Finding the volume of the cap of a sphere

Question

Let's say I have a sphere of radius 2 and we drill a hole with the shape of a sphere of radius 1. How shall we find the volume of the cap of the sphere?

What I did was finding the spherical coordinates for the surface:

$\Phi(\theta, \phi)=2(\cos\theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$ with $\theta \in [0, 2\pi]$

To find the boundaries for $\phi$, I used the Pythagorean theorem:

To find $h$: $(\frac{h}{2})^2=2^2+1^2$, so $h=\sqrt{20}$

So I end up with this triangle:

Then $\sin \phi = \frac{1}{2}$ so, $\phi = \frac{\pi}{6}$.

Setting up the integral:

$\int _{0} ^{\pi/6} \int _{0} ^{2\pi} \int _{0} ^{2} \rho^2 \sin \phi d\rho d\theta d\phi=-\frac{8}{3}(\sqrt{3}-2)\pi$

Which is wrong... I've done a lot of steps, so I can't really tell where I made a mistake

Answer 1

Applying Pythagoras, $h^2 = 2^2 - 1^2$. You instead did addition. (I am using $h$ instead of $h/2$ to keep things simpler).

So you know $h = \pm \sqrt3$.

You calculated the angle correctly $(0 \leq \phi \leq \frac{\pi}{6})$ for the spherical cap above $z$ axis. You can multiply by two if you need to find the volume for both spherical caps.

Now coming to your integral, the volume that you found is of the spherical cone and not of the spherical cap. As you need to find the volume of the spherical cap, you need to make sure that $\rho$ is only between $\frac{\sqrt{3}}{\cos \phi}$ and $2$. So your integral becomes -

$V = \displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{\sqrt{3}/cos{(\phi)}}^{2} r^2 \sin \phi \, dr \, d\phi \, d\theta \approx 0.431$. This is for one spherical cap.

EDIT: see a 2D diagram.

Answer 2

We need to find what is the height of the spherical cap. By the Pythagorean theorem, half of the height of the cylinder is $$2^2=1^2+\left(\frac{h}{2}\right)^2$$ so $\tfrac{h}{2}=\sqrt{3}$. Then, the height of the spherical cap is $$H=2-\sqrt{3}.$$ Plugging this value into the formula (see https://en.wikipedia.org/wiki/Spherical_cap), we get $$V = \frac{\pi H^2}{3}(3R-H)=\frac{\pi(2-\sqrt{3})^2}{3}(4+\sqrt3)=\frac{\pi}{3}(16-9\sqrt{3})$$