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I'm trying to find the integral of a chord length of a circle, divided by the surface area of a sphere. Imagine a circle, that is stuck between two layers of a spherical shell, and then take discrete slices of the circle and spherical shell. The integral would look like: $$\int_{a-r}^{a+r} \frac{2\sqrt{r^2-(a-x)^2}}{4\pi x^2}\,dx$$

where "r" is the radius of the circle and "a" is the radius of the sphere, and "x" is any given distance from the center of the sphere to between the two shell layers. I have not been able to find this integral. Why isn't the integral simply the area of a circle divided by the volume of a spherical shell: $$\frac{\pi r^2}{\frac43 \pi ((a+r)^3-(a-r)^3)}$$

The numerator and denominator of the integral each individually integrate to these values, so why is the combined integral equivalent to taking the integrals separately? What is the real integral?

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Hint: To evaluate the integral, you may set $a-x=r\sin\phi.$ Then the integrand becomes a constant multiple of $$\left(\frac{\cos\phi}{a-r\sin\phi}\right)^2,$$ which may be done by setting $t=\tan\phi.$

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  • $\begingroup$ Thanks for the hint, Allawonder. I’ll try to get that to work. By the way, this isn’t homework. It’s for research. So if someone is able to fully integrate this for me, I would appreciate it. $\endgroup$ – astrodyn Sep 9 at 3:24
  • $\begingroup$ @astrodyn I see your point, but isn't research just extended homework? ;) $\endgroup$ – Allawonder Sep 9 at 8:08

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