Is it possible to calculate sphere surface area with circles?

Question

Imagine cutting a sphere into circles(the distance between the two circles is almost zero).
Then is it correct to say that the sum of the circumference of all the circles is the surface area of the sphere? (Please describe why not)

$S_{sphere}=2\sum_{h=0}^{r}{2\sqrt{r^2-h^2}\pi}$       (Where $\sqrt{r^2-h^2}$ is radius in each circle with distance $h$ to center, Also multiplied by 2 because it's sum of circles in semisphere)

Thanks in advance.

Answer 1

Sphere $$x^2+y^2+z^2=r^2$$

Circle $$x^2+y^2=r^2-h_1^2,\;z=h_1$$

Next circle $$x^2+y^2=r^2-h_2^2,\;z=h_2$$

Thickness of sphere area bands located between circles $$w=\sqrt{(h_2-h_1)^2+(\sqrt{r^2-h_1^2}-\sqrt{r^2-h_2^2})^2}=\sqrt{2r^2-2h_1h_2-2\sqrt{(r^2-h_1^2)(r^2-h_2^2)}}$$

$$h_2=h_1+\Delta h, \Delta h \ll h_1 \Rightarrow w=\frac{r}{\sqrt{r^2-h_1^2}}\Delta h+O(\Delta h)^2$$

When $w\to 0$: $\Delta h \to \frac{w}{r} \sqrt{r^2-h_1^2}$

So you need to take $h$ in such way that difference of to subsequent $h$ values is $\frac{w}{r} \sqrt{r^2-h^2}$ with some small $w$ tending to zero.

Answer 2

Distance between intersecting planes is very small, you wish to say?

The circumference is that of a polygon. So you get a smaller boundary length.

However the disc volumes sum up correctly.

Answer 3

If you take $\theta$ to be the angle between $r$ and the north pole then your $h$ is just $r\sin{\theta}$ and at each value of theta the surface area of the ribbon is $$rd\theta(2\pi r\sin{\theta})=2\pi r^2\sin{\theta}d\theta$$

The $rd\theta$ term is the arc length corresponding to the width of the ribbon and $2\pi r\sin{\theta}$ is its circumference.

You end up with the integral $$2\pi r^2\int_{0}^{\pi}\sin{\theta}d\theta=4\pi r^2$$