Points are represented as $(x,y)$.

• The plane is divided into four quadrants based on the signs of $x$ and $y$.

• Points are represented as $(x,y,z)$.
• The space is divided into eight octants based on the signs of $x$, $y$, and $z$.

• Points are represented as $(r,θ)$ where $r$ is the radius and $θ$ is the angle.

• A point $M⃗$ in polar coordinates is given by $(r,θ)$, where:

  • $r = \sqrt{x^2 + y^2}$ 　is the distance from the origin.
  • $θ = tan^{−1}(\dfrac{y}{x})$ 　is the angle formed with the positive x-axis.

• Conversion to Cartesian:

  • $x = r·cos⁡θ$
  • $y = r·sin⁡θ$

Applications:

• Used for problems with circular symmetry.
• Equations of circles and spirals.

• Points are represented as $(r,θ,z)$.
• Relationship with Cartesian coordinates:
  • $x = r·cos⁡θ$
  • $y = r·sinθ$
  • $z = z$

• Points are represented as $(ρ,ϕ,θ)$ where $ρ$ is the radius, $ϕ$ is the polar angle, and $θ$ is the azimuthal (horizontal) angle.
• Relationship with Cartesian coordinates:
  • $x = ρ·sin⁡ϕ·cos⁡θ$
  • $y = ρ·sin⁡ϕ·sin⁡θ$
  • $z = ρ·cos⁡ϕ$

• Vectorial form:

  • $ℓ⃗(t) = r⃗_0 + t·d⃗$
    • where:
      • $r⃗_0$​ is a point on the line
      • $d⃗$ is the direction vector

• Parametric form:

  • $ℓ_x(t) = x_0 + t_x·d_x$
  • $ℓ_y(t) = y_0 + t_y·d_y$
  • $ℓ_z(t) = z_0 + t_z·d_z$

• Vector form

  • $r⃗_0⋅n̂ = d$
    • where
      • $n̂ = (n_x,n_y,n_z)$ is the normal vector
      • $r⃗_0 = (x_0,y_0,z_0)$ is a point vector in the plane

• Scalar Equation

  • $n_x·(x−x_0​) + n_y·(y−y_0​) + n_z·(z−z_0​) = 0$
  • $n_x·x + n_y·y + n_z·z = d$
    • where $d = n_x·x_0 + n_y·y_0 + n_z·z_0$

• Generated by rotating a curve around an axis.
  • Example: A paraboloid generated by rotating $y = x^2$ around the $z$-axis.

• Formed by extending a curve along a straight line parallel to an axis, tipically $z$.
  • Example: A cylinder with radius $r$ and height $h$, represented by $x^2 + y^2 = r^2$

• General second-degree surfaces in three dimensions.
  • Standard forms include:
    • Ellipsoid:
      • $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
    • Hyperboloid of one sheet:
      • $\frac{x^2}{a^2} + \frac{y^2}{b^2} − \frac{z^2}{c^2} = 1$
    • Hyperboloid of two sheets:
      • $\frac{x^2}{a^2} - \frac{y^2}{b^2} − \frac{z^2}{c^2} = 1$
    • Paraboloid:
      • Elliptic:
        • $z = x^2 + y^2$
      • Hyperbolic
        • $z = x^2 − y^2$

A vector function is a function that maps a real number $t$ to a vector $r(t)$ in space.

• $r⃗❨t❩=⟨f❨t❩,⠀g❨t❩,⠀h❨t❩⟩$
  • where $f❨t❩$, $g❨t❩$, and $h❨t❩$ are scalar functions.

• Addition
  • $r⃗_1❨t❩ + r⃗_2❨t❩ = ⟨f_1❨t❩+f_2❨t❩,⠀g_1❨t❩+g_2❨t❩,⠀h_1❨t❩+h_2❨t❩⟩$
• Scalar Multiplication
  • $𝑐·r⃗❨t❩ = ⟨𝑐·f❨t❩,⠀𝑐·g❨t❩,⠀𝑐·h❨t❩⟩$
• Dot Product
  • $r⃗_1❨t❩ ⋅ r⃗_2❨t❩=⟨f_1❨t❩⋅f_2❨t❩,⠀g_1❨t❩⋅g_2❨t❩,⠀h_1❨t❩⋅h_2❨t❩⟩$
• Cross Product

$$r⃗_1❨t❩ ⨯ r⃗_2❨t❩ = \begin{vmatrix} x̂ & ŷ & ẑ \\\ f_1❨t❩ & g_1❨t❩ & h_1❨t❩ \\\ f_2❨t❩ & g_2❨t❩ & h_2❨t❩ \end{vmatrix}$$ $$r⃗_1❨t❩ ⨯ r⃗_2❨t❩ = \left( g_1❨t❩ h_2❨t❩ - h_1❨t❩ g_2❨t❩ \right) \mathbf{i} - \left( h_1(t) f_2❨t❩ - f_1❨t❩ h_2❨t❩ \right) \mathbf{j} + \left( f_1❨t❩ g_2❨t❩ - g_1❨t❩ f_2❨t❩ \right) \mathbf{k}$$ $$\mathbf{r}_1(t) \times \mathbf{r}_2(t) = \left\langle \begin{matrix} g_1(t) h_2(t) - h_1(t) g_2(t) \\\ h_1(t) f_2(t) - f_1(t) h_2(t) \\\ f_1(t) g_2(t) - g_1(t) f_2(t) \end{matrix} \right\rangle$$

A vector function $r⃗❨t❩$ is continuous at $t = t_0$ if

$$\lim_{t \to t_0} r⃗❨t❩ = r⃗❨t_0❩$$

The derivative of $r⃗❨t❩$ with respect to $t$ is defined as:

$$r⃗̇̇❨t❩ = r⃗'❨t❩ =\frac{dr⃗}{dt} = \lim_{\Delta t \to 0} \frac{r⃗❨t + \Delta t❩ - r⃗❨t❩}{\Delta t}$$ $$\frac{dr⃗̇̇}{dt} = ⟨ \frac{df❨t❩}{dt}, \frac{dg❨t❩}{dt}, \frac{dh❨t❩}{dt} ⟩$$

The tangent line to the curve defined by $r⃗❨t❩$ at $t=t_0$​ is given by:

$$\$𝐓⃗❨t❩ = r⃗❨t_0❩ + (t - t_0) r⃗'(t_0)\$$$

The integral of a vector function is computed component-wise:

$$∫r⃗❨t❩dt=⟨∫f❨t❩dt,∫g❨t❩dt,∫h❨t❩dt⟩$$

Definition:

• A parametric curve in space is given by a vector function

$$ r⃗❨t❩=⟨x❨t❩,y❨t❩,z❨t❩⟩ $$

$$r⃗❨t❩ = ⟨ cos❨t❩ , sin❨t❩ , 0 ⟩$$

The process of reparametrization involves identifying a novel parameter $u$ that is related to the original parameter $t$ by means of a function $u = g❨t❩$, thereby enabling the rewriting of $rt$ as

$$r⃗( g^{-1}❨u❩)$$

The arc length $s$ of a curve $r⃗❨t❩$ from $t = a$ to $t = b$ is given by:

$$s = ∫_a^b ∥ \frac{r⃗❨t❩}{dt} ∥ dt$$

• Where $∥ x❨t❩ ∥$ is the magnitude of $x❨t❩$

The curvature $κ$ of a curve at a point measures how quickly the direction of the curve changes at that point.

For a curve $r⃗❨t❩$, the curvature is given by:

$$κ = \frac{∥r⃗′(t)×r⃗′′❨t❩∥​}{∥r⃗′❨t❩∥^3} = \frac{bending}{speed_{normalization}}$$

$Curvature$ measures how sharply a curve bends at a particular point. It is the rate of change of the tangent vector's direction with respect to arc length.

A scalar field is a function that assigns a scalar value to each point in space. For example, a temperature distribution in a room can be represented as a scalar field $T(x,y,z)$.

Mathematically, a scalar field is written as

$$f(x,y,z)$$

where $(x,y,z)$ are the coordinates in space.

Partial derivatives measure how the scalar field changes as each coordinate changes independently.

The partial derivative of $f$ with respect to $x$ is denoted as

$$\frac{∂f}{∂x}​$$

and represents the rate of change of $f$ along the $x$-direction, keeping $y$ and $z$ constant.

Similarly, the partial derivatives with respect to $y$ and $z$ are

$$\frac{∂f}{∂y}$$ $$\frac{∂f}{∂z}$$

Example:

For a scalar field $f(x,y,z)=x^2+y^2+z^2$:

• $\frac{∂f}{∂x} = 2x$
• $\frac{∂f}{∂y} = 2y$
• $\frac{∂f}{∂z} = 2z$

A tangent plane to a surface at a given point is a plane that just "touches" the surface at that point, approximating the surface near the point.

For a surface given by $z = f(x,y)$, the tangent plane at the point $(x_0,y_0,z_0)$ can be found using the partial derivatives of $f$.

$$z − z_0 ​= \frac{∂x}{∂f}​(x_0​,y_0​) · (x−x_0​) + \frac{∂y}{∂f}​(x_0​,y_0​)·(y−y_0​)$$

Example:

For a scalar field $z =x^2+y^2$:

• $\frac{∂f}{∂x} = 2x$
• $\frac{∂f}{∂y} = 2y$

The tangent plane at the point $(1,1,2)$ is:

• ⚪️ $z - 2 = 2·1·(x-1) + 2·1·(y-1)$
• ⚪️ $z = 2x + 2y - 2$

A normal line to a surface at a given point is a line that is perpendicular to the tangent plane at that point.

The direction of the normal line is given by the gradient vector of $f$, denoted as $∇f$.

The gradient vector $∇f$ is a vector consisting of the partial derivatives of $f$:

$$∇f = \left\langle \frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z} \right\rangle$$

The equation of the normal line at the point $r_0 = (x_0,y_0,z_0)$ is:

$$ℓ⃗_{f_0}(t) = r_0 + t·∇f_0$$ $$ℓ⃗_{f_0}(t) = r(x_0,y_0,z_0) + t·∇f(x_0,y_0,z_0)$$

Example:

For the field $z = x^2 + y^2$ at the point $(1,1,2)$, the gradient vector is:

• ⚪️ $\frac{∂f}{∂x} = 2x$
• ⚪️ $\frac{∂f}{∂y} = 2y$
• ⚪️ $\frac{∂f}{∂z} = -1$
  • 🔴 $∇f = ⟨ 2x, 2y, -1 ⟩$
  • 🟠 $∇f(1,1,2) = ⟨ 2, 2, -1 ⟩$
• 🟡 $ℓ⃗_{f_0}(t) = r(x_0,y_0,z_0) + t·∇f(x_0,y_0,z_0)$
• 🟢 $ℓ⃗_{f_0}(t) = (1,1,2) + t·⟨ 2, 2, -1 ⟩$
• 🟢 $ℓ⃗_{f_0}(t) = (1 + 2t, 1 + 2t, 2 - t)$

So, the equation of the normal line is:

A function $f(x,y,z)$ is differentiable at a point $(x_0,y_0,z_0)$ if it can be approximated by a linear function at that point.

Formally, $f$ is differentiable at $(x_0,y_0,z_0)$ if there exist constants $A$,$B$ and $C$ such that:

$$f(x,y,z) ≈ f(x_0,y_0,z_0) + A(x−x_0) + B(y−y_0) + C(z−z_0)$$

A function $f$ of several variables is differentiable at a point if it can be well-approximated by a linear function at that point.

Formally, $f$ is differentiable at $a⃗ = ⟨a_1,a_2,…,a_n⟩ if there exists a linear map $L$ such that:

$$\lim_{\mathbf{h} \to 0} \frac{f(\mathbf{a} + \mathbf{h}) - f(\mathbf{a}) - L(\mathbf{h})}{\|\mathbf{h}\|} = 0$$

The chain rule is used to differentiate composite functions.

If $z = f(x,y,z)$ where $x = f(t)$, $y = g(t)$ and $z = h(t)$, then:

$$\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}$$

Suppose $(F: \mathbb{R}^{n+m} \to \mathbb{R}^m )$ is continuously differentiable. Let $(\mathbf{z} = (\mathbf{x}, \mathbf{y}) \in \mathbb{R}^{n+m})$ with $(\mathbf{x} \in \mathbb{R}^n)$ and $(\mathbf{y} \in \mathbb{R}^m)$. Assume that:

1. $(F(\mathbf{a}, \mathbf{b}) = 0)$ for some $((\mathbf{a}, \mathbf{b}) \in \mathbb{R}^n \times \mathbb{R}^m)$.
2. The Jacobian matrix $(\frac{\partial F}{\partial \mathbf{y}})$ at $((\mathbf{a}, \mathbf{b}))$ is invertible.

Then, there exist neighborhoods $( U )$ of $(\mathbf{a})$ in $(\mathbb{R}^n)$ and $( V )$ of $(\mathbf{b})$ in $(\mathbb{R}^m)$, and a continuously differentiable function $( \mathbf{g}: U \to V )$ such that for every $(\mathbf{x} \in U)$:

$$ [ F(\mathbf{x}, \mathbf{g}(\mathbf{x})) = 0 ] $$

Additionally, the partial derivatives of $( \mathbf{g} )$ can be expressed as:

$$ [ \frac{\partial \mathbf{g}}{\partial \mathbf{x}} = - \left( \frac{\partial F}{\partial \mathbf{y}} \right)^{-1} \frac{\partial F}{\partial \mathbf{x}} ] $$

The directional derivative of a function $( f )$ at a point $( \mathbf{a} )$ in the direction of a vector $( \mathbf{u} )$ is the rate at which $( f )$ changes at $( \mathbf{a} )$ in the direction of $( \mathbf{u} )$.

It is given by:

$$ [ D_{\mathbf{u}} f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u} ] $$

where $( \nabla f(\mathbf{a}) )$ is the gradient vector of $( f )$ at $( \mathbf{a} )$.

The gradient vector of a scalar field $( f )$ is a vector that points in the direction of the greatest rate of increase of $( f )$ and whose magnitude is the rate of increase.

It is given by:

$$ \nabla f = \left\langle \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \ldots, \frac{\partial f}{\partial x_n} \right\rangle $$

The gradient vector of a scalar field $( f )$ is a vector that points in the direction of the greatest rate of increase of $( f )$ and whose magnitude is the rate of increase.

It is given by:

$$ \nabla f = \left\langle \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \ldots, \frac{\partial f}{\partial x_n} \right\rangle $$

1. Direction of Maximum Increase: The gradient vector points in the direction of the steepest ascent of the function.

2. Orthogonality: The gradient vector at any point is orthogonal (perpendicular) to the level surface passing through that point.

3. Rate of Increase: The magnitude of the gradient vector gives the rate of the steepest increase of the function.

Consider the scalar field $( f(x, y, z) = x^2 + y^2 + z^2 )$:

1. Compute the partial derivatives:

$$ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y, \quad \frac{\partial f}{\partial z} = 2z $$

2. The gradient vector is:

$$ \nabla f = \left\langle 2x, 2y, 2z \right\rangle $$

At the point $( (1, 1, 1) )$:

$$ \nabla f(1, 1, 1) = \left\langle 2 \cdot 1, 2 \cdot 1, 2 \cdot 1 \right\rangle = \left\langle 2, 2, 2 \right\rangle $$

This vector points in the direction of the greatest increase of ( f ) and its magnitude is:

$$ | \nabla f | = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3} $$

For a curve $(\mathbf{r}(t) = (x(t), y(t), z(t)))$, the tangent line at $( t = t_0 )$ is given by:

$$ \mathbf{T}(t) = \mathbf{r}(t_0) + (t - t_0) \mathbf{r}'(t_0) $$

where $(\mathbf{r}'(t_0))$ is the derivative of $(\mathbf{r}(t))$ at $( t = t_0 )$.

Example:

Consider the curve $(\mathbf{r}(t) = \langle t, t^2, t^3 \rangle)$. To find the tangent line at $( t = 1 )$:

1. Compute the derivative:

$$ \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle $$

2. At $( t = 1 )$: $$ \mathbf{r}'(1) = \langle 1, 2, 3 \rangle $$

3. The point on the curve at $( t = 1 )$ is: $$ \mathbf{r}(1) = \langle 1, 1, 1 \rangle $$

4. The equation of the tangent line is: $$ \mathbf{T}(t) = \langle 1, 1, 1 \rangle + (t - 1) \langle 1, 2, 3 \rangle $$

For a surface $( z = f(x, y) )$, the tangent plane at $( (x_0, y_0, z_0) )$ can be found using the gradient vector of $( f )$.

The equation of the tangent plane to the surface $( z = f(x, y) )$ at the point $( (x_0, y_0, z_0) )$ is given by:

$$ z - z_0 = \frac{\partial f}{\partial x} (x_0, y_0) (x - x_0) + \frac{\partial f}{\partial y} (x_0, y_0) (y - y_0) $$

Example:

Consider the surface $( z = x^2 + y^2 )$. To find the tangent plane at the point $( (1, 1, 2) )$:

1. Compute the partial derivatives:

$$ \frac{\partial f}{\partial x} = 2x, \quad \frac{\partial f}{\partial y} = 2y $$

2. At $( (1, 1) )$:

$$ \frac{\partial f}{\partial x} (1, 1) = 2, \quad \frac{\partial f}{\partial y} (1, 1) = 2 $$

3. The equation of the tangent plane is:

$$ z - 2 = 2(x - 1) + 2(y - 1) $$

4. Simplifying, we get:

$$ z = 2x + 2y - 2 $$

• The gradient vector provides the direction of steepest ascent for a function.
• The tangent line to a curve is found using the derivative of the vector function describing the curve.
• The tangent plane to a surface is found using the gradient vector of the scalar field defining the surface.

Unconstrained optimization involves finding the maximum or minimum values of a function $( f(x, y, \ldots) )$ without any restrictions on the variables.

• Local Extrema:

  • A function $( f )$ has a local maximum at $( \mathbf{a} )$ if $( f(\mathbf{a}) \geq f(\mathbf{x}) )$ for all $(\mathbf{x})$ near $(\mathbf{a})$.
  • A function $( f )$ has a local minimum at $( \mathbf{a} )$ if $( f(\mathbf{a}) \leq f(\mathbf{x}) )$ for all $(\mathbf{x})$ near $(\mathbf{a})$.

• Global Extrema: A function $( f )$ has a global maximum at $( \mathbf{a} )$ if $( f(\mathbf{a}) \geq f(\mathbf{x}) )$ for all $(\mathbf{x})$ in the domain. A function $( f )$ has a global minimum at $( \mathbf{a} )$ if $( f(\mathbf{a}) \leq f(\mathbf{x}) )$ for all $(\mathbf{x})$ in the domain.

Constrained optimization involves finding the extrema of a function subject to constraints, often given in the form of equations or inequalities.

• Lagrange Multipliers:
  • If we want to maximize or minimize $( f(x, y) )$ subject to the constraint $( g(x, y) = 0 )$, we use the method of Lagrange multipliers.
  • Define the Lagrangian function:

$$ \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda (g(x, y) - c) $$

• Solve the system of equations:

$$ \nabla \mathcal{L} = 0 $$

Critical points are points where the gradient of the function is zero or undefined.

• Finding Critical Points:
  • To find critical points of $( f(x, y, \ldots) )$:

$$ \nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \ldots \right\rangle = \mathbf{0} $$

• Classifying Critical Points:
  • Use the second partial derivative test. For a function $( f(x, y) )$:
  1. Compute the second partial derivatives to form the Hessian matrix $( H )$:

$$ H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{bmatrix} $$

2. Evaluate the determinant of $( H )$ at the critical point:

$$ D = f_{xx} f_{yy} - (f_{xy})^2 $$

3. Determine the nature of the critical point:
   • If $( D > 0 )$ and $( f_{xx} > 0 )$, local minimum.
   • If $( D > 0 )$ and $( f_{xx} < 0 )$, local maximum.
   • If $( D < 0 )$, saddle point.
   • If $( D = 0 )$, the test is inconclusive.

Optimization problems involve finding the best solution from all feasible solutions.

Example:

Maximize $( f(x, y) = xy )$ subject to $( x^2 + y^2 = 1 )$.

1. Define the Lagrangian:

$$ \mathcal{L}(x, y, \lambda) = xy + \lambda (1 - x^2 - y^2) $$

2. Compute the partial derivatives:

$$ \frac{\partial \mathcal{L}}{\partial x} = y - 2\lambda x = 0 $$

$$ \frac{\partial \mathcal{L}}{\partial y} = x - 2\lambda y = 0 $$

$$ \frac{\partial \mathcal{L}}{\partial \lambda} = 1 - x^2 - y^2 = 0 $$

3. Solve the system of equations to find the critical points and evaluate $( f )$ at these points to determine the maximum value.

This provides a comprehensive overview of the key concepts in optimization, critical points, and optimization problems. Let me know if you need further details or specific examples!

The double integral of a function $( f(x, y) )$ over a rectangular region $( R = [a, b] \times [c, d] )$ is defined as:

$$ \iint_R f(x, y) , dA = \int_a^b \int_c^d f(x, y) , dy , dx $$

For general regions, the double integral is given by:

$$ \iint_R f(x, y) , dA = \int_{x=a}^{x=b} \int_{y=g(x)}^{y=h(x)} f(x, y) , dy , dx $$

1. Linearity:

$$ \iint_R [af(x, y) + bg(x, y)] , dA = a \iint_R f(x, y) , dA + b \iint_R g(x, y) , dA $$

2. Additivity:
   • If $( R = R_1 \cup R_2 ) and ( R_1 \cap R_2 = \emptyset )$, then:

$$ \iint_R f(x, y) , dA = \iint_{R_1} f(x, y) , dA + \iint_{R_2} f(x, y) , dA $$

3. Non-negativity:
   • If $( f(x, y) \geq 0 )$ for all $( (x, y) \in R )$, then:

$$ \iint_R f(x, y) , dA \geq 0 $$

The volume $( V )$ under the surface $( z = f(x, y) )$ over the region $( R )$ is given by:

$$ V = \iint_R f(x, y) , dA $$

The average value $( f_{avg} )$ of a function $( f(x, y) )$ over the region $( R )$ is given by:

$$ f_{avg} = \frac{1}{\text{Area}(R)} \iint_R f(x, y) , dA $$

For a surface $( z = f(x, y) )$ over a region $( R )$, the surface area $( A )$ is given by:

$$ A = \iint_R \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} , dA $$

The triple integral of a function $( f(x, y, z) )$ over a region $( V )$ is defined as:

$$ \iiint_V f(x, y, z) , dV $$

The volume $( V )$ of a region $( E )$ is given by:

$$ V = \iiint_E 1 , dV $$

If the density of a solid region $( E )$ is given by $( \rho(x, y, z) )$, then the mass $( M )$ of the solid is:

$$ M = \iiint_E \rho(x, y, z) , dV $$

A vector field $( \mathbf{F} )$ in $( \mathbb{R}^3 )$ is a function that assigns a vector to each point in space:

$$ \mathbf{F}(x, y, z) = \langle P(x, y, z), Q(x, y, z), R(x, y, z) \rangle $$

The line integral of a scalar field $( f )$ along a curve $( C )$ parameterized by $( \mathbf{r}(t) )$ is:

$$ \int_C f , ds = \int_a^b f(\mathbf{r}(t)) |\mathbf{r}'(t)| , dt $$

The line integral of a vector field $( \mathbf{F} )$ along a curve $( C )$ parameterized by $( \mathbf{r}(t) )$ is:

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

• Work done by a force field:

$$ W = \int_C \mathbf{F} \cdot d\mathbf{r} $$

• Circulation and flux:

$$ \text{Circulation} = \int_C \mathbf{F} \cdot d\mathbf{r}, \quad \text{Flux} = \int_S \mathbf{F} \cdot d\mathbf{S} $$

If $( \mathbf{F} )$ is a conservative vector field, and $( C )$ is a curve from point $( A )$ to point $( B )$, then:

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

where $( \mathbf{F} = \nabla f )$.

For a vector field $( \mathbf{F} = \langle P, Q \rangle )$ in the plane:

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

For a vector field $( \mathbf{F} = \langle P, Q, R \rangle )$ in $( \mathbb{R}^3 )$:

$$ \iiint_V (\nabla \cdot \mathbf{F}) , dV = \iint_S \mathbf{F} \cdot d\mathbf{S} $$

For a vector field $( \mathbf{F} )$ and a surface $( S )$ with boundary curve $( \partial S )$:

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

A surface $( S )$ can be parameterized by $( \mathbf{r}(u, v) )$ where $( (u, v) )$ are parameters.

The surface integral of a scalar function $( f )$ over a surface $( S )$ parameterized by $( \mathbf{r}(u, v) )$ is:

$$ \iint_S f , dS = \iint_D f(\mathbf{r}(u, v)) |\mathbf{r}_u \times \mathbf{r}_v| , du , dv $$

The surface integral of a vector field $( \mathbf{F} )$ over a surface $( S )$ parameterized by $( \mathbf{r}(u, v) )$ is:

$$ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u, v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v) , du , dv $$