Beads for a Bracelet

The volume of the 5mm sphere is $\frac{4\pi (5)^3}{3} = \frac{500\pi}{3}$ .

Envision the bead as the area between the curves $y = \sqrt{25 - x^2}$ and $y = 3$ revolved around the $x$ -axis. Using the Washer Method , this volume will be given by

$\pi\int_{-4}^{4}\left(\sqrt{25 - x^2}\right)^2 - (3)^2dx = \pi\int_{-4}^{4}16 - x^2dx = \frac{256\pi}{3}$

Subtracting this result from the total volume gives the amount of material removed by the drill:

$\frac{500\pi}{3} - \frac{256\pi}{3} = \frac{244\pi}{3} \approx 255.516 \text{ mm}^3 .\square$

We want to integrate over the bottom of the cylinder, so our goal is something like $\int_0^r \int_0^{2\pi} \! f(r,\phi)\ \mathrm{d}\phi \mathrm{d}r$ with the radius of the cylinder r and the angle $\phi$ .

So, first off we need a value for the height z. The parameterisation of a sphere is $\{(x,y,z) \in \mathbb{R}^3: x^2 + y^2 + z^2 \leq R^2\}$ with R bing the radius of the sphere. From this parameterisation follows $z(x,y) = \sqrt{R^2-(x^2+y^2)}$ .

Now we have everything we need: $V=2 \int_A\! z(x,y) \mathrm{d}x \mathrm{d}y$ with the over all volume V (the factor 2 needs to be added because we only calculate half of the volume with the integral - the origin of our coordinate system is in the centre of the sphere).

Now we can change the integral to cylindical coordinates according to the conventions of Wikipedia and we get: $V=2 \int_0^r \int_0^{2\pi} \! r\sqrt{R^2-r^2}\ \mathrm{d}\phi \mathrm{d}r$

This calculation can be solved by substituting (for example) $R^2-r^2=u$ .

Solving everything we get the value $V=255.516mm^2$ .