FINALS

Dude... I had 3 tests today, and my candidate fitness assessment for the Air Force Academy. I'm so tired. I also have an interview with MIT tomorrow. Hopefully it will go well... haha of course it will. Time for more math.

Suggestions

I need some good acoustic guitar songs. Any that would be awesome, you should tell me. Gracias.

ick-Clay on the Ad-ays

Yep. That's Pig Latin :P Seriously though. ...I think it's time for bed. Procrastination owned me again. I need more followers. Night everybody.

Math and Physics ventures...

Well most likely no one will want to see this, but if anyone is interested... Today in Economics I wasn't paying attention to the teacher, but rather I was formulating and deriving the expression for moment of inertia of a hollow sphere. For those of you who don't know, moment of inertia is how hard it is to rotate something about a point or axis.

Well here we go. This probably will be really hard to understand because of the lack of symbols... sorry for the inconvenience.

int = integral; pi = pi; d = density; m = mass; r = radius; I = moment of inertia; variables = phi, theta, p;

I = mr^2

Converting to spherical coordinates and integrating we get:

I = int[int[int[mr^2,V]]] = int[int[int[(p*sin(phi))^2*p^2*sin(phi),p],phi],theta]

p = r in this case.

The tricky part was the innermost integral. I realized after thinking for a while that you can't integrate with respect to p because it is infinitesimally small and I don't need the whole radius. Therefore, I converted it into a trivial sum in order to just use the very edge. I also set up my bounds in order to account for the entire sphere.

int[int[sum[d*p(i)^4*(sin(phi))^3,i,1,1],phi,0,pi],theta,0,2*pi] where p(1) = r

Simplifying to a double integral...

int[int[d*r^4*(sin(phi))^3,phi,0,pi],theta,0,2*pi]

now some trig subs to make this integrable.

d*r^4*int[int[sin(phi)*(1-(cos(phi))^2),phi,0,pi],theta,0,2*pi]

simplifies down to:

d*r^4*int[4/3,theta,0,2*pi] = (8/3)*pi*d*r^4

We can assume a hollow sphere to only have an area: which is 4*pi*r^2
And planar density is mass over area
And now we substitute

(8/3)*pi*(m/(4*pi*r^2))*r^4 = (2/3)*m*r^2

Hooray calculus!

Pick Me!

Well, I'm home again. I need people to follow me and stuff. That'd be cool. Or not. Whatever. I think it's time for some recreational multi-variable calculus. Who else loves math?

Procrastination

I'm supposed to be reading right now... Well, that's not happening. I've planned on doing so much but find myself in front of the computer when I need something to be done. If anyone ever reads this: procrastinate, yes or no? (Question is meant to be general and is open for interpretation.)

Introduction

I've never really blogged before and this is kind of just a test run. I don't even know what a blog is really supposed to be. I'm just going to write down random thoughts I have... It's cold outside.