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

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• ZARATHUSTRA
In summary: The gradient field ##\vec{V}=-\vec{\nabla} \phi## is conservative everywhere, because the line integral along any closed curve vanishes.(b) The integral in (c) is zero, because the line integral along the unit circle ##\vec{x}(t)=\cos t \vec{i} + \sin t \vec{j}## is zero.
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?

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

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

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##.

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

vanhees71 said:
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

vanhees71 said:
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

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

vanhees71 said:
The line integral vanishes, but obviously not all line integrals. Try
$$\vec{x}(t)=\cos t \vec{i} + (1+\sin t) \vec{j}$$
what do you mean it vanishes? is my line integral zero?

vanhees71 said:
The line integral vanishes, but obviously not all line integrals. Try
$$\vec{x}(t)=\cos t \vec{i} + (1+\sin t) \vec{j}$$
but this doesn't describe the circle, does it?

ZARATHUSTRA said:
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).

vanhees71 said:
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)

ZARATHUSTRA said:
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 said:
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
vanhees71 said:
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?

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!

vanhees71 said:
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.

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!

Last edited:
vanhees71 said:
Which textbook? If it really claims what you say, find a better one!
hmm maybe I interpreted wrong

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

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

ZARATHUSTRA said:
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!

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

## 1. What is a line integral?

A line integral is a mathematical concept used in vector calculus to calculate the total value of a function along a specified path or curve. It takes into account both the magnitude and direction of the function, making it a powerful tool for analyzing physical systems.

## 2. How is a line integral evaluated?

To evaluate a line integral, you need to first parameterize the path or curve in terms of a single variable. Then, you must determine the appropriate limits of integration and use the fundamental theorem of calculus to integrate the function along the path. This will result in a numerical value that represents the total value of the function along the specified path.

## 3. What is the significance of line integrals in science?

Line integrals have many applications in science. For example, they are used in physics to calculate work done by a force along a specific path, and in electromagnetism to calculate the electric potential along a path. They are also used in fluid mechanics to calculate the flow of a fluid along a certain path.

## 4. What are some common types of line integrals?

The most common types of line integrals are the path integral, where the path is specified explicitly, and the line integral of the first and second kind, where the path is determined by a vector field. Other types include the contour integral, which is used in complex analysis, and the surface integral, which is used in multivariable calculus.

## 5. Are there any practical applications of line integrals?

Line integrals have many practical applications in fields such as physics, engineering, and economics. They are used to calculate quantities such as work, force, electric potential, and fluid flow, making them essential tools in understanding and analyzing real-world systems. Additionally, line integrals are also used in computer graphics to render 3D images and in geographic information systems to analyze terrain and elevation data.

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