Integration: Area Enclosed By Graph Of $X^4 + Y^4 = 1
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$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
Kirthi RamanKirthi Raman
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$endgroup$
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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
CommunityBot
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answered Apr 28, 2012 at 23:59
Pedro♦Pedro
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$endgroup$
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