I need to find the moment of inertia of the body $(x^2+y^2+z^2)^2=a^2 xy, \ x > 0, y > 0$ about the $xOy$ plane. The density of the body is $\rho$ (constant) So, it means I have to calculate $\rho \underset{V}{\int \int \int} z dx dy dz$. I moved to spherical coordinates, where the condition $xy > 0$ (in this case it's equivalent to $x,y > 0$ gave me that $\sin (2\theta) > 0 \rightarrow \phi \in [0, \pi/2]$.
So I had an integral $\rho \int_{0}^{\pi/2} d\theta \int_{0}^{\pi} d\psi \int_{0}^{a\cos (\psi) \sqrt{\cos\phi \sin \phi}} r^3 \sin^2 (\psi) dr$, which is $\dfrac{\rho a^4 \pi^2}{1024}$. The answer in the book is $\dfrac{3}{10} 2^{-10} \sqrt{\pi} \rho a^5$. Could someone give a hint what went wrong? Also it's possible the mistake is in the book.
Maybe there is a way to say right off the bat whether the answer is correct or not, using the physical meaning of the moment of inertia
Moment of inertia sounds wrong. This is just the moment, which means it’s computing the $z$-coordinate of the center of mass. Isn’t this volume symmetric about the $xy$-plane, in which case the answer should be $0$? What is $\psi$? This is all a jumbled mess.
@TedShifrin Sorry again for inconvenience of the translation, Ted!. In Russian mathematics we have spherical coordinates $(r, \phi, \psi)$, I knew that Europeans use $\theta$, so I changed it.
Gosh. You're right, it's all messed up. Let me use the comfortable notation for me, spherical coordinates -- $(r, \phi, \psi). \rho $ -- density of the body.
And here. We use spherical sub that looks like that: \begin{eqnarray} x=r\cos(\phi)\cos(\psi) \\ y=r\sin(\phi) \cos(\psi) \\ z= r \sin(\psi) \end{eqnarray}
@TedShifrin I see your point, but to me it's not obvious that the integral is going to be zero, if I'm looking at the formula. The book says: "Static moment about xOy is $I_{xOy}^{(1)} = \underset{V}{\int\int\int} z \cdot \rho (x,y,z) dxdydz$
@TedShifrin Well, from the equations it seems like $\phi$ is not limited at all, so it circles all the way around and changed from $0$ to $2\pi$. And from the condition $x,y > 0$ I got that, first, it's equivalent here to $x\cdot y > 0$. So $r^2 cos^2 (\psi) cos(\phi)sin(\phi) > 0$. So, $sin(2\phi)>0 \rightarrow \phi \in [0, \pi/2]$.
I mean, you wrote the answer yourself, I just informed you I was calculating it at the moment. I was quite sure it was what it was, but decided to check it via calculation. It's 5am here :)
I remember encountering a geometry problem, around 10 months ago. I began solving it at exactly 1:15AM and by the time I was done, the birds were chirping and I could see the sunlight start to seep in
I have a research task, but at the moments when I get stuck with spherical coordinates for no reason, it feels like the only thing I should do is to clean the streets with a shovel
Almost every math professor I meet who talks about their career always reveals how they dedicated weeks, even months, to a single task/problem, like Ted himself.
I'm not sure about Europe/USA, but here you get some hard problem from prof. and you work on it for 1 year and you write a paper on how you tried to crack it
@MagnusAlexander TokyoU is kinda like that. I've even been given the opportunity, along with 2 other guys, to take graduate level courses even before completing my first year.
@冥王Hades I take some courses in the beginning of the semester, but somewhere in the middle I realize I won't be able to pass it and that I have no time for it so I give up
Now I only attend p-adic calculus course, bc my task that I'll be trying to solve this summer is about this
Uh. We have English, but I know it better than the teacher. And I'm being objective. 80% of the info I consume is english, so this course takes me zero effort. Apart from that, we have no such subjects. I mean, the ones which are not related to maths
@MagnusAlexander I'm good at PE, but it's hard because everyone here is good at it because of how Japanese high schools work. For English, English classes in Japan are actually very hard, and I'm a native speaker. They dive way too deep and far beyond what is needed to have a conversation
@冥王Hades That's the problem with academic English, mostly it's overlearning. But I guess this is good to have in your arsenal when you have to write an article of some kind.
I have never ever really learned any grammar rule for English tbh. I just listened to Ed Sheeran songs and apart from trying Ed's patience with Russian notation, I think I communicate pretty well
@冥王Hades That's actually cool. As long as it all doesn't turn into grammar tests and other crap, I'm okay with such a class
@MagnusAlexander it does turn into tests. We share those classes with students from other departments like humanities and social sciences, and they're very good at this kind of stuff, so they always score the highest on it, even though I'm a native English speaker
Well, we have an opportunity not to care about our grades at all. You just need to get the min score to pass, and forget about it. I think it's not the case in Japan, you guys are having kind of an education cult or smth.
In Russia even if your avg is 2.5/5 but you're smart you can go anywhere (I mean the job)
you'll like this, ted. i took munchkin to the third of a group of ice skating classes. very elementary stuff, maybe too elementary, and the instructor is not paid enough to care. she essentially boycotted the lesson by standing in one place, ignoring all instructions, and skating only during the 10 minute 'free skate' at the end.
on the way home i said, we'll see how it goes next week. munchkin said: "i'm not skating. next week i'll do the same thing. you'll see and you'll learn." calmly, like a murderer taunting the police.
user: the only problem is i think she would enjoy that too much. it's a weird game of chicken that we're playing. i didn't exactly want to drive her to ice skating, either, but 4 is too early to learn "if i just complain enough i can avoid it."
yes, it's pretty bad. her other thing is sneaking contraband (usually toys) into places where they are not allowed (e.g. portions of day care not devoted to playing) or advisable (e.g. in her hand, on the ice, at ice skating, where she almost lost it).
i think some of the other kids at day care look up to her as this badass of rule-breaking.
she knows enough skating to (1) sneak a toy into ice skating, (2) misplace and drop the toy during ice skating, (3) later skate over to the toy on the ice and pick it up without going down on one knee or falling over, without her instructor seeing. that's what's annoying about all of this.
anyway, i'm sorry if she ends up destroying the planet.
Hey everyone. I'm trying to prove that it's safe to add product objects to a category, and I'm finding it surprisingly difficult. It seems like it should be pretty straightforward.
Specifically, suppose we have any category C. Define the category C' as the category which is generated by following: the objects, arrows, and identities of C; the operator that, given any two objects a and b, produces an object called $a \times b$; and the operators and identities comprising the definition of a product object. Then C is a full subcategory of C'.
I managed to prove that C' doesn't have any arrows between the objects of C that C itself does not have. What I have yet to prove is that distinct arrows in C are still distinct in C'.
What I did was to define a normal form for arrows in C', and then use several different induction proofs to eventually prove that every arrow in C' can be written in this normal form.
Is there an easier way to prove that it's possible to add product objects to a category, while preserving the original category as a full subcategory? Or is that just how you do it--define a normal form and do a bunch of induction?
Let (X, d) be a metric space and A\subset X and x_0\in X. Define d(x_0, A) =\inf {d(x_0, a) :a\in A} . Under what condition on A , d(x_0, A)=d(x_0, a) for some a\in A?
I have found that if A\subset X compact the minimizing element exists.
d_{x_0} : X\to \Bbb{R} is continuous and A\subset X compact implies d(x_0, A) is attained.
@AlessandroCodenotti
I know this property is interesting in the setting of Hilbert space.
In this setting, we need closed linear subspace (more specifically closed convex set) and the minimizing vector is UNIQUE.
ah okay, perhaps completeness is insufficient, we need some version of what is basically convexity but in the metric space setting , plus completeness of $A$
the total boundedness works as a stand in
@user977780 I know. I'm trying to think about whether total boundedness is replaceable by something weaker.
and something more reasonable than tautological
but as was said it seems hard
yeah maybe sequential compactness is the obvious choice of weakening, but in the metric space setting that isnt any different to compactness, so its only useful to consider something weaker than sequential compactness in spaces with more structure, like normed vector spaces
d(x_0, A) =\inf {d(x_0, A) :a\in A} then \exists (a_n) \subset A ( we need only first countability of X) such that d( x_0,a_n) \to d(x_0, A) . Now by joint continuity of d, we can interchange lim and d.
i dont know what you are trying to do @user977780 , are you claiming you only need first countability to ensure lim_n a_n exists (or a subsequence thereof)? because if you are i dont understand your argument at all. You cant just assume lim_n a_n exists, if you have compactness of A then you can find a convergent subsequence..
it doesnt work, basically im trying to think of a good condition that ensures d(a_n,x_0)-d(a_m,x_0) -> 0 for large n,m also ensures a_n,a_m are eventually in a compact set, which is either a neighbourhood of x_0 or something away from arbitrarily small neighbourhoods of x_0
but I think this , if it turns into anything meaningful, will require a condition on the whole space X, not just A
Let (X, T) be a topological space such that no continuous real valued map is one-to-one.
Quick examples: (X, \tau_p) particular point space, (\Bbb{S}^1, \tau_{euclidean})
The above two examples are not particularly interesting as both are pseudo compact spaces, every real valued continuous maps are CONSTANT. Extremely injective :)
yes, by Rudin, you can prove existence of P-points to prove it
I just thought that this "sum of ultrafilters" which is now part of what is called the Rudin-Frolik order, is an interesting construction of ultrafilters from others
The proof based on the book you mention is defined differently, so I believe it'll be detached from that, but I still found some explicit descriptions in a paper by Booth
Basically, $p\leq q $ iff $q = \sum(X, p)$ iff $p = \Omega(X, q)$ for some discrete family of ultrafilters $X$
so besides this ordering itself, there are those two operations $\Sigma$ and $\Omega$ on ultrafilters
and they can be explicitly described
Depending on that family of ultrafilters $X$
countable family*
So Frolik basically uses this relation $\leq$, defining it by explicitly mentioning what $\Sigma(X, p)$ is
then by counting argument, he shows $\mathbb{N}^*$ is not homogeneous
well it's more complicated than that I guess, since in all this he uses this ordering up to types of ultrafilters, which is sort of the equivalence classes from the quasi-order $\leq$ above
@AlessandroCodenotti f:S^1\to \Bbb{R} continuous implies f(S^1) is compact and connected. Hence f(S^1) =[a, b] and a, b is attained. Then by Darboux property f : S^1\to[a, b] onto map. If f is injective then \forall x\in S^1 f(x) =a or b
@AlessandroCodenotti My bad. Continuous one-to-one map
The last two lines are totally incorrect. Pseudo compact iff every real valued continuous maps are bounded ( need not be constant)
In case of particular point topology, every continuous map is constant.
yeah, you can just use that a continuous map $S^1 \rightarrow [a,b]$ lifts to a continuous map $f: [0,1] \rightarrow [a,b]$, satisfying $f(0) = f(1)$, what you want to show is that there necessarily exist $x_1,x_2 \in (0,1)$ mapping to the same point, so you just apply the extreme/intermediate value theorem (there is a global maximum attained at some point $c_0 \in (0,1)$, then consider $f$ on $[0,c_0]$ and on $[c_0,1]$, use $f(0) = f(1)$ and the intermediate value theorem for non-injectivity)
if $f(0) = f(1)$ is equal to the global max, use the global min instead
@porridgemathematics Given any countable ordinal \alpha , Borel sets of type G_{\apha} is of type F_{\alpha+1}.
This is true in \Bbb{R}
But not in general topological spaces.
For an example, (X, \tau_p) particular point space then \{p\} \in G_0 but \{p\} isn't in F_{1} or F_{\sigma} as no closed set can contain the particular point p.
G_0= open sets, F_0= closed sets, G_1=G_{\delta}, F_1=F_{\sigma}, G_2=G_{\delta\sigma} etc.
@MagnusAlexander I would like to clarify. I guess that there is a difference between every student in Moscow State University (or Moscow School of Economics, say) and "every student in Russia".
@Yai0Phah Oh yes. But MSU is just like 99% of other universities in Russia prefers soviet system and it's all the same everywhere. HSE has a pretty modern and European view on the matter, of course
@AlessandroCodenotti how exactly does one estimate cardinality of $\{p : p\leq q\}$ where $\leq$ is the Rudin-Frolik order on $\beta\mathbb{N}$ i.e. $p\leq q$ if there is a injective $f:\mathbb{N}\to\beta\mathbb{N}$ with discrete image such that $f^*(p) = q$ where $f^*$ is the continuous extension to $\beta\mathbb{N}$?
I have a question about some elementary Euclidean geometry. The proof belove (see screenshot) is a lot of text, but none of it is deep. I am wondering why $(\epsilon,\epsilon)=1$, where $\epsilon=\sum_i \epsilon_i$ with $\{\epsilon_i\}$ an "admissible" set of linearly independent unit vectors (meaning that $(\epsilon_i,\epsilon_j)\leq 0$ for $i\neq j$, and $4(\epsilon_i,\epsilon_j)^2=0,1,2,3$ for $i\neq j$).
Previously it was determined that $0<(\epsilon,\epsilon)=n+2\sum_{i<j}(\epsilon_i,\epsilon_j)$, so I don't know how we suddenly know that $(\epsilon,\epsilon)=1$. It should be true then that $(\epsilon,\eta_0)=1$, and it isn't clear to me why that would be the case.
