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

Respuesta :

Answer:

The answer is "6".

Step-by-step explanation:

Given:

[tex]C_1[/tex] line segment [tex](0,0) \ to \ (-3,-1)[/tex]

[tex]C_2[/tex] line segment [tex](-3,1)\ to\ (-6,0)[/tex]

[tex]C_1[/tex] line equation:

[tex]\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\\\\[/tex]

[tex]\bold{\int _{C_1} -3 \ dy - dx=0}\\\\[/tex]

[tex]C_2[/tex] line equation:

[tex]\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[/tex]

[tex]\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}\\\\[/tex]

Let [tex]C=C_1+C_2[/tex]

[tex]\to \bold{\int _{C} -3 \ dy - dx= \int _{C_1} -3 \ dy - 1\ dx +\int _{C_2} -3 \ dy - dx}\\\\[/tex]

                             [tex]=0+6\\\\=6[/tex]

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