Calculating Work with Line Integrals: Solving a Force Field Problem

[JulienB]

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.
$$\implies \dot{\vec{r}}(\tau) = (-1,1,1) \\<br /> \implies A = \int_{0}^{1} \vec{F} \cdot \vec{r} \cdot d\tau \\<br /> = \int_{0}^{1} (a (1 - \tau) \tau^3, b(1-\tau)^2\tau^2, c\tau^3)(-1,1,1)d\tau \\$$
Here again, I have a little doubt about what I did with the force field vector. Is that correct?
$$\int_{0}^{1} (-a\tau^3 + a\tau^4 + b\tau^2 - 2b\tau^3 + b\tau^4 + c\tau^3) d\tau \\<br /> a[\frac{\tau^5}{5} - \frac{\tau^4}{4}]_0^1 + b[\frac{\tau^5}{5} + \frac{\tau^3}{3} - \frac{\tau^4}{2}]_0^1 + c[\frac{\tau^4}{4}]_0^1 \\<br /> A = -\frac{a}{20} + \frac{b}{30} + \frac{c}{4}$$
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?

Thanks a lot in advance for your answers and suggestions, I appreciate it.Julien.

 

[TSny]

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

 

[JulienB]

@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.

 

[JulienB]

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

 

[TSny]

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).

 

[JulienB]

@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]

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

 

[JulienB]

@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]

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

 

[JulienB]

@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]

@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?

 

[JulienB]

@TSny I guess I could, but I'll first give it a go. See you in 20 hours

 

[TSny]

(It's not THAT bad.)

 

[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]

@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:

$\vec{F} \mbox{ is conservative, if } \vec{\nabla} \times \vec{F} = 0\\ \vec{\nabla} \times \vec{F} = (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z})\vec{e_x} + (\frac{\partial F_x}{\partial x} - \frac{\partial F_z}{\partial x})\vec{e_y} + (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y})\vec{e_z} \\ = 0 \vec{e_x} + 0 \vec{e_y} + (2bxy^2 - 3axy^2) \vec{e_z} \\ = xy^2 (2b - 3a) \vec{e_z} \\ \implies \vec{F} \mbox{ is conservative, if } 2b - 3a = 0 \\ \implies b = \frac{3}{2} a$
Is that a correct answer?

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.

 

[JulienB]

@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?
$$\phi = - \int \vec{F} d\vec{r} = - \int (axy^3, \frac{3}{2} ax^2y^2, c z^3) d\vec{r} \\<br /> = - (\frac{a}{2} x^2 y^3, \frac{a}{2} x^2 y^3, \frac{c}{4} z^4) + \vec{c}$$

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

Thanks a lot in advance for your answer.Julien.

 

[TSny]

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 $\phi(x, y, z)$?

 

[JulienB]

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:

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

What do you think? Did I do that right now?Thanks a lot for your answers.

Julien.

 

[TSny]

$$\vec{F} = -\nabla \phi = -(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}) \\<br /> \implies axy^3 = - \frac{\partial \phi}{\partial x} \implies d\phi = - \int axy^3 dx \implies \phi = -\frac{a}{2} x^2y^3 + g(y,z) \\<br /> \implies \frac{\partial \phi}{\partial y} = \frac{\partial g(y,z)}{\partial y} = -\frac{3}{2} a x^2 y^2 \\$$

Is the middle expression of the last line correct if you use the expression for $\phi$ at the end of the previous line?

But your method is good.

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.

 

[JulienB]

@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]

$$\vec{F} = -\nabla \phi = -(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}) \\<br /> \implies axy^3 = - \frac{\partial \phi}{\partial x} \implies d\phi = - \int axy^3 dx \\<br /> \implies \phi = -\frac{a}{2} x^2y^3 + g(y,z)$$

OK to here.

$$\implies \frac{\partial \phi}{\partial y} = \frac{\partial g(y,z)}{\partial y}$$

Shouldn't it be $$\frac{\partial \phi}{\partial y} = -\frac{\partial \left(\frac{a}{2} x^2y^3 \right)}{\partial y} + \frac{\partial g(y,z)}{\partial y}$$

 

[JulienB]

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

Thanks a lot.Julien.

 

[JulienB]

Okay new answer:

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

Hopefully that is better. What d'you think?Ju.

 

[haruspex]

@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?
$$\phi = - \int \vec{F} d\vec{r} = - \int (axy^3, \frac{3}{2} ax^2y^2, c z^3) d\vec{r} \\<br /> = - (\frac{a}{2} x^2 y^3, \frac{a}{2} x^2 y^3, \frac{c}{4} z^4) + \vec{c}$$

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

Thanks a lot in advance for your answer.Julien.

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

 

[TSny]

Okay new answer:

$$g(y,z) = ax^2 \int -\frac{3}{2} y^2 + y dy = \frac{a}{2} x^2 (y^2 - y^3) + h(z)$$

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 $g(y, z)$.

 

[JulienB]

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

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

@haruspex Now trying to do the integral properly:

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

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.

 

[JulienB]

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]

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]

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

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).