Using Stokes law, calculate the work done along a curve

[skrat]

Homework Statement

Using Stokes law, calculate the work done along a curve $\Gamma$ which is defined as edge of a spherical triangle in first octant of a sphere $x^2+y^2+z^2=R^2$. Vector field is $\vec{F}=(z^2,x^2,y^2)$.

Homework Equations

Stokes law: $\int _{\partial \Sigma }\vec{F}d\vec{r}=\int \int _{\Sigma }\triangledown \times \vec{F}dA$

The Attempt at a Solution

firstly $\triangledown \times \vec{F}=(2y,2z,2x)$.

Now I should somehow parametrize that triangle but anything i try... comes out nasty. For example:
$x=t$ for $t\in \left [ 0,R \right ]$ than also $y=t$ which would from $x^2+y^2+z^2=R^2$ mean that $z^2=r^2-2t^2$

Now i have parameterization as function of t only $r(t)=(t,t,\sqrt{r^2-2t^2})$ but... to calculate dS in stokes integral, I need parametrization of two paramateres... so... How do I continue?

Is it ok if I just say that $r(t,z)=(t,t,z)$ where $t\in \left [ 0,R \right ]$ and $z\in \left [ -\sqrt{r^2-2t^2},\sqrt{r^2-2t^2} \right ]$

 

[LCKurtz]

Homework Statement

Using Stokes law, calculate the work done along a curve $\Gamma$ which is defined as edge of a spherical triangle in first octant of a sphere $x^2+y^2+z^2=R^2$. Vector field is $\vec{F}=(z^2,x^2,y^2)$.

Homework Equations

Stokes law: $\int _{\partial \Sigma }\vec{F}d\vec{r}=\int \int _{\Sigma }\triangledown \times \vec{F}dA$

The Attempt at a Solution

firstly $\triangledown \times \vec{F}=(2y,2z,2x)$.

Now I should somehow parametrize that triangle but anything i try... comes out nasty. For example:
$x=t$ for $t\in \left [ 0,R \right ]$ than also $y=t$ which would from $x^2+y^2+z^2=R^2$ mean that $z^2=r^2-2t^2$

Now i have parameterization as function of t only $r(t)=(t,t,\sqrt{r^2-2t^2})$ but... to calculate dS in stokes integral, I need parametrization of two paramateres... so... How do I continue?

Is it ok if I just say that $r(t,z)=(t,t,z)$ where $t\in \left [ 0,R \right ]$ and $z\in \left [ -\sqrt{r^2-2t^2},\sqrt{r^2-2t^2} \right ]$

Do you have to use Stoke's theorem or does that just seem like the obvious choice to you? It is much easier to just work out the three line integrals. Lots of the variables are 0 on each curve of your spherical triangle.

 

[skrat]

Yes, I HAVE to.

 

[LCKurtz]

Then I would try spherical coordinates.$$
\vec r(\theta,\phi)=\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\rangle$$Calculate $\vec r_\theta$ and $\vec r_\phi$, and work out the integral$$
\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}(\nabla \times \vec F) \cdot \vec r_\phi \times \vec r_\theta~
d\phi d\theta$$where the $\pm$ sign is chosen to agree with the orientation which, by the way, you didn't specify.

 

[skrat]

I think you meant $\vec r(\theta,\phi)=\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\rangle$

But you kinda lost me here already... Why would I want to integrate over a sphere (or one eight of it) if the problems says to integrate $\vec{F}$ along curve where this curve consists of three lines that together form a triangle?

 

[LCKurtz]

I think you meant $\vec r(\theta,\phi)=\langle R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi\rangle$

Yes. I corrected it.

But you kinda lost me here already... Why would I want to integrate over a sphere (or one eight of it) if the problems says to integrate $\vec{F}$ along curve where this curve consists of three lines that together form a triangle?

That's why I asked you if you have to use Stoke's theorem. It let's you change the line integral around the boundary to a surface integral over the bounded surface. But that is no advantage in this problem. The line integrals are easy.

And I assume you don't literally mean the boundary curve is a triangle. It is three circular arcs that form what you might call a spherical triangle.

 

[vela]

Now I should somehow parametrize that triangle but anything i try... comes out nasty. For example:
$x=t$ for $t\in \left [ 0,R \right ]$ than also $y=t$ which would from $x^2+y^2+z^2=R^2$ mean that $z^2=r^2-2t^2$

Now i have parameterization as function of t only $r(t)=(t,t,\sqrt{r^2-2t^2})$ but... to calculate dS in stokes integral, I need parametrization of two paramateres... so... How do I continue?

Is it ok if I just say that $r(t,z)=(t,t,z)$ where $t\in \left [ 0,R \right ]$ and $z\in \left [ -\sqrt{r^2-2t^2},\sqrt{r^2-2t^2} \right ]$

How did you get y=t?

 

[skrat]

And I assume you don't literally mean the boundary curve is a triangle. It is three circular arcs that form what you might call a spherical triangle.

Hmm, well ... The problems says to integrate along a positively oriented spherical triangle cut by $x^2+y^2+z^2=R^2$ .. To be honest with you, I never even thought about circular arcs.

How did you get y=t?

Yes, I was the most worried about that part. I simply imagined that since it is a sphere the triangle is symmetrical. I guess, that's not ok.

 

[skrat]

Ok, I get it now. Now idea of how spherical triangle looks like was completely wrong! Thanks, LCKurtz!

Now I agree with this:
$\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}(\nabla \times \vec F) \cdot \vec r_\phi \times \vec r_\theta~ d\phi d\theta=\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}2R(\sin\theta \sin \phi,\cos \theta,\sin \theta \cos \phi )R^2(\sin^2\theta \cos \phi,\sin^2 \theta \sin \phi, \sin \theta cos \theta)$

Which gives me some nasty integrals:

$\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}2R^3(\sin^3\theta \sin\phi \cos\phi +\sin^2\theta \sin\phi \cos\theta + \sin^2\theta \cos\theta \cos\phi)$

Which of them is 0 and why?

 

[skrat]

Is the result $2R^3$ ?

 

[vela]

That's what I got. As a check, you could try calculating the line integrals and see if you get the same result.

 

[LCKurtz]

That's what I get too, with a $\pm$ sign. You still have to give the orientation and choose the correct sign.

 

[skrat]

I thought $2R^3$ including the sign.. But since you are asking about the sign, i am guessing it's -...

 

[LCKurtz]

Guesses don't count. In problems like this, it matters which way you go around the curve. You haven't told us which way to go around the curve so the problem is incompletely stated. The direction around the curve determines the orientation of the corresponding bounded surface. You have to make sure your vector $\vec r_\phi \times \vec r_\theta$ agrees with that orientation to decide whether to use that vector or its negative in the formula. There is no way to know which is correct until you tell us which way around the curve to consider.

 

[skrat]

There's nothing mentioned about the orientation of spherical triangle. So both signs should be ok.

However, let's say that spherical triangle in positively oriented. Than the overall sign is positive (that's IF I understand this right).

 

[LCKurtz]

To say the spherical triangle is "positively oriented" doesn't mean anything to me. The direction is usually specified with descriptions like "clockwise or "counterclockwise" as viewed from the origin" or more literally something like "moving from $(R,0,0)$ to $(0,R,0)$ along the curve in the xy plane then to $(0,0,R)$ in the yz plane, then etc... Then you use the right-hand rule to determine the orientation of the surface.

 

[skrat]

Ok, our definition of positively oriented surface:

If you stand on the edge of your surface (spherical triangle) and start walking, than the surface has positive orientation if it is on your left side and negative if it is on your right side.

Does this make any sense?

 

[skrat]

edit: opened another thread...

 

[LCKurtz]

You started with a curve (spherical triangle). It is true that the direction you go around the curve determines the orientation of the surface with respect to that curve. The right hand rule will tell you which normal to use for positive orientation. But you have to know which way you are going around the curve to do that. Just like you can't say if the surface is on your right or left if you don't know which direction you are moving. If your book doesn't describe the direction are you sure it doesn't show the direction with a picture with little arrows on the curves to indicate which direction?

 

[skrat]

You started with a curve (spherical triangle). It is true that the direction you go around the curve determines the orientation of the surface with respect to that curve. The right hand rule will tell you which normal to use for positive orientation. But you have to know which way you are going around the curve to do that. Just like you can't say if the surface is on your right or left if you don't know which direction you are moving. If your book doesn't describe the direction are you sure it doesn't show the direction with a picture with little arrows on the curves to indicate which direction?

Aaa, ok, thanks for this explanation. I'm 100% sure that there is no direction mentioned and believe me that there is no picture - otherwise I wouldn't have mistaken the curved lines for straight lines. :)

 

[LCKurtz]

Ok, since we are already talking about orientation... I have another problem, and I know that the solution is ok but I am not sure about the sign...
I hope you don't mind that I did NOT open another topic because of this:

I do mind. Please post that in a different thread. It will mess up the context in this one.

 

[vela]

Aaa, ok, thanks for this explanation. I'm 100% sure that there is no direction mentioned and believe me that there is no picture - otherwise I wouldn't have mistaken the curved lines for straight lines. :)

This is really unlikely if the problem is from a textbook. Can you please provide us the problem statement exactly as given?

 

[skrat]

Well, that's the problem. There is no textbook. This is what we got as "Homework" in school. I used quotes because it was given more as a home exercise and not as MUST to do.

The professor never wrote the problem down. There of course is an option that he mentioned the orientation and I forgot to write it down or he simply forgot to do that...

 

[LCKurtz]

A couple of final suggestions. Edit your post #18 and delete it if it isn't too late, since you reposted it.

Also, to wrap it up, suppose we specify that the spherical triangle is traversed clockwise as viewed from the origin. Then would you use the plus or minus sign in your answer and why? (Of course, we have been assuming all along we are calculating the work done by the force).

 

[skrat]

Also, to wrap it up, suppose we specify that the spherical triangle is traversed clockwise as viewed from the origin. Then would you use the plus or minus sign in your answer and why? (Of course, we have been assuming all along we are calculating the work done by the force).

$\nabla \times \vec{F}$ has all three components positive so does $r_{\phi }\times r_{\theta }=R^2(\sin^2\theta \cos \phi,\sin^2 \theta \sin \phi, \sin \theta cos \theta)$

So... I think that the sign is ok. It's +.

 

[LCKurtz]

A couple of final suggestions. Edit your post #18 and delete it if it isn't too late, since you reposted it.

Also, to wrap it up, suppose we specify that the spherical triangle is traversed clockwise as viewed from the origin. Then would you use the plus or minus sign in your answer and why? (Of course, we have been assuming all along we are calculating the work done by the force).

$\nabla \times \vec{F}$ has all three components positive so does $r_{\phi }\times r_{\theta }=R^2(\sin^2\theta \cos \phi,\sin^2 \theta \sin \phi, \sin \theta cos \theta)$

So... I think that the sign is ok. It's +.

Actually, I should re-phrase that. The origin isn't a good reference since it is in the plane of all three curves. Just so there is no doubt, I mean to go from $(R,0,0)\rightarrow (0,R,0)\rightarrow (0,0,R)\rightarrow (R,0,0)$. The signs of the components of $\nabla \times \vec F$ have nothing to do with it. Tell me why you like that the three components of $r_{\phi }\times r_{\theta}$ are all positive.

 

[skrat]

Tell me why you like that the three components of $r_{\phi }\times r_{\theta}$ are all positive.

Because this means that $r_{\phi }\times r_{\theta}$ doesn't have the opposite direction as $\vec{F}$ does?

 

[LCKurtz]

Actually, I should re-phrase that. The origin isn't a good reference since it is in the plane of all three curves. Just so there is no doubt, I mean to go from $(R,0,0)\rightarrow (0,R,0)\rightarrow (0,0,R)\rightarrow (R,0,0)$. The signs of the components of $\nabla \times \vec F$ have nothing to do with it. Tell me why you like that the three components of $r_{\phi }\times r_{\theta}$ are all positive.

Because this means that $r_{\phi }\times r_{\theta}$ doesn't have the opposite direction as $\vec{F}$ does?

No. You are missing the point. The orientation of the curve vs the surface has nothing to do with the force field. The force field will determine whether the final answer is positive or negative, but it doesn't affect the orientation. Think about the right hand rule. That determines which way the normal must point.

 

[skrat]

Ok, I think I got it now...

So, right hand rule for this orientation $(R,0,0)\rightarrow (0,R,0)\rightarrow (0,0,R)\rightarrow (R,0,0)$ ... z component is positive at all times. (normal pointed away from the origin)
And because $r_{\phi }\times r_{\theta}$ is positive in all three components, so z direction of the two matches.

... ?

 

[LCKurtz]

Ok, I think I got it now...

So, right hand rule for this orientation $(R,0,0)\rightarrow (0,R,0)\rightarrow (0,0,R)\rightarrow (R,0,0)$ ... z component is positive at all times. (normal pointed away from the origin)
And because $r_{\phi }\times r_{\theta}$ is positive in all three components, so z direction of the two matches.

... ?

That's right. If you went around the curve the other way you would take the opposite vector. The thing about these problems is you have a 50-50 chance of getting them right even if you don't understand that.

 

[skrat]

Wow, it feels so good to understand that! :D

So, the final sign, the result, basically depends on weather the $r_{\phi }\times r_{\theta}$ and normal vector determined by right hand rule match. If they do not, than the result must be multiplied by -1.

Please correct me if I am wrong.

 

[LCKurtz]

Wow, it feels so good to understand that! :D

So, the final sign, the result, basically depends on weather the $r_{\phi }\times r_{\theta}$ and normal vector determined by right hand rule match. If they do not, than the result must be multiplied by -1.

Please correct me if I am wrong.

$$\pm\int_0^{\frac \pi 2} \int_0^{\frac \pi 2}(\nabla \times \vec F) \cdot \vec r_\phi \times \vec r_\theta~
d\phi d\theta$$
We are talking about the choice of the $\pm$ sign in the statement of Stoke's theorem. Whether the final answer of the problem is positive or negative is a different question and it depends on what the integrand is.