# Calling all Oxbridge Physicists!

Scroll to see replies I'm a 2nd year Maths and Physics at Hild Bede. I'm very much looking forward to third year, when I can drop all the nasty practical modules and concentrate on the lovely mathsy theoretical type subjects . you not have to do a practical project as part of your finals? I'm guessing AMM is going to do a theoretical project.

Another joyous vector calculus proof for you guys to assist me with! Our lovely tutor has set us some extra exam questions on vector calculus and I thought they were going well until I saw this one! HELP!!!

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Q. A soolid of uniform density occupies a volume V bounded by a surface S. Show that the z-component of its centre of mass,

Zo = (Integral z dV)/(Integral dV)

can be written in the alternative form:

Zo = (3 * Integral zr.dS)/(4 * Integral r.dS)

I'm guessing it's do with the divergence (or Gauss') theorem:

Integral DivA dV = Integral F.dS

and that we have to say z is equivalent to DivA...but I really have no idea how to proceed! any ideas? I know something you don't know  shiny
I know something you don't know I don't doubt that! You're a filthy tab which means you know how to cheat  I'd work backwards.

div(zr) = 4z
div(r) = 3

Enough? EDIT: Not enough? OK...

Using those two things we just calculated...

z0 = INT[z dV]/INT[dV]
z0 = INT[div(zr)/4 dV] / INT[div(r)/3 dV]
z0 = 3*INT[div(zr) dV] / 4*INT[div(r) dV] Worzo
you not have to do a practical project as part of your finals?

Nope, I can choose to do a theoretical physics project, or a mathematical dissertation. Ok, that makes more sense. So...

Worzo
z0 = 3*INT[div(zr) dV] / 4*INT[div(r) dV]

Then by divergence theorem:

z0 = 3*INT zr.dS / 4*INT r.dS QED!!! That right? ANDDDDDDDDDDDDDDDDDDDD another vector calculus question:

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where r is the position vector and a is a constant vector
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I wrote r=(x,y,z) and a=(a,b,c) and tried expanding it, but I get something horrible which doesn't appear to simplify!!! Ideas?!

*****

Can someone also confirm that I'm right in that curl{(a.r)a} = 0 You can just use the chain rule can't you
Unparseable latex formula:

\fontsize{5}\frac{\partial}{\partial x}(ax+by+cz)^2 = 2(ax+by+cz)\times a

then the other bits are just times b and c so you get
Unparseable latex formula:

\fontsize{5}\nabla (\bf{a}.\bf{r})^2 = 2(\bf{a}.\bf{r})\bf{a} Hoofbeat
Can someone also confirm that I'm right in that curl{(a.r)a} = 0

Given what Bezza showed that is the curl of a gradient, therefore yes. Bezza
You can just use the chain rule can't you
Unparseable latex formula:

\fontsize{5}\frac{\partial}{\partial x}(ax+by+cz)^2 = 2(ax+by+cz)\times a

then the other bits are just times b and c so you get
Unparseable latex formula:

\fontsize{5}\nabla (\bf{a}.\bf{r})^2 = 2(\bf{a}.\bf{r})\bf{a}

Ooooo that sounds like a nifty way of doing it! Thanks I'll have a go a bit later (now onto circuit theory!). AntiMagicMan
Given what Bezza showed that is the curl of a gradient, therefore yes.

thanks  Just for your interest, thought I might put up the formal proof of that....(OK, I admit it I like using the crazy suffix ****....)

EDIT: What the frig? It automatically stars things? Worzo
Just for your interest, thought I might put up the formal proof of that....(OK, I admit it I like using the crazy suffix ****....)

EDIT: What the frig? It automatically stars things?

You're not allowed to swear andy, tut tut And you are SERIOUSLY obsessed with Einstein's Summation principle...but then again, I do agree that it is v.cool (I MUST teach myself it!) Y'know, it's not all Einstein's work. All he did was to decide to remove the sigma sign when you repeat an index twice. He didn't think of using the suffix notation in the first place. That question was a perfect example for it really. All Bezza did was say, "do this for the x-component, and then look, it's the same for the rest". The suffix notation puts the "look, it's the same for the rest" into a handy generalisation.

EDIT: Not allowed to swear? What a load of sh|tty b0||0cks. Einstein once said that the summation convention was his only contribution to mathematics. He was so modest  Worzo
EDIT: Not allowed to swear? What a load of sh|tty b0||0cks.

LOL! Brilliantly done! Boo! shiny
Boo!

ARGHHH!!!! IT'S THE PINK LEGEND! <runs away in fear>

why aren't u on msn?