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I am wondering what the formula for the surface area of a 4 D sphere is. I noticed recently that in a circle, the derivative of area with respect to the radius is its circumference, and that in a sphere the derivative of volume with respect to the radius is its surface area. My intuition tells me you should be able to extend this to 4 D. the formula for the volume of a 4d sphere is $V=\frac12 \pi^2 r^4$ so then just taking the derivative $$\frac d{dr} \frac12 \pi^2 r^4 = 2 \pi^2 r^3$$ if you're curious where I got this formula you can set up a quadruple integral over the region bounded by $r^2 = x^2 + y^2 + z^2 + w^2$. Is this correct? and is there any proof that this relationship between surface area and volume holds for Dimensions greater than R3.

Con
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Christopher Moore
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1 Answers1

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Your intuition is correct. If an $n$-ball of radius $n$ has volume $V_nr^n$, increasing $r$ by $dr$ adds a thickness-$dr$ shell of volume $dr$ times the ball's surface, i.e. the surface is $d(V_nr^n)/dr=nV_nr^{n-1}$. This $(n-1)$-dimensional measure is of what we call the $(n-1)$-sphere, so you've proven the fact that the $4$-ball has $4$-measure (i.e. hypervolume) $\tfrac12\pi^2r^4$ implies the $3$-sphere has $3$-measure (i.e. volume, in this case a surface volume) $2\pi^2r^3$.

J.G.
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  • Note that it's important that the surface moves outward at equal rates in every direction as you increase $r.$ This wouldn't work for (say) a non-spherical ellipsoid, but it's fine for a sphere. – David K Oct 14 '20 at 17:56
  • @DavidK What's more, even when this strategy works, it can require care, such as in explaining the factor of $2$ needed for a hypercube. – J.G. Oct 14 '20 at 18:23
  • Is that a hypercube with one corner fixed in space? If so, you have four out of eight cells moving outward at rate zero. – David K Oct 14 '20 at 23:03