Line Integral: A from (1,0,0) to (3,0,0) on Semi-Circular Path

[msd213]

Homework Statement

Calculate the line integral of A between to points x=1 and x=3 along the semicircular path with a center at x=2

Homework Equations

A=kx in the x hat direction.

\int A \bullet dl

dl = ds s hat +s d\phi \phi hat +dz z hat

The Attempt at a Solution

My biggest problem seems to be getting that A into cylindrical coordinates so I can take the dot product with the dl.

 

[gabbagabbahey]

HI msd213, welcome to PF!

Homework Equations

A=kx in the x hat direction.

\int A \bullet dl

dl = ds s hat +s d\phi \phi hat +dz z hat

You can write this more clearly as follows (click on the \LaTeX images to see the code used to generate them!):

\textbf{A}=kx\mathbf{\hat{x}}

\int \textbf{A}\cdot d\textbf{l}

d\textbf{l}=ds\mathbf{\hat{s}}+s d\phi\mathbf{\hat{\phi}}+dz\mathbf{\hat{z}}

My biggest problem seems to be getting that A into cylindrical coordinates so I can take the dot product with the dl.

In cylindrical coordinates, centered at x=2 (since that is the center of your curve, you might as well choose the origin of your cylindrical coordinates to be there) x-2=s\cos\phi and \mathbf{\hat{x}}=\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}

 

[msd213]

HI msd213, welcome to PF!



You can write this more clearly as follows (click on the \LaTeX images to see the code used to generate them!):

\textbf{A}=kx\mathbf{\hat{x}}

\int \textbf{A}\cdot d\textbf{l}

d\textbf{l}=ds\mathbf{\hat{s}}+s d\phi\mathbf{\hat{\phi}}+dz\mathbf{\hat{z}}



In cylindrical coordinates, centered at x=2 (since that is the center of your curve, you might as well choose the origin of your cylindrical coordinates to be there) x-2=s\cos\phi and \mathbf{\hat{x}}=\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}

Thanks for your response but I'm still confused. How did you derive that \mathbf{\hat{x}}?

 

[gabbagabbahey]

I didn't derive it...I looked it up!

The derivation is done in most Multi-variable calculus texts, and I think is done in an Appendix in Griffiths' Introduction to Electrodynamics as-well. But it is really just a matter of drawing a picture and doing some very basic trig.

 

[msd213]

I didn't derive it...I looked it up!

The derivation is done in most Multi-variable calculus texts, and I think is done in an Appendix in Griffiths' Introduction to Electrodynamics as-well. But it is really just a matter of drawing a picture and doing some very basic trig.

I think I see now. But when you take the dot product between the A (now converted into cylindrical coordinates, and dl, aren't you going to have terms multiplied by differentials such as (some term)*ds +(another term)*d\phi...etc. How can you take the integrals with that? Where does the volume element fit in?

 

[gabbagabbahey]

I think I see now. But when you take the dot product between the A (now converted into cylindrical coordinates, and dl, aren't you going to have terms multiplied by differentials such as (some term)*ds +(another term)*d\phi...etc. How can you take the integrals with that? Where does the volume element fit in?

Well, for your particular curve (the semicircle), s=1 and z are both constant, so ds=dz=0 and hence d\textbf{l}=d\phi\mathbf{\hat{\phi}}

In general, any 3D curve can be described by a single parameter. So, in cylindrical coordinates, for example, you would have something like s=s(u), \phi=\phi(u) and z=z(u) and so all the differentials (ds, d\phi and dz) can be written in terms of a single differential du. (For your particular curve, the easiest way is just to use \phi as your parameter)

If instead you wanted to integrate over a surface, you would express each coordinate in terms of two independent parameters, and your surface area element in terms of their differentials.

 

[msd213]

Well, for your particular curve (the semicircle), s=1 and z are both constant, so ds=dz=0 and hence d\textbf{l}=d\phi\mathbf{\hat{\phi}}

In general, any 3D curve can be described by a single parameter. So, in cylindrical coordinates, for example, you would have something like s=s(u), \phi=\phi(u) and z=z(u) and so all the differentials (ds, d\phi and dz) can be written in terms of a single differential du. (For your particular curve, the easiest way is just to use \phi as your parameter)

If instead you wanted to integrate over a surface, you would express each coordinate in terms of two independent parameters, and your surface area element in terms of their differentials.

I'm still a little confused about how this clears up the problem with the repeated differentials. For example, in my problem, isn't the integral going to look like.

\int -k s^2\\\ cos\phi\\ sin\phi\\ d\phi\\\ s\\ ds\\ d\phi\\ dz

Using \mathbf{\hat{x}}=\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}
and kx=kscos\phi

I'm not sure how I can write these in terms of one parameter.

 

[gabbagabbahey]

I'm still a little confused about how this clears up the problem with the repeated differentials. For example, in my problem, isn't the integral going to look like.

\int -k s^2\\\ cos\phi\\ sin\phi\\ d\phi\\\ s\\ ds\\ d\phi\\ dz

No,

\begin{aligned} \textbf{A} & = kx\mathbf{\hat{x}} \\ & = k(s\cos\phi+2)\left(\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}\right) \\ & = k(\cos^2\phi+2\cos\phi)\mathbf{\hat{s}}-k(\sin\phi\cos\phi+2\sin\phi)\mathbf{\hat{\phi}} \end{aligned}

(Since s=1 along your curve)

And d\textbf{l}=d\phi\mathbf{\hat{\phi}}, so

\begin{aligned} \textbf{A}\cdot d\textbf{l} & = \left(k(\cos^2\phi+2\cos\phi)\mathbf{\hat{s}}-k(\sin\phi\cos\phi+2\sin\phi)\mathbf{\hat{\phi}}\right)\cdot(d\phi\mathbf{\hat{\phi}}) \\ & = -k(\sin\phi\cos+2\sin\phi)\phi d\phi\end{aligned}

 

[msd213]

No,

\begin{aligned} \textbf{A} & = kx\mathbf{\hat{x}} \\ & = k(s-2)\cos\phi\left(\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}\right) \\ & = -k\cos^2\phi\mathbf{\hat{s}}+k\sin\phi\cos\phi\mathbf{\hat{\phi}} \end{aligned}

(Since s=1 along your curve)

And d\textbf{l}=d\phi\mathbf{\hat{\phi}}, so

\begin{aligned} \textbf{A}\cdot d\textbf{l} & = \left(-kcos^2\phi\mathbf{\hat{s}}+k\sin\phi\cos\phi\mathbf{\hat{\phi}}\right)\cdot(d\phi\mathbf{\hat{\phi}}) \\ & = k\sin\phi\cos\phi d\phi\end{aligned}

Thanks for your patience and prompt responses.

Whenver I use cylindrical coordinates, I suppose I just reflexively put in that s\\ ds\\ d\phi\\ dz whenever I convert to cylindrical coordinates and integrate. Why exactly isn't it needed here?

 

[gabbagabbahey]

d\tau=s ds d\phi dz is the differential volume element...you're not performing a volume integral here, so d\tau is pretty useless to you.

 

[msd213]

No,

\begin{aligned} \textbf{A} & = kx\mathbf{\hat{x}} \\ & = k(s-2)\cos\phi\left(\cos\phi\mathbf{\hat{s}}-\sin\phi\mathbf{\hat{\phi}}\right) \\ & = -k\cos^2\phi\mathbf{\hat{s}}+k\sin\phi\cos\phi\mathbf{\hat{\phi}} \end{aligned}

(Since s=1 along your curve)

And d\textbf{l}=d\phi\mathbf{\hat{\phi}}, so

\begin{aligned} \textbf{A}\cdot d\textbf{l} & = \left(-kcos^2\phi\mathbf{\hat{s}}+k\sin\phi\cos\phi\mathbf{\hat{\phi}}\right)\cdot(d\phi\mathbf{\hat{\phi}}) \\ & = k\sin\phi\cos\phi d\phi\end{aligned}

Thanks, I guess I've just been conditioned from calc to put that there.

Also, I think the whole idea of this section is path independence.

So, when you just take the path just along the x-axis, you have

\int_1^3 k\\ x\\ dx

This integral yields 4k. But the integral you derive yields zero, unless I'm way off. The bounds should be 0 to \pi correct? Shouldn't the integrals yield the same answer?

 

[gabbagabbahey]

Thanks, I guess I've just been conditioned from calc to put that there.

Also, I think the whole idea of this section is path independence.

So, when you just take the path just along the x-axis, you have

\int_1^3 k\\ x\\ dx

This integral yields 4k. But the integral you derive yields zero, unless I'm way off. The bounds should be 0 to \pi correct? Shouldn't the integrals yield the same answer?

Right, sorry my post contained an error (substituted x=(s-2)\cos\phi instead of x-2=s\cos\phi)...i'll edit it to fix the error...

 

[msd213]

Right, sorry my post contained an error (substituted x=(s-2)\cos\phi instead of x-2=s\cos\phi)...i'll edit it to fix the error...

Thank you, that should clear everything up. You've been a great help.