# Find the value of the line integral. F · dr (Hint: If F is conservative, the integration may be easier on an alternative path.) (2x − 8y + 6) dx − (8x + y − 6) dy?

## part c:

Jul 28, 2018

(c) $q \quad \frac{21 - 20 {e}^{2} - {e}^{4}}{2}$

(d) $q \quad 0$

#### Explanation:

If you mean that the line integral is:

int_C (2x − 8y + 6) dx − (8x + y − 6) dy qquad = triangle

Then bbF = (:2x − 8y + 6,-(8x + y − 6) :)

And "curl " bbF = det[(del_x, del_y),(2x − 8y + 6,-(8x + y − 6))] = 0

So $\boldsymbol{F}$ is a conservative vector field.

You can only see (c) and (d) in the screengrab

• (c)

Along $y = {e}^{x} q \quad \therefore \mathrm{dy} = {e}^{x} \mathrm{dx}$:

triangle = int_(0,2) (2x − 8bbe^x + 6) dx − (8x + bbe^x − 6) bb(e^x dx)

That looks ugly so just use a different 2-step path:

• $\left\{\begin{matrix}\mathbb{A} : q \quad \mathrm{dx} = 0 & x = 0 & y : \setminus 1 \to {e}^{2} \\ \mathbb{B} : q \quad \mathrm{dy} = 0 & y = {e}^{2} & x : \setminus 0 \to 2\end{matrix}\right.$

• Path $\mathbb{A}$

triangle = int_C cancel((2x − 8y + 6) dx) − (8*0 + y − 6) dy

$= {\int}_{1 , {e}^{2}} - y + 6 \setminus \mathrm{dy} = {\left[6 y - {y}^{2} / 2\right]}_{1}^{{e}^{2}}$

$= 6 {e}^{2} - \frac{{e}^{4} + 11}{2}$

• Path $\mathbb{B}$

triangle = int_C (2x − 8e^2 + 6) dx − cancel((8*0 + y − 6) dy)

 = int_(0,2) 2x − 8e^2 + 6 \ dx = [ x^2 − (8e^2 - 6)x]_0^2

 = 4 − 16e^2 + 12

${C}_{3} = \mathbb{A} + \mathbb{B}$

$= \frac{21 - 20 {e}^{2} - {e}^{4}}{2}$

• (d)

Zero, because this is a closed path and the field is conservative