Line Integral: Solving for \ointr.dr=0

[rado5]

Homework Statement

What is the result of this? \ointr.dr=?

Homework Equations

The Attempt at a Solution

\ointr.dr =\ointrdr=\int^{a}_{a}rdr = \frac{r^2}{2} \left| ^{a}_{a} = 0

Is it correct?

 

[HallsofIvy]

What argument do you have for asserting that \vec{r}\cdot d\vec{r}= r dr? Generally, the dot product of two vectors is equal to the product of their lengths when the vectors are parallel. Are \vec{r} and d\vec{r} parallel?

In Cartesian coordinates, \vec{r}= x\vec{i}+ y\vec{j} and d\vec{r}= dx\vec{i}+ dy\vec{j} so your integral becomes \oint xdx+ ydy around some closed path.

A circle of radius R can be written as \vec{r}= Rcos(\theta)\vec{i}+ R sin(\theta)\vec{j} and then d\vec{r}= -Rsin(\theta)d\theta\vec{i}+ Rcos(\theta)d\theta\vec{j} so your integral becomes R^2\int_{\theta= 0}^{2\pi} -sin(\theta)cos(\theta)+ cos(\theta) sin(\theta) d\theta which is even simpler.

 

[rado5]

Dear "HallsofIvy" thank you very much for your wonderful help. I thought the angle between r and dr is always zero because the angle between two curves are defined as the angle between their tangent lines.

 

[HallsofIvy]

Okay. And what are the tangent lines of "r" and "dr"?

 

[rado5]

Okay. And what are the tangent lines of "r" and "dr"?

Well now I think that the tangent lines are different because we have two different vectors, but I didn't notice it!

 

[HallsofIvy]

I may be pulling your chain a little too hard here! Yes, r and dr are parallel (they both point directly away from the origin) and so their dot product is just rdr. My point was that you have to show that, not just assert it.

But it is simpler to do it as R^2\int_{\theta= 0}^{2\pi} -sin(\theta)cos(\theta)+ cos(\theta) sin(\theta) d\theta because, of course, -sin(\theta)cos(\theta)+ cos(\theta)sin(\theta)= 0.

 

[rado5]

Okay. And what are the tangent lines of "r" and "dr"?

The tangent line of \vec{r} is placed on d\vec{r} and the tangent line of d\vec{r} is placed on itself because it is a straight line (because \vec{r}= x\vec{i} + y\vec{j}) so the angle between \vec{r} and d\vec{r} is zero.
Is it correct? If there is a better interpretation please let me know. And thank you very much in advance for your great help.

 

[lanedance]

have you looked at the definition of conservative vector fields at all?