# Find the value of the line integral. F · dr C (Hint: If F is conservative, the integration may be easier on an alternative path.) F(x, y) = yi − xj?

## (a) r1(t) = ti + tj, 0 ≤ t ≤ 1 (b) r2(t) = ti + t2j, 0 ≤ t ≤ 1 (c) r3(t) = ti + t3j, 0 ≤ t ≤ 1

Jul 27, 2018

See below

#### Explanation:

bbF(bbr) = y bbi − x bbj

Testing for conservative field:

$\text{curl } \boldsymbol{F} = \det \left[\begin{matrix}{\partial}_{x} & {\partial}_{y} \\ y & - x\end{matrix}\right] = {\partial}_{x} \left(- x\right) - {\partial}_{y} \left(y\right) = - 2 \boldsymbol{k} \textcolor{red}{\ne \boldsymbol{0}}$

This is not conservative , which is fine.

To use the parameterisation in $t$, note that:

• ${\int}_{C} \boldsymbol{F} \left(\boldsymbol{r}\right) \cdot d \boldsymbol{r} \equiv {\int}_{\Delta t} \boldsymbol{F} \left(\boldsymbol{r} \left(t\right)\right) \cdot \boldsymbol{r} ' \left(t\right) \setminus \mathrm{dt}$

(a)

• bbr_1(t) = underbrace(t)_(x)bbi + underbrace(t)\_(y)bbj, qquad 0 ≤ t ≤ 1 qquad bbr_1^'(t) = bbi + bbj

$\implies {\int}_{\Delta t} \boldsymbol{F} \left(\boldsymbol{r} \left(t\right)\right) \cdot \boldsymbol{r} ' \left(t\right) \setminus \mathrm{dt}$

= int_(0,1) (t bbi − t bbj) * (bbi + bbj) \ dt = 0

(b)

• bbr_2(t) = underbrace(t)_(x)bbi + underbrace(t^2)\_(y)bbj, qquad 0 ≤ t ≤ 1 qquad bbr_2^'(t) = bbi + 2t bbj

implies int_(0,1) (t^2 bbi − t bbj) * (bbi + 2t bbj) \ dt

$= - {\int}_{0 , 1} {t}^{2} \setminus \mathrm{dt} = - \frac{1}{3}$

(c)

• bbr_3(t) = underbrace(t)_(x)bbi + underbrace(t^3)\_(y)bbj, qquad 0 ≤ t ≤ 1 qquad bbr_3^'(t) = bbi + 3t^2 bbj

implies int_(0,1) (t^3 bbi − t bbj) * (bbi + 3t^2 bbj) \ dt

$= - 2 {\int}_{0 , 1} {t}^{3} \setminus \mathrm{dt} = - \frac{2}{3} = - \frac{1}{2}$