Recent content by naptor

Nope. But thinking about that integral was helpful I think I got it, at least according to wikipedia. Thanks.

Itorus=\int \frac{(a^{2}M)}{8\pi} d\theta + \int \frac{R^{2}M}{2\pi} d\theta, from 0 to 2\pi and got : \frac{(a^{2}M)}{4} + R^{2}M Well the substitution was silly because all I had was a bunch of constants and no functions of theta .

Sorry I didn't write the equations correctly. a is the radius of the little disk and R is the distance from the center of the torus to the center of the little disk, that is : Inner Radius = R-a.

Homework Statement To calculate the moment of inertia of a solid torus through the z axis(the torus is on the xy plane), using the parallel and perpendicular axis theorem. Homework Equations The Attempt at a Solution Well, first I divided the torus into tiny little disks and...

Welcome to the forum kxk010! You need to solve for F. The problem is of course that you don't know T so you have two unknowns. You have to figure out a way to get rid of one of the two variables (T) .One way to go is to divide one by the other, as your solutions manual did, a substitution...

Hi tim :smile thanks for the reply ok so :P(A \cup \bar{B})=P(A)+P(\bar{B})-P(A \cap \bar{B}) I can get rid off A \cap \bar{B} using P(A)=P(A\cap B )+P(A \cap \bar{B}) \Rightarrow P(A \cap \bar{B}) =P(A)-P(A\cap B ) now I can plug this in my first eq and use the hypothesis I got:P(A \cup...

Homework Statement Let E= A \cup \bar{B} and F= \bar{D} \cup C Assuming that A,B,C,D are independent show that F and E are independent Homework Equations By definition A and by are independent if and only P(AB)=P(A)P(B). The Attempt at a Solution I tried to use set theory to...

So, what you're saying is that I can show that : a is an n-ple solution \Rightarrow f(a)=f'(a)=f''(a)=...=f^n(x)=0 ?

I was justifying that based on an result on algebra that, as far as I can remember, says that any n-polynomial (n\geq1) has exactly(not sure about this "exactly") n roots. Therefore I thought that the only real root "a" found should have multiplicity 3. But I'm pretty sure now that the theorem...

Thank you Metallic. This approach definitely looks elegant, I'll try to use it to extend to the general case x^n-a^n .

Thank you, makes perfect sense now :) .

There is a theorem in algebra, whose name I don't recall, that states that given a polynomial and its roots I can easily factor it so for instance : p(x)=x^2-36 , assuming that p(x) is a real function, p(0)=0 \Leftrightarrow x=6,-6 then p(x) can be written as : P(x)=(x-6)(x+6) I...