Arc Length and Line Integrals

1. Apr 1, 2009

IniquiTrance

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!

2. Apr 2, 2009

HallsofIvy

Staff Emeritus
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)||?

3. Apr 2, 2009

IniquiTrance

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?

4. Apr 3, 2009

IniquiTrance

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

5. Apr 3, 2009

HallsofIvy

Staff Emeritus
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.

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: Apr 3, 2009
6. Apr 3, 2009

IniquiTrance

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.

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

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

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

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

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

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

7. Apr 3, 2009

HallsofIvy

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

8. Apr 3, 2009

IniquiTrance

By constructing a piece-wise vector valued function around the rectangle.