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Beta - special functions - manipulation

by DigitalSwitch
Tags: beta, special
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DigitalSwitch
#1
May5-12, 07:51 PM
P: 4
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.
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Dick
#2
May5-12, 08:05 PM
Sci Advisor
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P: 25,228
How about trying the substitution y=a*sqrt(t)?
DigitalSwitch
#3
May5-12, 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}(1-t)}[/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}(1-t)^{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.

Dick
#4
May5-12, 11:19 PM
Sci Advisor
HW Helper
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P: 25,228
Beta - special functions - manipulation

Quote Quote by DigitalSwitch View Post
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}(1-t)}[/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}(1-t)^{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.
It will. You did it wrong. How can you wind up with a y outside of the integral? y is a function of t. Do it more carefully.
DigitalSwitch
#5
May6-12, 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 = a2/a2 = 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}(1-t)^{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!
Dick
#6
May6-12, 06:57 AM
Sci Advisor
HW Helper
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P: 25,228
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 (1-t) into the integral.
benz31345
#7
May7-12, 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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