# Find the value of the line integral.F · dr (Hint: If F is conservative, the integration may be easier on an alternative path.) F(x,y) = ye^xyi + xe^xyj (a) r1(t) = ti − (t − 8)j, 0 ≤ t ≤ 8 ?

## (b) the closed path consisting of line segments from (0, 8) to (0, 0), from (0, 0) to (8, 0), and then from (8, 0) to (0, 8)

Jul 27, 2018

Zero in both cases

#### Explanation:

$\boldsymbol{F} \left(\boldsymbol{r}\right) = y {e}^{x y} \boldsymbol{i} + x {e}^{x y} \boldsymbol{j}$

A field $\boldsymbol{F} \left(\boldsymbol{r}\right)$ is conservtive if there exists a potential function $\phi \left(\boldsymbol{r}\right)$ such that:

• $\boldsymbol{F} = - \nabla \phi$

Or, in 2-D:

• $\boldsymbol{F} = - \left\langle{\phi}_{x} , {\phi}_{y}\right\rangle$

This can be solved directly here, because:

$\left\{\begin{matrix}{\phi}_{x} = - y {e}^{x y} \\ {\phi}_{y} = - x {e}^{x y}\end{matrix}\right. \implies \phi = - {e}^{x y} + {\phi}_{o}$

So for:

Path a)

• bbr_1(t) = t bbi − (t − 8) bbj, qquad 0 le t le 8

This is along line $\left(0 , 8\right) \to \left(8 , 0\right)$

$\Delta \phi = - {e}^{0 \times 8} + {e}^{8 \times 0} = 0$

Path b)

Regardless of path, this is closed path from $\left(0 , 8\right) \to \left(0 , 8\right)$

$\therefore \Delta \phi = 0$

Going instead with the hint , the curl of a conserved vector field is zero. This follows from the existence of a potential function and the gradient theorem.

Curl of the Field: $= \boldsymbol{\nabla} \times \boldsymbol{F} = \det \left(\begin{matrix}{\partial}_{x} & {\partial}_{y} \\ y {e}^{x y} & x {e}^{x y}\end{matrix}\right)$

$= \left({\partial}_{x} \left(x {e}^{x y}\right) - {\partial}_{y} \left(y {e}^{x y}\right)\right) \boldsymbol{k}$

$= \left({e}^{x y} + x y {e}^{x y} - \left({e}^{x y} + x y {e}^{x y}\right)\right) \boldsymbol{k} = \boldsymbol{0}$

The field is conservative / path-independent

Path a)

bbr_1(t) = t bbi − (t − 8) bbj, qquad 0 le t le 8

This is along line $\left(0 , 8\right) \to \left(8 , 0\right)$ ie $y = 8 - x$

Following the hint, break this into $\boldsymbol{1}$ followed by path $\boldsymbol{2}$:

• $\left\{\begin{matrix}\boldsymbol{1} & x = 0 & \mathrm{dx} = 0 & y \in \left[8 & 0\right] \\ \boldsymbol{2} & y = 0 & \mathrm{dy} = 0 & x \in \left[0 & 8\right]\end{matrix}\right.$

Path $\boldsymbol{1}$:

$\implies {\int}_{C} \boldsymbol{F} \cdot d \boldsymbol{r}$

$= {\int}_{C} \left(y {e}^{x y} \boldsymbol{i} + x {e}^{x y} \boldsymbol{j}\right) \cdot \left(\boldsymbol{i} \mathrm{dx} + \boldsymbol{j} \mathrm{dy}\right)$

$= {\int}_{C} \left(y \boldsymbol{i} + 0 \boldsymbol{j}\right) \cdot \left(\boldsymbol{i} 0 + \boldsymbol{j} \mathrm{dy}\right) = 0$

Path $\boldsymbol{2}$:

$= {\int}_{C} \left(0 \boldsymbol{i} + x \boldsymbol{j}\right) \cdot \left(\boldsymbol{i} \mathrm{dx} + \boldsymbol{j} 0\right) = 0$

$\boldsymbol{1} + \boldsymbol{2} = 0$

Path b)

This is clearly zero as it is a closed path in a conserved field.

FINALLY:

Noting that $\boldsymbol{F} \cdot \mathrm{db} b r \equiv \boldsymbol{F} \cdot \boldsymbol{r} ' \setminus \mathrm{dt}$, the full integral using the parameterisation is:

• $\boldsymbol{r} \left(t\right) : \left\{\begin{matrix}x = t \\ y = 8 - t\end{matrix}\right. q \quad \boldsymbol{r} ' \left(t\right) : \left\{\begin{matrix}x ' = 1 \\ y ' = - 1\end{matrix}\right.$

$\boldsymbol{F} \left(\boldsymbol{r} \left(t\right)\right) \cdot \boldsymbol{r} ' \left(t\right) = \left(\left(8 - t\right) {e}^{t \left(8 - t\right)} \boldsymbol{i} + t {e}^{t \left(8 - t\right)} \boldsymbol{j}\right) \cdot \left(\boldsymbol{i} - \boldsymbol{j}\right)$

$= \left(8 - t\right) {e}^{t \left(8 - t\right)} - t {e}^{t \left(8 - t\right)}$

$= \left(8 - 2 t\right) {e}^{t \left(8 - t\right)}$

$\implies {\int}_{0}^{8} \left(8 - 2 t\right) {e}^{8 t - {t}^{2}} \setminus \mathrm{dt}$