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Nov
5
comment How do I determine the measure for a volume integral?
Sorry, this was the integral of $r^2=x^2+y^2+z^2$. If I calculate the correct $r_z^2$, the squared distance from the $z$-axis, it's only $x^2+y^2$ so I only get two terms out of three. Due to the spherical symmetry, it's $2/3$ of the result above, so that the moment of inertia is $(2/5)MR^2$ for a solid ball.
Nov
5
comment How do I determine the measure for a volume integral?
Let me calculate the moment of inertia of a ball of radius $R$ here. I choose spherical coordinatets so $dV=r^2\,dr\,d\Omega$. The integral over $d\Omega$ gives $4\pi$, the integral of $r^4dr$ (there was $r^2$ from the moment) from $0$ to $R$ gives $R^5/5$: I can factorize them here. So the result is $4\pi \rho R^5/5=3/5 MR^2$. Less symmetric shapes are more complicated but it's pure maths.
Nov
5
comment How do I determine the measure for a volume integral?
Also, your question as well as your comments below my answer make it spectacularly self-evident that you don't understand what either of these symbols mean, in a striking contradiction with your assertion.
Nov
5
comment How do I determine the measure for a volume integral?
Dear @jnm2, the first equation in your comment above isn't a "magic transformation": it's a totally trivial identity. $\int$ is the integral sign but it's still a sum of infinitesimal pieces. The infinitesimal pieces are proportional to $dV$ and one may choose cubic shapes of $dV$ to make it clear that it may be written as $dx\,dy\,dz$. I don't understand what's your problem. If you can't understand this simple thing about the volume integral, you should give up studying physics quantitatively. Also, I wrote you how you can calculate the triple integral for any shape you mentioned.
Nov
5
answered How do I determine the measure for a volume integral?
Aug
17
awarded  Nice Answer
Jul
21
answered IMO 2011 Problem 5 - Show that $f(m) \mid f(n)$ if $f(m) \leq f(n)$
Jul
9
comment True, false, or meaningless?
I see, so $\forall i\in \{\}: T(i)$ holds for any $T(i)$ because there is no counterexample in an empty set for which $T(i)$ would fail - it holds for everyone (all zero of them). On the other hand, $\exists i\in\{\}: T(i)$ is always untrue because there doesn't exist any $i$ in an empty set that has a property - whatever property - because there's nothing in an empty set even without adjectives.
Jul
9
answered True, false, or meaningless?
Jul
9
answered Closed-form Expression for $\sum_{j=0}^{k-1}(2j+2)\sum_{i=1}^j \frac 1 {i^2}$? (problem with Mathematica)
Jul
9
answered Compositions of prime numbers
Jun
27
comment Validity of $\sum_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$?
Oh, I see, you're right.
Jun
27
comment Puzzle: $(\Box @)+(\Box @) = (\Box\bigstar\Box$)
Thanks for your care and implicit compliments, @JTL and @Jyrki, but I don't think it's a big issue. Let's be generous, people have the right to vote the way they want.
Jun
27
awarded  Nice Answer
Jun
26
revised Validity of $\sum_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$?
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Jun
26
revised Puzzle: $(\Box @)+(\Box @) = (\Box\bigstar\Box$)
added 123 characters in body
Jun
26
revised Validity of $\sum_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$?
added 454 characters in body; added 39 characters in body; added 55 characters in body; added 47 characters in body
Jun
26
comment Validity of $\sum_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$?
Interesting, I jumped on the same intuition as you did - it couldn't be right, I thought, because the RHS didn't have the minimal $\lambda$ anymore. Of course, the catch is that the "generalization" has almost nothing to do with the original inequality.
Jun
26
answered Validity of $\sum_{i=1}^n(a_i^2+b_i^2+c_i^2+d_i^2)\lambda_i\geq\lambda_1+\lambda_2+\lambda_3+\lambda_4$?
Jun
26
comment How to take the inverse of this signal?
You were faster, I erased my answer. ;-)