Thread: please help me the problem using beta and gamma

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))

Could someone please help me

tnanks .

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))

Could someone please help me

tnanks .

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

Originally Posted by HallsofIvy

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