Amanufacturer drills a round hole of radius r through the center of a metal sphere of radius r. find the volume of the remaining metal “bead.”

The media attached below is the bottom/top view of drilled sphere. If you look it from sideways, you'll see that the drilled hole is in the form of cylindrical shape. So,
Volume of the drilled shape= volume of cylinder = πr²h, where h is 2r.
∴Volume of cylinder(V1) = πr²(2r) = 2πr³.
Volume of sphere(V2) = 4πr³/3.

Therefore volume of remaining metal bead = V1 - V2
                                                                         =   2πr³ - 4πr³/3 
                                                                         =   (6πr³ - 4πr³)/3
                                                                         =   2πr³/3.

Hence volume of remaining bead is 2πr³/3

Solution: (-Infinite, -8/3] U (4, Infinite)

Using that a fraction is greater than or equal to zero when the numerator and denominator have the same sign:
a/b>=0. Then we have two cases:
Case 1) If the numerator is positive, the denominator must be positive too (at the same time):
if a>=0 ∩ b>0

Or (U)

Case 2) If the numerator is negative, the denominator must be negative too (at the same time):
if a<=0 ∩ b<0

In this case a=3x+8 and b=x-4, then:

Case 1):
if 3x+8>=0 ∩ x-4>0
Solving for x:
3x+8-8>=0-8 ∩ x-4+4>0+4
3x>=-8 ∩ x>4
3x/3>=-8/3 ∩ x>4
x>=-8/3 ∩ x>4
Solution Case 1: x>4 = (4, Infinite)

Case 2):
if 3x+8<=0 ∩ x-4<0
Solving for x:
3x+8-8<=0-8 ∩ x-4+4<0+4
3x<=-8 ∩ x<4
3x/3<=-8/3 ∩ x<4
x<=-8/3 ∩ x<4
Solution Case 2: x<=-8/3 = (-Infinite, -8/3]

Solution= Solution Case 1 U Solution Case 2
Solution = (4, Infinite) U (-Infinite, -8/3]
Solution: (-Infinite, -8/3] U (4, Infinite)