Integration: area enclosed by graph of $x^4 + y^4 = 1$

Question

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

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you are wrong. I dont think $\int (1-x^{4})^{1/4} = \frac{4}{5}(1-x^{4})^{5/4} +C$ — , Apr 28 '12 at 19:35
BTW: what you have is a special case of a Lamé curve. See here for the true area formula. The gamma function is involved. — J. M. ain't a mathematician, Apr 28 '12 at 19:41
@Chandrasekhar I would be very grateful if you could advise me where I have gone wrong in my integration? — Rob D, Apr 28 '12 at 20:13
Your supposed indefinite integral is just wrong. Differentiate it and compare to the original. Meanwhile, Dirichlet gave a pretty general recipe for this kind of calculation. I understand it is given in Whittaker and Watson, a book from the early 1900's that seems to be perpetually reprinted. — Will Jagy, Apr 28 '12 at 20:38
@NSDigital, you will not be able to do this integral easily; either look up the answers suggested above or make the change of variable $t=x^4$, and learn about beta functions. — Chris K. Caldwell, Apr 28 '12 at 22:55

score 8 · Accepted Answer · answered May 05 '12 at 00:31

8

Essentially the solution you have is good, but I would like to have this one anyways.

The required area is by symmetry

$$ A = 4 \int_0^1 (1-x^4)^{\frac{1}{4}} \hspace{4pt} \mathrm{d}x$$

Substitute $u = x^4 \hspace{4pt} \Rightarrow x^3 = u^{\frac{3}{4}} \hspace{4pt}, \mathrm{d}u = 4 x^3 \mathrm{d}x $

$$ \begin{align*} A &= 4 \int_0^1 \frac{(1-u)^{\frac{1}{4}} \hspace{4pt}\mathrm{d}u}{4 u^{\frac{3}{4}}}\\ &= \int_0^1 u^{\frac{-3}{4}} (1-u)^{\frac{1}{4}} \hspace{4pt}\mathrm{d}u\\ &= \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)}\\ &= \frac{2 \times 3.287}{\sqrt{\pi}} \\ &\approx 3.708 \end{align*} $$

answered May 05 '12 at 00:31

Kirthi Raman

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For those wondering about the jump from the second to the third line: the second integral is in the form of the integral definition for the beta function, which KV proceeded to directly turn to the gamma function definition. – J. M. ain't a mathematician May 05 '12 at 04:05
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Thanks, that's great @J.M. Could I ask how the gamma function works? i.e. how gamma(1/4) becomes 2? Help much appreciated. – Rob D May 09 '12 at 23:27
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@NS: actually, the $2$ came from $\Gamma(3/2)=(1/2)!=(\sqrt{\pi})/2$. The numerator, on the other hand, is $\Gamma(5/4)\Gamma(1/4)=\Gamma(1/4)^2/4\approx 3.286$ – J. M. ain't a mathematician May 09 '12 at 23:35

score 6 · Answer 2 · edited Apr 13 '17 at 12:20

In general, the curve

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

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

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

Here $\Gamma$ is Euler's Gamma function, defined for $\Re(\alpha)>0$ as $$\Gamma(\alpha)= \int_0^{\infty}e^{-\mu}\mu^{\alpha}\frac{d \mu}{\mu}$$ A closely related function is Euler's Beta function, given by

$$B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

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

Integration: area enclosed by graph of $x^4 + y^4 = 1$

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