Why Does the Line Integral of a Square Path's Perimeter Equal Zero?

[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!

 

[HallsofIvy]

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)||?

 

[IniquiTrance]

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?

 

[IniquiTrance]

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

 

[HallsofIvy]

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
[tex]\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
$$\int_4^0 (-dx)= -(-4)= 4$$

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.

 

[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.

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

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

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

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

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

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

 

[HallsofIvy]

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

 

[IniquiTrance]

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