@BernardoMeurer hello?

do you have specific questions

@0celouvsky Yes

But it's a really shit one, ready?

Like chub n' tuck integral level shitty

ok

Okay

Given a set $S=\{(x,y,z)\in\Bbb R^3 \colon x^2 + y^2 < 1; 0 < 2z < 3+y\}$ Write the expression that yields the volume of $S$ using $\int\int\int dxdydz$ as an integration order.

How do I even draw that set?

@0celouvsky

lol

$x^2+y^2<1$ is an open cylinder

jesus

ye

And you don't have the answer?

I do

You do have it?

Yep

I don't know how to get there though

starting by the bounds on z

where do they come from

I think you need to work this from the insie out

do you understand the $x$ integral?

Yeah, he just solved the inequality for x

Now y is easy

It's bound is just the radius of the circle defining the cylinder

it's not clear to me why he needs two integrals

now dafuk do I do for dz

ah, do you want me to draw the picture

0 to 2?

Indeed, why does he need two integrals?

Yeah, draw it please

So the straight dotted line is where he breaks up the integrals

The first integral is the lower half

that's just a plain old cylinder

the second is the part that's cut off at an angle

How did you conclude that the second eq. forms that plane?

Because $2z=3+y$ is linear, hence a plane

so you're cutting off at a plane

and leaving everything below it

(where the $z$ direction is "up")

Got it

so with that picture I think it should be clear

And we know it splits at 1 because?

splits at 1?

So we know that $y$ goes from $-1$ to $1$ in the uncut region

How do I know at which z to split the integral

Yeah

Ok, let's look at the first integral

$y$ goes from $-1$ to $1$

So then we look at what $z$ runs over

Yep

you get that from the inequality

the lower bound on $z$ is $0$ because that's the base of the cone

for $y=-1$ z goes up to 1

and for y=1 it goes to 2

that's right

Ah

I see

Alright

ah, but for $y=1$ your $z$ equation takes you into the cut part

Cool

you have to stop at $y=-1$

Yeah

But I see why it cuts off at 1 now

so now for the second part

the integral that is

now you're being cut off by the plane

Yep

so your lower bound on $y$ is now $2z-3$

But you're free to go to $y=1$ with the upper bound

Because we solve the plane boundary for y

right

so then $z$...well, you're picking up where the first one left off. So the lower bound is 1.

Yeah, it all makes sense now

But the largest $z$ is $2z=2+2$

Moral of the story: draw a picture

Okay, another one

Using the appropriate coordinate shift compute the mass of the solid $$B=\{(x,y,z)\in\Bbb R^3 \colon 0 < y < x^2 + z^2 ; y < \sqrt{2-x^2-z^2}\}$$. The function for mass density is given by $$\sigma(x,y,z) = 2y$$

what's a coordinate shift?

Changing the coordinate basis, to cylindrical or spherical or ...

Bad translation I guess?

It's called a coordinate change in English

You want to use cylindrical coordinates where $y$ is the height coordinate.

because then $x^2+z^2=r^2$

Ah, I hadn't seen that

I was about to ask how do I detect the best coordinate system

how many do you know?

polar, cylindrical, spherical

and cartesian

ok, so that's pretty limited

polar is when I'm in 2d and see circles

cylindrical I try to see if the Z variable is independent somewhat

cylindrical is for 3d and circles

or one of them is

spherical is when I have no more hope

$$\bar\partial^*=-\bar\star_{E^*}\bar\partial\bar\star_E$$

Christ

What the fuck is that

adjoint of the antiholomorphic differential on a Hermitian vector bundle

Oh dude come on

the notation is getting a bit sill

@BernardoMeurer I'm reading a book

I'll answer questions if you ask

@0celouvsky i know half of those words but i don't know the meaning of the sentence

Okay, so I detected a chance for a coordinate change

I write $r = x^2 + z^2$

@JoshuaLin page 168 of "Differential Analysis on Complex Manifolds"

@0celouvsky Shouldn't I try and draw this first?

To figure out the theta

Uh

the theta is actually irrelevant

the equations don't depend on it

you're just going to integrate it out

this isn't chub and tuck level, come on :P

I'm confused

What do I do now?

I just figured r out

but so what?

what are your constraints?

$0 < y < r$ and $y < \sqrt{2-r}$

so pick $\theta=\arctan(z/x)$ or whatever

it doesn't really matter because $\theta$ doesn't appear anywhere

What? Why would I do that?

you wanted a theta

can't I just integrate theta from 0 to 2 pi and be quiet about it

It doesn't show so who cares

I swear to christ

does anyone ever fucking read what I say

:^)

Now y

goes from 0

to r

no

to

it's $y<\sqrt{2-r^2}$ btw

@0celouvsky Oh, yeah correct

so, how the heck do I get the bounds of y?

They depend on $r$

I mean the lower bound is 0

so $0<y<r^2$ and $y<\sqrt{2-r^2}$

are you sure you copied the problem right?

Yes

yeah, so solve the constrains for $\rho$

well, $r$ in our notation

$r>\sqrt y$

$r<\sqrt{2-y^2}$

exactly

now figure out the $y$ bounds

the lower one is zero

Yes

That's trivial

the upper one is the trick

you're not gonna be happy

?

you need to solve $r^2=\sqrt{2-r^2}$

Goddamit

Why?

you need to find out at which $r$ the two upper bounds for $y$ agree

yeah

that's the largest one

makes sense

sigh

if $y$ gets larger, it violates one of them

so that's $r^4+r^2=2$

write $k=r^2$

so $k^2+k=2$

that factors

$(k+2)(k-1)$

So you get $r^2=-2,1$

So $r=\pm \sqrt{2}i,\pm 1$

the only reasonable one is $r=1$, so $y<1$

good?

Got it

Yeah

Yeesh

Okay

and we integrate our density function

yes, and don't forget the $r$ from the Jacobian!!

$$\int_0^{2\pi}\int_0^1\int_{\sqrt{y}}^{\sqrt{2-y^2}}2yr drdyd\theta$$

Oh, yeah, almost forgot that

looks right

I can just shuf the 2 straight out of there, right?

yes

and do the $\theta$ integral

And the y our of the r integral

yes

"doing the $\theta$ integral" means just removing it?

Since it has no power here

well you just write a $2\pi$ out front

okay, so the primitive of $r$ is $r^2/2$

yes

$$4\pi\int_0^1y-y^2dy$$

Getting there

just do the integral

Hm?

I'll just separate them

then it's trivial

you should be able to write down the primitive by inspection

$y^2/2 - y^3/3$

yes

so the integral is $1/2-1/3$

$1/6$

yes

so $4\pi/6$

which is wrong

lol

dammit

You didn't do the $r$ integral correctly

and you forgot about the $y$ you took out

I didn't

you still did the $r$ integral incorrectly

$$4\pi\int_0^1 y(\frac{2-2y}{2})dy$$

write down the $r$ integral

isn't the r integral just $\frac{r^2}{2}$ limited from $\sqrt{y}$ to $\sqrt{2-y}$

Ah

oh, you have a $2y$

well

y^2

yes

It's y^2

Well whatever

Okay, it makes sens to me now

I'm going to head to bed, exhausted

I think I'll do okay at the exam

good night

good luck

Goodnight man, thanks for the help!

np

@Slereah help

Jesus Christ

@Slereah back in 1916, the energy momentum tensor was $G_{\mu\nu}$

@Slereah The original Schwarzschild paper uses +---

Are we using the wrong convention?

hey

@Slereah I don't understand the derivation of the Schwarzschild interior solution

there's an ode $$\frac{\partial}{\partial\psi}(\sin^2(\psi)\frac{\partial}{\partial\psi}\xi)=-3‌​\sin^2(\psi)\xi$$

Are we talking black holes or not black holes

stars

the solution to that is clearly $A\cos\psi$, but I'm pretty sure there should be a two-parameter family of solutions

it's a second order equation after all

Who even uses Maple

That's below Mathlab and Mathematica

I use it and it works pretty well

Speaking of which

I plugged in that equation into Wolfram

Apparently two families of solutions are $A e^{-2i\psi} \csc (\psi)$ and $B \dfrac{e^{3i\psi}}{-1 + e^{2i\psi}}$

Not too sure how you can construct $A \cos(\psi)$ from those

I tried $B \sin(\psi)$ because that felt right but obviously not at all a solution

huh

Stephani doesn't seem to include that ODE

does Stephani have the solution?

Well the solution to what, exactly

where do you get that equation

Frankel "Gravitational Curvature"

You know what's mysterious?

Bigfoot?

Does anyone actually check that the double cover $\mathrm{Spin}\to\mathrm{SO}$ is actually a covering map?

Probably somewhere?

I think it's just that $\lambda, -\lambda \in \textrm{Spin}$ maps to $\Lambda \in \textrm{SO}$

A covering map is more than that.

i'm gonna guess none

Since most swords are forged with no blood at all

> Hemoglobin count in dragons is similar to large land mammals, or, more precisely, giraffes.

insane

Who is that dragon vet that checks hemoglobin count in dragons

super clifford algebra

you have got to be kidding me

what

is that one of those supersymmetric shenanigans

I don't think anyone actually gives a shit about it being a covering map

isn't it a covering map just by virtue of the quotient construction?

quotient construction?

$\mathrm{Spin}(V):=\mathrm{Pin}(V)\cap \mathrm{C}\ell^\mathrm{ev}(V)$.

I mean more the $Spin = SO / \sim$ sort of thing

that doesn't even make sense

spin is larger than so

Oh right, other way around

and...it might

You might be right

idk I'm too damn tired

I guess the hard part is proving the two definition equivalent

that should follow from uniqueness of universal covers.

@JohnRennie I need another $200 book

I can't keep this up

Also what is $C\ell^{ev}$

Even degree clifford things

Also how do you define the pin group without reference to the rotation group

isn't it also a Clifford thing

$\mathrm{Pin}(V)$ is the group in $\mathrm C\ell(V)$ consisting of elements of the form $a_1\cdots a_n, a_j\in V, ||a||_j=1$.

Ok I gotta sleep

Cheers

night

@Slereah here that would be closed as insufficient effort since a simple Google suffices to answer it.

what's the site for blood forging

@Kaumudi.H Morning. I thought you'd be on a train by now.

If I haven't already said so, good luck with the exam!

Will you get a chance to explore the town where you're staying?

Ah, I didn't realise the exam was in the place where you used to live.

At least you don't have the added stress of finding your way ina strange town.

Are you staying with your grandparents?

Can anyone clear me on the fact why a charge doesn't feel force due to its own electric field

What will you do tomorrow before the exam. Just chill, or do some exploring?

It seems to me it can distinguish between it's ownfiel and field produced by other charges?

@Xasel what sort of force would a charge feel from its own field? In what direction would the force be?

I meant to say Electrostatic force

IF I place charge in other charges electic field it feels a force

then why it doesn't feel a force in its own field?

@Xasel In what direction would the force be?

@Kaumudi.H Sit on the porch and watch the fireflies - wow, that sounds idyllic :-) In my case I would also drink beer though I suggest you don't want to sit the exam with a hangover :-)

Well...I don't know at this moment ...Did multiple readings and some google and still unable to resolve the confusion

@Xasel Hint: the field around a point charge is spherically symmetric.

???....a picture is forming in my brain but I'm pretty much still in haze(and yesterday's discussion wasn't less confusing :(

@Kaumudi.H In the UK you can't buy alcohol if you're under 18, but there's no restriction on drinking alcohol e.g. with meals at home.

@Kaumudi.H : You can prove to law enforcement that that alcohol content in your blood stream is due to strenuous exercise :P

@Xasel the force on a charge $q$ due to an electric field $\mathbf E$ is $\mathbf F = q\mathbf E$. So the force and the field are in the same direction. Yes?

yes.

@Kaumudi.H I have a certain sympathy for that point of view, though it's more extreme than I would be. Drinking by young people does cause problems in the UK, and that applies to girls as well as boys, but it's a small minority who cause the problems.

@Xasel and spherical symmetry means the field is the same in all directions, so the net field sums to zero.

Boys can't get pregnant.

NB that's just an observation and doesn't imply any moral stance.

I don't mean to defend sexism. I wouldn't, and don't, take that approach with female members of my family.

But I guess it's understandaable. Go back a few decades and securing a good marriage was key to a woman's future security.

Even in the UK that was the case in the 19th century.

I agree.

I've read articles claiming that the male-female ratio is rising in India, and that now some families are requiring the boy supplies the dowry.

My father once told me that he's harsher on my sister than he was on me about curfew timings, contactable friends while going out, knowing where she was going and reminding her more often about the dangers of drinking because it's more perilous to be a drunk girl than it is to be a drunk guy. I understand your rant about impurities and learning to make rotis else the in-laws be angered, though @Kaumudi.H