Question

A solid metallic cylinder of radius \( 6 \, \text{cm} \) and height \( 20 \, \text{cm} \) is melted and recast into a large sphere. What is the radius of the sphere?

Hint:

Before reaching for a calculator, try factoring the number under the cube root into primes and pulling out any group of three identical factors, since that group can be taken straight out of the cube root as a whole number, leaving only a smaller cube root to estimate.

Answer 1

Step 1: Formula for volume of the cylinder and sphere. 
The volume of a cylinder is given by: \[ V_{\text{cylinder}} = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height. The volume of a sphere is given by: \[ V_{\text{sphere}} = \frac{4}{3} \pi R^3 \] where \( R \) is the radius of the sphere. 

Step 2: Using the given data. 
The volume of the cylinder is melted and recast into the sphere, so: \[ \pi r^2 h = \frac{4}{3} \pi R^3 \] Substitute \( r = 6 \, \text{cm} \) and \( h = 20 \, \text{cm} \) into the formula: \[ \pi (6)^2 (20) = \frac{4}{3} \pi R^3 \] Simplifying: \[ \pi \times 36 \times 20 = \frac{4}{3} \pi R^3 \] \[ 720 \pi = \frac{4}{3} \pi R^3 \] Canceling \( \pi \) from both sides: \[ 720 = \frac{4}{3} R^3 \] Multiplying both sides by 3: \[ 2160 = 4 R^3 \] Dividing both sides by 4: \[ R^3 = 540 \] Taking the cube root: \[ R = \sqrt[3]{540} \approx 8.18 \, \text{cm} \] 

Step 3: Conclusion. 
The radius of the sphere is approximately 8.14 cm.

Answer 2

Concept:

• Instead of computing the cube root as a decimal directly, the number under the cube root can first be broken into its prime factors, pulling out any perfect cube factor to leave a simpler cube root behind.
• This gives the radius as an exact simplified radical, and only the last small remaining cube root needs to be approximated.

Step 1: Set up the equal-volume equation, same as before.
$\pi r^2 h = \dfrac{4}{3} \pi R^3$, with $r = 6$ cm and $h = 20$ cm, gives $R^3 = \dfrac{3 r^2 h}{4} = \dfrac{3 \times 36 \times 20}{4} = 540$.

Step 2: Break 540 into prime factors.
$540 = 2^2 \times 3^3 \times 5$.

Step 3: Pull out the perfect cube factor from under the cube root.
$3^3$ is a perfect cube, so $\sqrt[3]{540} = \sqrt[3]{3^3 \times 2^2 \times 5} = 3\sqrt[3]{20}$.

Step 4: Approximate only the small remaining cube root.
$\sqrt[3]{20} \approx 2.71$, so $R \approx 3 \times 2.71 = 8.14$ cm.

Final Answer: The radius of the sphere is approximately 8.14 cm (exactly $3\sqrt[3]{20}$ cm).