2
$\begingroup$

This question might be similar to MSE Question but I didn't really get the answer.

Suppose I want to evaluate the integral $$W = \int \underline{F}\cdot \underline{dr}$$ which actually has the physical meaning of giving the work done by the force $F$ along $r$. Now, Imagine I have a non conservative vector field (or force), i.e. $F$ is non conservative, it is a resistive force which resists with constant value $R$, suppose it acts for a distance $d$. Then can I do the following? $$W = \int \underline{F}\cdot\underline{dr} =\underline{F}\cdot\int\underline{dr} = \underline{F}\cdot\underline{\triangle r} = -Rd$$

and if so

why can I do that? When can I take out one of the factors of a dot product inside an integral?

I.e. can I take out the force? This happens in the last line of this page of my lecture notes: Image 1

image 2

  • 0
    You can take out F if it is a constant vector, otherwise not. – 2017-01-03
  • 0
    @Paul what I don't understand is, how do we know it works? Of course we know it for standard multiplication, but here we have a dot product, it's not trivial that we can take it out, or is it? – 2017-01-03

1 Answers 1

2

If the force is constant, \begin{align*} \int {\bf F}\cdot d{\bf r} &= \int \sum_i F_i dr_i \\ &= \sum_i F_i \int dr_i \\ &= {\bf F}\cdot \int d{\bf r} \\ &= {\bf F}\cdot \Delta{\bf r}. \end{align*} This is allowed due to the definition of the dot product and the fact that for constant $c$, $\int c f(x) dx = c\int f(x) dx$.

  • 0
    but how can you do the step in the second line? Recall that you have$$\int \sum_i(F_idr_i)$$, whereas you are treating it as if it was $$\int (\sum_iF_i)dr$$. For example if $F = (a,b)$ and $dr = (dx,dy)$ then we would have $$\int adx+bdy$$ but you can't factor out $a$ and $b$ like you did. You'd have to do $$a\int dx + b\int dy$$ – 2017-01-03
  • 0
    @Euler_Salter: We have \begin{align*} \int{\bf F}\cdot d{\bf r} &= \int \sum F_i d r_i \\ &= \int(F_x dx + F_y dy) \\ &= \int F_x dx + \int F_y dy \\ &= F_x \int dx + F_y \int dy \\ &= \sum_i F_i \int dr_i. \end{align*} – 2017-01-03
  • 0
    so in the last line you mean $$\sum_i \left(F_i\int dr_i\right)$$? – 2017-01-03
  • 0
    That is correct! – 2017-01-03