Aimee packs ice cream into an ice cream cone. She then puts a perfect hemisphere of ice cream on top of the cone that has a volume of 4 in.3 The diameter of the ice cream cone is equal to its height. What is the total volume of ice cream in and on top of the cone? Use the relationship between the formulas for the volumes of cones and spheres to help solve this problem. Show your work and explain your reasoning.
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The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height.
Given that the diameter of the ice cream cone is equal to its height, we can let the height of the cone be h and the radius of the base be r. Therefore, the diameter of the base is 2r.
The volume of the cone is V_cone = (1/3)πr^2h.
The volume of the hemisphere is V_hemisphere = (2/3)πr^3.
We are given that the volume of the hemisphere is 4 in^3, so (2/3)πr^3 = 4.
Simplifying this equation, we have πr^3 = 6.
Dividing both sides by π, we get r^3 = 6/π.
Taking the cube root of both sides, we have r = (6/π)^(1/3).
Since the diameter of the base is 2r, the diameter of the base is 2 * (6/π)^(1/3).
The height of the cone is also equal to the diameter of the base, so the height of the cone is 2 * (6/π)^(1/3).
Now, we can calculate the volume of the cone. The volume is V_cone = (1/3)πr^2h.
Plugging in the value of r and h, we have V_cone = (1/3)π[(6/π)^(1/3)]^2 * 2 * (6/π)^(1/3).
Simplifying this expression, we have V_cone = (1/3)π *(6/π)^(2/3) * 2 * (6/π)^(1/3).
V_cone = (2/3) * π * (36/π)^(1/3) * (6/π)^(1/3).
V_cone = (2/3) * π * (216/π^2)^(1/3).
V_cone = (2/3) * π * (216)^(1/3) * (π^(-2/3)).
V_cone ≈ (2/3) * (6) * (0.770302) * (π^(-2/3)).
V_cone ≈ 3.080364 * (π^(-2/3)).
The total volume of the ice cream in and on top of the cone is the sum of the volume of the cone and the volume of the hemisphere.
V_total = V_cone + V_hemisphere.
V_total ≈ 3.080364 * (π^(-2/3)) + 4.
Therefore, the total volume of ice cream in and on top of the cone is approximately 7.080364 in^3.