Line Integrals

If $\vec{F}$ is a vector field and $\vec{\sigma}(t)$ , $a \le t \le b$ , is a path, the line integral of $\vec{F}$ along $\vec{\sigma}(t)$ is

$$\int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds} = \int_a^b \vec{F}(t) \cdot \vec{\sigma}\,'(t)\,dt.$$

Here is what this looks like in two dimensions.

$$\hbox{\epsfysize=2in \epsffile{line-integrals-1.eps}}$$

The vector field gives a vector at each point of the plane. As you move along the path, at each point you compute the dot product of the velocity vector $\vec{\sigma}'(t)$ with the vector from the field at that point. Integrate to "add up" the results; the total is the line integral.

Notice that I'm writing $\bvec{ds}$ instead of $ds$ , the differential for a path integral. Here's the difference:

$$\bvec{ds} = \vec{\sigma}\,'(t)\,dt, \quad\hbox{while}\quad ds = \|\vec{\sigma}\,'(t)\|\,dt.$$

One is a vector, the other is a scalar: $\bvec{ds}$ uses the velocity vector, while $ds$ uses the length of the velocity vector.

Example. Compute $\displaystyle \int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds}$ for $\vec{F}(x,y) = (y, x)$ :

(a) Along the path $\vec{\sigma}(t) = (\cos t, \sin t)$ for $0 \le t \le \dfrac{\pi}{2}$ .

(b) Along the path $\vec{\tau}(t) = (\cos 2 t, \sin 2t)$ for $0 \le t \le \dfrac{\pi}{4}$ .

(a) The path is the part of $x^2 + y^2 = 1$ lying in the first quadrant --- a quarter circle. $\vec{\sigma}(0) = (1, 0)$ while $\vec{\sigma}\left(\dfrac{\pi}{2}\right) = (0, 1)$ , so the quarter circle is traversed counterclockwise.

The velocity is

$$\vec{\sigma}\,'(t) = (-\sin t, \cos t).$$

Write the vector field in terms of t:

$$\vec{F}(t) = (y, x) = (\sin t, \cos t).$$

Then

$$\vec{F}(t) \cdot \vec{\sigma}\,'(t) = -(\sin t)^2 + (\cos t)^2 = \cos 2 t.$$

(I used a double angle formula from trigonometry.) Therefore,

$$\int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds} = \int_0^{\pi/2} \cos 2 t\,dt = \left[\dfrac{1}{2} \sin 2 t\right]_0^{\pi/2} = 0.\quad\halmos$$

(b) $\tau$ is the same quarter circle as in (a), but

$$\vec{\tau}\,'(t) = (-2 \sin 2 t, 2 \cos 2 t).$$

The extra factor of 2 means the path is traversed twice as fast as (a). The field is

$$\vec{F}(t) = (y, x) = (\sin 2 t, \cos 2 t).$$

So

$$\vec{F}(t) \cdot \vec{\tau}\,'(t) = -2(\sin 2 t)^2 + 2(\cos 2 t)^2 = 2 \cos 4 t.$$

Therefore,

$$\int_{\vec{\tau}} \vec{F} \cdot \bvec{ds} = \int_0^{\pi/4} 2 \cos 4 t\,dt = \left[\dfrac{1}{2} \sin 4 t\right]_0^{\pi/4} = 0.$$

Traversing the path twice as rapidly made no difference. This is true in general: If you traverse the same path in the same direction at different speeds, the line integral does not change.

Example. Compute $\displaystyle \int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds}$ for $\vec{F}(x, y) = (x + y, x - y)$ :

(a) Along the path $\vec{\sigma}(t) = (t, t^2)$ for $0 \le t \le 1$ .

(b) Along the path $\vec{\tau}(t) = (1 - t, (1 - t)^2)$ for $0 \le t \le 1$ .

(a) The path is the part of the parabola $y = x^2$ from $(0, 0)$ to $(1, 1)$ . I have

$$\vec{\sigma}'(t) = (1, 2 t) \quad\hbox{and}\quad \vec{F}(x, y) = (x + y, x - y) = (t + t^2, t - t^2).$$

Hence,

$$\vec{F}(t) \cdot \vec{\sigma}'(t) = t + 3 t^2 - 2 t^3.$$

The integral is

$$\int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds} = \int_0^1 (t + 3 t^2 - 2 t^3)\,dt = \left[\dfrac{1}{2} t^2 + t^3 - \dfrac{1}{2} t^4\right]_0^1 = 1.\quad\halmos$$

(b) If I change the path to $\vec{\tau}(t) = (1 - t, (1 - t)^2)$ , I am now traversing $y = x^2$ from $(1, 1)$ to $(0, 0)$ --- the same path as in (a), but in the opposite direction.

I have

$$\vec{\tau}\,'(t) = (-1, -2 + 2 t).$$

$$\vec{F}(t) = ((1 - t) + (1 - t)^2, (1 - t) - (1 - t)^2) = (t^2 - 3 t + 2, t - t^2).$$

Hence,

$$\vec{F}(t) \cdot \vec{\tau}\,'(t) = -2 t^3 + 3 t^2 + t - 2.$$

Therefore,

$$\int_{\vec{\tau}} \vec{F} \cdot \bvec{ds} = \int_0^1 (-2 t^3 + 3 t^2 + t - 2)\,dt = \left[-\dfrac{1}{2} t^4 + t^3 + \dfrac{1}{2} t^2 - 2 t\right]_0^1 = -1.$$

This is true in general: If you traverse the same path in the opposite direction, the line integral is multiplied by -1.

Differential notation.

If $\vec{F} = (F_1, F_2, F_3)$ is a vector field and $\vec{\sigma}(t)$ for $a \le t \le b$ , is a path, then

$$\vec{F} \cdot \vec{\sigma}'(t) = (F_1, F_2, F_3) \cdot \left(\der x t, \der y t, \der z t\right) = F_1 \der x t + F_2 \der y t + F_3 \der z t.$$

So

$$\int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds} = \int_a^b \left(F_1 \der x t + F_2 \der y t + F_3 \der z t\right)\,dt.$$

If I formally multiply the $dt$ in, I get

$$\int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds} = \int_a^b (F_1\,dx + F_2\,dy + F_3\ dz).$$

You will sometimes see line integrals written in this form.

You can do the computation converting $F_1$ , $F_2$ , and $F_3$ to functions of t using $\vec{\sigma} = (x(t),y(t),z(t))$ . Replace $dx$ , $dy$ , $dz$ by $\der x t$ , $\der y t$ , $\der z t$ , and integrate the whole thing with respect to t.

Alternatively, you can use one of the coordinate variables x, y, or z as the parameter.

Example. Let $\vec{\sigma}(t)$ be the segment joining $(1, 1, 1)$ to $(2, 3, 4)$ . Compute

$$\int_{\vec{\sigma}} x\,dx + y\,dy + (x + y - 1)\,dz.$$

The segment is

$$\vec{\sigma}(t) = (1 - t) \cdot (1, 1, 1) + t \cdot(2, 3, 4) = (1 + t, 1 + 2 t, 1 + 3 t).$$

That is,

$$x = 1 + t, \quad y = 1 + 2 t, \quad z = 1 + 3 t.$$

The parameter range is $0 \le t \le 1$ .

Now

$$\der x t = 1, \quad \der y t = 2, \quad \der z t = 3.$$

Plugging everything into the integral, I get

$$\int_{\vec{\sigma}} x\,dx + y\,dy + (x + y - 1)\,dz = \int_0^1 \left[(1 + t)(1) + (1 + 2 t)(2) + (1 + 3 t)(3)\right]\,dt = \int_0^1 (6 + 14 t)\,dt =$$

$$\left[6 t + 7 t^2\right]_0^1 = 13.\quad\halmos$$

Example. Compute $\displaystyle \int_C (x + 8 y)\,dx + 5 x^2\,dy$ where C is the part of the curve $y = x^3$ going from $(0, 0)$ to $(1, 1)$ .

I'll do everything in terms of x. I have $\der y x = 3 x^2$ , so $dy$ becomes $3 x^2\,dx$ . The curve extends from $x = 0$ to $x = 1$ . Substituting $y = x^3$ and $dy = 3x^2\,dx$ , I get

$$\int_C (x + 8 y)\,dx + 5 x^2\,dy = \int_0^1 [(x + 8 x^3) + 5 x^2 \cdot 3 x^2]\,dx = \int_0^1 (x + 8 x^3 + 15 x^4)\,dx =$$

$$\left[\dfrac{1}{2} x^2 + 2 x^4 + 3 x^5\right]_0^1 = \dfrac{11}{2}.\quad\halmos$$

If $\vec{F}$ represents a force field and $\vec{\sigma}$ is the tranjectory of an object moving in the field, the work done by the object in moving along the path through the field is

$$\int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds}.$$

Example. Suppose a force field is given by

$$\vec{F}(x, y) = (x + y, 2 y).$$

Find the work done by a particle moving along a path consisting of the segment from $(0, 0)$ to $(1, 0)$ , followed by the arc of the circle $x^2 + y^2 = 1$ from $(1, 0)$ counterclockwise to $(0, 1)$ .

$$\hbox{\epsfysize=1.75in \epsffile{line-integrals-2.eps}}$$

The segment from $(0, 0)$ to $(1, 0)$ may be parametrized by

$$\vec{\sigma}(t) = (t, 0) \quad\hbox{for}\quad 0 \le t \le 1.$$

Thus,

$$x = t, \quad y = 0.$$

Hence,

$$\vec{\sigma}\,'(t) = (1, 0).$$

On this segment,

$$\vec{F}(t) = (t, 0).$$

So

$$\vec{F} \cdot \vec{\sigma}\,'(t) = (t, 0) \cdot (1, 0) = t.$$

The work done is

$$\int_{\vec{\sigma}} \vec{F} \cdot \bvec{ds} = \int_0^1 t\,dt = \left[\dfrac{1}{2} t^2\right]_0^1 = \dfrac{1}{2}.$$

The arc of the circle from $(1, 0)$ to $(0, 1)$ may be parametrized by

$$\vec{\tau}(t) = (\cos t, \sin t) \quad\hbox{for}\quad 0 \le t \le \dfrac{\pi}{2}.$$

Thus,

$$x = \cos t, \quad y = \sin t.$$

Hence,

$$\vec{\tau}\,'(t) = (-\sin t, \cos t).$$

On this arc,

$$\vec{F}(t) = (\cos t + \sin t, 2 \sin t).$$

So

$$\vec{F} \cdot \vec{\tau}\,'(t) = (\cos t + \sin t, 2 \sin t) \cdot (-\sin t, \cos t) = -\cos t \sin t - (\sin t)^2 + 2 \sin t \cos t = \sin t \cos t - (\sin t)^2.$$

The work done is

$$\int_{\vec{\tau}} \vec{F} \cdot \bvec{ds} = \int_0^{\pi/2} (\sin t \cos t - (\sin t)^2)\,dt = \int_0^{\pi/2} \left(\dfrac{1}{2} \sin 2 t - \dfrac{1}{2} (1 - \cos 2 t)\right)\,dt =$$

$$\left[-\dfrac{1}{4} \cos 2 t - \dfrac{1}{2} t + \dfrac{1}{4} \sin 2 t\right]_0^{\pi/2} = \dfrac{1}{2} - \dfrac{\pi}{4} = -0.28539 \ldots.\quad\halmos$$

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