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We had a question which goes like this:

A particle is displaced from $(0,0)$ to $(1,1)$ along $\rm y=x$. The force $F$ on the particle is $\rm ( x^2 \hat j + y \hat i)$. Find Work done during displacement.

What I did:

$$\rm W = \int F\cdot dx$$

$$\rm = \int ( y \hat i + x^2 \hat j)\cdot( dx \hat i + dy \hat j)$$

$$\rm = \int (y \ dx + x^2 \ dy)$$

The problem is that we can't integrate with respect to one variable keeping other as constant because both variables change.

What to do in this case? Also can we put $y = x$ since it was moved along this path? Also if we were not given this condition, how we find the integral then?

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    $\begingroup$ This is a line integral (Google it), you have to parametrize the path to get a parametrization $\gamma(t)$ and evaluate the integral of $F(\gamma(t))\cdot \gamma'(t)$. $\endgroup$ Jan 30, 2016 at 6:27
  • $\begingroup$ Just set $y=x$, as you suggested, and do the integrals from 0 to 1. $\endgroup$ Jan 30, 2016 at 6:27
  • $\begingroup$ Parametrize the path $\gamma$. Here the parametrization $x(t)=t$, $y(t)=t$ with $t$ ranging over $[0,1]$ suggests itself. $\endgroup$ Jan 30, 2016 at 6:27
  • $\begingroup$ A simple parametrization that works is $\gamma:[0,1]\to \Bbb R, t\mapsto(t,t)$ (here you use the fact that $y=x$ along the path. $\endgroup$ Jan 30, 2016 at 6:29
  • $\begingroup$ What if we were not given $y = x$, then how I would solve it? $\endgroup$
    – integer
    Jan 30, 2016 at 6:32

1 Answer 1

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Correct. As you mentioned in post and others mentioned in the comments, we can find the parameter $t$ which relates $y$ and $x$.

The particle was moved along $ x = y$, therefore this relation exists between all $ x$ and $\rm y$.

So we can replace $x$ with $y$ and vice versa, or replace both x and y with parameter $t$

$$ \therefore W = \int_{(0,0)}^{(1,1)}(x \ dx+y^2\ dy) = \frac{y^2}{2} + \frac{x^3}{3}\ \Big{|}^{(1,1)}_{(0,0)} = \frac{5}{6}$$

What if the particle wasn't moved along $ x= y$ ? How would we solve the integral:

$$ = \int (y \ dx + x^2 \ dy)$$

I think is the integral would have been unsolvable. The reason could be that the force $ F = (y \hat i + x^2 \hat j)$ is non conservative since $\nabla \times F \neq 0$. So work done would depend on curve along which particle was moved, and that path equation would need to be givenm

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