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Beta  special functions  manipulation 
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#1
May512, 07:51 PM

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1. The problem statement, all variables and given/known data
I have this incomplete Beta function question I need to solve using the Beta function. [itex]\int^{a}_{0}y^{4}\sqrt{a^{2}y^{2}}dy[/itex] 2. Relevant equations Is there an obvious substitution which will help convert to a variant of Beta? Beta function and variants are in Beta_function Wikipedia article 3. The attempt at a solution Every time I look at the question I just start trying to integrate it as if the Beta function is irrelevant. I can't grasp how the beta function is applied to the 'incomplete' questions. Thanks. 


#2
May512, 08:05 PM

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How about trying the substitution y=a*sqrt(t)?



#3
May512, 11:07 PM

P: 4

That did it thanks. Great skills! (I want them)
I've still got something wrong. Subbing [itex]y=a\sqrt{t}[/itex] into integrand: [itex]a^{4}t^{2}\sqrt{a^{2}(1t)}[/itex] Changing bounds and var (wrongly?): [itex]t=\frac{y^{2}}{a^{2}}\rightarrow \frac{a^{2}}{2y}dt=dy[/itex] [itex]\frac{a^{2}}{2y}a^{5}\int^{1}_{0}t^{2}(1t)^{1/2}dt[/itex] which gives (by subbing indices + 1 into Beta function): [itex]\frac{a^{7}}{2y}B(3,3/2)[/itex] but [itex]B(3,3/2)=\frac{16}{105}[/itex] Mathematica says the answer by integration is [itex]\frac{\pi a^{6}}{32}[/itex] Pi is missing! I have a feeling that the beta function should of produced it. 


#4
May512, 11:19 PM

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Beta  special functions  manipulation



#5
May612, 12:25 AM

P: 4

To solve: [itex]\int^{a}_{0}y^{4}\sqrt{a^{2}y^{2}}dy[/itex] (Utilising the Beta special function)
Use the substitution [itex]y=a\sqrt{t}[/itex] This implies [itex]t = \frac{y^{2}}{a^{2}}[/itex] Change bounds and variable for y=0, t=0; for y=a, t = a^{2}/a^{2} = 1 [itex]t=\frac{y^{2}}{a^{2}}\rightarrow \frac{a^{2}}{2y}dt=dy[/itex] Changing variable, bounds and subbing for y gives equivalent integrals: [itex]\int^{a}_{0}y^{4}\sqrt{a^{2}y^{2}}dy \Leftrightarrow \int^{1}_{0}\frac{a^{6}t^{2}(a^{2}a^{2}t)^{1/2}}{2a\sqrt{t}}dt[/itex] Simplifying (which gives the desired Beta function form) [itex]\frac{a^{6}}{2}\int^{1}_{0}t^{3/2}(1t)^{1/2}dt[/itex] Solution [itex]\frac{a^{6}}{2} B(5/2,3/2) = \frac{a^{6}}{2}\cdot\frac{\pi}{16} = \frac{\pi a^{6}}{32}[/itex] Thanks for the help Dick. I obviously need to work more carefully and practice far more calculus as I keep making the simplest mistakes. As for choosing the substitution, I hope that comes with practice. I hope my working will help somebody, I end up on these forums from google searches often! 


#6
May612, 06:57 AM

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That looks great. If you look back at it, the choice of y=a*sqrt(t) isn't really all that clever. It's just about the only thing you can do to get a (1t) into the integral.



#7
May712, 06:15 AM

P: 5

I can't not post on forum homework I want to know so fast please help me



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