# Find the value of the line integral F · dr C (Hint: If F is conservative, the integration may be easier on an alternative path.) (x2 + y2)dx + 2xy dy?

## Jul 29, 2018

(a) $\frac{57344}{3}$

(b) $- \frac{125}{3}$

#### Explanation:

$\boldsymbol{F} \left(\boldsymbol{r}\right) = \left\langle{x}^{2} + {y}^{2} , 2 x y\right\rangle$

$\text{curl } \boldsymbol{F} = \det \left[\begin{matrix}{\partial}_{x} & {\partial}_{y} \\ {x}^{2} + {y}^{2} & 2 x y\end{matrix}\right] = \boldsymbol{0}$

Hence $\boldsymbol{F}$ is conservative and line integrals are path-independent.

• (a)

${\boldsymbol{r}}_{1} = {t}^{5} \boldsymbol{i} + {t}^{4} \boldsymbol{j} q \quad q \quad 0 \le t \le 2$

So that path is: $\left(0 , 0\right) \to \left({2}^{5} , {2}^{4}\right)$

An alternative path is:

$\boldsymbol{r} = t \boldsymbol{i} + \frac{t}{2} \boldsymbol{j} q \quad q \quad 0 \le t \le {2}^{5} q \quad q \quad {\boldsymbol{r}}^{'} = \boldsymbol{i} + \frac{1}{2} \boldsymbol{j}$

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

$= {\int}_{0 , {2}^{5}} \left\langle{t}^{2} + {t}^{2} / 4 , {t}^{2}\right\rangle \cdot \left\langle1 , \frac{1}{2}\right\rangle \setminus \mathrm{dt}$

$= {\int}_{0 , {2}^{5}} \frac{7}{4} {t}^{2} \setminus \mathrm{dt} = \frac{57344}{3}$

• (b)

${\boldsymbol{r}}_{2} = 5 \cos t \boldsymbol{i} + 4 \sin t \boldsymbol{j} q \quad 0 \le t \le \frac{\pi}{2}$

Which is: $\left(5 , 0\right) \to \left(0 , 4\right)$

As much simpler calculation will follow from this 2 step approach:

• $\left\{\begin{matrix}\mathbb{A} q \quad \left(5 0\right) \to \left(0 0\right) & \mathrm{dy} = y = 0 \\ \mathbb{B} q \quad \left(0 0\right) \to \left(0 4\right) & \mathrm{dx} = x = 0\end{matrix}\right.$

$\triangle = {\int}_{\Delta x} {x}^{2} + {y}^{2} \setminus \mathrm{dx} + {\int}_{\Delta y} 2 x y \setminus \mathrm{dy}$

• $\mathbb{A}$

$\triangle = {\int}_{5 , 0} \left({x}^{2} + \cancel{{y}^{2}}\right) \mathrm{dx} + \cancel{\int 2 x y \setminus \mathrm{dy}}$

$= {\left[{x}^{3} / 3\right]}_{5 , 0} = - \frac{125}{3}$

• $\mathbb{B}$

$\triangle = \cancel{\int \left({x}^{2} + {y}^{2}\right) \mathrm{dx}} + {\int}_{0 , 4} 2 \cancel{x} y \setminus \mathrm{dy} = 0$

${C}_{2} = \mathbb{A} + \mathbb{B} = - \frac{125}{3}$