SOLUTION:
Step 1:
In this question, we are given the following:
Step 2:
Part A:
The volume of the holding tank=
[tex]\begin{gathered} \text{Volume of the cylinder}=\text{ }\pir^2h \\ \text{where }\pi\text{ = }\frac{22}{7} \\ r\text{ = 6 f}eet \\ \text{h = 16 f}eet \end{gathered}[/tex]
Volume of the cylinder=
[tex]\begin{gathered} \frac{22}{7}\text{ X 6 X 6X 16} \\ =\text{ }\frac{12672}{7} \\ =\text{ 1810.2857} \\ \cong1810.29ft^3(\text{ to 2 decimal places)} \end{gathered}[/tex]
Step 3:
Part B:
The volume of the containment tank=
[tex]\begin{gathered} \pi r^2h \\ \text{where }\pi=\frac{22}{7} \\ r\text{ = 8 f}eet \\ h\text{ = 4 fe}et \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} \frac{22}{7}\text{ X 8 X 8 X 4} \\ =\text{ }\frac{5632}{7} \\ =\text{ 804.5714} \\ \cong804.57ft^3\text{ ( to 2 decimal places)} \end{gathered}[/tex]
Step 4:
Part C:
If the center tank is full of heating fuel and develops a leak at the bottom, will the containment be able to hold all the heating fuel that leaks out?
Answer: NO