A hole in the sphere

Commented solution code is attached. Results are printed at the end. The answer comes out to around $28.8 \%$

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68

import math

Num = 500   # spatial resolution parameter

Rs = 20.0   # sphere radius
Rh = 10.0   # hole radius

dr = Rs/Num   # infinitesimal quantities - spherical coordinates
dtheta = 2.0*math.pi/Num
dphi = (math.pi/2.0)/Num

Vref = (4.0/3.0)*math.pi*(Rs**3.0)  # total sphere volume for reference

#######################################

V = 0.0   # initialize hole volume to zero

r = 0.0    # triple integral over sphere volume

while r <= Rs:

    theta = 0.0

    while theta <= 2.0*math.pi:

        phi = 0.0

        while phi <= math.pi/2.0:

            x = r*math.cos(theta)*math.sin(phi)   # coordinates of point within sphere
            y = r*math.sin(theta)*math.sin(phi)
            z = r*math.cos(phi)

            if x**2.0 + (z-Rh)**2.0 <= Rh**2.0:      # if point in hole, add to volume

                dV = (r**2.0)*math.sin(phi)*dr*dtheta*dphi
                V = V + dV

            phi = phi + dphi

        theta = theta + dtheta

    r = r + dr

#######################################

# print results

print Num
print (100.0*V/Vref)

#>>> ================================ RESTART ================================
#>>> 
#100
#29.1753658406
#>>> ================================ RESTART ================================
#>>> 
#200
#28.7037956416
#>>> ================================ RESTART ================================
#>>> 
#400
#28.7417287715
#>>> ================================ RESTART ================================
#>>> 
#500
#28.8323134583
#>>> 

First, let's scale down by a factor of $10$ in each direction. This makes the algebra neater and doesn't affect the percentage result.

By placing the centre of the sphere at the origin and the axis of the cylinder on $x=1$ , we have the equations $x^2+y^2+z^2=4$ for the sphere and $x^2-2x+y^2=0$ for the cylinder.

It's easier to work in cylindrical coordinates $r,\phi,z$ , in which $x=r \cos \phi$ and $y=r \sin \phi$ ( $z$ is unchanged).

The sphere now has equation $r^2+z^2=4$ , and the cylinder $r^2-2r\cos \phi=0$ , or (almost) equivalently $r=2\cos \phi$ .

We have $\begin{aligned} V &=2\int_{0} ^ {\pi} \int_0 ^{2\cos \phi} \sqrt{4-r^2} \; r \; dr d\phi \\ &=\int_{0} ^ {\pi} \frac{16}{3} \left(1- \sin^3 \phi \right) d\phi \\ &=\frac{16}{9}\left( 3\pi-4\right) \end{aligned}$

The volume of the sphere is $\frac{32\pi}{3}$ ; so the required percentage is $100 \times \frac{\frac{16}{9}\left( 3\pi-4\right)}{\frac{32\pi}{3}} = 100 \times \frac{3\pi-4}{6\pi} \approx \boxed{28.779\%}$