Hein's Egg

Calculus Level 5

The poet and scientist Piet Hein created an egg that will stand on its oblong edge. He called it the superegg, which is a solid of revolution (about the y-axis) characterized by the equation ( x a ) 2.5 + ( y b ) 2.5 = 1. { \left( \frac { x }{ a } \right) }^{ 2.5 }+{ \left( \frac { y }{ b } \right) }^{ 2.5 } =1.

Calculate the volume of the superegg if a = 2 a = 2 and b = 3 b = 3 .


The answer is 56.5508.

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2 solutions

Michael Mendrin
Jul 24, 2014

The volume formula for Piet Hein's superellipse (not to be confused with superellipsoid) is

2 π a b 2 Γ ( 1 + 1 n ) Γ ( 1 + 2 n ) Γ ( 1 + 3 n ) 2\pi a{ b }^{ 2 }\dfrac { \Gamma (1+\dfrac { 1 }{ n } )\Gamma (1+\dfrac { 2 }{ n } ) }{ \Gamma (1+\dfrac { 3 }{ n } ) }

so that for n = 5 2 n=\dfrac { 5 }{ 2 } , a = 3 a=3 , and b = 2 b=2 (as suggested by the photograph of Hein's Egg), the volume works out to 56.5508... 56.5508... . For a = 2 a=2 and b = 3 b=3 , it works out to 84.8262... 84.8262... .

The volume of the superellipsoid, however, is

8 a b c ( Γ ( 1 + 1 n ) ) 3 Γ ( 1 + 3 n ) 8abc\dfrac { { (\Gamma (1+\dfrac { 1 }{ n } )) }^{ 3 } }{ \Gamma (1+\dfrac { 3 }{ n } ) }

so that for n = 5 2 n=\dfrac { 5 }{ 2 } , a = 3 a=3 , and b = c = 2 b=c=2 it works out to 60.8591... 60.8591... .

The difference between Piet Hein's superellipse and the general superellipsoid is that the latter is not a solid of revolution.

So I wasn't wrong the first time (except you're right, when revolved around the y-axis it looks like a curling stone). Somebody reported that the egg was not a solid of revolution, so I changed it to a general superellipsoid. I think I will create a problem called "The Curling Stone" and have the answer be 84.8262.

Steven Zheng - 6 years, 10 months ago

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Thanks for clarifying. I've updated the answer accordingly.

Calvin Lin Staff - 6 years, 10 months ago

Yeah, Stephen, even I can be wrong sometimes. It's a very confusing subject, but, hey, it pushed me to spend time looking more closely at this and I've learned a few things.

Michael Mendrin - 6 years, 10 months ago

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Since you are the only person who answered, it is not too late to change the equation for this problem (and ask the moderators to change the answer).

Steven Zheng - 6 years, 10 months ago

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@Steven Zheng Sure, but I wonder why nobody else tried. I thought it was a nice problem.

Michael Mendrin - 6 years, 10 months ago

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@Michael Mendrin Did you use the Dirichlet Integral for the superellipsoid? I had to learn that to re-solve this problem.

Also, I think I did not type the correct axis of revolution. It's been a while since I solved such problems. I think I wrote revolve around the x-axis, while the correct axis is y. So the answer is 56.5508 despite a =2 and b=3.

Steven Zheng - 6 years, 10 months ago

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@Steven Zheng No, I said I couldn't do it, in the case of the superellipsoid. Or at least I didn't try very hard, after I consulted Wikipedia on exactly what is a superellipsoid, and how it's different from a "superellipse". The Wikipedia article provide the formula for the volume, and I was struck by the similarity between the two, even though I couldn't get anywhere with the former.

Michael Mendrin - 6 years, 10 months ago

Excuse me. How did you integrated the function? Can you give me a reference to the method? Thanks!

Carlos David Nexans - 6 years, 10 months ago
Haroun Meghaichi
Jul 31, 2014

We can calculate the volume on the side x > 0 x>0 then multiply with two (by symmetry).

For x 0 x\geq 0 the equation reduces to : y = f ( x ) = 3 ( 1 ( x 2 ) 5 / 2 ) 2 / 5 y=f(x)= 3\left(1-\left(\frac{x}{2}\right)^{5/2}\right)^{2/5} We know that : V = 4 π 0 2 x f ( x ) d x = x = 2 t 8 π 0 1 t f ( 2 t ) d t = 48 π 0 1 t ( 1 t 5 / 2 ) 2 / 5 d t . V=4\pi \int_0^{2} x |f(x)|\ \mathrm{d}x \overset{x=2t}{=} 8\pi \int_0^1 tf(2t)\ \mathrm{d}t=48\pi \int_0^1 t(1-t^{5/2})^{2/5}\ \mathrm{d}t . Now set z = t 5 / 2 z=t^{5/2} then t d t = 2 5 z 1 / 5 t\mathrm{d}t=\frac{2}{5}z^{-1/5} , and then : V = 96 π 5 0 1 z 1 / 5 ( 1 z ) 2 / 5 d z = 96 π 5 B ( 4 5 , 7 5 ) V=\frac{96\pi}{5} \int_0^1 z^{-1/5}(1-z)^{2/5} \ \mathrm{d}z=\frac{96\pi}{5} \mathrm{B} \left(\frac{4}{5},\frac{7}{5}\right) Where B \mathrm{B} is the Beta function , use the Beta-Gamma relation and then Γ ( x + 1 ) = x Γ ( x ) \Gamma(x+1)=x\Gamma(x) to get : V = 96 π 5 Γ ( 7 5 ) Γ ( 4 5 ) Γ ( 11 5 ) = 32 π Γ ( 4 5 ) Γ ( 2 5 ) Γ ( 1 5 ) V= \frac{96\pi }{5} \cdot\frac{\Gamma\left(\frac{7}{5}\right) \Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{11}{5}\right) }=32\pi \frac{\Gamma\left(\frac{4}{5}\right)\Gamma\left(\frac{2}{5}\right) }{\Gamma\left(\frac{1}{5}\right)} Use WolframAlpha to get that the answer is 56.5508 \approx 56.5508 .

Remark 1: We can use the reflection formula to get V = 32 π 2 sin π 5 Γ ( 2 5 ) Γ 2 ( 1 5 ) V=\frac{32\pi^2}{\sin \frac{\pi}{5}} \cdot\frac{\Gamma\left(\frac{2}{5}\right)}{\Gamma^2 \left(\frac{1}{5}\right)} You can even use python but you need more work.

Remark 2: Using the same method we can generalize the result like what @Michael Mendrin has found.

So that's how it's done, in the case of the superellipisoid.

Michael Mendrin - 6 years, 10 months ago

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I believe so.

Haroun Meghaichi - 6 years, 10 months ago

That's how I did it too (including the W|A for evaluating Gamma functions).

Steven Zheng - 6 years, 10 months ago

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