Is a Line Integral Zero if the Vector Field is Not Conservative?

[ZARATHUSTRA]

calculate the line integral for a vector field F= -xy⋅j over a circle which is c(t)=costi+sintj,
so I used x=cost y=sint and ∫(0 to 2pi) -(sintcost)(cost)dt=(cos^3(2pi)-cos^3(o))/3=0 now here is the problem, if this enclosed line integral is zero then why is the vector field not conservative?

 

[ZARATHUSTRA]

dr(t)/dt=(-sint,cost) then dot with F gets -(sintcos^2t)

 

[ZARATHUSTRA]

cant find anything wrong with my calculation process
and the textbook tells me that if enclosed line integral is zero then vector field must be conservative which means there is something with my work but I checked my work so many times and I just can't find it

 

[vanhees71]

Why should it be conservative, if just the integral along one closed line vanishes? It's definitely not conservative, not even locally, because $\vec{\nabla} \times \vec{F} \neq 0$.

 

[ZARATHUSTRA]

Thanks in advance for anyone who takes time to read it and would like to help me with the problem

 

[ZARATHUSTRA]

Why should it be conservative, if just the integral along one closed line vanishes? It's definitely not conservative, not even locally, because $\vec{\nabla} \times \vec{F} \neq 0$.

right, but the enclosed line integral is zero, which means either I am wrong with the integral calculation, or the vector field is conservative

 

[ZARATHUSTRA]

Why should it be conservative, if just the integral along one closed line vanishes? It's definitely not conservative, not even locally, because $\vec{\nabla} \times \vec{F} \neq 0$.

but I just can't find where I calculated wrong

 

[vanhees71]

The line integral vanishes, but obviously not all line integrals. Try
$$\vec{x}(t)=\cos t \vec{i} + (1+\sin t) \vec{j} $$
instead.

 

[ZARATHUSTRA]

The line integral vanishes, but obviously not all line integrals. Try
$$\vec{x}(t)=\cos t \vec{i} + (1+\sin t) \vec{j} $$
instead.

what do you mean it vanishes? is my line integral zero?

 

[ZARATHUSTRA]

The line integral vanishes, but obviously not all line integrals. Try
$$\vec{x}(t)=\cos t \vec{i} + (1+\sin t) \vec{j} $$
instead.

but this doesn't describe the circle, does it?

 

[vanhees71]

I think your calculation is right, although it's hard to read. Use LaTeX:

https://www.physicsforums.com/help/latexhelp/

 

[ZARATHUSTRA]

I think your calculation is right, although it's hard to read. Use LaTeX:

https://www.physicsforums.com/help/latexhelp/

sorry about that, I will, I was very anxious.

 

[vanhees71]

but this doesn't describe the circle, does it?

It describes a unit circle with its center in $(0,1,0)$. You'll see that it's a closed line for which the line integral does not vanish, and that explains, why the field is not conservative. For a field to be conservative (in some region) all line integrals along any closed line must vanish (in this region).

 

[ZARATHUSTRA]

It describes a unit circle with its center in $(0,1,0)$. You'll see that it's a closed line for which the line integral does not vanish, and that explains, why the field is not conservative. For a field to be conservative (in some region) all line integrals along any closed line must vanish (in this region).

right, but I was calculating the line integral of a circle centered at (0,0,0)

 

[vanhees71]

what do you mean it vanishes? is my line integral zero?

Yes, your line integral vanishes, as you have correctly evaluated yourself. If you try the other line integral instead, you'll see this one doesn't vanish, and thus the field cannot be conservative!

 

[vanhees71]

Please think about what I try to explain with my answers!

 

[ZARATHUSTRA]

It describes a unit circle with its center in $(0,1,0)$. You'll see that it's a closed line for which the line integral does not vanish, and that explains, why the field is not conservative. For a field to be conservative (in some region) all line integrals along any closed line must vanish (in this region).[/QUOT

Yes, your line integral vanishes, as you have correctly evaluated yourself. If you try the other line integral instead, you'll see this one doesn't vanish, and thus the field cannot be conservative!

and if The region I calculated is a closed curve and it is zero, then it should be conservative, right?

 

[vanhees71]

NO! The field is conservative in a region, if the line integrals along ALL closed lines in that region vanish. It's not sufficient that one special line integral along a special closed curve vanishes!

 

[ZARATHUSTRA]

Please think about what I try to explain with my answers!

yes of course, the integral I calculated from your region is definitely NOT zero. But the region I calculated IS zero, and the region is still closed curve. But the textbook tells me that if the Line integral of an enclosed curve is zero, then the vector field must be conservative.

 

[vanhees71]

Which textbook? If it really claims what you say, find a better one!

Just to give you another example for something really puzzling! Consider the scalar field
$$\phi(\vec{x})=\arctan(y/x).$$

(a) calculate the gradient field $\vec{V}=-\vec{\nabla} \phi$.
(b) Integrate this vector field along the unit circle $\vec{x}(t)=\cos t \vec{i} + \sin t \vec{j}$.
(c) Explain the (hopefully) surprising result!

 

[ZARATHUSTRA]

Which textbook? If it really claims what you say, find a better one!

hmm maybe I interpreted wrong

 

[ZARATHUSTRA]

so I guess the statement "if the Line integral of an enclosed curve is zero, then the vector field must be conservative." is false

 

[ZARATHUSTRA]

Thank you so much sir, I think I get it now.

 

[vanhees71]

so I guess the statement "if the Line integral of an enclosed curve is zero, then the vector field must be conservative." is false

Which book is it? Maybe it's good to avoid it in the future!

 

[ZARATHUSTRA]

It was my bad and hasty interpretation. But the Book name is Calculus Early Transcendentals(2nd E) by Briggs Cochran Gillett