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Gauss' Theorum and curl of a vector field

  • Thread starter Haths
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  • #1
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Two problems one that I have some idea about solving, the other I have no idea at all about where to start.

1. Find the surface integral of E . dS where E is a vector field given;

E = yi - xj + 1/3 z3 and S is the surface x2 + z2 < r2 and 0 < y < b

Well Gauss' theorum would be the place to start to make this easy for me...Already having calculated the divergance of the field to be;

div E = z2

But my problems came to the paramaterisation of the integral. Assuming that my co-ord. transfer is;

x = r cos(a)
y = y
z = r sin(a)

z2 = r2sin2(a)

The jacobian for the cylindrical co-ords I believe being r2

Hence my integral becomes;

[tex]
$ int int int r^{4} sin^{2}(a) dr da dy$
[/tex]

The limits being b -> 0 / 2PI -> 0 / r -> 0 respectfully

...and I end up with;

1/5 PI b r5

I have no idea if this is right or not, it 'feels' right, but I'm not confident in the answer, nor if I have made some mistake along the way.

2. Finding the curl of;

[tex]
A(r,t) = a e^{(ip \dot r - i \omega t)}
[/tex]

where a and p are constant vectors...

The only idea I had was to convert it into sines and cosines, but that didn't help me that much.

Cheers,
Haths
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
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6
and S is the surface x2 + z2 < r2 and 0 < y < b
These inequalities describe a volume (a cylinder of length [itex]b[/itex] and radius [itex]r[/itex]) not a surface. Surely you meant to say "S is the surface that bounds the volume x2 + z2 < r2 and 0 < y < b", right?

Well Gauss' theorum would be the place to start to make this easy for me...Already having calculated the divergance of the field to be;

div E = z2

But my problems came to the paramaterisation of the integral. Assuming that my co-ord. transfer is;

x = r cos(a)
y = y
z = r sin(a)

z2 = r2sin2(a)

The jacobian for the cylindrical co-ords I believe being r2
Your parameterization is correct. However since you are integrating one of your coordinates from 0 to [itex]r[/itex], it is bad notation to call that variable [itex]r[/itex]. You should use [itex]r'[/itex] or a different letter like [itex]u[/itex] to make clear the meaning of the integration. Also, the determinant of the Jacobian (the Jacobian itself is a matrix, not a scalar) is incorrect; using [itex]r'[/itex] as your variable, you should find that the determinant is just [itex]r'[/itex] not [itex]r'^2[/itex] and so your integral becomes (click on the image to see the LaTeX code that generated it):

[tex]\int_0^r \int_0^{2\pi} \int_0^b (r')^3\sin(a)^2 dr' da dy[/tex]

2. Finding the curl of;

[tex]
A(r,t) = a e^{(ip \dot r - i \omega t)}
[/tex]

where a and p are constant vectors...

The only idea I had was to convert it into sines and cosines, but that didn't help me that much.

Cheers,
Haths
Here, there is a vector calc product rule which will help:

[tex]\vec{\nabla}\times(f\vec{A})=f(\vec{\nabla}\times\vec{A})-\vec{A}\times(\vec{\nabla}f)[/tex]

for any scalar function [itex]f[/itex] and vector function [itex]\vec{A}[/itex]....Have you seen this product rule before?....Do you see how it helps you here?
 
  • #3
33
0
Here, there is a vector calc product rule which will help:

[tex]\vec{\nabla}\times(f\vec{A})=f(\vec{\nabla}\times\vec{A})-\vec{A}\times(\vec{\nabla}f)[/tex]

for any scalar function [itex]f[/itex] and vector function [itex]\vec{A}[/itex]....Have you seen this product rule before?....Do you see how it helps you here?
Thanks on the first section, yeah the jacobian is wrong :p.

__________________________________________________________________

I've not seen the RHS of that equation before, but the first one appears to look just like the normal equation for calculating the curl of a field. So I'm not quite sure how to apply that to this senario. Because first and formost I don't understand/remember how to take the field equation I have been given and re-arrange it into some form that is useful to me.

I can see that;

[tex]
a(cos(pr-wt) + isin(pr-wt))
[/tex]

But the complex notation throws me, because I don't know what I could do from here, other than double angle formula and from there I don't know if indeed that is one way to aproach the solution Furthermore does anything unexpected happen with imaginary parts of the equation, like are they included in the solution for the curl, or excluded as imaginary parts.

Haths
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6
Thanks on the first section, yeah the jacobian is wrong :p.

__________________________________________________________________

I've not seen the RHS of that equation before, but the first one appears to look just like the normal equation for calculating the curl of a field. So I'm not quite sure how to apply that to this senario. Because first and formost I don't understand/remember how to take the field equation I have been given and re-arrange it into some form that is useful to me.

I can see that;

[tex]
a(cos(pr-wt) + isin(pr-wt))
[/tex]

But the complex notation throws me, because I don't know what I could do from here, other than double angle formula and from there I don't know if indeed that is one way to aproach the solution Furthermore does anything unexpected happen with imaginary parts of the equation, like are they included in the solution for the curl, or excluded as imaginary parts.

Haths
I would leave it in exponential notation, and make clear what quantities are vectors and which are scalars:

[tex]\vec{A}(\vec{r},t)=\vec{a}e^{i(\vec{p}\cdot\vec{r}-\omega t)}[/tex]

If you were to use the vector product rule I posted, you would note that [itex]\vec{a}[/itex] is a vector function and [tex]e^{i(\vec{p}\cdot\vec{r}-\omega t)}[/tex] is a scalar. The product rule tells you how to calculate the curl of a product between any scalar function and any vector function, and so you could apply it here as follows:

[tex]\vec{\nabla}\times(\vec{A}(\vec{r},t))=\vec{\nabla}\times(\vec{a}e^{i(\vec{p}\cdot\vec{r}-\omega t)})=e^{i(\vec{p}\cdot\vec{r}-\omega t)}(\vec{\nabla}\times\vec{a})-\vec{a}\times(\vec{\nabla}e^{i(\vec{p}\cdot\vec{r}-\omega t)})[/tex]

And since [itex]\vec{a}[/itex] is a constant vector, [itex]\vec{\nabla}\times\vec{a}=0[/itex], so all you would need to do is calculate the gradient of [itex]e^{i(\vec{p}\cdot\vec{r}-\omega t)}[/itex] (which you could do by letting [itex]\vec{p}=p_x\hat{i}+p_y\hat{j}+p_z\hat{k}[/itex] and, as always, [itex]\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}[/itex] and calculating the gradient using Cartesian coordinates).

However(!), if you haven't learned this product rule yet, you may not be permitted to use it. If not, you have no choice but to say [itex]\vec{a}=a_x\hat{i}+a_y\hat{j}+a_z\hat{k}[/itex],[itex]\vec{p}=p_x\hat{i}+p_y\hat{j}+p_z\hat{k}[/itex] and, as always, [itex]\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}[/itex] and calculate the curl of [itex]\vec{A}(\vec{r},t)[/itex] directly.

For example,

[tex]\frac{\partial A_z}{\partial y}=\frac{\partial}{\partial y}\left(a_ze^{i(p_x x+p_y y+p_z z-\omega t)}\right)=i p_y a_ze^{i(p_x x+p_y y+p_z z-\omega t)}=i p_y a_ze^{i(\vec{p}\cdot\vec{r}-\omega t)}[/tex]

Since [itex]\vec{a}[/itex] and [itex]\vec{p}[/itex] are constant vectors, you know that their components are independent of [itex]x[/itex], [itex]y[/itex] and [itex]z[/itex].
 
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