Master of Science in Mathematics

Lecture Notes. Accessed on: 2019-10-21 17:50:03

## Independence of Path

**Definition**. The line integral is called ** independent of path** if in a region $D$, if for any two points $P$ and $Q$ in $D$ then the line integral along every piecewise smooth curve in $D$ from $P$ to $Q$ has the same value.

**Theorem**. (** Independence of Path**) If $V$ is a continuous vector field on the open connected set $D,$ then the following three conditions are either all true or all false:

$(1) \quad$ $V$ is conservative on $D$

$(2) \quad$ $\int_C c{V}\cdot d{R}=0$ for every piecewise smooth closed curve $C$ in $D.$

$(3) \quad$ $\int_C {V}\cdot d{R}$ is independent of path within $D.$

**Example**. Let $F=\langle y,-x\rangle$ and let $C_1$ and $C_2$ be the following two paths joining $(0,0)$ to $(1,1)$; $C_1$: $y=x$ for $0\leq x\leq 1$ and $C_2$: $y=x^2$ for $0\leq x\leq 1.$ Show that $$ \int_{C_1}{F}\cdot d {R}\neq \int_{C_2}{F}\cdot d {R}. $$ Explain what this means? On $C_1:$ $x=t,$ $y=t$ with $0\leq t\leq 1,$ $$ \int_{C_1} y \, dx-x \, dy=\int_0^1 (t-t) \, dt=0. $$ On $C_2:$ $x=t,$ $y=t^2,$ with $0\leq t\leq 1,$ $$ \int_{C_2}y \,dx-x\,dy=\int_0^1 \left(t^2-t(2t)\right) \, dt=-\frac{1}{3}. $$

**Solution**. Since these two line integrals do not have the same value we see that not all line integrals are independent of path.

## Fundamental Theorem of Line Integrals

Now we are ready for a generalization of the Fundamental Theorem of Calculus.

**Theorem**. (** Fundamental Theorem of Line Integrals**) Let $C$ be a piecewise smooth curve that is parametrized by the vector function ${R}(t)$ for ${a\leq t\leq b}$ and let ${V}$ be a vector field that is continuous on $C.$ If $f$ is a scalar function such that ${V}=\nabla f,$ then $$ \int_C {V}\cdot d{R}=f(Q)-f(P)$$ where $Q=R(b)$ and $P=R(a)$ are the endpoints of $C.$

**Proof**. Suppose ${R}(t)=x(t) {i}+y(t) {j}+z(t) {k}$ and let $G$ be the composite function $G(t)=f(x(t),y(t),z(t)).$ We have \begin{align} \int_C {F}\cdot d{R} & =\int_C\nabla f\cdot d{R} =\int_C \left[\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\right] \\ & =\int_a^b \left[\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}\right] \, dt \\ & =\int_a^b \frac{d G}{d t} \, dt =G(b)-G(a) =f(Q)-f(P). \end{align}

**Example**. Show that the vector field $$ {F}(x,y,z)=\left\langle \frac{y}{1+x^2}+\tan ^{-1}z,\tan ^{-1}x,\frac{x}{1+z^2}\right\rangle, $$ is conservative and find a scalar potential $f$ for $F.$ Then evaluate the line integral $$ \int_C {F}\cdot d {R}. $$ where $C$ is any piecewise smooth path connecting $A(1,0,-1)$ to $B(0,-1,1).$

**Solution**. Since $\mathop{curl} {V}={0},$ ${V}$ is conservative. Now we set out to find $f.$ Since $$ \frac{\partial f}{\partial x}=\frac{y}{1+x^2}+\tan ^{-1}z $$ we set $$ f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c(y,z). $$ Since $$ \frac{\partial f}{\partial y}=\tan ^{-1}z=\frac{\partial }{\partial y}\left(y \tan ^{-1}x+x \tan ^{-1}z+c\right)=\tan ^{-1}x+\frac{\partial c}{\partial y}, $$ so $\frac{\partial c}{\partial y}=0$ and $c=c_1(z)$ and so we set $f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z+c_1(z).$ Since $$ \frac{\partial f}{\partial z}=\frac{x}{1+z^2}=\frac{\partial }{\partial z}\left[y \tan ^{-1}x+x \tan ^{-1}z+c_1(z)\right]=\frac{x}{1+z^2}+c’_1(z),$$ so $c_1′(z)=0,$ $ c_1=0$ and so we set $ f(x,y,z)=y \tan ^{-1}x+x \tan ^{-1}z.$ \begin{equation} \int_C{F}\cdot d {R}=f(1,0,-1)-f(0,-1,1) =\frac{3\pi }{4}+\frac{\pi }{4} =\pi . \end{equation}

**Example**. Show that the vector field $$ {V}(x,y,z)=\left(x y^2+y z\right){i}+\left(x^2y+x z+3y^2z\right){j}+\left(x y+y^3\right){k} $$ is conservative and find a scalar potential $f$ for $F.$ Then evaluate the line integral $$ \int_C {F}\cdot d {R}. $$ where $C$ is any piecewise smooth path joining $A(8,6,1)$ to $B(1,3,5).$

**Solution**. First we determine $\mathop{curl} {V}.$ We find \begin{align} \mathop{curl} {V} & =\left| \begin{array}{cc} \begin{array}{c} {i} \\ \begin{array}{c} \partial / \partial x \\ x y^2+y z \end{array} \end{array} & \begin{array}{cc} {j} & {k} \\ \begin{array}{c} \partial / \partial y \\ x^2y+x z+3y^2z \end{array} & \begin{array}{c} \partial / \partial z \\ x y+y^3 \end{array} \end{array} \end{array} \right| \\ & =\left(x+3y^2-x-3y^2\right){i}-(y-y){j}+(2x y+z-2x y-z){k} ={0}\end{align} and so the vector field is conservative and we can find the scalar potential function. Now we set out to find $f$ with $\nabla f={V}.$ Since $f_x=x y^2+y z$ we know that $$ f(x,y,z)=\frac{x^2}{2}y^2+x y z+C_1(y,z). $$ Then we find $$ f_y=x^2y+x z+\frac{ \partial C_1}{\partial y}; $$ and so comparing this with the given $x^2y+x z+3y^2z$ we determine that $$ \frac{ \partial C_1}{\partial y}=3y^2z. $$ So $C_1=y^3z+C_2(z).$ So far we have $$ f=\frac{x^2}{2}y^2+x y z+y^3z+C_2(z). $$ Also since $f_z=x y+y^3+\frac{dC_2}{d z}$ and comparing this to the given $x y+y^3$ we determine $\frac{dC_2}{dz}=0.$ So $C_2$ is a constant with respect to $x,$ $y,$ and $z.$ Therefore a scalar potential function is $$ f(x,y,z)=\frac{x^2}{2}y^2+x y z+y^3z$$ (taking the constant to be zero). Finally \begin{equation} \int_C {V}\cdot d{R}=f(8,6,1)-f(1,3,5)=\frac{2523}{2}. \end{equation}

**Example**. Show that the vector field is conservative and find a scalar potential $f$ for $F.$ Then evaluate the line integral $$ \int_C {F}\cdot d {R} = {F}(x,y)=\frac{(y+1){i}-x {j}}{(y+1)^2} $$ where $C$ is any smooth path connecting $A(0,0)$ to $B(1,1).$

**Solution**. Let ${F}(x,y)=u(x,y){i}+v(x,y){j}$, then since $$ \frac{\partial u}{\partial y}=\frac{-1}{(y+1)^2}=\frac{\partial v}{\partial x}, $$ we know that $F$ is conservative. By definition, we know $$ f_x(x,y)=\frac{1}{y+1} \qquad \text{and}\qquad f_y(x,y)=\frac{-x}{(y+1)^2}. $$ Integrating with respect to $x,$ $f(x,y)=\frac{x}{y+1}+c(y).$ Since, $$ f_y(x,y)=-\frac{x}{(y+1)^2}+c'(y)=-\frac{x}{(y+1)^2}, $$ so $c'(y)=0.$ For $c(y)=0;$ and then $f(x,y)=\frac{x}{y+1}$ is a scalar potential function for ${F}.$ By the fundamental theorem of line integrals, \begin{equation} \int_C {F}\cdot d{R}=f(1,1)-f(0,0)=\frac{1}{2}-0=\frac{1}{2}. \end{equation}

## Work in a Conservative Vector Field

**Example**. Find the work done an object moves in the force field $$ {F}(x,y,z)=x {i}+y {j}+(x z-y){k} $$ along the curve $C$ defined parametrically by ${R}(t)=t^2{i}+2t {j}+4t^3 {k}$ for $0\leq t\leq 1.$

**Solution**. We determine \begin{align} & d{R}=\left(2t {i}+2{j}+12t^2{k}\right)dt, \qquad {F}(t)=t^2{i}+2t{j}+\left(4t^5-2t\right){k}, \end{align}

(from $x(t)=t^2, y(t)=2t,$ and $z=4t^3$) and $$ {F}\cdot d{R}=\left(2t^3+4t+48t^7-24t^3\right)dt. $$ Thus, \begin{equation} W=\int_C {F}\cdot d{R}=\int_0^1 \left(2t^3+4t+48t^7-24t^3\right) \, dt=\frac{5}{2}. \end{equation}

**Example**. Find the work done when an object moves along a closed path in a connected domain where the force field is conservative.

**Solution**. In such a force field ${F},$ where $f$ is a scalar potential of ${F}$ and because the path of motion is closed, it begins and ends at the same point $P.$ Thus, the work is given by $$ W=\oint _C {F}\cdot d{R}=f(P)-f(P)=0. $$

**Example**. Let $$ {F}(x,y)=\frac{-y {i}+x {j}}{x^2+y^2}. $$ Evaluate the line integral $$ \int_{C_1}{F}\cdot d {R} $$ where $C_1$ is the upper semicircle $y=\sqrt{1-x^2}$ transversed counterclockwise. What is the value of $$ \int_{C_2}{F}\cdot d {R} $$ if $C_2$ is the lower semicircle $y=-\sqrt{1-x^2}$ also transversed counterclockwise?

**Solution**. On the upper semi-circle $C_1$, we let $x=\cos (t)$ and $y=\sin (t)$ for $0\leq t\leq \pi ,$ and then $$ \int_{C_1} \frac{-y \, dx+x \, dy}{x^2+y^2}=\int_0^{\pi } \frac{-\sin (t)(-\sin (t))+\cos (t)(\cos (t))}{\sin ^2(t)+\cos ^2(t)} \, dt=\int_0^{\pi }dt=\pi . $$ For $\pi \leq t\leq 2\pi ,$ we have the lower semi-circle $C_2$: \begin{equation} \int{C_2}\frac{-y\, dx+x\, dy}{x^2+y^2}=\int_{\pi }^{2\pi }dt=\pi . \end{equation}

## Finding Area with Line Integral

**Theorem**. If $R$ is a region bounded by a piecewise smooth simple closed curve $C$ oriented counterclockwise, then the area of $R$ is given by $$ A=\oint _C x dy=-\oint _C y dx=\frac{1}{2}\oint _C x dy-y dx. $$