Question

A snow cone is a tasty treat with flavored ice and a spherical bubble gum ball at the bottom. The radius of the cone is 1.25 inches, and its height is 2.75 inches. If the diameter of the bubble gum ball is 0.5 inches, what is the closest approximation of the volume of the cone that can be filled with flavored ice?

Answer 1

Volume of the cone=1/3πr^2h
V=1/3π(1.25)^2×2.75
V≈4.5 cubic inches

Volume of the sphere=4/3πr^3
V≈0.07 cubic inches

4.5-0.07=4.43. As a result, the volume of the cone that can fipp with ice is about 4.43 cubic inches. Hope it help!

Answer 2

The original volume V₁ of the cone = 1/3(πR²₁).H₁

So V₁ = 1/3(π,1.25²).2.75 ==> V₁ = 4.5 in³

Now to calculate the lower cone (excluding the bubble) we should know
its height that is H₂.
You notice that the largest cone is similar to the smaller one, the we ca write:

R₁/R₂ =H₁/H₂==> 1.25/0.25 = 2.75/H₂ ===> H₂ = 0.55 in

Hence the volume V₂ of the smaller cone = V₂=0.036 in³

Now the volume of the trunk is V₁ - V₂ = 4.5 in³ - 0.036 in³ => V₁-V₂=4.464 in³

The last thing is to calculate V₃ the volume of the spherical bulbe. Once found we will deduct it from the remaining volume that is (V₁-V₂=4.464 in)
Volume of the bulbe which is half of a sphere
Volume sphere= 4/3(πR₂³).H₂ =>V₃=4/3(π.0.25³)(0.55)=0.036=>V₃=0.036 in³
So the bulbe = 1/2 V₃ ==> 0.036/2=0.018 in==> V₃ =0.018 in³

Now the total volume remaining for the ice cream :
Remaining Volume = V₁ - V₂ -V₃ =====> 4.446 in³