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 (Flower)

tnanks .

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- Jan 18th 2010, 03:57 AMwesamplease 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 (Flower)

tnanks . - Jan 18th 2010, 05:50 AMHallsofIvy
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? - Jan 18th 2010, 06:10 AMwesam