Question

Let C be the curve which is the union of two line segments, the first going from (0,0) to (−3,−1) and the second going from (−3,−1) to (−6,0). Compute the line integral ∫C−3dy 1dx

Answer 1

Answer:

The answer is "6".

Step-by-step explanation:

Given:

$C_1$ line segment $(0,0) \ to \ (-3,-1)$

$C_2$ line segment $(-3,1)\ to\ (-6,0)$

$C_1$ line equation:

$\to \frac{x-0}{-3-0}=\frac{y-0}{1-0}=t\\\\\to x=-3t\\\\\to y= t\\\\ \longrightarrow \int _{C_1} -3 \ dy - 1 dx\\\\=\int^1_{0} -3 \ dt -(-3 \ dt)\\\\=\int^1_{0} 0 \ dt \\\\=0\\\\\therefore\\\\\to x=-3t\ \ \ \ \ \ \ \ \ \to y=t\\\\\to dx= -3 dt\ \ \ \ \ \ \ \ \to dy = dt$

$\bold{\int _{C_1} -3 \ dy - dx=0}$

$C_2$ line equation:

$\to \frac{x-(-3)}{-6-(-3)}=\frac{y-1}{0-1}=t\\\\\to x=-3t-3 \ \ \ \ \ \ \ \ \ \ \to dx= -3 \ dt \\\\\to y= 1-t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \to dy= -dt$

$\int _{C_2} -3 \ dy - dx\\\\=\int^1 _{0} -3 \ (-dt) - (-3\ dt)\\\\=\int^1 _{0} 6 \ dt\\\\=6\\\\\to \bold{\int _{C_2} -3 \ dy - dx=6}$

Let $C=C_1+C_2$

$\to \bold{\int _{C} -3 \ dy - dx= \int _{C_1} -3 \ dy - 1\ dx +\int _{C_2} -3 \ dy - dx}$

                             $=0+6\\\\=6$