Integration: Area Enclosed By Graph Of $X^4 + Y^4 = 1

     
I"m attempting lớn find the area enclosed by the graph $x^4 + y^4 = 1$ as shown below.My approach was to lớn rearrange the equation so it is in terms of $y = f(x)$ và integrate one of the đứng đầu two quadrants with respect to $x$ & then multiply by $4$ lớn get the area for the whole shape. I"ve never tried to lớn integrate this kind of graph before và I"m not sure If I"ve done it correctly. Any input đầu vào or assistance would be much appreciated. Thanks.

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$egingroup$ you are wrong. I dont think $int (1-x^4)^1/4 = frac45(1-x^4)^5/4 +C$ $endgroup$
Essentially the solution you have is good, but I would lượt thích to have this one anyways.

The required area is by symmetry

$$ A = 4 int_0^1 (1-x^4)^frac14 hspace4pt gamize.vnrmdx$$

Substitute $u = x^4 hspace4pt Rightarrow x^3 = u^frac34 hspace4pt, gamize.vnrmdu = 4 x^3 gamize.vnrmdx $

$$ eginalign*A &= 4 int_0^1 frac(1-u)^frac14 hspace4ptgamize.vnrmdu4 u^frac34\ &= int_0^1 u^frac-34 (1-u)^frac14 hspace4ptgamize.vnrmdu\ &= fracGammaleft(frac14 ight)Gammaleft(frac54 ight)Gammaleft(frac32 ight)\&= frac2 imes 3.287sqrtpi \&approx 3.708endalign*$$


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answered May 5, 2012 at 0:31
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Kirthi RamanKirthi Raman
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$egingroup$
In general, the curve

$$y^1/a+x^1/a=1$$

is called a superellipse, and it"s area is given by

$$4cdot Gamma(alpha+1)^2 over Gamma(2alpha +1)$$

Here $Gamma$ is Euler"s Gamma function, defined for $Re(alpha)>0$ as $$Gamma(alpha)= int_0^inftye^-mumu^alphafracd mumu$$A closely related function is Euler"s Beta function, given by

$$B(a,b)=int_0^1 t^a-1(1-t)^b-1dt=fracGamma(a)Gamma(b)Gamma(a+b)$$

for $Re(a),Re(b) > 0$.For the details, you can see this question.


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edited Apr 13, 2017 at 12:20
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answered Apr 28, 2012 at 23:59
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Pedro♦Pedro
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