Calling all Oxbridge Physicists!

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.

Reply 322

OP

ok thanks, I'll try projecting it in a second when I've done the last question on this paper (waves)

Reply 323

OP

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!

Reply 324

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.

Reply 325

OP

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!

Reply 326

RichE

15

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.

Reply 327

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!

Reply 328

RichE

15

Do you guys only ever work out the flux across planar surfaces then?

Reply 329

it's either planar or has some kind of cylindrical or spherical symmetry....at least in the first year that is!

Reply 330

OP

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!!! :biggrin:

Reply 331

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?)

Reply 332

OP

LennonMcCartney

(Can we use LaTeX markup on these forums?)

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

Reply 333

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

Reply 334

Bezza

2

just tex in square brackets

Unparseable latex formula:

\fontsize{4}\sec\gamma = \frac{\nabla\vec F}{|\nabla\vec F . \vec n|}

Reply 335

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!

Reply 336

\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).

Reply 337

Bezza

2

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

Reply 338