Question

what is the volume of a hemisphere with a diameter of 51.8 in, rounded to the nearest tenth of a cubic inch?

Explanation:

Step1: Find the radius

The diameter \( d = 51.8 \) in, so the radius \( r=\frac{d}{2}=\frac{51.8}{2} = 25.9 \) in.

Step2: Recall the volume formula for a hemisphere

The volume of a sphere is \( V_{sphere}=\frac{4}{3}\pi r^{3} \), so the volume of a hemisphere \( V_{hemisphere}=\frac{1}{2}\times\frac{4}{3}\pi r^{3}=\frac{2}{3}\pi r^{3} \).

Step3: Substitute the radius into the formula

Substitute \( r = 25.9 \) into \( V_{hemisphere}=\frac{2}{3}\pi r^{3} \):
\( V=\frac{2}{3}\times\pi\times(25.9)^{3} \)
First, calculate \( (25.9)^{3}=25.9\times25.9\times25.9 = 25.9\times670.81=17374.079 \)
Then, \( \frac{2}{3}\times\pi\times17374.079=\frac{2\times\pi\times17374.079}{3} \)
\( \frac{34748.158\pi}{3}\approx\frac{34748.158\times3.1415926535}{3} \)
\( \approx\frac{109150.5}{3}=36383.5 \) (rounded to the nearest tenth)

Answer:

\( 36383.5 \)