# Line Integral Example - mistake or am I missing something?

• kostoglotov
In summary: So, we need to get the scalar function of f, call it the potential function, we know that in a conservative vector field \vec{F} = \nabla f, no worries. However, shouldn't it be\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x} = -\frac{mMG|\vec{x}|}{|\vec{x}|^3}\vec{u} = -\frac{mMG}{|\vec{x}|^2}\vec{u}And so converting the vector form of F into a scalar field
kostoglotov
This is an example at the beginning of the section on the Fundamental Theorem for Line Integrals.

1. Homework Statement

Find the work done by the gravitational field

$$\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x}$$

in moving a particle from the point (3,4,12) to (2,2,0) along a piece wise smooth curve

Now, I think that there's a mistake in the solution given...but this textbook is pretty good, and plenty of times in the past I've thought it had made a mistake and really I was mistaken.

So, I understand all the concepts (edit: I obviously didn't), all good there. It's here where I'm scratching my head.

$$f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}$$

So, we need to get the scalar function of f, call it the potential function, we know that in a conservative vector field $\vec{F} = \nabla f$, no worries. However, shouldn't it be

$$\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x} = -\frac{mMG|\vec{x}|}{|\vec{x}|^3}\vec{u} = -\frac{mMG}{|\vec{x}|^2}\vec{u}$$

And so converting the vector form of F into a scalar field from which we can compute the grad vector, doesn't

$$|\vec{x}|^2 = x^2 + y^2 + z^2$$

not

$$\sqrt{x^2+y^2+z^2}$$

so shouldn't it be

$$f(x,y,z) = \frac{mMG}{|\vec{x}|^2} = \frac{mMG}{x^2+y^2+z^2}$$

how would one end up with

$$f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}$$

??

edit: both the 6th and 7th editions have the same example...so I'm guessing that I'm missing something.

Last edited:
Can you work out the gradient of the f(x,y,z) you suggest? Does it give the right force?

kostoglotov
DEvens said:
Can you work out the gradient of the f(x,y,z) you suggest? Does it give the right force?

No, it gives N not N m.

I'm still not sure what's going on here though. Is it the case that we're NOT trying to just find the magnitude of the gravity vectors on a given domain, but rather searching for some other related function, that will produce a grad vec that when dot producted with a path vector will result in the dimensions that we need in our answer?

I still don't know how they got to their answer.

edit: Oh wait, I think I see what's happening...f(x,y,z) is essentially an antiderivative ?? Thus, we need to be able to get to the grad vec from that f(x,y,z)...??

edit2: yep, that worked, and it made sense...thanks :)

Last edited:
kostoglotov said:
This is an example at the beginning of the section on the Fundamental Theorem for Line Integrals.

1. Homework Statement

Find the work done by the gravitational field

$$\vec{F}(\vec{x}) = -\frac{mMG}{|\vec{x}|^3}\vec{x}$$

in moving a particle from the point (3,4,12) to (2,2,0) along a piece wise smooth curve

Now, I think that there's a mistake in the solution given...but this textbook is pretty good, and plenty of times in the past I've thought it had made a mistake and really I was mistaken.

So, I understand all the concepts (edit: I obviously didn't), all good there. It's here where I'm scratching my head.

$$f(x,y,z) = \frac{mMG}{\sqrt{x^2+y^2+z^2}}$$Standard result: if ##\vec{r} = x\vec{e}_x + y\vec{e}_y + z \vec{e}_z## with magnitude ##r = \sqrt{x^2 + y^2 + z^2}##, then
$$\vec{ \nabla} \frac{1}{r} = -\frac{\vec{r}}{r^3}$$
Just look at ##\partial r^{-1} / \, \partial x ## for example.

## What is a line integral?

A line integral is a type of integral that is used to calculate the total value of a function along a certain curve or path. It involves breaking up the curve into small segments and adding up the values of the function at each segment.

## What is an example of a line integral?

An example of a line integral would be calculating the work done by a force along a certain path. This would involve integrating the dot product of the force and the displacement along the path.

## Why might there be a mistake in a line integral calculation?

There could be a mistake in a line integral calculation due to human error, such as a miscalculation or using the wrong formula. It could also be due to missing information or incorrect assumptions made during the calculation.

## How can I check if I made a mistake in my line integral calculation?

You can check your line integral calculation by double-checking your work and making sure you used the correct formula. You can also try solving the problem using a different method or approach to see if you get the same result.

## What should I do if I am still unsure about my line integral calculation?

If you are still unsure about your line integral calculation, it is best to seek help from a teacher, tutor, or colleague who is knowledgeable in the subject. They can review your work and help identify any mistakes or areas that need clarification.

• Calculus and Beyond Homework Help
Replies
3
Views
865
• Calculus and Beyond Homework Help
Replies
20
Views
775
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
10
Views
780
• Calculus and Beyond Homework Help
Replies
4
Views
981
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
1K