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

The answer is "6".

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 (x-0)/(-3-0)=(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 (x-(-3))/(-6-(-3))=(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`