Radius of the hemisphere and the cone base: R = 2 in
Height of the cone: H = 6 - 2 = 4 in
π ≈ 3,14
Ice inside cone:
[tex]V_1=\frac13\pi R^2\cdot H=\frac13\pi\cdot2^2\cdot 4=\frac{16}3\pi[/tex]
Ice outside cone:
[tex]V_2=\frac12\cdot\frac43\pi R^3=\frac23\pi\cdot2^3=\frac{16}3\pi[/tex]
Total of ice:
[tex]V=V_1+V_2=\frac{16}3\pi+\frac{16}3\pi=\frac{32}3\pi\approx\frac{32}3\cdot3,14=33,49(3)\approx33,5[/tex]
There are 33,5 cubic inches of ice in the cone.
