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my physics professor gave us this brain teaser in class for fun, I think i got the answer. Is this correct?

If the earth were a 100-dimensional hypersphere with radius 4000 miles rather than a three-dimensional sphere of that radius, what fraction of its volume would be within 40 miles of its surface?

I think the answer is $1/100^{100}$ I know that the volume of the hypersphere varies directly with r, meaning in 1d its $r^1$, in 2d its $r^2$... etc. and the constants will cancel out since it's a proportion.

so I took $40^{100}/4000^{100}$ = $1/100^{100}$. Is that correct?

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  • $\begingroup$ This would have been correct if you would have been $40$ miles from the center. For estimating $40$ miles from the surface, you have to subtract the volume of the sphere of radius $3960$ miles. $\endgroup$ – астон вілла олоф мэллбэрг Oct 6 '16 at 2:40
  • $\begingroup$ awwwwh, top commentor, you are correct. dang. should've caught that. $\endgroup$ – mac5 Oct 6 '16 at 2:43
  • $\begingroup$ @астон: Perhaps that comment would be better suited as an answer, since you essentially answered the question. $\endgroup$ – MvG Oct 6 '16 at 6:15
  • $\begingroup$ @MvG You are right, but it is too small to be an answer. I will leave it as a comment. $\endgroup$ – астон вілла олоф мэллбэрг Oct 6 '16 at 6:16
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As астон вілла олоф мэллбэрг already commented, your computation would have been correct if the question were asking about the proportion that is $40$ miles from the center. For $40$ miles from the surface, $40^{100}$ is not the correct formula, as the hypersphere shell will have larger volume than that. Instead you have to subtract the volume of a hypersphere of radius $3960$ miles from your total hypersphere to get at that shell volume.

$$\frac{4000^{100}-(4000-40)^{100}}{4000^{100}}= 1-\left(1-\frac{40}{4000}\right)^{100}\approx0.634$$

Here you can also see how increasing the dimension will make that proportion increase, since the parenthesis is less than one. For dimension $3$ you only get a result of $0.03$.

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