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Suppose I have a spherical segment like the one in the picture. enter image description here

I want to find the infinitesimal volume of such a segment. The angle between point A and B is $d\theta$. And the radius of the sphere is $R$. Here, the volume is stated to be $\frac{\pi}{6}h(3a^2+3b^2+h^2)$. Now I try to express the volume with $R$ and $d\theta$ only, and I am having trouble with it. Any help would be appreciated.

Another approach for this, I guess, is using the Jacobian in spherical coordinates: enter image description here

Integrating $dV$ from $\phi=0$ to $\phi=2\pi$:

$$\int_{\phi=0}^{\phi=2 \pi}r^2 \sin \theta dr d\theta d\phi$$ yeilds $2\pi \cdot r^2 \sin \theta dr d\theta$. Is that correct?

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  • $\begingroup$ No, because you must also integrate over $r$. $\endgroup$ – M. Wind Jul 6 '15 at 22:24
  • $\begingroup$ @ M. Wind: What are the limits of integration? $\endgroup$ – E Be Jul 7 '15 at 10:06
  • $\begingroup$ It is actually very simple. Just integrate $r$ from $0$ to $R$. $\endgroup$ – M. Wind Jul 7 '15 at 14:35
  • $\begingroup$ you need more than just $R$ and $d\theta$ - you can move $a$ and keep $d\theta$ the same and get different results. $\endgroup$ – JonMark Perry Jul 9 '15 at 14:52
  • $\begingroup$ So given a fixed radius $a$ which is simply $R \sin \theta$, what is the calculation? $\endgroup$ – E Be Jul 9 '15 at 14:55
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Yes, $\;dV = 2\pi \cdot r^2 \sin \theta\, dr\, d\theta \;$ is the correct answer. Simple and straightforward.
Nothing to be improved. It seems to me that the OP doesn't have any trouble at all.

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This is better handled in cylindrical coordinates.

The infinitesimal volume is the area of the circular section times the infinitesimal height, $\pi r^2(z)dz$.

The radius as the function of the height is given by $r^2(z)=R^2-z^2$, then

$$V=\int_{z_a}^{z_a+h}\pi(R^2-z^2)dz=\pi\left(R^2z-\frac{z^3}3\right)\Big|_{z_a}^{z_a+h}.$$

Now, we know that

$$R^2=a^2+z_a^2=b^2+(z_a+h)^2.$$ By subtraction, $$(z_a+h)^2-z_a^2=2z_ah+h^2=a^2-b^2,$$ and

$$z_a=\frac{a^2-b^2-h^2}{2h},R^2=a^2+\left(\frac{a^2-b^2-h^2}{2h}\right)^2,$$ and the rest will follow.

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It is not that simple with just the integration you suggested. Think about your limit of r and $\theta$. Let's talk about a simpler case first, say a spherical cap, ie. with only a lower base. If you use $\theta$ from 0 to $\theta_1 $corresponding to base, and r from 0 and $r_1$, you will actually end up with a cone shape! because r is integrated from 0 to $r_1$, so it will have an extra volume of the cone. However, if you minus what you obtained from above, ie. $2\pi R^2h/3$ with the volume of cone you will get exactly the same formula as given by wolfram alpha.

Check http://www.mathalino.com/reviewer/solid-mensuration-solid-geometry/spherical-sector and http://www.mathalino.com/reviewer/solid-mensuration-solid-geometry/spherical-segment too

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