Beta - special functions - manipulation

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.

$\int^{a}_{0}y^{4}\sqrt{a^{2}-y^{2}}dy$

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

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Beta - special functions - manipulation

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

 To solve: $\int^{a}_{0}y^{4}\sqrt{a^{2}-y^{2}}dy$ (Utilising the Beta special function) Use the substitution $y=a\sqrt{t}$ This implies $t = \frac{y^{2}}{a^{2}}$ Change bounds and variable for y=0, t=0; for y=a, t = a2/a2 = 1 $t=\frac{y^{2}}{a^{2}}\rightarrow \frac{a^{2}}{2y}dt=dy$ Changing variable, bounds and subbing for y gives equivalent integrals: $\int^{a}_{0}y^{4}\sqrt{a^{2}-y^{2}}dy \Leftrightarrow \int^{1}_{0}\frac{a^{6}t^{2}(a^{2}-a^{2}t)^{1/2}}{2a\sqrt{t}}dt$ Simplifying (which gives the desired Beta function form) $\frac{a^{6}}{2}\int^{1}_{0}t^{3/2}(1-t)^{1/2}dt$ Solution $\frac{a^{6}}{2} B(5/2,3/2) = \frac{a^{6}}{2}\cdot\frac{\pi}{16} = \frac{\pi a^{6}}{32}$ 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!
 Recognitions: Homework Help Science Advisor 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.