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This question is based a little on orbit mechanics, so the sphere has the Earth at the centre (but isn't the Earth) - not relevant but helps me explain it.

Given a hollow sphere I can calculate the volume as such:

V = 4/3*pi*(R^3-r^3)
Where: V = volume, R = outer radius, r = inner radius

Now suppose I want to calculate the volume of a segment of this hollow sphere. The segment is defined by two angles of inclination from the N-S direction. On a 2D plane these angle draw lines from the centre of the Earth to the outer edge of the hollow sphere, and have are symmetrical around the N-S axis. Here's a picture to explain:

enter image description here Here the red lines are the defined by the 2 angles of inclination, the blue lines are the projection of the red lines into 3 dimension.

My issue here is I can't work out how to calculate the volume of this halo structure. At first I thought I could calculate it from 2 spherical caps (larger - small), but that isn't right because a spherical cap is cut horizontally, and these cuts are clearly at an angle. Is there a name for this structure that I should be using in my research?

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    $\begingroup$ I would call it a spherical washer. The volume can be determined by revolving the cross sectional area around the N-S axis. You have 4 equations defining the cross section, 2 linear and 2 circular. Do you know how to determine the volume using integration via "volumes of revolution" ? You basically have 4 concentric discs generated about the y (N-S) axis which you combine and subtract to get the volume. $\endgroup$ – Phil H Apr 25 '18 at 15:59
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In spherical coordinates the region is described as $$ r\le\rho\le R,\quad \phi_1\le\phi\le\phi_2,\quad 0\le\theta\le2\,\pi. $$ It's volume is then $$ \int_r^R\int_0^{2\pi}\int_{\phi_1}^{\phi_2}\rho^2\sin\phi\,d\phi\,d\theta\,d\rho. $$

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  • $\begingroup$ Thanks for the answer. To my eye this resolves to -((c^3 - d^3) (Cos[a] - Cos[b]))/3 * 2*pi.... where a,b,c,d, = roe1,roe2,r,R respectively. I'm always cautious when an integral comes out negative, what does this mean practically (I can't have a negative volume of course). Have I incorrectly integrated? $\endgroup$ – FraserOfSmeg Apr 26 '18 at 6:26
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    $\begingroup$ Angles are measured from the north pole ($\phi=0$) to the south pole ($\phi=\pi$). $a$ should be the aperture of the wider cone, and $b$ of the narrower one; then $a>b$ and $-(\cos a-\cos b)>0$. $\endgroup$ – Julián Aguirre Apr 26 '18 at 15:10
  • $\begingroup$ Thanks for clearing that up! $\endgroup$ – FraserOfSmeg Apr 27 '18 at 0:04

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