# Thread: A question on line integral= urgent help please

1. ## A question on line integral= urgent help please

From the textbook conclusion formula => reversing the orientation of the curve changed the sign of the line integral.

eg. Evaluate integral sub c of 3xy dx, where c is the line segment joining (0,0) and (1,2) with the given orientation.

a) oriented from (0,0) to (1,2) .
b) oriented from (1,2) to (0,0).

the line integral of a is 4 and the line integral of b is -4.
therefore, reversing the orientation of the curve did change the sign of the line integral in this example.

However, I found another example that is against the above conclusion.
eg) Evaluate the integral sub c of 4x^3 ds where C
a) is the line segment from (-2,-1) to (1,2).
b) is the line segment from (1,2) to (-2,-1).

the answer of a and b are the same ---is -15sqrt(2) .
therefore, this question indicated reversing the orientation of the curve did not change the sign of the line integral.

Why?
Please teach me. Thank you.

2. Originally Posted by kittycat
From the textbook conclusion formula => reversing the orientation of the curve changed the sign of the line integral.

eg. Evaluate integral sub c of 3xy dx, where c is the line segment joining (0,0) and (1,2) with the given orientation.

a) oriented from (0,0) to (1,2) .
We can parametrize this curve as $\bold{r}(t) = t\bold{i}+2t\bold{j}$ for $0\leq t\leq 1$.
So,
$\int_C 3xy \ dx = \int_0^1 3 (t)(2t)(t)' dt$

Now to do (b) that is simply the negative of this answer.

3. Originally Posted by kittycat

the answer of a and b are the same ---is -15sqrt(2) .
therefore, this question indicated reversing the orientation of the curve did not change the sign of the line integral.

Why?
Please teach me.
No big deal here. the difference between the two was that in the first problem, you were integrating with respect to x. for the second problem you were integrating with respect s, the arc length. since arclength only has one sign, regardless of what orientation you are using ( $\Delta s_i$ is always positive), you get the same result going either way. but since $\Delta x_i$ and $\Delta y_i$ can change signs based on orientation, we have that, in general:

$\int_{-C} f(x,y)~dx = - \int_C f(x,y)~dx$ and $\int_{-C}f(x,y)~dy = - \int_C f(x,y)~dy$

while

$\int_{-C}f(x,y)~ds = \int_C f(x,y)~ds$