<h2>
Answer:</h2>
<u>First Part</u>
Given that
![Volume = \frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D)
We have that
![Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{Diameter}{2})^{3} = \frac{4}{3} \pi 9^{3} = 972\pi cm^{3} \approx 3053.63 cm^{3}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%28%5Cfrac%7BDiameter%7D%7B2%7D%29%5E%7B3%7D%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%209%5E%7B3%7D%20%3D%20972%5Cpi%20cm%5E%7B3%7D%20%5Capprox%203053.63%20cm%5E%7B3%7D)
<u>Second Part</u>
Given that
![Volume = \frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D)
If the Diameter were reduced by half we have that
![Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{r}{2}) ^{3} = \frac{\frac{4}{3} \pi r^{3}}{8}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%28%5Cfrac%7Br%7D%7B2%7D%29%20%5E%7B3%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%7D%7B8%7D)
This shows that the volume would be
of its original volume
<h2>
Step-by-step explanation:</h2>
<u>First Part</u>
Gather Information
![Diameter = 18cm](https://tex.z-dn.net/?f=Diameter%20%3D%2018cm)
![Volume = \frac{4}{3} \pi r^{3}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D)
Calculate Radius from Diameter
![Radius = \frac{Diameter}{2} = \frac{18}{2} = 9](https://tex.z-dn.net/?f=Radius%20%3D%20%5Cfrac%7BDiameter%7D%7B2%7D%20%3D%20%5Cfrac%7B18%7D%7B2%7D%20%3D%209)
Use the Radius on the Volume formula
![Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi 9^{3}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%209%5E%7B3%7D)
Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s
![Volume = \frac{4}{3} \pi 9^{3} = \frac{4}{3} \pi 9 * 9 * 9 = \frac{4}{3} \pi 9 * 9 * 3 * 3](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%209%5E%7B3%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%209%20%2A%209%20%2A%209%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%209%20%2A%209%20%2A%203%20%2A%203)
We can see now that one of the 3s can be already divided by the 3 in the denominator
![Volume = \frac{4}{3} \pi 9 * 9 * 3 * 3 = 4 \pi 9 * 9 * 3](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%209%20%2A%209%20%2A%203%20%2A%203%20%3D%204%20%5Cpi%209%20%2A%209%20%2A%203)
Finally, since we can't simplify anymore we just calculate it's volume
![Volume = 4 \pi 9 * 9 * 3 = 12 \pi * 9 * 9 = 12 * 81 \pi = 972 \pi cm^{3}](https://tex.z-dn.net/?f=Volume%20%3D%204%20%5Cpi%209%20%2A%209%20%2A%203%20%3D%2012%20%5Cpi%20%2A%209%20%2A%209%20%3D%2012%20%2A%2081%20%5Cpi%20%3D%20972%20%5Cpi%20cm%5E%7B3%7D)
![Volume \approx 3053.63 cm^{3}](https://tex.z-dn.net/?f=Volume%20%5Capprox%203053.63%20cm%5E%7B3%7D)
<u>Second Part</u>
Understanding how the Diameter reduced by half would change the Radius
![Radius =\frac{Diameter}{2}\\\\If \\\\Diameter = \frac{Diameter}{2}\\\\Then\\\\Radius = \frac{\frac{Diameter}{2} }{2} = \frac{\frac{Diameter}{2}}{\frac{2}{1}} = \frac{Diameter}{2} * \frac{1}{2} = \frac{Diameter}{4}](https://tex.z-dn.net/?f=Radius%20%3D%5Cfrac%7BDiameter%7D%7B2%7D%5C%5C%5C%5CIf%20%5C%5C%5C%5CDiameter%20%3D%20%5Cfrac%7BDiameter%7D%7B2%7D%5C%5C%5C%5CThen%5C%5C%5C%5CRadius%20%3D%20%5Cfrac%7B%5Cfrac%7BDiameter%7D%7B2%7D%20%7D%7B2%7D%20%3D%20%5Cfrac%7B%5Cfrac%7BDiameter%7D%7B2%7D%7D%7B%5Cfrac%7B2%7D%7B1%7D%7D%20%3D%20%5Cfrac%7BDiameter%7D%7B2%7D%20%2A%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%5Cfrac%7BDiameter%7D%7B4%7D)
Understanding how the Radius now changes the Volume
![Volume = \frac{4}{3}\pi r^{3}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20r%5E%7B3%7D)
With the original Diameter, we have that
![Volume = \frac{4}{3}\pi (\frac{Diameter}{2}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{2^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{2 * 2 * 2} = \frac{4}{3}\pi \frac{Diameter^{3}}{8}\\\\](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%28%5Cfrac%7BDiameter%7D%7B2%7D%29%20%5E%7B3%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B2%5E%7B3%7D%7D%5C%5C%5C%5C%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B2%20%2A%202%20%2A%202%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B8%7D%5C%5C%5C%5C)
If the Diameter were reduced by half, we have that
![Volume = \frac{4}{3}\pi (\frac{Diameter}{4}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{4^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 4 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 2 * 2 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{8 * 8} = \frac{\frac{4}{3}\pi\frac{Diameter^{3}}{8}}{8}](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%28%5Cfrac%7BDiameter%7D%7B4%7D%29%20%5E%7B3%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B4%5E%7B3%7D%7D%5C%5C%5C%5C%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B4%20%2A%204%20%2A%204%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B4%20%2A%202%20%2A%202%20%2A%204%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%5Cpi%20%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B8%20%2A%208%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%5Cfrac%7BDiameter%5E%7B3%7D%7D%7B8%7D%7D%7B8%7D)
But we can see that the numerator is exactly the original Volume!
This shows us that the Volume would be
of the original Volume if the Diameter were reduced by half.