a) Radius of cylinder, r = 6m
where Height is h and Volume is V
![Volume,V=\pi^{}r^2h](https://img.qammunity.org/2023/formulas/mathematics/college/a50o0wqdu8pfu0pr0wjh0pugqu7owo3voj.png)
c) if height is doubled
Replace the h with 2h
![h=2(h)=2h](https://img.qammunity.org/2023/formulas/mathematics/college/vrt9hvkd4hw4mki8xemrvkvh38mk7xetyn.png)
![V=\pi r^2h^{}](https://img.qammunity.org/2023/formulas/mathematics/college/a48fiiavt8qnzzpjilbkvj4gg5ciywwgc6.png)
![V=\pi r^2(2h)](https://img.qammunity.org/2023/formulas/mathematics/college/ex45sgwctmeaye91pk8f8b2u96tetjdp68.png)
From the equation, the volume will be doubled when the height is doubled.
d) if the height is multiplied by 1/3
Replace h with 1/3(h)
![h=(1)/(3)(h)](https://img.qammunity.org/2023/formulas/mathematics/college/91hskcsggwilr76knunf6chrdstbfi6x08.png)
![V=\pi r^2((1)/(3)h)](https://img.qammunity.org/2023/formulas/mathematics/college/f1k1pgrd2vqxeglpy1w7ym18nccsmpy38y.png)
From the equation, the volume of the cylinder will be 1/3 of it's original volume when height is multiplied by 1/3.
b) Where r= 3.14, below lies the graph
![\begin{gathered} V=\pi r^2h \\ \pi=3.14,\text{ r}=6m\text{ substitute for }\pi\text{ and r into the equation above} \\ V=(3.14)(6^2)h=(113.04h)m^3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r8ley225kio1r82xiygg2o2djdd1k6l2ar.png)