Work done along a quarter circle path

1. Feb 18, 2006

I wasnt really sure if I should post that question here since I actually didnt understand its math part.
this question actually already has its solution but I didnt understand this part, when I want to find the work done by a force along a quarter circle path, I define x and y with r and angle and which is for the first quarter of circle is x= r.cos(teta) and y=r.sin(teta) when it is the second quarter of the circle why the book took the x=rsin(teta) and y=r(1-cos(teta)) ?
do you have any idea?

2. Feb 18, 2006

Tide

The second x and y you wrote look like changes in x and y positions after integrating something over a quarter circle. Do you have any ideas why someone would do an integration for this kind of problem?

3. Feb 18, 2006

actually I have my force F=(y,2x) and I need to find the Work done along that curve so I need to integrate F.dr so it becomes integral ydx + 2xdy and I need to turn x,dx,y and dy into (r) and (theta)
there is something here but I cant see it :((

4. Feb 18, 2006

lightgrav

you can measure theta from the positive-x axis, or from the positive y-axis
(or anywhere else for that matter), but you have to be consistent throughout the entire solution.
In any case, the 2x.dy results in r^2 2.sin^2(t) or r^2 2.cos^2(t).
books will replace that 2.sin^2(t) = 1 - cos(2t) = 1 - cos^2(t) + sin^2(t) , or 2.cos^2(t) = 1 + cos(2t) ... maybe that's what they intended.
but YOU should do the integral in a way you're comfortable with!

5. Feb 18, 2006

lightgrav

you can measure theta from the positive-x axis, or from the positive y-axis
(or anywhere else for that matter), but you have to be consistent throughout the entire solution.
In any case, the 2x.dy results in r^2 2.sin^2(t) or r^2 2.cos^2(t).
books will replace that 2.sin^2(t) = 1 - cos(2t) = 1 - cos^2(t) + sin^2(t) , or 2.cos^2(t) = 1 + cos(2t) ... maybe that's what they intended.
but YOU should do the integral in a way you're comfortable with!