Line and surface integrals

This was probably my favorite chapter.

Types of Integrals

1. Scalar line integral along a curve $C$ given by $r(t)$ for $a \le t \le b$ (can be used to compute arc length, mass, electric potential):
   $$\int_C f(x,y,z) ds = \int^b_a f(r(t)) ||r'(t)|| dt$$
2. Vector line integral to calculate work along a curve $C$ given by $r(t)$ for $a \le t \le b$
   $$\int_C F \cdot dr = \int^b_a F(r(t)) \cdot r'(t)dt = \int_C F_1 dx + F_2dy + F_3 dz$$
3. Vector line integral to calculate flux across a curve $C$ given by $r(t)$ for $a \le t \le b$
   $$\int_C F \cdot n ds = \int^b_a F(r(t)) \cdot N(t) dt$$
4. Surface integral over a surface with parametrization $G(u,v)$ and parameter domain $D$ (can be used to calculate surface area, total charge, gravitational potential)
   $$\iint_S f(x,y,z) dS = \iint_D f(G(u,v)) ||N(u,v)|| dudv$$
5. Vector surface integral to calculate flux of a vector field $F$ across a surface $S$ with parameterization $G(u,v)$ and parameter domain $D$:
   $$\iint_S (F \cdot n) dS = \iint_S F \cdot dS = \iint_D F(G(u,v)) \cdot N(u,v)dudv$$

16.1 Vector Fields

Vector Field

A field to represent physical phenomena such as force fields, electric fields, magnetic fields and other large systems.

A vector field is typically represented in the form:
$$F(x,y,z) = \langle F_1(x,y,z), F_2(x,y,z), F_3(x,y,z) \rangle$$
To each point $P=(a,b,c)$ is associated the vector $F(a,b,c)$, which we also denote as $F(P)$ or $F=F_1 {\bf i} +F_2 {\bf j} + F_3 {\bf k}$

Example :: Finding vector at a point using a vector field:

Which vector corresponds to the point $P = (2,4,2)$ for the vector field $F(x,y,z) = \langle y-z, x, z- \sqrt{y} \rangle$?

1. Plug point $P$ into $F(x,y,z)$
   $$F(P) = F(2,4,2) = \langle (4)-(2), (2), (2)- \sqrt{(4)} \rangle = \langle 2, 2, 0 \rangle$$

Constant Vector Field

A sketch used to represent a complicated vector field as a uniform set of similar vectors

Unit Vector Field

A vector field, but all vectors are [[vectors#Unit Vector Normalizing a Vector|unit vector]] and represented through the equation $||F(P)||$

Radial Vector Field

If $F(P)$ is parallel to $\vec{OP}$ and $||F(P)||$ depends only on the distance $r$ from $P$ to the origin

Unit Radial Vector Field:
$$e_r = \Big \langle \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \Big \rangle = \Bigg \langle \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}} , \frac{z}{\sqrt{x^2 + y^2 + z^2}} \Bigg \rangle$$

Solving for 2D Unit Vector Field

Match each of the following planar vector fields with the corresponding plot. Note that the vectors are scaled to avoid overlap

$$F(x,y)= \langle \frac{-4y}{\sqrt{x^2 + y^2}}, \frac{4x}{\sqrt{x^2 + y^2}} \rangle$$

$$F(x,y)= \langle \frac{4x}{\sqrt{x^2 + y^2}}, \frac{4y}{\sqrt{x^2 + y^2}} \rangle$$

$$F(x,y)= \langle \frac{-4y}{\sqrt{x^2 + y^2}}, \frac{4y}{\sqrt{x^2 + y^2}} \rangle$$

$$F(x,y)= \langle \frac{4x}{\sqrt{x^2 + y^2}}, \frac{4x}{\sqrt{x^2 + y^2}} \rangle$$

Operations on Vector Fields

Three of the most important derivative operations in multivariable calculus are the [[multivariable functions#The Gradient|gradient]], divergence, and curl.

Divergence

Operations on a vector field using dot product
$$\text{div}(F) = \nabla \cdot F = \Bigg \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \Bigg \rangle \cdot \langle F_1, F_2, F_3 \rangle = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}$$
Divergence laws:

Curl

Operations on a vector field using cross product
$$\text{curl}(F) = \nabla \times F = \begin{bmatrix} {\bf i} & {\bf j} & {\bf k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ F_1 & F_2 & F_3 \end{bmatrix} =<br>\Bigg ( \frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\Bigg ){\bf i} - \Bigg (\frac{\partial F_3}{\partial x} - \frac{\partial F_1}{\partial z}\Bigg ){\bf j} + \Bigg ( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \Bigg )$$

Solving for Divergence and Curl

Calculate $\text{div}(F)$ and $\text{curl}(F)$
$F = \langle y, 2z, x \rangle$

1. Calculate the $\text{div}(F)$
   1. Solve for $\nabla$
      $$\nabla = \langle 0, 0, 0 \rangle$$
   2. Plug into the [[line and surface integrals#Divergence|divergence equation]]
      $$\text{div}(F) = \langle 0, 0, 0 \rangle \cdot \langle y, 2z, x \rangle = 0 + 0 + 0 = 0$$
2. Calculate the $\text{curl}(F)$
   1. plug in nabla solved for above into [[line and surface integrals#Curl|curl equation]]
3. $$\text{curl}(F) = \Bigg ( \frac{\partial }{\partial y}(x)-\frac{\partial}{\partial z}(2z)\Bigg ){\bf i} - \Bigg (\frac{\partial}{\partial x}(y) - \frac{\partial}{\partial z}(x)\Bigg ){\bf j} + \Bigg ( \frac{\partial}{\partial x}(2z) - \frac{\partial}{\partial y}(y) \Bigg ) = \langle -2, -1, -1\rangle$$

Let $F=\langle 5x-4zx^2, 4z-xy, 8z^2x^2 \rangle$. Calculate $div(F)$ and $curl(F)$

1. Solve $div(F)$
   1. Plug into the [[line and surface integrals#Divergence|divergence equation]]
      $$\text{div}(F) =\nabla F = \text{sum}\langle 5-4zx, -x, 16zx^2 \rangle = 5-4zx-x+16zx^2$$
2. Solve for $\text{curl}(F)$
   1. Plug in nabla solved for above into [[line and surface integrals#Curl|curl equation]]
      $$\text{curl}(F) = \nabla \times F = \begin{bmatrix} {\bf i} & {\bf j} & {\bf k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 5x-4zx^2 & 4z-xy & 8z^2x^2 \end{bmatrix}$$
      $$=\frac{\partial}{\partial y} (8z^2x^2) - \frac{\partial}{\partial z} (4z-xy) - \Bigg (\frac{\partial}{\partial x} (8z^2x^2) - \frac{\partial}{\partial z} (5x-4zx^2) \Bigg ) + \frac{\partial}{\partial x} (4z-xy) - \frac{\partial}{\partial y} (5x-4zx^2)$$
3. $$=(0 - 4)i - (16z^2x - (- 4x^2))j + (-y - 0)k = -4{\bf i} - (16z^2 x + 4x^2 ){\bf j} -y{\bf k}$$

Let $F= \langle 3x, 7x^2, 2 \rangle$ and $G=\langle 2, 7x, 3 \rangle$

1. Calculate $\text{curl}(F)$
   1. Evaluate using [[line and surface integrals#Curl|curl equation]]
      $$\text{curl}(F) = \nabla \times F = \begin{bmatrix} {\bf i} & {\bf j} & {\bf k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 3x & 7x^2 & 2 \end{bmatrix} = \langle 0, 0, 14x\rangle$$
2. Calculate $\text{curl}(G)$
   1. Evaluate using [[line and surface integrals#Curl|curl equation]]
      $$\text{curl}(G) = \nabla \times G = \begin{bmatrix} {\bf i} & {\bf j} & {\bf k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 2 & 7x & 3 \end{bmatrix} = \langle 0, 0, 7 \rangle$$
3. Calculate $\text{curl}(F + G)$
   1. Using the curl rule $Curl(F + G) = Curl(F) + Curl(G)$
      $$\text{curl}(F + G) = \langle 0, 0, 14x\rangle + \langle 0, 0, 7\rangle = \langle 0, 0, 14x + 7 \rangle$$

Conservative Vector Fields

A vector field F is conservative if there is a differentiable function $f(x,y,z)$ such that
$$F = \nabla f = \Big \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\Big \rangle$$

Example :: Finding the potential vector field function:

Consider the vector field $F = \langle 8xyz, 6x^2z, x^2yz \rangle$. Identify the potential function for the vector field F.

1. To check whhether the vector field $F = \langle F_1, F_2, F_3 \rangle$ is conservative, examine the cross-partials equation
   $$\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x} \quad \quad \frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y} \quad \quad \frac{\partial F_3}{\partial x} = \frac{\partial F_1}{\partial z}$$
2. Plug in values into cross-partials equation
   $$8xz \ne 12xz \quad \quad 6x^2 \ne x^2z \quad \quad 2xyz \ne 8xy$$
   If even just one didn’t equal the others we wouldn’t have a valid vectorfield

16.2 Line Integrals

Integrals can be separated into two categories, integrals over functions and integrals over vectorfields.

Scalar Line Integrals

$$\int_C f(x,y,z)ds = \lim_{[\Delta s_i]\rightarrow 0} \sum^N_{i=1}f(P_i)\Delta s_i$$

• This formula is stating how the line integral can be represented as a riemann sum

The line integral where $f(x,y,z)=1$ is equal to the length of the line shown below
$$\int_C 1ds = length(C)=\Delta s_i = \int^{t_i}<em>{t</em>{i-1}}||r'(t)||dt$$

Theorem 1: Computing a Scalar Line Integral

Let $r(t)$ be a parametrization that directly traverses $C$ for $a \le t \le b$. If $f(x,y,z)$ and $r'(t)$ are continuous, then
$$\int_C f(x,y,z)ds = \int^b_a f(r(t))||r'(t)||dt$$
$$ds=||r'(t)||dt \quad ||r'(t)||=\sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}$$

Computing Scalar Line Integral

Compute the integral of the scalar function $f(x,y,z)=3xy + 5z$ over $r(t)=\langle \cos(t), \sin(t), t \rangle$ for $0 \le t \le \pi$

1. Compute $ds$
   $$r'(t)=\langle -\sin(t), \cos(t), 1 \rangle$$
   $$ds = ||r'(t)||dt=\sqrt{(-\sin t)^2 + \cos^2t + 1}(dt)=\sqrt{2}dt$$
2. Write out the integrand and evaluate
   $$f(r(t)) = f(\cos t, \sin t, t) = 3\cos t \sin t + 5t$$
   $$f(x,y,z)ds = f(r(t)) ||r'(t)||dt = [3(\cos t)(\sin t) + (1)5 ]\sqrt{2}dt$$
3. $$\sqrt{2}\int^{\pi}_0 (3\cos t\sin t + 5t)dt = \sqrt{2} (-\frac{3\cos^2(t)}{2} + \frac{5t^2}{2})\Big |^\pi_0 = \frac{5\pi ^2}{\sqrt{2}}$$

Calculate the total mass of a metal tube in a helical shape $r(t)=\langle \cos t, \sin t, t^2 \rangle$ over $0 \le t \le 2\pi$ if the mass density is $\delta (x,y,z) = 8 \sqrt{z} \space g/cm$

1. Calculate $ds$
   $$ds=||r'(t)||dt =\sqrt{-(\sin t)^2 + (\cos t)^2 + (2t)^2} \space dt = \sqrt{4t^2 + 1}\space dt$$
2. Write out the integrand and evaluate
   $$\delta(x,y,z) = \delta(r(t))\sqrt{4t^2 + 1}\space dt = 8(\sqrt{t^2})\sqrt{4t^2 + 1}\space dt$$
   $$\int^{2\pi}_0 8t \sqrt{4t^2 + 1} \space dt = \:1334.854$$

Example :: Arc Length:

The rear wheel of a bicycle is following the path $\langle t, t^2 \rangle$ for $0 \le t \le 2$. The front wheel is always exactly 1 meter in front of the rear wheel.

1. Give a parametrization for the position of the front wheel of the tire.
   1. Find the direction of real wheel path
      1. $\frac{d}{dt}\langle t, t^2 \rangle = \langle 1, 2t \rangle$
   2. Find the unit vector of the directional vector
      1. $$||\langle 1, 2t \rangle|| = \langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}} \rangle$$
2. [C] Which wheel travels a longer distance? Set up arclength integrals for both paramtrization, and then use a computer to calculate the integrals and determine what distance each wheel travels.
   1. Arclength integral
      1. $$\text{Front Wheel =}\int^2_0 \sqrt{1^2 + (2t)^2}dt = 4.6467$$
      2. $$\text{Back Wheel =}\int^2_0 \sqrt{ \Big (\frac{d}{dt}( \frac{1}{\sqrt{1 + 4t^2}}) \Big )^2 + \Big (\frac{d}{dt}( \frac{2t}{\sqrt{1 + 4t^2}}) \Big )^2} = 5.13$$
3. Consider the function $f(x,y) = x^2 e^y - xy^2$, and calculate $\int_{C_1} \nabla f \cdot dr$, where $C_1$ is the path taken by the rear wheel.
   1. Find $f(r(t)) = t^2e^{t^2} - t^5$
   2. $f(r(2)) - f(r(0)) = 2^2e^{2^2} - 2^5 - e^0 = 4e^4 - 32 - 1 = 4e^4 - 33$

Vector Line Integrals

The difference between vector and scalar line integrals is that vector line integrals depend on the direction along the curve. This can be analgous to travelling up and down a mountain being very important to distinguish when regarding work.

Vector Line Integral Formula

The line integral of a vector field $F$ along an oriented curve $C$ is the integral of the tangential component of $F$.

$$\int_C (F \cdot T) \space ds = \int_C F \cdot dr\quad \quad dr = T \cdot ds$$

Orientation

The specified direction along curve $C$ is known as the orientation. Positive is movement in the chosen direction while negative is opposing said direction.

Theorem 2: Computing a Vector Line Integral

If $r(t)$ is a positively oriented regular paramtrization of an oriented curve $C$ for $a \le t \le b$, then
$$\int_C F \cdot dr = \int_C F \cdot Tds = \int_a^b F(r(t)) \cdot r'(t) dt$$
$$dr = r'(t)dt = \langle x'(t), y'(t), z'(t) \rangle \space dt$$

Example :: Computing Vector Line Integral:

Calculate the work done by the field $F(x,y,z) = \langle e^x, e^y, xyz \rangle$ when the object moves along the path $r=\langle t^2, t, \frac{t}{3} \rangle$ for $0 \le t \le 1$

1. Using [[line and surface integrals#Theorem 2 Computing a Vector Line Integral|Theorem 2]], we will follow and use the formula
   $$W=\int_C F(r(t)) \cdot r'(t)dt$$
2. Find the value of $F(r(t))$
   $$F(r(t)) = \langle e^{t^2}, e^{t}, (t^2) (t)(\frac{t}{3}) \rangle = \langle e^{t^2}, e^{t},\frac{t^4}{3} \rangle$$
3. Find the value of $r'(t)$
   $$r'(t) = \langle 2t, 1, \frac{1}{3} \rangle$$
4. Plug into the equation
   $$F \cdot dr = F(r(t)) \cdot r'(t) dt = \langle e^{t^2}, e^{t},\frac{t^4}{3} \rangle \cdot \langle 2t, 1, \frac{1}{3} \rangle dt= (2te^{t^2} + e^{t} +\frac{t^4}{9})dt$$
5. Set up integral
6. $$\int^1_0 (2te^{t^2} + e^{t} +\frac{t^4}{9})dt = 2e-\frac{89}{45}$$

Evaluate the line integral:
$$\int_C 17ydx + 16zdy + xdz, \quad r(t) = (2 + t^{-1}, t^3, t^2) \text{ for } 0 \le t \le 1$$

1. Using [[line and surface integrals#Theorem 2 Computing a Vector Line Integral|Theorem 2]], we will follow and use the formula
   $$\int_a^b F(r(t)) \cdot r'(t) dt$$
2. Find $r'(t)$
   $$r'(t) = ( -\frac{1}{t^2}, 3t^2, 2t)$$
3. Plug in $r(t)$ to find $F(r(t))$
   $$F = 17ydx + 16zdy + xdz = \langle 17y, 16z, x \rangle$$
   $$F(r(t)) = \langle 17(t^3), 16(t^2), (2 + t^{-1}) \rangle$$
4. Plug both $F(r(t))$ and $r'(t)$ into [[line and surface integrals#Theorem 2 Computing a Vector Line Integral|Theorem 2]]
   $$\int_a^b F(r(t)) \cdot r'(t) dt = \int_0^1 \langle 17t^3, 16t^2, 2 + t^{-1} \rangle \cdot \langle -\frac{1}{t^2}, 3t^2, 2t \rangle dt$$
   $$=\int_0^1 -17t + 48t^4 + 4t + 2 =\frac{51}{10}$$

Theorem 3: Properties of Vector Line Integrals

Let $C$ be a smooth oriented curve, and let $F$ and $G$ be vector fields
$$\text{i. Linearity: }\int_C (F+G) \space dr = \int_C F \space dr + \int_C G \space dr$$
$$\int_C kF \space dr = k \int_C F \space dr \quad \text{(k a constant)}$$
$$\text{ii. Reversing Orientation: } \int_{-C} F \space dr = -\int_C F \space dr$$
$\text{iii.}$ Additivity: If $C$ is a union of $n$ smooth curves $C_1, …, C_N$, then
$$\int_C F \space dr = \int_{C_1} F \space dr + … + \int_{C_N} F \space dr$$

Example :: Additivity Integration:

Calculate the line integral of $F(x,y,z) = (e^z, e^{x-y}, e^y )$ over the closed path ABCA, where $A=(2,0,0), B=(0,4,0), C=(0,0,6)$. Solve for $\int_C F \cdot dr$

1. To solve this problem we will use [[line and surface integrals#Theorem 3 Properties of Vector Line Integrals|additivity theorem]] to find the area under a smooth curve
   $$\int_C F \space dr = \int_{{\overline{AB}}} F \space dr + \int_{{\overline{BC}}} F \space dr + \int_{{\overline{CA}}} F \space dr$$
2. First compute $r(t)$ for $\overline{AB}$ .
   $$r(t) = B-A = (0, 4t, 0) - (-2 + 2t, 0,0) = \langle 2-2t, 4t, 0 \rangle$$
3. Set up integral of $\overline{AB}$ .
   $$\int_{{\overline{AB}}} F(r(t)) \cdot r'(t) = (e^0, e^{2-2t-4t}, e^{4t} ) \cdot (-2, 4, 0) = \int_0^1 -2 + 4e^{2-6t}$$
4. Compute $r(t)$ for $\overline{BC}$ .
   $$r(t) = C - B = (0,0,6t) - (0, -4 + t, 0) = (0, 4 - 4t, 6t)$$
5. Set up integral of $\overline{BC}$
   $$\int_{{\overline{BC}}} F \space dr = (e^{6t}, e^{0-4+4t}, e^{4-4t} ) \cdot (0, - 4, 6) = \int^1_0 -4e^{4t-4} + 6e^{4-4t} dt$$
6. Find $r(t)$ of $\overline{CA}$
   $$r(t) = A - C = (2t, 0, 0) - (0 ,0 ,-6 + 6t) = (2t, 0, 6-6t)$$
7. Set up integral for $\overline{CA}$
   $$\int_{{\overline{CA}}} F \space dr = (e^{6-6t}, e^{2t-0}, e^0 ) \cdot (2, 0, -6) = \int^1_0 2e^{6-6t} -6 dt$$
8. Sum all the integrals together
   $$\int_C F \space dr = \int_0^1 -2 + 4e^{2-2t}dt + \int^1_0 -4e^{4t-4} + 6e^{4-4t}dt + \int^1_0 2e^{6-6t} -6 dt$$
   $$\int_0^1 ( 4e^{2-2t} -4e^{4t-4} + 6e^{4-4t}dt + 2e^{6-6t} -8 )dt$$
   $$= -\frac{65}{6} + \frac{1}{3}e^{-4} + \frac{2}{3}e^2 + \frac{3}{2}e^4 + \frac{1}{3}e^6$$

Work Equation

Work refers to the energy expended when a force is applide to an object as it moves along a path.
$$W = \text{tangential component of F} \times \text{distance}=(||F||\cos\theta) \times ||\overrightarrow{PQ}||$$

Example :: Integration for total charge:

Find the total charge on the curve $y=x^{4/3}$ for $8 \le x \le 27$ assuming a charge density of $\delta (x,y) = \frac{x}{y}$

1. We can equate
   $$y = x^{4/3}, \quad r(t) = \langle t, t^{4/3}, 0 \rangle \quad 8 \le t \le 27$$
   $$r'(t) = \langle 1, \frac{4}{3}t^{1/3}, 0 \rangle$$
2. Solve for ds
   $$ds = ||r'(t)||dt = \sqrt{1^2 + \left( \frac{4x^{(1/3)}}{3}\right)^2 + 0^2 } = \sqrt{1^2 + \frac{16}{9}t^{2/3}}dt$$
   $$u = \sqrt{1^2 + \frac{16}{9}t^{2/3}}, du = \frac{32}{27}t^{-1/3}dt$$
   $$u(8) = \frac{73}{9} \quad u(27) = 17$$
3. Set up integral
   $$\int^{27}<em>8 t^{-1/3} \sqrt{1 + \frac{16}{9}t^{2/3}}dt = \frac{27}{32 \int^{17}</em>{73/9}} \sqrt{u}du = 26.43$$

Example :: Integration for work done by a field:

Calculate the work done by the field F when the object moves along the given path from the initial ponit to the final point.
$$F(x,y,z) = \langle x,y,z \rangle, \quad r(t) = \langle \cos t, \sin t, t \rangle \quad \frac{\pi}{2} \le t \le \frac{7\pi}{2}$$

1. Using the [[line and surface integrals#Theorem 2 Computing a Vector Line Integral|theorem 2 formula]], we can find the answer
2. We find $r'(t)$
   $$r'(t) = \langle -\sin t, \cos t, 1 \rangle$$
3. Find $F(r(t))$
   $$F(r(t)) = \langle \cos t, \sin t, t \rangle$$
4. Plug it into the formula
   $$\int_C = \langle \cos t, \sin t, t \rangle \cdot \langle -\sin t, \cos t, 1 \rangle = \int^{7\pi/2}_{\pi/2} (-\cos t \sin t + \sin t \cos t + t)dt = 6\pi ^2$$

16.3 Conservative Vector Fields

A conservative vector field is one that satisfies the equation
$$\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x} \quad \quad \frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y} \quad \quad \frac{\partial F_3}{\partial x} = \frac{\partial F_1}{\partial z}$$

• The circulation around a closed path is denoted as $\oint_C F \cdot dr$

Path Independence

Theorem indicating that the line integral of $F$ along a path from $P$ to $Q$ depends only on the endpoints $P$ and $Q$

Theorem: Fundamental Theorem for Conservative Vector Fields

Assume that $F = \nabla f$ on a domain $D$

1. If $r$ is a path along a curve $C$ from $P$ to $Q$ in $D$, then
   $$\int_C F \cdot dr = f(Q) - f(P)$$
   where $F$ is path independent
2. In circulation around a closed curve $C$, where $P=Q$, we get 0
   $$\oint_C F \cdot dr = 0$$

Example :: Fundamental Theorem for Conservative Vector Field

Let $F(x,y,z) = \langle 2xy + z, x^2, x \rangle$

1. Verify that $f(x,y,z) = x^2 y + xz$ is a potential function for $F$
   1. Find the partial derivative of $f(x,y,z)$
      1. $$\text{Potential function} = \nabla f = \langle 2xy + z, x^2, x \rangle$$
2. Evaluate $\int_C F \cdot dr$, where $C$ is a curve from $P = (1, -1, 2)$ to $Q = (2, 2, 3)$
   1. Use the [[line and surface integrals#Fundamental Theorem for Conservative Vector Fields|fundamental theorem for conservative vector fields]] to find $\int_C F \cdot dr$
   2. $$\int_C F \cdot dr = f(Q) - f(P) = f(2,2,3) - f(1, -2, 2)$$

Let $f(x,y) = 7x \cos(y)$. Find the conservative vector field $F$, which is the gradient of f.

1. Find the gradient of $f(x,y)$
   1. $$\nabla f = F = \langle 7 \cos (y) , - 7x \sin (y)\rangle$$
      Evaluate the line integral of $F$ over the upper half of the unit circle centered at the origin, oriented clockwise
2. Since the orgin is oriented clockwise we get $P=(-1, 0)$ and $Q = (1, 0)$

$$\int_C F \cdot dr = f(Q) - f(P) = 7(1)\cos(0) - 7(-1)\cos (0) = 7 + 7 = 14$$

Let $f(x,y,z) = 3y + 3z\ln(x)$. Find the conservative vector field $F$, which is the gradient of $f$.

1. $$F = \nabla f = \langle \frac{3z}{x}, 3, 3\ln(x) \rangle$$
   Evaluate the line integral of $F$ over the circle $(x - 2)^2 + y^2 = 1$ in the counter clockwise direction. Find $\int_C F \cdot dr = ?$
2. Since the line is a circle, we know that the integral is equal to zero

A vector field $F$ and contour lines of a potential function for $F$ are shown in the figure

Calculate the common value of $\int_C F \cdot dr$ for the curves in the direction from $P$ to $Q$, where $\int_C F \cdot dr = ?$

1. Find $f(Q)$ and $f(P)$
   1. $f(Q) = 22 + 3.5 = 25.5$
   2. $f(P) = 1 + 3.5 = 4.5$
   3. $$\int_C F \cdot dr = f(Q) - f(P) = 25.5 - 4.5 = 21$$

Calculate the following line integrals using the fundamental theorem of line integrals; be sure to check that the vector fields are conservative

1. $\int_C F \cdot dr$, where $F(x,y) = \langle 4xye^{x^2}, 2e^{x^2} -3y^2 \rangle$ where $C$ is the path along $(5t - 4t^2 + t^3, 3 + 3t - 2t^2)$ for $0 \le t \le 2$
   1. Calculate the [[line and surface integrals#Curl|curl]] of $F$
      1. $$Curl(F) = \begin{bmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \ 4xye^{x^2} & 2e^{x^2} - 3y^2\end{bmatrix} = x4e^{x^2} - x4e^{x^2} = 0$$
      2. $t=0: 5(0) - 4(0)^2 + (0)^3, 3 + 3(0) - 2(0)^2) = (0,3)$
      3. $t=3: 5(2) - 4(2)^2 + (2)^3, 3 + 3(2) - 2(2)^2) = (2, 1)$
   2. Find $\nabla F$
      1. $\nabla F = (8xye^{x^2}, -6y)$
   3. Plug into equation
      1. $F(0, 3) = 2(0)e^{(0)^2} - ()^3o = -25$
      2. $F(2,1) = 2(1)e^{(2)^2} - (1)^3 = 2e^4 - 5$
      3. $F(2, 1) - F(3, 0) = 2e^4 - 5 - (-25) = 2e^4 + 20$
      4. ANS: $2e^4 + 20$
2. $\int_C (2xy-3z, x^2 + 8y^3 z, 2y^4 - 3x)$ where $C$ is the path $(\cos (\pi t), 3t^2 - 5t, t \sin^2 (\pi t/4)$ from $(1, 0,0)$ to $(1,2,2)$
3. $\oint_C A \cdot dr$, where $A(x,y,z) = \langle \frac{y^2}{z^2 + 1}, \frac{2xy}{z^2 + 1}, \frac{-2xy^2z}{(z^2 + 1)^2} \rangle$ and $C$ is the path along the ellipse $4x^2 + 3xy + y^2 = 10$ oriented counter-clockwise
   1. Since $z^2 + 1$ in each component enables the shape of $A$ to be a circle, we get the answer of $0$
   2. ANS: 0

Theorem 2

A vector field $F$ on an open connected domain $D$ is path independent if and only if it is conservative

Potential Function

Consider the vector field $F = \langle y, x, z^3 \rangle$. Choose a potential function for $F = \langle y, x, z^3 \rangle$ if it exists.

1. Identify whether F is conservative. So, determine whether the field $F = \langle y, x, z^3 \rangle$ satisfies the cross-partials condition.
   1. $\frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y}(y) = 1$
   2. $\frac{\partial F_1}{\partial z} = \frac{\partial}{\partial z}(y) = 0$
   3. $\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}(x) = 1$
   4. $\frac{\partial F_1}{\partial z} = \frac{\partial}{\partial z}(x) = 0$
   5. $\frac{\partial F_1}{\partial x} = \frac{\partial}{\partial x}(z^3) = 0$
   6. $\frac{\partial F_1}{\partial y} = \frac{\partial}{\partial y}(z^3) = 0$
2. Now compare values to check
   1. $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x} = 1$
   2. $\frac{\partial F_2}{\partial z} = \frac{\partial F_3}{\partial y} = 0$
   3. $\frac{\partial F_3}{\partial x} = \frac{\partial F_1}{\partial z} = 0$
3. Thus, $F$ satisfies the cross-partial condition everywhere. Hence, $F$ is conservative. If$f(x,y,z)$ is a potential function, then it satisfies the following equation
   1. $f(x,y,z) = \int ydx = xy + f(y,z)$
   2. $f(x,y,z) = \int xdy = xy + g(x,z)$
   3. $f(x,y,z) = \int z^3 dz = \frac{1}{4}z^4 + h(x,y)$
4. These three ways of writing $f(x,y,z)$ must be equal so
   1. $xy + f(x,y) = xy + g(x,z) = \frac{z^4}{4} + h(x,y)$ for any constant $C$
5. Therefore, there are two potential functions for $F$:
   1. $f(x,y,z) = xy + \frac{z^4}{4} + 2\pi$
   2. $f(x,y,z) = xy + \frac{z^4}{4} - 67$

Evaluate $\int_C 2xyzdx + x^2 z dy + x^2 y dz$ over the path $r(t) = \Big( \frac{t^2}{6}, \sin(\frac{\pi t}{2}), e^{t^2 - 6t} \Big )$ over the path $0 \le t \le 6$

1. Find the points $P$ and $Q$
   1. $P = r(0) = \Big( \frac{(0)^2}{6}, \sin(\frac{\pi (0)}{2}), e^{(0)^2 - 6(0)} \Big ) = r(t) = \Big( 0, 0, 1\Big )$
   2. $Q = r(6) = \Big( \frac{(6)^2}{6}, \sin(\frac{\pi (6)}{2}), e^{(6)^2 - 6(6)} \Big ) = r(t) = \Big( 6, 0, 1 \Big )$
2. Find the gradient of
   1. $f = \langle 2xyz, x^2z, x^2y \rangle$
   2. $\nabla f = F = \langle 2yz, 0, 0 \rangle$
3. Integrate
   1. $\int_C F \cdot dr = f(Q) - f(P)$
   2. $f(P) = 2(0)(0)(1)dx + (0)^2 (1) dy + (0)^2 (0) dz = 0$
   3. $f(Q) = 2(6)(0)(1)dx + (6)^2 (1) dy + (6)^2 (0) dz = 36$