# spherical segment volume

Suppose I have a spherical segment like the one in the picture.

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:

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?

• No, because you must also integrate over $r$. Commented Jul 6, 2015 at 22:24
• @ M. Wind: What are the limits of integration?
– E Be
Commented Jul 7, 2015 at 10:06
• It is actually very simple. Just integrate $r$ from $0$ to $R$. Commented Jul 7, 2015 at 14:35
• you need more than just $R$ and $d\theta$ - you can move $a$ and keep $d\theta$ the same and get different results.
– JMP
Commented Jul 9, 2015 at 14:52
• So given a fixed radius $a$ which is simply $R \sin \theta$, what is the calculation?
– E Be
Commented Jul 9, 2015 at 14:55

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.

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.

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.