Help With The Maths Please!

[JKPolk]

Re: Help With The Maths Please!

You DO spy a Jacobian. Nice eye. I wrote the expression MSWord using Equation Editor, and just SSd it and imageshacked it.

The equation is written correctly, it is solvable. It is a double integral, so one has to evaluate the inner integral first with respect to r, and then evaluate the result with respect to theta. I don't think trig substitutions would really help, as this is already in polar coordinates and using trig would probably only create a small catastrophe.

I've tried using u substitution and integration by parts but I just can't seem to crack the code.

Oh! All the variables equal zero! That appears to work, since those "less-than-or-equal-to" signs are there.

The less than or equal to signs give the range of integration, so the integral with respect to r goes from 0 to a/sqrt(2) and the integral with respect to theta goes from 0 to 2Pi.

And to those who are asking what to solve for, it should come out to be a number with respect to the variable 'a,' I'm not really trying to solve for anything, seeing as this isn't an equation.

Thanks for your input, I'm gonna keep plugging away at this until I pass out probably!

[LifereaperX]

Re: Help With The Maths Please!

Message removed due to hi19hi19

[Chrissi]

Re: Help With The Maths Please!

They wouldn't take precedence over the negative sign... It would be -2*-2, which equals positive 4...

No, actually it would be - (2*2).

Which = -4.

What you are saying is (-2)^2, rather than -2^2.

[JKPolk]

Re: Help With The Maths Please!

OK so I eventually got it. I split it up into two integrals, distributing the r at the end into each factor and integrating the first and second sections separately. Thanks for the suggestions on how to tackle it!

[Could_Strife007]

Re: Help With The Maths Please!

If this was differential calc we probably would have been of a bit more help, but then of course you probably wouldn't be asking for help

[otis1228]

Re: Help With The Maths Please!

2 pie=360 degrees of a circle and there is a theta thrown in there. So im going to assume you are trying to figure out something about the triangle. Umm well the variables could be the length of the radius, generally 1 in this case. I think. So I dunno if that was any help but anyway, just babbling. good luck.

[GuidoHunter]

Re: Help With The Maths Please!

Guys, really, if you don't know Cal II, just don't try. All of the responses in this thread besides Kit's have just been laughable. The thought is appreciated, but it's really getting him nowhere...

Pie? It's a tasty dessert.
"There is a theta thrown in there"? In where?
Triangle? Nobody mentioned a triangle. Just because there's a root(a^2 + r^2) doesn't mean that you're dealing with the Pythagorean theorem.
Radius? Sorry, no radius here, just a radial parameter, which is, as you said, a variable, and therefore isn't exactly a radius.
Generally 1? Well, we might just ignore that given range of possibilities, then. =)

On the plus side, you gave me quite a good laugh today. No offense meant, of course; it was just humorous.

--Guido

http://andy.mikee385.com