# Vector calculus -line integral

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Vector calculus ---line integral

## Homework Statement

If vector F(r)= (x^2)y i + 2yz j + 9(z^2)x k , find ∫ vector F dot vector dr between (0,0,0) and (1,2,3)

## The Attempt at a Solution

If I want to find work done, then I just use F dot dr in this case , in F act in the direction of r, then I will get the answer in term of x y z, then substitute x=1, y=2, z=3 to get the work done.
But for this case, it didnt say it is work done, so still have to do ∫F dot dr , ∫Fdxi+∫Fdyj+∫Fdzk , then ∫Fdxi from x=0 to x=1 by keeping y and z constant, then what should I substitute to my y and z ?

Thank you.

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If the F here is force, then is it a conservative or non conservative force??

Are you sure you wrote everything in the problem statement? You can determine whether F is conservative or not by taking the curl of F and seeing if it vanishes.

Whether F is "conservative" or not, even whether it is a force, is unknown and irrelevant. If you knew this were a "conservative force" (in physics language- a "total differential" in mathematics) you would not have to do the line integral but here you are told to do it. The line from (0, 0, 0) to (1, 2, 3) (how original!) is given by x= t, y= 2t, z= 3t, as t goes from 0 to 1. The vector differential is $(\vec{i}+ 2\vec{j}+ 3\vec{k}) dt$. Take the dot product of that with $x^2y \vec{i} + 2yz \vec{j}j + 9z^2x \vec{j}$ and integrate.

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Are you sure you wrote everything in the problem statement? You can determine whether F is conservative or not by taking the curl of F and seeing if it vanishes.

I forget to write straight line c from (0,0,0) to (1,2,3)!
then I can define x=t, y=2t, z=3t, I know how to do already, thank you.

∇ cross F ,then I get -2yi-(3z^2)j-(x^2)k, so it is non conservative force.
Why it is conservative force will vanish?

Whether F is "conservative" or not, even whether it is a force, is unknown and irrelevant. If you knew this were a "conservative force" (in physics language- a "total differential" in mathematics) you would not have to do the line integral but here you are told to do it. The line from (0, 0, 0) to (1, 2, 3) (how original!) is given by x= t, y= 2t, z= 3t, as t goes from 0 to 1. The vector differential is $(\vec{i}+ 2\vec{j}+ 3\vec{k}) dt$. Take the dot product of that with $x^2y \vec{i} + 2yz \vec{j}j + 9z^2x \vec{j}$ and integrate.

If the path was closed and F conservative, then they wouldn't ask to do the integral. However, the question gives no information about what path we take to (1,2,3) from O. For example, it may be a helix or it may be a straight line. In the case that F was conservative then you could take any path, however it turns out F is not conservative so that is why I wondered if something was missing from OP.

EDIT: The OP has clarified this issue in the above post.

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Why is conservative force will vanish?

If F is a conservative force, we may associate a potential, that is ##\mathbf{F} = \nabla f##, where ##f## is the potential function for ##\mathbf{F}##. Now take the curl: $$\nabla \times \mathbf{F} = \nabla \times ( \nabla f ) = \mathbf{0}$$ since we have a curl-grad which is always zero.

Whether F is "conservative" or not, even whether it is a force, is unknown and irrelevant. If you knew this were a "conservative force" (in physics language- a "total differential" in mathematics) you would not have to do the line integral but here you are told to do it. The line from (0, 0, 0) to (1, 2, 3) (how original!) is given by x= t, y= 2t, z= 3t, as t goes from 0 to 1. The vector differential is $(\vec{i}+ 2\vec{j}+ 3\vec{k}) dt$. Take the dot product of that with $x^2y \vec{i} + 2yz \vec{j}j + 9z^2x \vec{j}$ and integrate.

Thank you. if it is conservative force, then the path of the work done will not important already, so that is why you said no need to do line integral.
can you please explain more about the total differetial in math mean? total differential of the force? normally if total differential will get what ?

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If F is a conservative force, we may associate a potential, that is ##\mathbf{F} = \nabla f##, where ##f## is the potential function for ##\mathbf{F}##. Now take the curl: $$\nabla \times \mathbf{F} = \nabla \times ( \nabla f ) = \mathbf{0}$$ since we have a curl-grad which is always zero.

##\mathbf{F} = \nabla f##, where ##f## is the potential function for ##\mathbf{F}##.
F is the rate of change of the potential energy ,
then $$\nabla \times ( \nabla f ) = \mathbf{0}$$
if it equal to zero ,will prove that the F have potential function , so it is conservative force. is it??

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##\mathbf{F} = \nabla f##, where ##f## is the potential function for ##\mathbf{F}##.
F is the rate of change of the potential energy ,
then $$\nabla \times ( \nabla f ) = \mathbf{0}$$
if it equal to zero ,will prove that the F have potential function , so it is conservative force. is it??

Yes, you can easily prove that curl-grad is always zero for general ##f##.