Question

question
what is the volume of a hemisphere with a radius of 42.1 ft, rounded to the nearest tenth of a cubic foot?

Explanation:

Step1: Recall the formula for the volume of a sphere

The volume of a sphere is \(V_{sphere}=\frac{4}{3}\pi r^{3}\). A hemisphere is half of a sphere, so the volume of a hemisphere \(V_{hemisphere}=\frac{2}{3}\pi r^{3}\).

Step2: Substitute the radius \(r = 42.1\) into the formula

\(V=\frac{2}{3}\pi(42.1)^{3}\). First, calculate \((42.1)^{3}=42.1\times42.1\times42.1 = 74509.661\). Then, \(V=\frac{2}{3}\pi\times74509.661\).
\(\frac{2}{3}\times74509.661 = 49673.1073\). So \(V = 49673.1073\pi\).

Step3: Calculate the numerical value

Using \(\pi\approx3.14159\), \(V=49673.1073\times3.14159\approx156076.4\)

Answer:

\(156076.4\) cubic - feet