_C \sqrt{1 + y’^2(x)} \,\mathrm{d}x$
  • $A = \iint_D |\mathbf{r}_u(u, v) \times \mathbf{r}_v(u, v)| \,\mathrm{d}A = \iint_D \sqrt{1 + (\partial z/\partial x)^2 + (\partial z/\partial y)^2}\,\mathrm{d}A$

# HOW: How to do surface integral

Scalar field: $\iint_S \rho(x, y, z) \,\mathrm{d}S = \iint _D \rho(\mathbf{r}(u, v)) |\mathbf{r}_u \times \mathbf{r}_v| \,\mathrm{d}A$

Vector field: $\iint_S \mathbf{F} \cdot \mathrm{d}\mathbf{S} = \iint_S \mathbf{F} \cdot \hat{n} \,\mathrm{d}S = \iint_S \mathbf{F} \cdot \mathbf{n} \,\mathrm{d}A = \iint_D \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v)\,\mathrm{d}A$

For $\hat{n}$ and $\mathbf{n}$: $\displaystyle \mathbf{n} = \hat{n}\mathrm{(1)} \cdot |\mathbf{r}_u \times \mathbf{r}_v|\mathrm{(2)} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{|\mathbf{r}_u \times \mathbf{r}_v|}\mathrm{(1)} \cdot |\mathbf{r}_u \times \mathbf{r}_v|\mathrm{(2)}$, for $\mathrm{(2)}$ is related to Jacobian Matrix

# HOW: How to switch the integral from vector integral to iterative integral under a closed surface/curve

Greens’ Theorem to Stokes’ Theorem, and Divergence Theorem

$$\int_{\partial D}\omega = \int_D \partial\omega$$

$$\Rightarrow \oint_{C(\partial{S})}\mathbf{F} \cdot \mathrm{d}\mathbf{r} = \iint_S \vec\nabla \times \mathbf{F} \cdot \,\mathrm{d}\mathbf{S} = \iint_S \vec\nabla \times \mathbf{F} \cdot \mathbf{n} \,\mathrm{d}A = \iint_S |\vec\nabla \times \mathbf{F}| \,\mathrm{d}A \text{ in $\mathbb{R}^2$}$$

$$\Rightarrow \oint_C \mathbf{F} \cdot \mathrm{d}\mathbf{S} = \iint_D \vec\nabla \cdot \mathbf{F} \,\mathrm{d}A, \oiint_S \mathbf{F} \cdot \mathrm{d}\mathbf{S} = \iiint_E \vec\nabla \cdot \mathbf{F} \,\mathrm{d}V$$