Superb Circular Shapes

Calculus Level 3

In this problem I asked your attention for the curve $S:\ \sqrt{1-|x|} + \sqrt{1-|y|} = 1,$ which looks a lot like a circle but bulges just a little. It is easy to check that this curve runs through the points $(0,\pm 1),\ (\pm 1,0)$ , as well as $\big(\pm\tfrac34,\pm\tfrac34\big)$ .

As someone pointed out, this has to do with superellipses --but curve $S$ is not actually a superellipse. However, there exists a superellipse that is very similar to our shape, given by $T:\ |x|^n + |y|^n = 1$ (with $n$ a real parameter), which runs through the same eight points $(0,\pm 1),\ (\pm 1,0)$ , as well as $\big(\pm\tfrac34,\pm\tfrac34\big)$ .

A section of both curves is shown in the diagram above: the blue graph is curve $S,$ and the orange graph the superellipse $T$ .

Clearly, the nearly circular area included by $T$ is slightly larger than that included by $S$ . How much bigger is it, expressed as a percentage?

Notes:

Give your answer as a percentage with three decimals precision.
This problem requires numerical integration. (If you can find a different solution, I am very interested!)

The answer is 0.367.

1 solution

Arjen Vreugdenhil
Feb 9, 2017

We limit the areas to the first quadrant.

The area included by curve $S$ can be expressed in closed form; it is equal to $A_S = 5/6$ . See my solution here for the details.

The area included by curve $T$ is found by integration, $A_T = \int_0^1 \left(1-x^n\right)^{1/n}\:dx.$ To find the value of $n$ , note that the graph passes through $(x,y) = (\tfrac34,\tfrac34)$ , which requires that $\left(\frac34\right)^n = \frac12\ \ \ \therefore\ \ \ n = \frac{\log 1/2}{\log 3/4} \approx 2.409421.$ Performing the integration numerically we get $A_T \approx 0.836391.$ This is about $\boxed{0.367}\%$ larger than $A_S$ .

I got AT ~ 0.854178, with 2.504% difference. I used the online free version of Wolfram alpha to compute this value.

Siva Bathula - 4 years, 4 months ago

Interesting... that's not what I get.

Arjen Vreugdenhil - 4 years, 4 months ago

Yeah I was wrong, messed up the integral somewhere.

Siva Bathula - 4 years, 4 months ago

Nice problem,approx 11/30 %

Nikola Djuric - 4 years, 3 months ago

(If you can find a different solution, I am very interested!)

Not sure if this helps, but.... The integral simplifies to a Beta function , $\dfrac1n \cdot \dfrac{(\Gamma( n+1) )^2}{\Gamma(2n+2)}$ . Apply the recurrence formula for the Gamma function , and we just need to approx $\Gamma(Q)$ , where $0<Q<1$ , which is a much simpler approximation to deal with.

Pi Han Goh - 4 years, 3 months ago

Almost. The area is $\int_0^1 \big(1 - x^n\big)^{\frac{1}{n}}\,dx \; = \; \tfrac{1}{n}B\big(\tfrac1n,\tfrac1n+1\big) \; = \; \frac{\Gamma(\frac1n)^2}{2n\Gamma(\frac2n)}$

Mark Hennings - 4 years, 3 months ago

Oh! Whoops!!!

Pi Han Goh - 4 years, 3 months ago

Superb Circular Shapes

The answer is 0.367.

1 solution

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