# Line integral in a uniform force field

1. Jul 22, 2014

### Born

I have had some trouble with Kleppner and Kollenkow's derivation of work in a uniform force field. As the attached image shows, all three integrals (with respect to dx, dy, dz) are evaluated as follows: $$\int_{x_a, y_a, z_a} ^ {x_b, y_b, z_b}$$ . I am not sure how to proceed with such limits.

Any help is more than welcome. Thanks!

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• ###### Kleppner and Kollenkow - Work Done by a Uniform Force.png
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2. Jul 22, 2014

### Mr-R

Welcome to PF Born

So the three integrals each calculate the contribution from each axis to the resultant force. I don't know why they put the limits like that, but you can see it like this: dx ($i$ direction) must go from some x to another x right? In this case $x_{b}-x_{a}$ and the same for he other directions.

Now $(x,y,z)$ is just r

Hope that helps.

3. Jul 23, 2014

### Born

Thank you for the quick response. It does intuitively make sense to me in the way you've said it, yet it feels weird since I know that after my integrations I would get something that will look like this:

$$x(x_b, y_b, z_b) - x(x_a, y_a, z_a) \text{ (same with}\ y \text{ and}\ z \text{)}$$

$$\text{How could I show (mathematically) that:}\ x(x_b, y_b, z_b)=x(x_b), x(x_a, y_a, z_a)=x(x_a), \text{ etc?}$$

4. Jul 23, 2014

### Mr-R

Something like this actually, $$r_b(x_b, y_b, z_b) - r_a(x_a, y_a, z_a)$$

I am afraid I don't get you here. Seems you got mixed up? see my above respond. Hopefully it will clear things up.

5. Jul 23, 2014

### Born

I got what you said, it just isn't satisfactory. There must be a reason why the limits were given thusly.

6. Jul 23, 2014

### HallsofIvy

Staff Emeritus
$$\int_{x_a, y_a, z_a}^{x_b, y_b, z_b} dzdydx$$
is simply shorthand for
$$\int_{x_a}^{x_b}\int_{y_a}^{y_b}\int_{z_a}^{z_b} dzdydx$$

7. Jul 23, 2014

### vanhees71

It's not a volume integral as suggested in the previous posting but a line integral in a pretty awful notation. The correct way is to say we have a curve $C$, defined, e.g., parametrically as a function $\vec{r}:[0,1] \rightarrow \mathbb{R}^3$. Then the line integral of a vector field along the curve $C$ is defined as a simple integral over the parameter:
$$\int_{C} \mathrm{d} \vec{r} \cdot \vec{V}(\vec{r}):=\int_0^1 \mathrm{d} t \; \frac{\mathrm{d} \vec{r}(t)}{\mathrm{d} t} \cdot V[\vec{r}(t)].$$
Now, sometimes the vector field is "conservative", i.e., the integral doesn't depend on the curve (perhaps restricted to cuves within a certain region of space) but only on the boundary points of the curve. Then you may write the integral in the sloppy way as in the scan of the book.

The specific form of the parametrization chosen is to first integrate from $(x_a,y_a,z_a)$ along a straight line parallel to the $x$-axis to $x_b,y_a,z_a$ and so on.

This is however, unnecessarily complicated. Since the force field is homogeneous in the example, i.e., $\vec{F}(\vec{r})=\text{const}$ you can integrate along any path connecting the two points, because the line integral is path independent anywhere, because it's analytic everywhere and $\vec{\nabla} \times \vec{F}=0$ everywhere. Then according to the Poincare lemma, the line integral is path independent. So we can chose just a straight line connecting the points $\vec{a}$ and [/itex]\vec{b}[/itex],
$$\vec{r}(t)=\vec{a} + t (\vec{b}-\vec{a}) ,\quad t \in [0,1].$$
This gives
$$W_{ba}=\int_0^1 \mathrm{d} t (\vec{b}-\vec{a}) \cdot \vec{F}=(\vec{b}-\vec{a}) \cdot \vec{F},$$
because $\vec{F}=\text{const}$.

8. Jul 23, 2014

### Born

Thank you vanhees71! Now I believe I'm beginning to understand the reasoning behind the notation.

Does this mean that it's just notation as well when the integral is broken up (i.e. dr becomes dx, dy, and dz) and therefore each integral should be treated as a straight line through each of the component's endpoints?

So; $$\int_{x_a} ^ {x_b}\, dx$$

9. Jul 24, 2014

### vanhees71

To stress it again: You should really use the definition of the line integral of a vector field as I've written down in my previous posting. The way the integral is done in the book is very misleading. It only works out right, because here you integrate over a constant force field!