# Arc Length and Line Integrals

When I take the line integral around a square shape path "C" as follows:

From A to B to C to D to A

C1 = A(0, 0) to B (4, 0)

t i
0 <= t <= 4

C2 = B (4, 0) to C (4, 7)

4 i + (t - 4) j
4 <= t <= 11

C3 = C (4, 7) to D (0, 7)

(15 - t) i + 7 j

11 <= t <= 15

C4 = D (0, 7) to A (0, 0)

(22 - t) j

15 <= t <= 22

Why is that when I take the line integral around this path using $$\int_{C }||r'(t)|| dt$$ and the above parameterization, I end up with 0, when I should be getting the arc length of the path, which is the perimeter of the square?

Thanks!

HallsofIvy
Homework Helper
When you take the integral of what around that path?

If f(x,y)dx+ g(x,y)dy is an exact differential, that is if there exist F(x,y) such that dF= f(x,y)dx+ g(x,y)dy, then the integral from any point $(x_0,y_0)$ to $(x_1,y_1)$,
$$\int_{(x_0,y_0)}^{(x_1,y_1)} f(x,y)dx+ g(x,y)dy= F(x_1,y_1)- F(x_0,y_0)$$
and, in particular, if $(x_1,y_1)= (x_0,y_0)$ then $F(x_1,y_1)= F(x_0,y_0)$ and that difference is 0.

Now, what is ||r'(t)||?

When you take the integral of what around that path?

If f(x,y)dx+ g(x,y)dy is an exact differential, that is if there exist F(x,y) such that dF= f(x,y)dx+ g(x,y)dy, then the integral from any point $(x_0,y_0)$ to $(x_1,y_1)$,
$$\int_{(x_0,y_0)}^{(x_1,y_1)} f(x,y)dx+ g(x,y)dy= F(x_1,y_1)- F(x_0,y_0)$$
and, in particular, if $(x_1,y_1)= (x_0,y_0)$ then $F(x_1,y_1)= F(x_0,y_0)$ and that difference is 0.

Now, what is ||r'(t)||?
I was taking the arc length of f(x, y) = 1 along ds.

Since "x" is a potential function of 1, it seems I should then be getting 0.

Yet I thought taking a line integral where f(x, y) = 1 should be giving me the arc length of the path, yet this turns out to be 0.

Where am I mistaken?

Also $$||r'(t)||= \frac{ds}{dt}$$

HallsofIvy
Homework Helper
I was taking the arc length of f(x, y) = 1 along ds.
This makes no sense. You don't find the "arc length" of a function, you find the length of an arc. In particular, integrating the dervative of any differentiable function around a closed path will give 0.

Since "x" is a potential function of 1, it seems I should then be getting 0.

Yet I thought taking a line integral where f(x, y) = 1 should be giving me the arc length of the path, yet this turns out to be 0.

Where am I mistaken?
You are not taking direction into account. On the line (0,0) to (4,0), ds= dx and you have
$$\int_0^4 dx= 4[/itex] On the line segment (4, 0) to (4, 7), ds= dy and you have [tex]\int_0^7 dy= 7[/itex] But on the line segment (4, 7) to (0, 7), x is decreasing from 4 to 0 so ds= -dx and you integrate [tex]\int_4^0 (-dx)= -(-4)= 4$$

On the line integral (0,7) to (0,0), y is decreasing from 7 to 0 so ds= - dy and you integrate
$$\int_7^0 (-dy)= -(-7)= 7$$

The integral around the entire rectangle is 4+ 7+ 4+ 7= 22.

Last edited by a moderator:
Hmm, I see your point. I meant we are taking the line integral along $$ds$$ with $$f(x,y) =1$$

I was computing it wrong, as the perimeter can be calculated with such a line integral.

$$\int_C f(x, y)||r'(t)|| dt =$$

$$C_{1}=\int_0^4 dt = 4$$

$$C_{2}=\int_4^{11} dt = 7$$

$$C_{3}=\int_{11}^{15} ||-1|| dt = 4$$

$$C_{4}=\int_{15}^{22} ||-1|| dt = 7$$

$$C_{1} + C_{2} + C_{3} + C_{4} = 22$$

HallsofIvy
How did you get the values t= 11, t= 15, t= 22? 