1. ## Work

Calculate the work done in moving the particle under the action of a force $F(x,y,z) = (2x+3y)i + xyj$ along the arc of the circle $x^2 + (y-1)^2 = 1$ of (0,0) to (1,1).

I am confused on the parameterization

2. Originally Posted by Apprentice123
Calculate the work done in moving the particle under the action of a force $F(x,y,z) = (2x+3y)i + xyj$ along the arc of the circle $x^2 + (y-1)^2 = 1$ of (0,0) to (1,1).

I am confused on the parameterization

A standard parameterization of the unit circle centered at (0,0), $x^2+ y^2= 1$ is $x= cos(\theta)$, $y= sin(\theta)$ because $x^2+ y^2= cos^2(\theta)+ sin^2(\theta)= 1$.

That can be shifted to center (0, 1) using $x= cos(\theta)$, $y- 1= sin(\theta)$ so $y= 1+ sin(\theta)$. To go from from (0,0) to (1,1), which is counter clockwise, integrate from $\theta= 3\pi/2$ to $2\pi$.

3. Originally Posted by HallsofIvy
A standard parameterization of the unit circle centered at (0,0), $x^2+ y^2= 1$ is $x= cos(\theta)$, $y= sin(\theta)$ because $x^2+ y^2= cos^2(\theta)+ sin^2(\theta)= 1$.

That can be shifted to center (0, 1) using $x= cos(\theta)$, $y- 1= sin(\theta)$ so $y= 1+ sin(\theta)$. To go from from (0,0) to (1,1), which is counter clockwise, integrate from $\theta= 3\pi/2$ to $2\pi$.
Ok. Thank you