A silo is shaped like a cylinder with half of a sphere on top. In this problem we will combine our knowledge of volume formulas for cylinders and spheres to solve silo volume problems.
a. Find a general equation to find the volume of a silo in terms of the radius \(r\) and the height of the cylinder \(h\).
b. What is the volume of a silo with radius \(15\) meters and height \(18\) meters?
c. If the volume of the silo is \(864 \pi\) cubic meters and the radius is \(6\) meters, what is its height?
Solution:
a. The volume of this complex shape is the sum of the volumes of its parts. Thus the volume formula is the sum of the formulas for the volume of a cylinder and the volume of half of a sphere. As you recall, the formula for the volume of a cylinder is
$$V_C = \pi r^2h$$
Similarly, the volume of a sphere is
$$V_P = \frac{4}{3} \pi r^3$$
However, the final volume includes only half of a sphere, so when combining the volumes, we use \(\frac{1}{2}V_P\) rather than \(V_P\). Thus the total volume for a silo is
$$V_S = V_C + \frac{1}{2}V_P = \pi r^2h + \frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \pi r^2h + \frac{2}{3}\pi r^3$$
b. We can use our new formula to solve this problem. Simply plug in \(15\) for \(r\) and \(18\) for \(h\):
$$V_S = \pi \cdot 15^2 \cdot 18 + \frac{2}{3} \pi \cdot 15^3 = 4050 \pi + 2250 \pi = 6300 \pi \; m^3$$
c. We can use our equation again, but this time we are solving for \(h\) using \(V_S\) and \(r\). Solve for the target variable before plugging in any values:
$$V_S = \pi r^2 h + \frac{2}{3} \pi r^3 \Rightarrow$$
$$\pi r^2 h = V_S - \frac{2}{3} \pi r^3$$
Our goal is to isolate the variable \(h\), and that is exactly what we are doing. Divide both sides of the equation by \(\pi r^2\):
$$h = \frac{V_S}{\pi r^2} - \frac{2}{3} r$$
As we know \(V_S\) and \(r\), we can now plug and chug to find \(h\):
$$h = \frac{864 \pi}{\pi \cdot 6^2} - \frac{2}{3} \cdot 6 = \frac{864}{36} - 4 = \frac{864 - 144}{36} = \frac{720}{36} = 20 \; m$$