# Work and line integral

## Homework Statement

Hi everybody! I would like to make sure I properly solved that problem because I find the result strange:

Given a force field ##F_x = axy^3##, ##F_y = bx^2y^2##, ##F_z = cz^3##.
Calculate the work with the line integral ##\int_{C} \vec{F} \cdot d\vec{r}## from point ##P_1(1,0,0)## to ##P_2(0,1,1)## along:
a) a line;
b) an helix by the z-axis.

## The Attempt at a Solution

Okay so first I parametered the line:
$$\vec{r}(\tau) = (1-\tau, \tau, \tau) \mbox{ with } \tau \in [0,1]$$
Is the first component of my vector correct? I have a little doubt about it.
$$\implies \dot{\vec{r}}(\tau) = (-1,1,1) \\ \implies A = \int_{0}^{1} \vec{F} \cdot \vec{r} \cdot d\tau \\ = \int_{0}^{1} (a (1 - \tau) \tau^3, b(1-\tau)^2\tau^2, c\tau^3)(-1,1,1)d\tau \\$$
Here again, I have a little doubt about what I did with the force field vector. Is that correct?
$$\int_{0}^{1} (-a\tau^3 + a\tau^4 + b\tau^2 - 2b\tau^3 + b\tau^4 + c\tau^3) d\tau \\ a[\frac{\tau^5}{5} - \frac{\tau^4}{4}]_0^1 + b[\frac{\tau^5}{5} + \frac{\tau^3}{3} - \frac{\tau^4}{2}]_0^1 + c[\frac{\tau^4}{4}]_0^1 \\ A = -\frac{a}{20} + \frac{b}{30} + \frac{c}{4}$$
So yeah here is my result for (a). I'd like to know if I use the right method before I attempt to solve (b).

What do you guys think?

Julien.

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TSny
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Looks good. (In your integral for A you meant to type a dot over the r vector.)

@TSny Ah yes, thanks a lot. I leave it as it is so that your message doesn't look strange in the feed :)

Great that it is correct. I'm struggling to find the equation for (b) though, I don't even get how to search for it. There is a picture with the problem, I attached it.

Thanks for all your help, I'm always making progress here.

Julien.

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Is the equation ##x(\tau) = \cos \tau, y(\tau) = \sin \tau, z = \tau## actually?

TSny
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Is the equation ##x(\tau) = \cos \tau, y(\tau) = \sin \tau, z = \tau## actually?
Close. Make sure the path goes through the final point (0, 1, 1).

@TSny Oh yeah I said that kind of randomly but I think I get it now. What about that: ##x(\tau) = \cos( \tau), y(\tau) = \sin (\frac{\tau}{2}), z(\tau) = \sin(\frac{\tau}{2})## with ## \tau \in [0,\pi]##?

TSny
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This is no longer a helix. You were much closer before.

@TSny Oops. The integral was also not that nice anymore. Okay another try: ##x(\tau) = \cos(\pi \tau), y(\tau) = sin(\frac{\pi}{2} \tau) , z(\tau) = \tau ## with this time ##\tau \in [0,1]##.

TSny
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This is not a helix either and it doesn't pass through P2. But it's pretty close!

@Ah I think I was thinking too much about the path. This should be it: ##x(\tau) = cos(\frac{\pi}{2} \tau) , y(\tau) = sin(\frac{\pi}{2} \tau), z(\tau) = \tau ##, right?

TSny
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@Ah I think I was thinking too much about the path. This should be it: ##x(\tau) = cos(\frac{\pi}{2} \tau) , y(\tau) = sin(\frac{\pi}{2} \tau), z(\tau) = \tau ##, right?
Right. Good luck with the integration. A little tedious. Can you use software?

@TSny I guess I could, but I'll first give it a go. See you in 20 hours TSny
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• JulienB
@TSny At the end I used a calculator because I must go to sleep. Did you also calculate it? I get ##A = \frac{-a}{5} + \frac{2b}{15} + \frac{c}{4}##.

I have an extra question: in the next exercise, they ask for which values a, b, c does this force field have a potential, and what is it. I answered the first one like that:

##\vec{F} \mbox{ is conservative, if } \vec{\nabla} \times \vec{F} = 0\\
\vec{\nabla} \times \vec{F} = (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z})\vec{e_x} + (\frac{\partial F_x}{\partial x} - \frac{\partial F_z}{\partial x})\vec{e_y} + (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y})\vec{e_z} \\
= 0 \vec{e_x} + 0 \vec{e_y} + (2bxy^2 - 3axy^2) \vec{e_z} \\
= xy^2 (2b - 3a) \vec{e_z} \\
\implies \vec{F} \mbox{ is conservative, if } 2b - 3a = 0 \\
\implies b = \frac{3}{2} a##
Is that a correct answer? I'm very unexperienced with gradients. What does it mean physically that z doesn't play a role in the "conservativeness" of the field?

Thanks a lot.

Julien.

TSny
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@TSny At the end I used a calculator because I must go to sleep. Did you also calculate it? I get ##A = \frac{-a}{5} + \frac{2b}{15} + \frac{c}{4}##.
Yes, that's what my computer got.
I have an extra question: in the next exercise, they ask for which values a, b, c does this force field have a potential, and what is it. I answered the first one like that:

##\vec{F} \mbox{ is conservative, if } \vec{\nabla} \times \vec{F} = 0\\
\vec{\nabla} \times \vec{F} = (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z})\vec{e_x} + (\frac{\partial F_x}{\partial x} - \frac{\partial F_z}{\partial x})\vec{e_y} + (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y})\vec{e_z} \\
= 0 \vec{e_x} + 0 \vec{e_y} + (2bxy^2 - 3axy^2) \vec{e_z} \\
= xy^2 (2b - 3a) \vec{e_z} \\
\implies \vec{F} \mbox{ is conservative, if } 2b - 3a = 0 \\
\implies b = \frac{3}{2} a##
Yes, very good.
What does it mean physically that z doesn't play a role in the "conservativeness" of the field?
I'm not sure what you are asking here. In this particular problem, only the z-component of F depends on z. Thus, the curl of F does not involve z. But, I don't think that addresses your question.

@TSny Thanks again for your answer. Nevermind my question, I think it was senseless. :)

To calculate the potential, can I do the integral of the force field?
$$\phi = - \int \vec{F} d\vec{r} = - \int (axy^3, \frac{3}{2} ax^2y^2, c z^3) d\vec{r} \\ = - (\frac{a}{2} x^2 y^3, \frac{a}{2} x^2 y^3, \frac{c}{4} z^4) + \vec{c}$$

If that's correct, I guess I could also determine the constant with the help of the points ##P_1## and ##P_2##, right?

Julien.

TSny
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Note that you ended up with a vector expression for something that should be a scalar function. Suppose you integrated the force along a path from the origin to some arbitrary point (x, y, z)? What would the integral represent in terms of ##\phi(x, y, z)##?

Okay, I am not sure to properly understand what you said but I give it a go using some resources from Internet too. It's kind of tedious, so after the 1st part I'll skip the steps:

$$\vec{F} = -\nabla \phi = -(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}) \\ \implies axy^3 = - \frac{\partial \phi}{\partial x} \implies d\phi = - \int axy^3 dx \\ \implies \phi = -\frac{a}{2} x^2y^3 + g(y,z) \\ \implies \frac{\partial \phi}{\partial y} = \frac{\partial g(y,z)}{\partial y} = -\frac{3}{2} a x^2 y^2 \\ \implies g(y,z) = -\frac{3}{2}a \int x^2y^2 dy = -\frac{a}{2}x^2y^3 + h(z) \\ \implies \frac{\partial h(z)}{\partial z} = -cz^3 \\ \implies h(z) = -c \int z^3 dz = -\frac{c}{4} z^4 \\ \implies \phi = -ax^2y^3 - \frac{c}{4} z^4$$

What do you think? Did I do that right now?

Julien.

TSny
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$$\vec{F} = -\nabla \phi = -(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}) \\ \implies axy^3 = - \frac{\partial \phi}{\partial x} \implies d\phi = - \int axy^3 dx \implies \phi = -\frac{a}{2} x^2y^3 + g(y,z) \\ \implies \frac{\partial \phi}{\partial y} = \frac{\partial g(y,z)}{\partial y} = -\frac{3}{2} a x^2 y^2 \\$$
Is the middle expression of the last line correct if you use the expression for ##\phi## at the end of the previous line?

My earlier suggestion was to integrate F along a path from the origin to an arbitrary point (x, y, z). Up to a sign, this will give you the change in ##\phi##: ##\phi(x, y, z) - \phi(0,0,0)##. Thus, you will have an expression for ##\phi(x, y, z)##. The result is independent of path, so you can choose a path that makes it easy to do the integral. But this method is not very different than your method.

@TSny Thanks for your answer (so quick!). I don't find the error though. I get it that the partial derivative of my final expression of Φ with respect to x doesn't match with the line you pointed out, but I didnt find yet why.

TSny
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$$\vec{F} = -\nabla \phi = -(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}) \\ \implies axy^3 = - \frac{\partial \phi}{\partial x} \implies d\phi = - \int axy^3 dx \\ \implies \phi = -\frac{a}{2} x^2y^3 + g(y,z)$$
OK to here.
$$\implies \frac{\partial \phi}{\partial y} = \frac{\partial g(y,z)}{\partial y}$$
Shouldn't it be $$\frac{\partial \phi}{\partial y} = -\frac{\partial \left(\frac{a}{2} x^2y^3 \right)}{\partial y} + \frac{\partial g(y,z)}{\partial y}$$

Oh yeah right. I'll come back to you soon with a corrected version. Thanks a lot.

Julien.

$$g(y,z) = ax^2 \int -\frac{3}{2} y^2 + y dy = \frac{a}{2} x^2 (y^2 - y^3) + h(z) \\ \frac{\partial h(z)}{\partial z} = -cz^3 \\ h(z) = -\frac{c}{4} z^4 \\ \implies \phi = \frac{a}{2}x^2y^2 - \frac{c}{4}z^4$$

Hopefully that is better. What d'you think?

Ju.

haruspex
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@TSny Thanks again for your answer. Nevermind my question, I think it was senseless. :)

To calculate the potential, can I do the integral of the force field?
$$\phi = - \int \vec{F} d\vec{r} = - \int (axy^3, \frac{3}{2} ax^2y^2, c z^3) d\vec{r} \\ = - (\frac{a}{2} x^2 y^3, \frac{a}{2} x^2 y^3, \frac{c}{4} z^4) + \vec{c}$$

If that's correct, I guess I could also determine the constant with the help of the points ##P_1## and ##P_2##, right?

Julien.
That was fine, except that you did not take the dot product. (f(), g(), h()).(dx,dy,dz)=?

TSny
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$$g(y,z) = ax^2 \int -\frac{3}{2} y^2 + y dy = \frac{a}{2} x^2 (y^2 - y^3) + h(z)$$
I don't see how you are getting this. You should have found that ##\frac{\partial g(y, z)}{\partial y} = 0##, which leads to a much simpler expression for ##g(y, z)##.

@TSny I had a mistake in the power of y. Now I get:

$$g(y,z) = \int 0 dy + h(z) = 0 + h(z) \\ \implies \frac{\partial \phi}{\partial z} = \frac{\partial h(z)}{\partial z} = -cz^3 \\ \implies h(z) = -c \int z^3 dz = \frac{-c}{4} z^4 + C \\ \implies \phi = -\frac{a}{2}x^2y^3 - \frac{c}{4} z^4 + C$$

@haruspex Now trying to do the integral properly:

$$\phi = - \int \vec{F} d\vec{r} = - (\frac{a}{2} x^2y^3 + \frac{a}{2} x^2y^3 + \frac{c}{4} z^4 + C) \\ = - ax^2y^3 - \frac{c}{4} z^4 + C$$

Now that's only a little different from the other one, but I'm still searching for the mistake.

Thanks a lot to both of you for helping me.

Julien.

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What I see is that only the first solution in my last post satisfies the equation ## \vec{F} = -(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z})##. It's very frustrating for me to not find the mistake in the integration though.

haruspex
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What I see is that only the first solution in my last post satisfies the equation ## \vec{F} = -(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z})##. It's very frustrating for me to not find the mistake in the integration though.
I was wrong to say what you were doing in post #20 was ok. That integral is not a valid way to find the potential, as you discovered.

• JulienB
@haruspex Ah okay! No problem, the most I learn comes from the mistakes I make. Thanks for your advices!

TSny
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##\phi = - \int \vec{F} \cdot d\vec{r}##

This will get you ##\phi## as long as ##d\vec{r}## represents a displacement along a path that takes you from some reference point to the point (x, y, z) where you want to know ##\phi(x, y, z)##. So, the components of ##d\vec{r}## are not independent during the integration. But you can pick any path you want. For example, you can take the origin as the reference point and then integrate along the x axis to (x, 0, 0), then integrate parallel to the y axis to (x, y, 0), and then integrate parallel to the z axis to (x, y, z).