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I would like to make sure I understand where the following integration fails (note -- I'm aware of how the surface area is typically computed etc. but I'm trying to understand the intuition better by testing where other seemingly appealing alternatives fail).

If we know the circumference of the circle is $2\pi r$ given radius $r$, then the surface area of the unit sphere is twice the integral of the the circle's circumference (one for each hemisphere):

$$ 2\int_{0}^12\pi r dr = 2\pi$$

This result is clearly incorrect.

My intuition says that this fails because the integral, being with respect to the Lebesgue measure, does not take into account the 'density' of the circles somehow -- in a sense treating the integration as though we have flattened out the half-hemispheres, much like if we stretch a rubber band to bulge outwards to create a hemisphere it would seem 'less dense' around the bulge?). Is that intuition correct?

Thanks.

Anon
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    One per hemisphere -- I'm imagining horizontal circles passing from the top (r=0) to the middle (r=1) and then again from there to the bottom (r=0). – Anon Jun 21 '25 at 13:54
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    Note that if you simply computed $\int_0^12\pi r,dr=\pi$, you would correctly (and for correct reasons) get the area of the unit disk. This has to do with the “$dr$” being orthogonal to the circumferences being integrated. So I guess you could ascribe this as a “density” issue where each $2\pi r$ contribution must be counted differently. On a somewhat related matter, this reminds me of my old answers: Behaviour of density function when flattening the hemisphere and at a lower level, this. – peek-a-boo Jun 21 '25 at 14:18
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    and of course more generally, if $A_{n-1}(r)$ is the surface area of the $(n-1)$-dimensional sphere of radius $r$, then $\int_0^RA_{n-1}(r),dr=V_n(R)$ gives the volume of the $n$-Ball of radius $R$, again due to orthogonality of the $dr$ from the spheres. btw in case you want to see a full-blown generalization, see the latter half of my answer here just to get a feel for the fact that indeed, we have to take into account how the various submanifolds sit inside the bigger manifold when computing integrals. – peek-a-boo Jun 21 '25 at 14:30

1 Answers1

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You are not accounting for the shape of the sphere. If you apply this failed reasoning to a cone or a cylinder you would get the same formula.

So, to account for the sphere, the radius should follow a circular shape: this would suggest we integrate $$ 2\int_0^12\pi\sqrt{1-r^2}\,dr. $$ But this is too naive (and wrong), because as you move along the curve the "$dr$" slants. To account for this slanting the contribution of the $dr$ depends on the length of the curve. The length differential for a function $f$ is $$ ds=\sqrt{1+[f' (x)]^2}\,dx, $$ and the integral becomes $$ 2\int_0^12\pi f(r)\,\sqrt{1+[f' (r)]^2}\,dr. $$ For the sphere this is $$ 2\int_0^12\pi\sqrt{1-r^2}\,\frac1{\sqrt{1-r^2}}\,dr=4\pi. $$

Disclaimer: I've always found these deductions super tricky, and mistakes likes yours I've made plenty in my life.

Martin Argerami
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  • Thanks! So essentially what that wrong solution ignored was how the radius changes as the circles move; all it did was account for the radius value itself. Would that be the correct way to sum this up? – Anon Jun 21 '25 at 14:12
  • Yes, but like I said, this is still tricky (to my intuition, at least) and I had to look the formula for "surface of a solid of revolution" to see what the correct form was. – Martin Argerami Jun 21 '25 at 14:17