# Calling all Oxbridge Physicists!

Scroll to see replies Bezza
That will only give you the flux through that 1 side of the tetrahedron that's bounded by the plane 2x + y + 2z = 4 whereas you need to sum the flux through all the sides.

If you put in x=y=0 then you see the plane crosses the z-axis at 2, then do similar things to get it crosses the x-axis at 2 and the y-axis at 4 then do a sketch.

I'm not too sure what he's done with that cross product either! He must be doing it to find a normal vector to the plane, but personally I would just use the fact that a plane can also be written as r.n=d so the normal vector is just (2,1,2)

One question I have is why don't you have to use a unit normal vector?

Thanks Bezza, I edited my post while you were obviously replying and realised my mistake with my initial question. Thanks for the explanation of the limits.

I'm glad someone else is confused with his last method. I used the unit normal vector method but I wasn't entirely sure about the limits and ended up with 64/9 which isn't right! grrrr he has projected the area onto the xz plane...so he can use the simple limits. To project it onto the plane you use: dxdz = |dS|U.j/|U| where U is some normal vector to the plane of interest.

|dS| = dxdz|U|/U.j

or dS = |dS|U/|U| = dxdzU/U.j

and because U.j in this case = 1, he has validly used (2,1,2)dxdz

that make any sense?

you can not use (unit vector)dxdz because dxdz is not the valid area element size....you must take into account it is a projection. He has projected the plane onto the xz-plane and used the suitable limits. ok thanks, I'll try projecting it in a second when I've done the last question on this paper (waves) Right now I'm so confused! I tried projecting it, the way I know how and got the annoying answer of 8/3...I'm getting closer! Could someone point out where I've gone wrong in my working? Thanks:

Project onto xy plane.

F=(0,2x,z) but z=0 therefore F=(0,2x,0)

Sec gamma = (2,1,2)/[|(2,1,2).(0,0,1)|]
sec gamma = (1,1/2,1)

Now F.sec gamma = x

Then we know the limits are:

x: 0 -> (2-y/2)
y: 0 -> 4

So Integral (0->4) Integral (0->(2-y/2)) x dxdy

Which when I evaluate it gives me 8/3!

HELP! have you remembered to evaluate the other sides of the tetrahedron? you have successfully found the surface integral of the side defined by the plane: 2x+y+2z=4
you must now find the surface integrals of the remaining sides. Since they are all on the xy-xz-yz planes you dont have to worry about any projections, just use the suitable limits. Willa
have you remembered to evaluate the other sides of the tetrahedron? you have successfully found the surface integral of the side defined by the plane: 2x+y+2z=4
you must now find the surface integrals of the remaining sides. Since they are all on the xy-xz-yz planes you dont have to worry about any projections, just use the suitable limits.

Yeah I've done all the other sides and agree with the workings Rich E put up earlier and get a total of -8/3 for the other 3 sides. Therefore, when I sum those 3 sides with this last side I get zero!!! which I know is wrong as a) it disagrees with everyone else, b) it disagrees with the next bit of the question! ARGHHH!!!!!!!!!!!!!

EDIT: I'm guessing it's my limits when I'm projecting into the xy plane that is causing the trouble...but I'm soo confused now! Hoofbeat
Thanks Rich. But a few questions:

1) How did you know when you were working in the z=0 (or xy) plane that the limits of z are 0->2? I understand how you got the limits for x.

2) On the second page, I'm not entirely sure what you're doing! Sorry! Are you doing a cross product to get an area? Can't you just define say G = 2x + y + 2z and then find the unit normal from grad G/|grad G|? but then I'm not sure how the limits would work?

1) For the z=0 plane then you don't need the limits do you? As the field.normal is zero anyway. Not sure which part of my answer you're talking about.

2) The way I did the upper face's flux was to use

dS = r_u x r_v du dv

for a normal vector with magnitude of infinitesimal area. I guess it all depends how you've been taught. It being a plane it's obvious that (2,1,2) is normal (from the r.n=a.n formula), but I was taught to take dS in F.dS as the above.

The limits for the upper face depend on what parametrisation you take. You need to cover that face with two co-ordinates and the limits are whatever they need to be to achieve that. I'm using the co-ordinates (x,z) and these need to vary over the range x,z>=0, x+z <=2. no you cant set z=0.....you only project the area element onto the plane not the value of the flux.

you must evaluate x+z dxdy

so to deal with the z use: z = 2 - .5y - x

plug that in and it should work! Do you guys only ever work out the flux across planar surfaces then? it's either planar or has some kind of cylindrical or spherical symmetry....at least in the first year that is! Willa
no you cant set z=0.....you only project the area element onto the plane not the value of the flux.

you must evaluate x+z dxdy

so to deal with the z use: z = 2 - .5y - x

plug that in and it should work!

Ok thanks, let me try that!

EDIT: Yay, thanks ever so much, it works!!!    RichE
Do you guys only ever work out the flux across planar surfaces then?

Yes usually using a suitable co-ordinate system it can be reduced to a fairly simple integral (such as sphere or cylinder, as Willa said [I think]). But we would generally use the grad f/(|gradf|.n) to find sec gamma.

(Can we use LaTeX markup on these forums?) LennonMcCartney
(Can we use LaTeX markup on these forums?)

Yup...I just don't know how to! Hoofbeat
Yup...I just don't know how to!

So what are the tags?

EDIT: Or are you saying you don't know? Let me try.

$\hat{H} | \psi \rangle = i \hbar \frac{\partial}{\partial t} | \psi \rangle$ just tex in square brackets
Unparseable latex formula:

\fontsize{4}\sec\gamma = \frac{\nabla\vec F}{|\nabla\vec F . \vec n|} do [ tex ] and tags (without the spaces) and use symbols as given by: http://www.fi.uib.no/Fysisk/Teori/KURS/WRK/TeX/symALL.html

so to do an alpha sign use: \alpha

also it's useful to know how to do superscript...which is just: ^{this is superscript}
subscript is: _{this is subscript}
fractions are useful as well: \frac{enumerator}{denominator}

it's best if you just experiment to see what works! $\hat{H} | \psi \rangle = i \hbar \frac{\partial}{\partial t} | \psi \rangle$

And that was the time-independent Schrödinger equation, ladies and gents. It's not a pretty font though. They have a better engine on www.physicsforums.com (hint hint mods). LennonMcCartney
And that was the time-independent Schrödinger equation, ladies and gents. It's not a pretty font though. They have a better engine on www.physicsforums.com (hint hint mods).

yeah, that does look a lot nicer. theres a thread in about on the latex so try posting in there Bezza
yeah, that does look a lot nicer. theres a thread in about on the latex so try posting in there

I am a member of physicsforums. Unfortunately, it's home to a lot of crackpots who try and prove collective consciousness stuff, and refute relativity. Fortunately it's also home to a couple of geniuses. http://www.thestudentroom.co.uk/t91246.html is the thread I was talking about. It was pig who implemented it so if you ask him nicely he might be able to change it.