Question 1:
For this case we have that by definition, the volume of a cylinder is given by:
![V = \pi * r ^ 2 * h](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9bbr1apmwdfg5ekvyuzuhy0h4dyqymtih6.png)
Where:
A: It's the radio
h: It's the height
In this case we have to:
Since the diameter is 10ft, then
![r = 5 \ ft](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nqrbbi4vo4rf8x0aetivso8wyl4ync0irz.png)
![h = 21 \ ft](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8zerqsqm7a8l8s2vxpayvsbgbfmk2it102.png)
Substituting:
![V = \pi * (5) ^ 2 * 21\\V = 3.14 * 25 * 21\\V = 1648.5 \ ft ^ 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gvjmk9xt92dgl3bn8k09efdtfyef42qkjw.png)
Thus, the volume of the cylinder is
![1648.5 \ ft ^ 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6h7uzizkwyqchupkws9txdnn6225zakdju.png)
Answer:
![1648.5 \ ft ^ 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6h7uzizkwyqchupkws9txdnn6225zakdju.png)
Question 2:
For this case we have that by definition, the volume of a sphere is given by:
![V = \frac {4} {3} *\pi * r ^ 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/mnc9m5s9an13cyst0f2yqqbvz4mxggen7e.png)
Where:
A: It's the radio
We have to:
![r = 20cm](https://img.qammunity.org/2020/formulas/mathematics/middle-school/atw9m68howkbtyhsfiyl831gxa8a1ctyoh.png)
Substituting:
![V = \frac {4} {3} * \pi * (20) ^ 3\\V = \frac {4} {3} * 3.14 * 8000\\V = 33493.33333](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t2uqi3tru8dj36q8eovbhntnzp3lj64bfa.png)
Rounding:
![V = 33493.3 \ cm ^ 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qfng3zztowq4jvlxe6bans3sejdylgenrc.png)
Answer:
![33493.3 \ cm ^ 3](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k0207oogadwdz65rziowz1hidcnv0jjarq.png)