Section 5.4 Fundamental Theorem of Line Integral

5.4 Fundamental Theorem of Line Integral

We extend the Fundamental Theorem of Calculus to the line integration of vector field in this section. Recall that if $g(x)=F^{\prime }(x)$ is a continuous function on $[a,b]$ then 

$${\int }_a^b{F}^{\prime }(x)dx=F(b)-F(a).$$

The idea is that we know $g(x)$ is the derivative of the function $F(x)$. From previous section, we say a vector field $\mathbf{F}(x,y)=<P(x,y),Q(x,y)>$ is conservative if there is a potential function $f(x)$ such that

$\mathbf{F}(x,y)=<P(x,y),Q(x,y)>=\mathrm{\nabla }f=<f_x,f_y>$. 

Theorem: Fundamental Theorem of Line integral

Let $C$ be a smooth curve given by vector function $\overrightarrow{r(t)}=<x(t),y(t)>$ or $\overrightarrow{r(t)}=<x(t),y(t),z(t)>$ with $a\le t\le b$. Let $f$ be a differential function of two or three variables whose

gradient vector $\mathrm{\nabla }f$ is continuous on $C$ then 

$${\int }_C\mathrm{\nabla }f\cdot d\overrightarrow{r}=f(\overrightarrow{r(b)})-f(\overrightarrow{r(a)}).$$

Example 1: Evaluate ${\int }_C\mathrm{\nabla }f\cdot d\overrightarrow{r}$ where $C$ is the curve $\overrightarrow{r(t)}=<t^2,-t,t^3>$ with $0\le t\le 2$ and $f(x,y,z)=x^{2}yz$.

Exercise 1: Evaluate ${\int }_C\mathrm{\nabla }f\cdot d\overrightarrow{r}$ where $C$ is the curve $\overrightarrow{r(t)}=<-2t,t^3,t>$ with $0\le t\le 1$ and $f(x,y,z)=x+y^2+z$.

When we apply Fundamental Theorem of Line Integral, we can see that we could have the integration, ${\int }_C\mathrm{\nabla }f\cdot d\overrightarrow{r}=0$ if the initial point and the end point coincide. This is not true if we integrate any vector field or any curve! We are going to see what assumptions we need in order to have vector filed line integral equal to $0$.

Definition 

Curve $C$ is a closed curve if there is a parameterization $\overrightarrow{r(t)}$, $a\le t\le b$ of $C$ such that the parameterization traverses the curve exactly once and $\overrightarrow{r(a)}=\overrightarrow{r(b)}$. Curve $C$ is a simple curve if $C$ does not cross itself. That is, $C$ is simple if there exists a parameterization $\overrightarrow{r(t)}$, $a\le t\le b$ of $C$ such that $\overrightarrow{r(t)}$ is one-to-one over $(a,b)$. It is possible for $\overrightarrow{r(a)}=\overrightarrow{r(b)}$, meaning that the simple curve is also closed.

Definition 

If $\mathbf{F}$ is a continuous vector field with domain $D$, we say that the line integral ${\int }_C\mathbf{F}\cdot d\overrightarrow{r}$ is independent of path if ${\int }_{C_1}\mathbf{F}\cdot d\overrightarrow{r}={\int }_{C_2}\mathbf{F}\cdot d\overrightarrow{r}$ for any two paths $C_1$ and $C_2$ in $D$.

Theorem

${\int }_C\mathbf{F}\cdot d\overrightarrow{r}$ is independent of path if and only if ${\int }_C\mathbf{F}\cdot d\overrightarrow{r}=0$ for every closed path $C$ in $D$. 

Theorem

Suppose $\mathbf{F}$ is a vector field that is continuous on an open connected region $D$. If${\int }_C\mathbf{F}\cdot d\overrightarrow{r}$ is independent of path in $D$, then $\mathbf{F}$ is conservative vector filed in $D$, that is there exists a function $f$ such that $\mathrm{\nabla }f=\mathbf{F}$.

With two theorems above, deciding a vector is conservative becomes essential. Recall we only have necessary condition for a conservative field $\mathbf{F}(x,y)=<P(x,y),Q(x,y)>$ then we must have $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$. But we cannot use $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ to conclude $\mathbf{F}$ is conservative. 

Theorem

Let $\mathbf{F}(x,y)=<P(x,y),Q(x,y)>$ be a vector field on an open simply connected region $D$. Suppose that $P$ and $Q$ have continuous first order partial derivatives and $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ through out the whole region $D$. Then $\mathbf{F}$ is conservative. 

Remark: ${\mathbb{R}}^2$ is always simply connected, hence in this content of this course, we just need $P$ and $Q$ have continuous first order partial derivatives.

Example 2: Decide if $\mathbf{F}$ is conservative. If it is conservative then find $f$ such that $\mathrm{\nabla }f=\mathbf{F}$. $\mathbf{F}(x,y)=<xy+y^2,\frac{1}{2}{x}^2+2xy>$.

Exercise 2: Decide if $\mathbf{F}$ is conservative. If it is conservative then find $f$ such that $\mathrm{\nabla }f=\mathbf{F}$. $\mathbf{F}(x,y)=<6xy+y^3+y,3x^2+3x{y}^2+x>$.

Example 3: Decide if $\mathbf{F}$ is conservative. If it is conservative then find $f$ such that $\mathrm{\nabla }f=\mathbf{F}$. $\mathbf{F}(x,y)=<y^2{e}^{xy},(1+xy)e^{xy}>$.

Exercise 3: Decide if $\mathbf{F}$ is conservative. If it is conservative then find $f$ such that $\mathrm{\nabla }f=\mathbf{F}$. $\mathbf{F}(x,y)=<(1+x{y}^2)e^{x{y}^2},2x^{2}y{e}^{x{y}^2}>$.

Example 4: Decide if $\mathbf{F}$ is conservative. If it is conservative then find $f$ such that $\mathrm{\nabla }f=\mathbf{F}$. $\mathbf{F}(x,y)=<2xy+y^2+2x,x^2+2xy+2y>$.

Exercise 4: Decide if $\mathbf{F}$ is conservative. If it is conservative then find $f$ such that $\mathrm{\nabla }f=\mathbf{F}$. $\mathbf{F}(x,y)=<y+y\text{cos}(x)+1,x+\text{sin}(x)+3y^2>$.

Example 5: Decide if $\mathbf{F}$ is conservative and find ${\int }_C\mathbf{F}\cdot d\overrightarrow{r}$ using the Fundamental Theorem of Line integral. $\mathbf{F}(x,y)=<2x{e}^{-y},2y-x^2{e}^{-y}>$ and $C$ is the curve on the $x^2+y^2=1$ traversed counter clockwise once. 

Exercise 5: Decide if $\mathbf{F}$ is conservative and find ${\int }_C\mathbf{F}\cdot d\overrightarrow{r}$ using the Fundamental Theorem of Line integral. $\mathbf{F}(x,y)=<y{e}^x+\text{sin}(y),e^x+x\text{cos}(y)>$ and $C$ is the curve on the $4=2x^2+y^2$ traversed counter clockwise once. 

Example 6: Decide if $\mathbf{F}$ is conservative and find ${\int }_C\mathbf{F}\cdot d\overrightarrow{r}$ using the Fundamental Theorem of Line integral. $\mathbf{F}(x,y)=<xy+y^2,\frac{1}{2}{x}^2+2xy>$ and $C[latex]isthecurveonthe[latex]y=x^2+\frac{1}{4}{x}^4$ from $(0,0)$ to $(1,\frac{5}{4})$. 

Exercise 6: Decide if $\mathbf{F}$ is conservative and find ${\int }_C\mathbf{F}\cdot d\overrightarrow{r}$ using the Fundamental Theorem of Line integral. $\mathbf{F}(x,y)=<xy+y^2,\frac{1}{2}{x}^2+2xy>$ and $C$ is the curve on the $y=x^2+x^3$ from $(0,0)$ to $(1,2)$.

Example 7: (a) Find the work done by $\mathbf{F}(x,y)=<x^3,y^3>$ on a particle that moves on a line segment from $P(1,0)$ to $Q(2,2)$.

(b) Find the work done by $\mathbf{F}(x,y)=<y^3,x^3>$ on a particle that moves on a line segment from $P(1,0)$ to $Q(2,2)$.

Group work:

1. Find the work done by $\mathbf{F}(x,y)=<x^2{y}^3,x^3{y}^2>$ on a particle that moves on a line segment from $P(1,0)$ to $Q(2,2)$.

2. Find the work done by $\mathbf{F}(x,y)=<3+2x{y}^2,2x^{2}y>$ on a particle that moves from $P(1,1)$ to $Q(4,\frac{1}{4})$ along the curve $y=\frac{1}{x}$.

3. Find the work done by $\mathbf{F}(x,y)=<y^3{e}^{xy},2y{e}^{xy}+x{y}^2{e}^{xy}>$ on a particle that moves along the curve $\overrightarrow{r(t)}=<\text{cos}(t),\text{sin}(t)>$, $0\le t\le 2\pi$.

4.  Find the work done by $\mathbf{F}(x,y)=<2x{y}^2,x^{2}y>$ on a particle that moves on a line segment from $P(0,1)$ to $Q(2,3)$.