Solve the problem.
-A bead is formed from a sphere of radius 3 by drilling through a diameter of the sphere with a drill bit of radius 1. Find the volume of the bead.
A) $\frac { 56 } { 3 } \pi$
B) $\frac { 28 } { 3 } \pi$
C) $18 \pi$
D) $36 \pi$

Question

Solve the problem.
-A bead is formed from a sphere of radius 3 by drilling through a diameter of the sphere with a drill bit of radius 1. Find the volume of the bead. 
A) $\frac { 56 } { 3 } \pi$
B) $\frac { 28 } { 3 } \pi$
C) $18 \pi$
D) $36 \pi$

Quizplus · Accepted Answer

The volume of the bead can be found by subtracting the volume of the cylindrical hole from the volume of the original sphere. The volume of the sphere is $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3)^3 = 36\pi$. The volume of the cylindrical hole is $\pi r^2h = \pi (1)^2(6) = 6\pi$, since the diameter of the sphere is 6, which is also the height of the cylinder. Subtracting the volume of the hole from the volume of the sphere gives $36\pi - 6\pi = 30\pi$. However, we also need to account for the volume of the two caps removed by the drilling, which are two spherical caps with height $h = 3 - 1 = 2$. The volume of each cap is $\frac{1}{3}\pi h^2(3R - h) = \frac{1}{3}\pi (2)^2(3*3 - 2) = \frac{4}{3}\pi(7) = \frac{28}{3}\pi$. Adding the volume of the two caps to the volume after drilling gives $30\pi + \frac{28}{3}\pi = \frac{90}{3}\pi + \frac{28}{3}\pi = \frac{118}{3}\pi$. This calculation is incorrect; the correct approach is to directly subtract the cylindrical volume and add the volume of the spherical caps correctly calculated. The mistake was in the calculation of the caps' volume and the final addition. The correct volume of the bead is found by subtracting the cylindrical hole's volume from the sphere's volume without incorrectly adding cap volumes, as initially described. The correct calculation for the bead's volume involves understanding the geometry of the bead and correctly applying the formula for the volume of a sphere ($\frac{4}{3}\pi r^3$) and the volume of a cylinder ($\pi r^2h$), but the explanation provided contains a mistake in the final computation and explanation of the caps' volume. The correct approach is to calculate the volume of the sphere and subtract the volume of the cylinder drilled through it. The volume of the sphere is $36\pi$, and the volume of the cylinder (including the caps, which was incorrectly added instead of being part of the cylinder's removal) is $6\pi$, leading to the correct subtraction for the bead's volume without the need to separately add cap volumes, as they are not added back but are part of the removed cylinder's volume.

Solve the Problem 563π\frac { 56 } { 3 } \pi356​π

Solve the Problem $\frac { 56 } { 3 } \pi$