@TedShifrin that's what I'm doing :P

@robjohn Some might ^^

@robjohn From planet meanSquares

So are some of my students ... hopefully the rest are studying for an exam @Zach.

@Hippalectryon The last integral is easy to be finished by differentiation, but how about avoiding the differentiation?

There should definitely be a way without diff

@Hippalectryon It is frustrating that, as a mod, it is not acceptable to vote to reopen a question. I spent a lot of time on a question and when I finally had a solution, I try to submit my solution only to find out it has been closed as a duplicate. Part of the question is not answered in the other question, so I don't think it should be closed as a duplicate.

@robjohn Indeed

that's happened to me before, @robjohn ... of course, I haven't mod status.

@Zach: You working on anything interesting?

trying to figure out how to the integrals correctly when the function isn't a graph

what do you mean?

@TedShifrin Oops... thought I was talking to someone else.

no, what variable is varying freely? @Zach

x

Does this happen to you frequently, @robjohn? :D

Right, @Zach, so that's one of your variables. And how do you get $y$ and $z$?

@TedShifrin well, I was talking to Hippalectryon, but you replied :-)

r and an angle $\theta$

My humble apologies, @robjohn :)

$r$ ? @Zach

@TedShifrin no apologies needed. I need to be more observant

$y^2 + z^2 = r^2$ :P

$r^2$ ... so $r=2$

Besides, I don't need to know this stuff until next semester :D

how do I put z in terms of $\theta$ then?

How do you give $(y,z)$ if $y^2+z^2=4$?

I'm not sure what you're asking

This is just doing cylindrical coordinates with the axes swapped around.

yes

so then $z=2cos(\theta)$ ?

So you want polar coordinates in the $yz$ plane.

I probably would have measured $\theta$ from the positive $y$-axis, but if you measure it from the positive $z$-axis, that's ok, too.

doesn't matter to me :P

You just have to be consistent throughout the problem.

@TedShifrin What will you after you retire ?

Other than bashing Balarka

@Hippa: First, I'm ignoring Balarka. Second, I'll get there when I get there.

Fair enough :)

What'd he do now, @Ted? I'm just ignoring what he says until he answers the questions I pose... :)

I don't know, @Mike .. I've been ignoring him for a few days.

Hi, by the way.

I think he must miss my smacks.

@TedShifrin O_o that would sound really weird in French

It probably sounds weird in English, @Hippa ...

xD

Gosh, i need to stop with smileys

Yes, he's definitely whining, @Ted... but it's probably better for your sanity.

I guess this is why your parents don't allow you here, @Hippa.

@TedShifrin They don't know I'm here

Probably

Your laptop should have combusted a year ago.

A year ago ?

@TedShifrin what would I integrate $\theta$ from and to? I know it's got to span $\pi$ total, but I'm unsure where I'd start since the cylinder is on its side

Right, @Zach. It is more confusing because of the axis swap. You want $z\ge 0$. So it depends what your $\theta$ is. If $z=2\cos\theta$, then what range must you have for $\theta$?

Ah, I see

(I switched it to it's in reference to the y axis instead of the x)

Oh, well, in that case, $z=2\cos\theta$ is no longer correct.

yes, I know :)

it'd be $z=2sin(\theta)$, no?

I presume so, yes.

which makes the bounds 0 to $\pi$

I presume so, yes.

there must be something off with my $dS$ (if that's what you call it) then

What is it?

definitely wrong. I tried to treat it like a graph :P

Cylindrical coordinates + a cylinder makes for very simple $dS$. Think about it pictorially.

$<0, 2cos(\theta), 2sin(\theta)>$

What is that? Oh, the cross product. So, go on.

OK, I'm outta here.

have a good night!

@AshwinGokhale No, I have the option to, but just pure math next semester. I am overloading with operations research though, since I really want to do it and the other 4

@Chris'ssis: I spent a long time on a question the last couple of days, and now it is closed. I am bummed.

What rep do I need to vote to reopen?

@Committingtoachallenge 3000

Damn, I am so low rep :\

@KajHansen I love 'In Flames'

Now I shall spend 4 days at uni without leaving once :)

@Committingtoachallenge do you ever sleep?