# B Integration of a quantity when calculating Work

1. May 25, 2017

### Gurasees

While integrating work we write dw = INT. F.ds. But why can't we write dw = INT. dF.s ?

2. May 25, 2017

### Staff: Mentor

The definition of work is $work = force \times distance$ and while the force can change over position and over time as we move along some path a distance s, we can generalize it as an integral over the path with ds being a very small distance to handle non-linear paths.

However dF isn't a force its a change in force and if used breaks the definition of work which is $work = force \times distance$ not $work = change in force \times distance$.

http://hyperphysics.phy-astr.gsu.edu/hbase/wint.html

3. May 25, 2017

First what do you mean by dF multiplied by s....give the physical interpretation...

Last edited by a moderator: May 25, 2017
4. May 25, 2017

### Buffu

Isn't by definition $\displaystyle w = \int_C f\cdot ds$ ?

5. May 25, 2017

Yes, it is so...

6. May 25, 2017

### Diegor

Check integration by parts. You will see that f.ds is not the same as df.s but there is a relation between them. That is math. then you will have to see if that has physical sense.

7. May 25, 2017

### nasu

8. May 25, 2017