Line integral of a vector field

[nmsurobert]

Homework Statement

Consider the vector field F(r) = Φ^
(a) Calculate ∫ F⋅dl where C is a circle of radius R (oriented counterclockwise) in the xy-plane centered on the origin.

Homework Equations

maybe
Φ^ = -sinΦx^ + cosΦy^

The Attempt at a Solution

not really a solution. i am just stuck at what "dl" should be. if i go by my notes the "dl" is equal ∂l/∂θ (θ). but in our example its in terms of θ. so i don't know if "dl" here is equal to -sinΦx^ + cosΦy^. but can i evaluate the integral from 0 to 2π with Φ and not θ.

thanks.

 

[Orodruin]

In order to perform a line integral, find a parametrisation of the curve you are integrating along. You can then express ##d\vec l## according to
$$ d\vec l = \frac{d\vec r}{dt} dt$$
where ##t## is the curve parameter.

So first order of business: Can you find a parametrisation of the curve?

 

[nmsurobert]

Well I think t is Φ

 

[Orodruin]

You are on the right track, but you must be much more specific. A given t should uniquely identify a point on the curve, you might have φ = t, but what are the other coordinates for a given t?

 

[nmsurobert]

I also know the radius, R.

 

[Orodruin]

So write down the following functions of t:
##\phi(t) = \ldots##, ##\theta(t)= \ldots##, ##r(t)= \ldots##
Please try to do things in a systematic and proper way, it will help you in the long run.

 

[nmsurobert]

ok i figured it out after some note digging. but i still don't quite understand how to solve for dr/dt dt.but i just i guess just had a "duh" moment. i shouldve realized that because it centered on the xy plane i can just use cylindrical coordinates.

i can just treat it like the base of a cylinder. so dl would equal ρΦ^dΦ evaluated from 0 to 2π and ρ = R.

so the solution to the problem is simply R2π.

you for the replies.