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 .

2. 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?

3. 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