1. ## Work

I have this vector field equation, the first part of the question is to find the potential equation for it, I found it.
The second part of the question is to find the work of the field through this path.
My idea is to plug t in the r equation, because I'm not sure but I think (x,y,z)=(component of r), is that right? and that way I find the start point of the path and the end point, plug it into the potential equation and that's it, is that right?

2. Originally Posted by asi123
I have this vector field equation, the first part of the question is to find the potential equation for it, I found it.
The second part of the question is to find the work of the field through this path.
My idea is to plug t in the r equation, because I'm not sure but I think (x,y,z)=(component of r), is that right? and that way I find the start point of the path and the end point, plug it into the potential equation and that's it, is that right?
If $F = \nabla \phi$ then $W = \int_{C} F \cdot dr = \phi(x_2, y_2, z_2) - \phi(x_1, y_1, z_1)$ where $(x_1, y_1, z_1)$ are the coordinates of the start of C and $(x_2, y_2, z_2)$ are the coordinates of the end of C.

To get the coordinates of the start and endpoints of C, substitute $t = 0$ and $t = 2 \pi$ into r.

So yes, that's right.

3. Originally Posted by mr fantastic
If $F = \nabla \phi$ then $W = \int_{C} F \cdot dr = \phi(x_2, y_2, z_2) - \phi(x_1, y_1, z_1)$ where $(x_1, y_1, z_1)$ are the coordinates of the start of C and $(x_2, y_2, z_2)$ are the coordinates of the end of C.

To get the coordinates of the start and endpoints of C, substitute $t = 0$ and $t = 2 \pi$ into r.

So yes, that's right.

10x, it seemed so easy, and they gave 10 points for it (it's a question from a test), so I got confuse...