Calculus 2

$\text{Chapter 9: Paramatric equation and polar coordinates}$

Parametric curves: $(x,y)=(f(t),g(t))$

• $x=f(t),y=f(t)$ for $a<t<b$

  • Then there are initial point $(f(a),g(a))$ and terminal point $(f(b),g(b))$
  • We can graph $x=g(t)$ and $y=t$ by graphing $x=g(y)$
    • eg. $x={\sin }^{2}t,y=\sin t⟹x=y^2$

• Cycloid:

  • Let the parameter be $\theta$
  • $|OT|=\text{arc}PT=r.\theta$

  $⟹x=|OT|-|PQ|=r.\theta -r.\sin \theta$

  $⟹y=|CT|-|CQ|=r-r.\cos \theta$

Calculus with parametric curves:

• Tangent:

  • By chain rule, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ when $dx/dt\ne 0$
  • $\frac{d^{2}y}{d{x}^2}=\frac{d(\frac{dy}{dx})}{dx}=\frac{d(dy/dx)/dt}{dx/dt}$

• Area under graph:

  • $y=g(t)$ and $dx=f^{\prime }(t)dt$
  • $a$ is $t$ of the leftmost point which has minimum $x$

• Arc length:
  • Curve C is traversed exactly once as $t$ increases from $\alpha$ to $\beta$, then arc length is
    • If $t$ traverses twice then $L$ twice times the actual arc length

$x=f(t),y=g(t),\alpha \le t\le \beta$ where $f^{\prime }(t)$ and $g^{\prime }(t)$ are continuous on $[\alpha ,\beta ]$,

Polar coordinates: $(r,\theta )$

• $O$, The pole is the origin
• The ray(half-line) is drawn $\theta$ angle horizontally corresponds to the positive $x$-axis
  • $r$, the distance from $O$ can be negative
  • $\theta$ is the angle

• Polar curve:
  • a collection of points satisfying $r=f(\theta )$
    • $x=r.\cos \theta$
    • $y=r.\sin \theta$
    • $r^2=x^2+y^2$
    • $\tan \theta =\frac{y}{x}$

• Tangent to polar curve:
  • Find the point where tangent is horizontal $⟹\frac{dy}{d\theta }=0$
    • Put $\theta$ and only take points where $r>0$

Area and lengths in polar coordinates:

• Area under graph:
  • Find area between 2 curves:
    • Find the intersection first, which are the limits
    • or cut the shape to parts in which each part is area under graph of 1 curve

$A={\int }_a^b\frac{1}{2}{r}^{2}d\theta ={\int }_a^b\frac{1}{2}{f}^2(\theta )d\theta$

$A=\frac{1}{2}[{\int }_a^b{f}^2(\theta )d\theta -{\int }_a^b{g}^2(\theta )d\theta ]$

• Arc length:
  • Distance traveled is difference

$L={\int }_a^b\sqrt{r^2+(\frac{dr}{d\theta }{)}^2}d\theta$

$\text{Chapter 10: Vectors and the geometry space}$

3D coordinates systems:

• Equation of a sphere: $(x-h{)}^2+(y-k{)}^2+(z-l{)}^2=r^2$ where the centre is $C(h,k,l)$

Vectors:

The Dot Product:

$⁍$

• $\theta$ is the angle between the 2 vectors both pointing toward the intersection
  • Find $\angle ABC⟹\overrightarrow{AB}\cdot \overrightarrow{CB}$, both vectors are pointing toward B
  • $\pi -\theta$ to find the angle if 1 vector is pointing toward and the other is pointing away
  • Or change direction of 1 vector by inverting it or
• $a\cdot b=|a|.|b|.Cos90=0$ when 2 vectors are perpendicular
• $a\cdot b=|a|.|b|.Cos0=|a{|}^2$ when 2 vectors are parallel and have same direction
• $a\cdot b\cdot c$ has no meaning

The Cross Product:

• Vector determinant:
  • results to another vector

$n=a\times b=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_i& a_j& a_k\\ b_i& b_j& b_k\end{array}\right|;\text{note that}-\mathbf{j}$

• if $\theta$ is the angle between $a$ and $b$ then: $|a\times b|=|a|.|b|.\sin \theta$
• Right-hand rule:
  • If the fingers of your right hand curl in the direction of a rotation (through an angle less than 180) from $a$ ****to $b$, then your thumb points in the direction of $a\times b$

• Properties:
  • $a\times a=0$ for all $a\in V_3$
  • The vector $a\times b$ is orthogonal, common perpendicular, to both $a$ and $b$
  • Two non-zero vectors $a$ and $b$ are parallel if and only if $a\times b=0$

• The area of parallelogram is $|a\times b|=|a|.(|b|.\sin \theta )$

• Triple products (scalar triple product):

  $a.(b\times c)=a\left|\begin{array}{ccc}{a}_i& a_j& a_k\\ b_i& b_j& b_k\\ c_i& c_j& c_k\end{array}\right|$

  • Volume of parallelepiped:
    • $$\begin{gathered} h=|a|.\cos \theta \\ ⟹V=|b\times c|.|a|.|\cos \theta | \\ V=|a.|b\times c|| \end{gathered}$$

Equation of lines and planes:

• Find the line intersection of 2 planes:
  • Let one variable be free, $(z=t)$
  • Substitute in others 2 equations $(x=f(t),y=g(t))$
  • Split it to equation in term $r=(a,b,c)+\lambda (x,y,z)$

Cylinders and quadric surfaces:

• Cylinders
  • Surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given plane curve

$z=x^2$ to $(x,y,x^2)$

$y^2+z^2=1$ to $(x,y^2,1-y^2)$

• Quadric surface
  • A second-degree equation in three variables x, y, and z
  • $A{x}^2+B{y}^2+C{z}^2+J=0\text{or}A{x}^2+B{y}^2+Iz=0$
  • Draw:

    • Let $z=k$, we have $x^2+(\frac{y}{3}{)}^2=1-\frac{k^2}{4};(k^2<4)$

      which is many level curves of ellipse on $x-y$ plane at $(0,0)$ with 3 units in $y$-direction

    • Let $y=k$ then $x=k$

$x^2+\frac{y^2}{9}+\frac{z^2}{4}=1$

Vector functions and space curves:

• Functions $fgh(t)$ are parametric equations

• As $t$ varies throughout the interval, it forms a space curve

  $\mathbf{r}(t)=⟨f(t),g(t),h(t)⟩,$ then $\underset{t\to a}{\lim }\mathbf{r}(t)=⟨\underset{t\to a}{\lim }f(t),\underset{t\to a}{\lim }g(t),\underset{t\to a}{\lim }h(t)⟩$

  Provides limits of the component functions exist

• A vector function $\mathbf{r}$ is continuous at $a$ if $\underset{t\to a}{\lim }=\mathbf{r}(a)$

A helix $\cos (t)+\sin (t)+t$

• Sketch:

  • $⟨t,2-t,2t⟩⟹(0,2,0)+t(1,-1,2)$ thus a line
  • $⟨\cos (t),\sin (t),t⟩⟹$

• Derivatives:

  $$\begin{gathered} \frac{d\mathbf{r}}{d\theta }={\mathbf{r}}^{\prime }(t)=\underset{h\to 0}{\lim }\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h} \\ =f^{\prime }(t)i+g^{\prime }(t)j+h^{\prime }(t)k \end{gathered}$$

  • Tangent vector of $\mathbf{r}={\mathbf{r}}^{\prime }(t)$

• Integral:

${\int }_a^b\mathbf{r}(t)dt=({\int }_a^{b}f(t)dt)i+({\int }_a^{b}g(t)dt)j+({\int }_a^{b}h(t)dt)k+C$

• $C$ is an arbitrary constant vector

Arc length and curvature:

• Arc length:

  $L={\int }_a^b|r^{\prime }(t)|dt={\int }_a^b\sqrt{(\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2+(\frac{dz}{dt}{)}^2}dt$

  • Note that $a$ and $b$ are from $t$
  • Distance traveled is difference

• Parametrizations:

  • A single curve $C$ can be represented by more than one vector functions, they are parametrizations of curve $C$

    

  • ex: $r_1(t)=⟨t,t^2,t^3⟩,1\le t\le 2$ is the same as $r_2(u)=⟨e^u,e^{2u},e^{3u}⟩,0\le u\le \ln 2$

    • $t=e^u$

  • Thus, arc length is independent of the parametrization that is used

  • Parametrize a curve with respect to arc length

    • $r(t(s))$: to find the coordinates knowing the arc length

    1. Find $s(t)=L$, the arc length function
    2. $s=s(t)={\int }_0^t|{\mathbf{r}}^{\prime }(u)|du$
    3. substitute $t$ as function of $s$ in

• Curvature:

  • A parametrization $\mathbf{r}(t)$ is smooth on an interval I if $\mathbf{r}$ is continuous and $\mathbf{r}(t)\ne 0$ in I

  • Unit tangent vector:

  $\mathbf{T}(t)=\frac{{\mathbf{r}}^{\prime }(t)}{|{\mathbf{r}}^{\prime }(t)|}$

  • Curvature of the curve:

  $\kappa =\left|\frac{d\mathbf{T}}{ds}\right|=\left|\frac{d\mathbf{T}/dt}{ds/dt}\right|=\frac{|{\mathbf{T}}^{\prime }(t)|}{|{\mathbf{r}}^{\prime }(t)|}$

  $\kappa =\frac{|{\mathbf{r}}^{\prime }(t)\times {\mathbf{r}}^{\prime \prime }(t)|}{|{\mathbf{r}}^{\prime }(t){|}^3}$ use this

  • For explicit equations: $y=f(x)$

  $\kappa =\frac{|f^{\prime \prime }(x)|}{[1+(f^{\prime }(x){)}^2{]}^{3/2}}$

• Normal and Binormal vectors:

  • Principal unit normal vector: $\mathbf{N}(t)=\frac{{\mathbf{T}}^{\prime }(t)}{|\mathbf{T}(t)|}$
    • Indication of which direction the curve is turning
  • Binormal vectors: $\mathbf{B}(t)=\mathbf{T}(t)\times \mathbf{N}(t)$
    • An unit vector, perpendicular to both $\mathbf{T}$ and $\mathbf{N}$

Chapter 11: Partial Derivative

Chapter 12: Multiple integral

$\text{Chapter 13:Vector calculus}$

Vector fields:

• Vector fields: $\mathbf{F}(\mathbf{x})=f(\mathbf{x})i+g(\mathbf{x})j+h(\mathbf{x})k$
  • $f(x,y,z)$ is a component function
  • For every point $({\mathbf{x}}_1)$ becomes $(f(\mathbf{x})i,g(\mathbf{x})j,h(\mathbf{x})k)$ then $\mathbf{F}(\mathbf{x})$ is a vector field

• Gradient fields: $\nabla f(\mathbf{x})=f_x(\mathbf{x})i+g_y(\mathbf{x})j+h_z(\mathbf{x})k$
  • Forms a gradient vector field
  • A vector field $\mathbf{F}$ is called a conservative vector field if there exists a function $f$ **such that $\mathbf{F}=\nabla f$
    • $f$ is a potential function of $\mathbf{F}$
    • Any path $C$ between two point have the same ${\int }_{C}F\cdot d\mathbf{r}$ (like work done by gravity), a fluid flows back to where it was with same speed

Line integrals:

• Integral over a curve where x=x(t),y=y(t), a\le t\le b$$x=x(t),y=y(t), a<t<b

  • Equivalent to $\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}$

• $\color{tomato}\displaystyle \int_C f(x,y)ds =\int^b_a f\big(x(t),y(t)\big)\sqrt{

\bigg( \frac{dx}{dt}\bigg)^2+ \bigg( \frac{dy}{dt}\bigg)^2 }

dt$-Use$x=\cos t,y=\sin t$ for circle

• $\int f(x,y)ds={\int }_{C1}f(x,y)ds+{\int }_{C2}f(x,y)ds+...+{\int }_{Cn}f(x,y)ds$
  • Define domain of $t$ for each piece

• Applications are similar to double and triple integral

• Line integral with respect to $x$ or $y$:

  • Vector representation of a line starts at ${\mathbf{r}}_0$ ends at ${\mathbf{r}}_1$:
    • $\mathbf{r}(t)=(1-t){\mathbf{r}}_0+t{\mathbf{r}}_1$ for $0\le t\le 1$
  • $dx=x^{\prime }(t)\to {\int }_{C}f(x,y)dx={\int }_a^{b}f(x(t),y(t)).x^{\prime }(t)dt$
  • $dy=y^{\prime }(t)\to {\int }_{C}f(x,y)dy={\int }_a^{b}f(x(t),y(t)).y^{\prime }(t)dt$
  • ${\int }_{-C}f(x,y)dx=-{\int }_{C}f(x,y)dx$

• ***Line integrals in space: ${\int }_C\mathbf{F}\cdot d\mathbf{r}$

  • $\displaystyle {\color{tomato}\int_C f(x,y,z)ds} =\int^b_a f\big(x(t),y(t),z(t)\big).\sqrt{

  \bigg( \frac{dx}{dt}\bigg)^2+ \bigg( \frac{dy}{dt}\bigg)^2 + \bigg( \frac{dz}{dt}\bigg)^2 }

  dt\ \color{tomato}=\int^b_a f\big(\bold r(t)\big) \ |\bold r’(t)| dt$

  • Line integral with respect to $z$:
    • $dz=z^{\prime }(t)\to {\int }_{C}f(x,y,z)dz={\int }_a^{b}f(x(t),y(t),z(t)).z^{\prime }(t)dt$

• ***Line integrals of vector fields:

  • Work done on moving an object on a curve in 3D space in a vector field (force/electric) is:
    • $W={\int }_C\mathbf{F}.d\mathbf{r}={\int }_C\mathbf{F}\cdot \mathbf{T}ds={\int }_a^b\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)dt$ (dot)
      • where $\mathbf{T}(x,y,z)=\frac{{\mathbf{r}}^{\prime }(t)}{|{\mathbf{r}}^{\prime }(t)|}$ is unit tangent vector
      • $\mathbf{F}(\mathbf{r}(t))$ means $\mathbf{F}(x(t),y(t),z(t))$

The fundamental theorem for line integrals:

• Net change theorem for line integral:
  • Let $C$ be smooth curve by vector function $\mathbf{r}(t)$ and gradient vector $\Delta f$ continuous on $C$.
    • Then ${\int }_C\Delta f\cdot d\mathbf{r}=f(\mathbf{r}(b))-f(\mathbf{r}(a))$, true for 2 and 3 variables

• Independence of path:
  • Line integrals of conservative vector fields are independent of path, ${\int }_{C_1}F\cdot d\mathbf{r}={\int }_{C_2}F\cdot d\mathbf{r}$ if $C_1$ and $C_2$ have the same initial and terminal points
  • Closed curve: ${\mathbf{r}}_1(a)={\mathbf{r}}_2(b)$ have ${\int }_{C}F\cdot d\mathbf{r}=0$

${\int }_{C_1}F\cdot d\mathbf{r}=-{\int }_{-C_2}F\cdot d\mathbf{r}$

• If $\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}$ is a conservative vector field, where $P$ and $Q$ have continuous first-order $\partial$ then though out $D$ we have $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$
• Let $\mathbf{F}(x,y)=P\mathbf{i}+Q\mathbf{j}$ be a vector field on an open simple-connected region $D$. Suppose $P$ and $Q$ have continuous first-order $\partial$ and $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ though out $D$ then $\mathbf{F}$ is conservative
• To test for conservative:
  • $\mathbf{F}=\Delta f=<\underset{M}{\underbrace{\frac{\partial f}{\partial x}}},\underset{P}{\underbrace{\frac{\partial f}{\partial y}}},\underset{Q}{\underbrace{\frac{\partial f}{\partial z}}}>$
  • Take pairs and test if $\frac{\partial M}{\partial y}=\frac{\partial P}{\partial x},\frac{\partial M}{\partial z}=\frac{\partial Q}{\partial x},\frac{\partial P}{\partial z}=\frac{\partial Q}{\partial y}$

• Find potential function:
  • Integrate $f_x(x,y,z)$ with respect to $x$ to have $f(x,y,z)=\int f_x(x,y,z)dx+g(y,z)$
    • $g(y,z)$ as arbitrary constant of integration
  • Compare $f_y(x,y,z)$ with $y$ partial derive of found $f(x,y,z)$ to have $g_y(y,z)$
  • Integrate $g_y(y,z)$ with respeect to $y$ to have $\int g_y(y,z)dy+h(z)$
    • $h(z)$ as arbitraty constant
  • Differentiate found $f$ with respect to $z$ to compare with $f_z$
• Conservation of energy:
  • Work done by the force field along $C$ is equal to the change in kinetic energy at the endpoints of $C$, $W=K(B)-K(A)$

Green’s theorem: (2d version of stoke)

• Relates a double integral over a plane region $D$ to a line integral around its plane boundary curve.
• Positive orientation refers to counter clockwise traversal of $C$
• Let $C$ be a positively oriented, piecewise-smooth, simple closed curve in the plane. If $P(\mathbf{F}i)$ and $Q$ have continuous partial derivatives on an open region that contains $D$
  • ${\oint }_C\mathbf{F}\cdot d\mathbf{r}={\oint }_{C}Pdx+Qdy=\int {\int }_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$
    • If $P(x,y)=Q(x,y)=0\to {\oint }_{C}Pdx+Qdy=0$
  • The curl along the line equals to total curl on the graph because inside, curls cancels out
  • Green’s theorem also work with regions with holes

• Since the area of $D=A=\int {\int }_{D}1dA$, we choose $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1$, then:
  • $A={\oint }_{C}xdy=-{\oint }_{C}ydx=\frac{1}{2}{\oint }_{C}xdy-ydx$

Curl and divergence:

• Curl: $\text{curl }\mathbf{F}=\nabla \times \mathbf{F}$

  • $\mathbf{F}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k}$, a vector field and the partial derivatives all exist

    • $\displaystyle \text {curl } \bold F=

    \bigg( \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z} \bigg)\bold i+ \bigg( \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x} \bigg)\bold j+ \bigg( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \bigg)\bold k$

  • Vector, represents tendency to spin around an axis with magnitude measures the strength of the rotation and direction indicates the axis of rotation

  • If a function $f$ of 3 variables that has continuous second-order partial derivatives, then $\text{curl }(\nabla f)=0$

    • Since $\mathbf{F}=\nabla f$ is conservative vector field, then if $\mathbf{F}$ is conservative then $\text{curl }\mathbf{F}=0$
    • $\text{curl }\mathbf{F}=0$ then $\mathbf{F}$ is conservative if $\mathbf{F}$ is simply connected

$\nabla =\{\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}\}$ treated as a variable

• Divergent: $\text{div }\mathbf{F}=\nabla \cdot \mathbf{F}$
  • If $\mathbf{F}=P\mathbf{i}+Q\mathbf{j}+R\mathbf{k}$, a vector field and the partial derivatives of $P,Q$ and $R$ **all exist, then the div of $\mathbf{F}$ ****is the vector field on 3 defined by $\text{curl }\mathbf{F}=\nabla \times \mathbf{F}$
    • $\text{div }\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$
    • $\text{div curl }\mathbf{F}=0$ always because $a\cdot (a\times b)=0$, otherwise $\text{curl }\mathbf{F}$ doesnt exist
  • Scalar, represents tendency for fluid to diverge away from that point, $\text{div }\mathbf{F}=0$ is said to be incompressible

• Vector form of Green’s theorem:
  • ${\oint }_C\mathbf{F}\cdot d\mathbf{r}=\int {\int }_D(\text{curl }\mathbf{F})\cdot \mathbf{k}dA$
  • Second form of Green’s theorem:
    • ${\oint }_C\mathbf{F}\cdot \mathbf{n}ds=\int {\int }_D\text{div }\mathbf{F}(x,y)dA$

Parametric surface and their areas: (parameterised when possible)

• Parametric surfaces: $S$
  • $\mathbf{r}(u,v)=x(u,v)\mathbf{i}+y(u,v)\mathbf{j}+z(u,v)\mathbf{k}$ with parametric equations, $x=x(u,v)$
  • $\mathbf{r}(u_0,v)$ is a grid curve

• Convert from to cylindrical, polar, spherical coordinates

  • Vector function that contains vector $\mathbf{a}$ and $\mathbf{b}$, and pass through ${\mathbf{r}}_0(u,v)$ is $\mathbf{r}(u,v)={\mathbf{r}}_0+u\mathbf{a}+v\mathbf{b}$

  

  • From the spherical coors:$\mathbf{r}(u,v)=a\sin \varphi \cos \theta \mathbf{i}+b\sin \varphi \sin \theta \mathbf{j}+c\cos \varphi \mathbf{k}$

  

• Surface of revolution:

  • Surface $S$ obtained by rotating the curve $y=f(x)$ about the $x$-axis is:
    • $x=x,y=f(x)\cos \theta ,z=f(x)\sin \theta$ where $u=x,v=\theta$
  • Convert to parametric equations by simplifying the equations when 2 of xyz variables can be replaces

• Tangent plane:
  • ${\mathbf{r}}_u=\frac{\partial x}{\partial u}(u_0,v_0)\mathbf{i}+\frac{\partial y}{\partial u}(u_0,v_0)\mathbf{j}+\frac{\partial z}{\partial u}(u_0,v_0)\mathbf{k}$
  • so normal of tangent plane, ${\mathbf{r}}_n={\mathbf{r}}_u\times {\mathbf{r}}_v$
    • 0 if it has corner

• Surface area:
  • $A(S)=\int {\int }_D|{\mathbf{r}}_u\times {\mathbf{r}}_v|dA$

• Surface area of the graph of a function:

  • For surface $S$ with equation $x=x,y=y,z=f(x,y)$
  • ${\mathbf{r}}_x=\mathbf{i}+\left(\frac{\partial f}{\partial x}\right)\mathbf{k}$ and ${\mathbf{r}}_y=\mathbf{j}+\left(\frac{\partial f}{\partial y}\right)\mathbf{k}$
  • $\displaystyle A(S)=\int\int_D {\color{tomato}\sqrt{1 + \bigg( \frac{\partial z}{\partial x}\bigg)

  +\bigg( \frac{\partial z}{\partial y}\bigg)^2}}dA$

Surface integrals:

• Parametric surfaces:
  • $\int {\int }_{S}f(x,y,z)dS=\int {\int }_{D}f(\mathbf{r}(u,v)).|{\mathbf{r}}_u\times {\mathbf{r}}_v|dA$

• Graph: for surface $S$ with equation $x=x,y=y,z=g(x,y)$

  • $\displaystyle\int\int_S f(x,y,z)\ dS=\int\int_D {\color{tomato} f\big(z,y,g(x,y)\big) \sqrt{ 1+\bigg( \frac{\partial z}{\partial x} \bigg)^2+

  \bigg( \frac{\partial z}{\partial y} \bigg)^2} \ }dA$

  • use $x=g(y,z)$ or $y=g(x,z)$ if necessary

• Oriented surface:

  • There are two possible orientations for any orientable surface

  • Positive orientation: normals point outward, concave inward

  

Mobius strip

Surface integral of vector fields:

• $\mathbf{F}$ is a continuous vector field on oriented surface $S$ with uni normal vector $\mathbf{n}$, then surface integral of $\mathbf{F}$ over $S$ (known as the flux of $\mathbf{F}$ across $S$) is

  • $\displaystyle \int\int\limits_S {\color{tomato}\bold F\cdotp dS}=\int\int\limits_S {\color{tomato}\bold F\cdotp \bold n\ dS}

  =\int\int\limits_D \color{tomato}\bold F\cdotp (\bold r_u\times \bold r_v) \ dA$

  • If surface $S$ is given by graph $z=g(x,y)$ then change to xy and:

    • $\displaystyle \int\int\limits_S {\color{tomato}\bold F\cdotp dS}=\int\int\limits_D

    {\color{tomato}\bigg(-P\frac{\partial g}{\partial x}-Q\frac{\partial g}{\partial y} + R\bigg)} \ dA$

*$PQR$ is from $\mathbf{F}$

• There might be 2 or more parts in a closed surface
• Electric flux of electric field $\mathbf{E}$ though surface $S$: $\int \underset{S}{\int }\mathbf{E}\cdot dS$
• The net rate of outflow of substance through outward oriented surface $S$: $\int \underset{S}{\int }\mathbf{F}\cdot dS$
  • where $\mathbf{F}=p(\mathbf{x})\mathbf{v}(\mathbf{x})$, density times velocity
• Temperate of any point is $u(x,y,z)$ the heat flow is defined as a vector field: $\mathbf{F}=-K\nabla u$
  • then the rate of heat flow across surface $S$ is $\int \underset{S}{\int }\mathbf{F}\cdot dS=-K\int \int \nabla u\cdot dS$

Stoke’s theorem: (3d version of green)

• Relates a surface integral over a surface $S$ to a line integral around the boundary space curve of $S$

• The orientation of surface $S$ is the positive orientation of boundary curve $C$

• Let $S$ be an oriented piece-wise smooth surface that is bounded by a simple, closed, piecewise-smooth boundary curve $C$ with positive orientation

  • $\displaystyle\color{tomato} \oint_C \bold F \cdotp d\bold r\equiv

  \int\int_S \text{curl }\bold F\cdotp dS$,curl$\bold F$ is treated as vector field

• If the surface $S$ is a graph function, $z=g(x,y): \displaystyle

\int\int_S \text{curl }\bold F\cdotp dS$

![Screenshot 2023-06-17 at 00.36.06.png](<https://s3-us-west-2.amazonaws.com/secure.notion-static.com/459fa1da-377f-4829-8af9-fbe55af5e0ea/Screenshot_2023-06-17_at_00.36.06.png>)

• If $S_1$ and $S_2$ are oriented surfaces with same oriented boundary curve $C$ and both satisfy the hypotheses of Stoke’s theorem: $\int {\int }_{S_1}\text{curl }\mathbf{F}\cdot dS={\int }_C\mathbf{F}\cdot d\mathbf{r}=\int {\int }_{S_1}\text{curl }\mathbf{F}\cdot dS$, Both curl F integral are equal

The divergence theorem:

• 3D vector version of green theorem: ${\int }_C\mathbf{F}\cdot \mathbf{n}ds$:

  • Let $E$ be a simple solid region, $S$ be the boundary surface of $E$, given with positive (outward) orientation. $\mathbf{F}$ is vector field whose component functions have continuous partial derivatives on an open region that contains $E$. Then:
    • $\int {\int }_S\mathbf{F}\cdot dS=\int {\int }_S\mathbf{F}\cdot \mathbf{n}dS=\int \int {\int }_E\text{div }\mathbf{F}dV$, use triple integral

• A region $E$ is inside $S_2$ but outside $S_1$:

  • $\int {\int }_S\text{div }\mathbf{F}dV=-\int {\int }_{S_1}\mathbf{F}\cdot dS+\int {\int }_{S_2}\mathbf{F}\cdot dS$

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