A cylindrical piece of steel has a diameter of 2" and a height of 2". It is to be melted down and re-molded into a spherical ball bearing.
1) Assuming no metal is lost in the melting/re-molding process, what will the volume of the ball bearing be?
1:
I began with the volume of the cylinder: $V=πr^2h$
—> $V=π((1)^2)(2)$ —> $V=2π$ —> Because this is the volume of the cylinder, I think $2π$ in.$^3$ is the volume of the sphere, too ("no metal is lost").
2) What will be the radius of the ball bearing to the nearest hundredth of an inch?
2:
I am not sure if I was supposed to find the diameter in (1). I would use the volume of 2π from the equation of the cylinder to equal the volume of the sphere: V=4/3π(r)^3=2π —> To the nearest hundredth, my answer for the ball's radius is 1.14 in.^3.
In this two-step process, was there error in my judgement? If yes, please explain why.
Best Answer
I don't know why you call $2\pi$ a diameter, but otherwise your reasoning seems to be sound. You calculated the volume of the cylinder and equated it to volume of the sphere, solving for $r$. That's correct.
$$V = \pi (1 \, \mathrm{inch})^2 \cdot 2 \, \mathrm{inch} = 2 \pi \, \mathrm{inch}^3$$ $$V = \frac 4 3 \pi r^3 \Rightarrow r = \left( \frac {3V} {4 \pi}\right)^{1/3}= \left(\frac 3 2 \right)^{1/3} \, \mathrm{inch} \approx 1.1447 \, \mathrm{inch}$$