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I am dealing with the following conundrum. Suppose you have a 3-sphere of radius $R$ in $\mathbb{R}^4$:

$$ \mathbb{S}^3 := \lbrace(x,y,z,w)\in \mathbb{R}^4: x^2 + y^2 + z^2 + w^2 = R^2 \rbrace $$

For any $r \in [0, R]$, the torus: $$ \mathbb{T} (r) := \lbrace(x,y,z,w)\in \mathbb{R}^4: z^2 + w^2 = r^2 \mbox{ and } x^2 + y^2 = R^2 - r^2 \rbrace $$ is a subset of $\mathbb{S}^3$. In fact: $$ \mathbb{S}^3 = \bigcup_{r\in [0, R]} \mathbb{T}(r) $$ I thought that this observation would allow me to compute the area of the hypersurface with only elementary calculus. The area of each torus equals: $$ (2\pi r) * (2\pi \sqrt{R^2-r^2}) $$ But when integrating from $0$ to $R$, i.e. $$ 4 \pi^2 \int_0^R r \sqrt{R^2-r^2} dr$$ does not give the right answer.

What mistake am I making here?

Regards,

Nithin

Nithin Govindarajan
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1 Answers1

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Compare to the analagous method for computing the area of the 2-sphere. We have a disjoint union of circles $$C(r) = \{x,y,z\in \mathbb{R}^3\mid x^2 + y^2 = R^2-r^2, z^2 = r^2\}.$$

each of whose length is $2\pi \sqrt{R^2-r^2}.$ The proper way to proceed is not to do $$ \int_0^R 2\pi \sqrt{R^2-r^2}dr = \frac{\pi^2}{2}R^2.$$

There are a few things wrong with this. First off, we should really be integrating over $z,$ not $r$ since this will give us the true range over the union of circles. However, this only accounts for a factor of two, and the answer is still wrong.

The other error was to not take into account the surface element. $dr$ (or $dz $ anyway) is not the proper line element to multiply by $2\pi\sqrt{R^2-z^2}$ to get the surface element. The correct line element is $$\sqrt{du^2+dz^2} = dz\sqrt{1+\left(\frac{du}{dz}\right)^2} = dz\sqrt{1+\frac{z^2}{R^2-z^2}} = \frac{Rdz}{\sqrt{R^2-z^2}}.$$ where $u^2 = x^2 + y^2 = R^2-z^2$

So the integral for the area of the 2-sphere is $$ \int_{-R}^R 2\pi\sqrt{R^2-z^2}\frac{Rdz}{\sqrt{R^2-z^2}} = 4\pi R^2.$$

So you're going to have to take into account those things when you try it for the 3-sphere.

spaceisdarkgreen
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