Recent content by caseyjay

Hi Tiny-Tim, May I know why do you square cosecθ + cotθ? Did I miss out any "squares" in my working? I do not know from where and why did you square cosecθ + cotθ.

Dear all, I would like to evaluate \int\frac{1}{1-2sin\left(x\right)}dx Firstly, I make use of the Weierstrass substitution method by letting: u=tan\left(\frac{x}{2}\right) and therefore sin\left(x\right)=\frac{2u}{1+u^{2}} and dx=\frac{2}{1+u^{2}}du Eventually I can...

Hi alxm, Thanks for your reply, but why multiply it with STPSQH [FONT="Courier New"][(delta t)^2]/2? What do you mean by "scaling factor" and to convert to "what" units?

Hi nbo10, I am not sure if there is something wrong with LATEX or not. Anyway I have duplicate the earlier question over here but this time without using LATEX in it. The question I am having is related to molecular dynamics. In the book "Molecular Dynamics Simulation: Elementary Methods"...

Is Latex still down? I can't seem to produce right Latex despite I thought what I typed was compliant to Latex code.

Dear all, I am not sure if this is the right place to ask the following. (Moderators, please move this thread to the right sub-forum if necessary.) The question I am having is related to molecular dynamics. In the book "Molecular Dynamics Simulation: Elementary Methods" by J. M. Haile, a...

Sorry for the late reply. Thank you very much for your help! Regards, Casey

Hi kof9595995, Thanks for your help. I more or less what is going on. Basically to find the upper bound, you let sin^2(x) and sin^2(y) be 0. But why do you not let sin^2(x) and sin^2(y) be 1 in order to find the lower bound? Why do we proceed to convert the integrand to polar coordinates...

Erm... please let me know if I am on the right track. the biggest value of sin^2(x) is 1. Hence, the biggest value of sqrt(4+sin^2(x)+sin^2(y)) is sqrt(6). But why did you compare it with sqrt(4)? Also I know that region A is defined by x^2+y^2=1, hence it's radius is 1. But why did...

Oh dear I think my calculus is really lousy. :( I'm so confused now. How is the integrand be smaller than 1/2 and how will that lead to upper bound being pi/2? I'm sorry I am really lost on this.

Thanks kof9595995. How will using x^2>=sin^2(x), I will get a new integrand smaller than the original one? I'm lost on this. Secondly, even if I manage to find the lower bound, how do I approach to find the upper bound? Finally if there is any website or book that discussed about...

Hi, So instead of letting x^2=sin^2(x), we can let x^2>sin^2(x)? can you explain to me in more details? Thanks!

Hi, I still don't quite get you. Basically you are letting x^2 > sin^2(x) and y^2 > sin^2(y). Why do you need to set it to greater and not let them be equal? And how can that lead to inequality? Is there some websites or book I can read about this? Thanks in advance.

Hi all, I am faced with this question. I am asked to show that 2(\sqrt{5}-2)\pi\leq\iint_{R}\frac{1}{\sqrt{4+sin^2x+sin^2y}}dA\leq\frac{\pi}{2} Noting that the double integral is to be performed on region R which is bounded by the circle x^2+y^2=1 From what I know, the double...

Hi all, I think I have solved the problem and thanks for all your help. I have made some fundamental mistakes earlier and hence I am unable to solve it correctly. The solution is as follows: Surface: x2+3y2+2z2=9 Let f(x,y,z)=x2+3y2+2z2 Therefore...