The volume we're looking for is the volume of both cones in the figure.
The volume of a cone is [tex]V=\pi r^2\frac{h}{3}[/tex] .
So, [tex]V_{tot} = V_{1} + V_2[/tex] .
Cone 1's variables:
r = 2.6
h = 5
Cone 2's variables:
r = 2.6
h = 3
Now we can just plug and chug!
[tex]V_{tot} = [(3.14)(2.6^2)(\frac{5}{3} )] + [(3.14)(2.6^2)\frac{3}{3} )]\\V_{tot} = [(3.14)(6.76)(1.67)] + [(3.14)(6.76)(1)]\\V_{tot} = (35.45) + (21.23)\\V_{tot} = 56.68[/tex]
V = c. 56.6 cubic units
