 # Imagine cutting a sphere into circles(the distance between the two circles is almost zero). Then i Leah Pope 2022-06-25 Answered
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.
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Sphere
${x}^{2}+{y}^{2}+{z}^{2}={r}^{2}$
Circle
${x}^{2}+{y}^{2}={r}^{2}-{h}_{1}^{2},\phantom{\rule{thickmathspace}{0ex}}z={h}_{1}$
Next circle
${x}^{2}+{y}^{2}={r}^{2}-{h}_{2}^{2},\phantom{\rule{thickmathspace}{0ex}}z={h}_{2}$
Thickness of sphere area bands located between circles
$w=\sqrt{\left({h}_{2}-{h}_{1}{\right)}^{2}+\left(\sqrt{{r}^{2}-{h}_{1}^{2}}-\sqrt{{r}^{2}-{h}_{2}^{2}}{\right)}^{2}}=\sqrt{2{r}^{2}-2{h}_{1}{h}_{2}-2\sqrt{\left({r}^{2}-{h}_{1}^{2}\right)\left({r}^{2}-{h}_{2}^{2}\right)}}$
${h}_{2}={h}_{1}+\mathrm{\Delta }h,\mathrm{\Delta }h\ll {h}_{1}⇒w=\frac{r}{\sqrt{{r}^{2}-{h}_{1}^{2}}}\mathrm{\Delta }h+O\left(\mathrm{\Delta }h{\right)}^{2}$
When $w\to 0$: $\mathrm{\Delta }h\to \frac{w}{r}\sqrt{{r}^{2}-{h}_{1}^{2}}$
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.

We have step-by-step solutions for your answer! Summer Bradford
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.

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