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.

enter image description here

  • $\begingroup$ you are wrong. I dont think $\int (1-x^{4})^{1/4} = \frac{4}{5}(1-x^{4})^{5/4} +C$ $\endgroup$ – user9413 Apr 28 '12 at 19:35
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    $\begingroup$ BTW: what you have is a special case of a Lamé curve. See here for the true area formula. The gamma function is involved. $\endgroup$ – J. M. is a poor mathematician Apr 28 '12 at 19:41
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    $\begingroup$ 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. $\endgroup$ – Will Jagy Apr 28 '12 at 20:38
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    $\begingroup$ @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 <a href=en.wikipedia.org/wiki/Beta_function>beta functions</a>. $\endgroup$ – Chris K. Caldwell Apr 28 '12 at 22:55
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    $\begingroup$ possible duplicate of The area of the superellipse $\endgroup$ – Pedro Tamaroff Apr 28 '12 at 23:59

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 $

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

  • $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician May 5 '12 at 4:05
  • $\begingroup$ 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. $\endgroup$ – Rob D May 9 '12 at 23:27
  • $\begingroup$ @NS: actually, the $2$ came from $\Gamma(3/2)=(1/2)!=(\sqrt{\pi})/2$. The numerator, on the other hand, is $\Gamma(5/4)\Gamma(1/4)=\Gamma(1/4)^2/4\approx 3.286$ $\endgroup$ – J. M. is a poor mathematician May 9 '12 at 23:35

In general, the curve


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.


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