ANSWER :
The volume of the prop rounded to the nearest tenth is 2882.5 in^3
EXPLANATION :
Recal the volume formulas :
Cone :
[tex]V=\frac{1}{3}\pi r^2h[/tex]
Hemisphere/Half sphere :
[tex]V=\frac{2}{3}\pi r^3[/tex]
From the problem, the radius of the cone and half sphere is the same.
The radius is r = 9 in
The height of the cone is 16 in
Using the formulas above :
Cone :
[tex]\begin{gathered} V=\frac{1}{3}(3.14)(9)^2(16) \\ V=1356.48 \end{gathered}[/tex]
Half sphere :
[tex]\begin{gathered} V=\frac{2}{3}(3.14)(9)^3 \\ V=1526.04 \end{gathered}[/tex]
The area of the prop is the sum of these two volumes :
[tex]1356.48+1526.04=2882.52[/tex]
