# Line integral over a vector field

#### says

1. The problem statement, all variables and given/known data
Evaluate ∫C < −y, x − 1 > dr where C is the closed piecewise continuous curve formed by the line segment joining the point A(− √ 2, √ 2) to the point B( √ 2, − √ 2) followed by the arch of the circle of radius 2, centered at the origin, from B to A.
2. The attempt at a solution
Vector field, F = < −y, x − 1 > = -y i + (x-1) j
F
(r(t)) = (-sin(t))i + (cos(t)-1)j
r
(t) = cos(t)i+sin(t)j
r
'(t) = -sin(t)i+cos(t)j
F
(r(t))⋅r'(t)=sin2(t)+cos2(t)-1
Now I need to evaluate that between the regions 0≤t≤π
I just wanted to check if I've set this up correctly. The arc from point B to A is = 180°= π, so that's why I've defined the region of t as going from 0≤t≤π.

Related Calculus and Beyond Homework News on Phys.org

#### LCKurtz

Homework Helper
Gold Member
1. The problem statement, all variables and given/known data
Evaluate ∫C < −y, x − 1 > dr where C is the closed piecewise continuous curve formed by the line segment joining the point A(− √ 2, √ 2) to the point B( √ 2, − √ 2) followed by the arch of the circle of radius 2, centered at the origin, from B to A.
2. The attempt at a solution
Vector field, F = < −y, x − 1 > = -y i + (x-1) j
F
(r(t)) = (-sin(t))i + (cos(t)-1)j
r
(t) = cos(t)i+sin(t)j
r
'(t) = -sin(t)i+cos(t)j
F
(r(t))⋅r'(t)=sin2(t)+cos2(t)-1
Now I need to evaluate that between the regions 0≤t≤π
I just wanted to check if I've set this up correctly. The arc from point B to A is = 180°= π, so that's why I've defined the region of t as going from 0≤t≤π.
Your parameterization is not a circle of radius $2$. And your $t$ range is incorrect. It needs to start at the right place when $t=0$.

#### says

Oh, I forgot the radius = 2. I think (below) is better!
r(t) = 2cos(t)i+2sin(t)j
r'(t) = -2sin(t)i+2cos(t)j
As for the range.
( √ 2, − √ 2)≤t≤(− √ 2, √ 2)
This is the same as the range 0≤t≤π.

#### LCKurtz

Homework Helper
Gold Member
Oh, I forgot the radius = 2. I think (below) is better!
r(t) = 2cos(t)i+2sin(t)j
r
'(t) = -2sin(t)i+2cos(t)j
As for the range.
( √ 2, − √ 2)≤t≤(− √ 2, √ 2)
This is the same as the range 0≤t≤π.
No it isn't. $\vec r(0) =\langle 2,0\rangle$.

#### says

edit: not sure if you were correcting my last statement or my r(t)?

7π/4≤t≤3π/4

Last edited:

#### Mark44

Mentor
As for the range.
( √ 2, − √ 2)≤t≤(− √ 2, √ 2)
This makes no sense. t is a real number that you're comparing to two points. In addition there is no ordering principle for points, so you can't say ( √ 2, − √ 2) ≤ (− √ 2, √ 2)
says said:
This is the same as the range 0≤t≤π.

edit: not sure if you were correcting my last statement or my r(t)?

7π/4≤t≤3π/4
This is saying, among other things, that 7π/4 ≤ 3π/4, which is clearly not true.

#### says

Ok. I see what you mean. I'm not sure where to go from here though with defining the range...

#### LCKurtz

Homework Helper
Gold Member
Your original post did not tell you which arch to take getting back from B to A. You have to decide that first and make sure your parameterization goes in the correct direction, either clockwise or counterclockwise. Then you need to think about what geometric angle $t$ represents in your parameterization. Make sure it starts at B and traverses the arch to A in the direction you want.

#### says

I'm having trouble understanding what you mean, sorry. I'm assuming we are calculating the integral from B to A, so it's counterclockwise. It start at an angle of 315 degrees, and finishes at an angle of 135 degrees, a difference of 180 degrees.

I understood the question to mean that we have a vector field, < −y, x − 1 >, and a path from point B to point A, which followed the arc of a circle that was centred at the origin and had a radius 2. i.e. I pictured it as a particle moving in a semi-circle from point B to point A. The particle's path is determined by the field vector, F, and it's own movement, r(t). (Sorry if I don't sound to mathematically rigorous here, I'm still trying to get the intuition of line integrals over vector fields.)

#### LCKurtz

Homework Helper
Gold Member
I'm having trouble understanding what you mean, sorry. I'm assuming we are calculating the integral from B to A, so it's counterclockwise. It start at an angle of 315 degrees, and finishes at an angle of 135 degrees, a difference of 180 degrees.
While there is nothing wrong with assuming it is counterclockwise, your original problem doesn't say that. It's OK with me to assume that, but understand that you could just as well go clockwise on the circle from B to A, and the answers are probably different. The direction should have been stated in the problem, or maybe you just left it out. Now, assuming you want to go counterclockwise and start at 315 degrees, you won't end at 135 degrees. Counterclockwise motion adds to the angle.

I understood the question to mean that we have a vector field, < −y, x − 1 >, and a path from point B to point A, which followed the arc of a circle that was centred at the origin and had a radius 2. i.e. I pictured it as a particle moving in a semi-circle from point B to point A.
Yes. You just didn't specify which semi-circle. See above.

The particle's path is determined by the field vector, F, and it's own movement, r(t). (Sorry if I don't sound to mathematically rigorous here, I'm still trying to get the intuition of line integrals over vector fields.)
The particles path is determined by $\vec r(t)$, not by the field vector $\vec F$. The vector field affects how much work is done in the process.

#### says

The particles path is determined by $\vec r(t)$, not by the field vector $\vec F$. The vector field affects how much work is done in the process.
Ok, I'm trying to understand that comment a bit more with a much simpler example. I jump from point B to point A. I only got to point A because I put in positive work, while the vector field (gravity) did negative work, which made me land at point A. The amount of work I did in getting from B to A is equal to the ∫F(r(t))⋅r'(t).

But back to the question:
If I start at 315 degrees, i.e. 315 degrees = 0, and I end at 135 degrees, that would mean I start at 0 degrees and end at 180 degrees. So 0≤t≤180 or 0≤t≤π. I calculated the integral with this range and got -π as a result.

#### LCKurtz

Homework Helper
Gold Member
Ok, I'm trying to understand that comment a bit more with a much simpler example. I jump from point B to point A. I only got to point A because I put in positive work, while the vector field (gravity) did negative work, which made me land at point A. The amount of work I did in getting from B to A is equal to the ∫F(r(t))⋅r'(t).
That integral calculates the work done by the vector field.

But back to the question:
If I start at 315 degrees, i.e. 315 degrees = 0,
$315\ne 0$. And it doesn't agree with $t=0$ either. See below

and I end at 135 degrees, that would mean I start at 0 degrees and end at 180 degrees. So 0≤t≤180 or 0≤t≤π. I calculated the integral with this range and got -π as a result.
You need to show your work. Are you still using $\vec r(t) = 2\cos t i + 2\sin(t) j$? If so, I have pointed out to you before that when $t=0$ it gives (2,0). And at $t=\pi$ it gives $(-2,0)$. That does not describe the correct semicircle.

Last edited:

"Line integral over a vector field"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving