# please help me the problem using beta and gamma

• January 18th 2010, 03:57 AM
wesam
hi:
have a problem solving this:
solved by using beta and gamma function

show that the area enclosed by the curve X^4+y^4=1
is {Gamma(1/4) }^2/2(sqrt(pi))

tnanks .
• January 18th 2010, 05:50 AM
HallsofIvy
Quote:

Originally Posted by wesam
hi:
have a problem solving this:
solved by using beta and gamma function

show that the area enclosed by the curve X^4+y^4=1
is {Gamma(1/4) }^2/2(sqrt(pi))

tnanks .

Well, one way to do that, since this is a closed curve, is to change to polar coordinates: let $x= r cos(\theta)$ and [tex]y= r sin(\theta). Now the equation of the curve is [tex]r^4 cos^4(\theta)+ r^4 sin^4(\theta)= 1 or $r= \frac{1}{\sqrt[4]{cos^4(\theta)+ sin^4(\theta)}}$ and then the area is given by
$\int_{\theta= 0}^{2\pi}\int_{r=0}^\frac{1}{\sqrt[4]{cos^4(\theta)+ sin^4(\theta)}} r drd\theta$ $= \int_{\theta= 0}^{2\pi}\frac{1}{2(\sqrt[4]{cos^4(\theta)+ sin^4(\theta)})^2} d\theta$ $= \int_{\theta= 0}^{2\pi}\frac{1}{2\sqrt{cos^4(\theta)+ sin^4(\theta)}} d\theta$.
.

Now, what are the definitions of the Gamma and Beta functions and how do they relate to that integral?
• January 18th 2010, 06:10 AM
wesam
Quote:

Originally Posted by HallsofIvy
Well, one way to do that, since this is a closed curve, is to change to polar coordinates: let $x= r cos(\theta)$ and [tex]y= r sin(\theta). Now the equation of the curve is [tex]r^4 cos^4(\theta)+ r^4 sin^4(\theta)= 1 or $r= \frac{1}{\sqrt[4]{cos^4(\theta)+ sin^4(\theta)}}$ and then the area is given by
$\int_{\theta= 0}^{2\pi}\int_{r=0}^\frac{1}{\sqrt[4]{cos^4(\theta)+ sin^4(\theta)}} r drd\theta$ $= \int_{\theta= 0}^{2\pi}\frac{1}{2(\sqrt[4]{cos^4(\theta)+ sin^4(\theta)})^2} d\theta$ $= \int_{\theta= 0}^{2\pi}\frac{1}{2\sqrt{cos^4(\theta)+ sin^4(\theta)}} d\theta$.
.

Now, what are the definitions of the Gamma and Beta functions and how do they relate to that integral?

thank you very much (Yes)