To solve the problem, we need to calculate the total volume of ice cream in the cone and the hemisphere placed on top.
Step 1: Understanding the Volume of the Cone
We know that Aimee packs ice cream into an ice cream cone that has a volume of \(4 , \text{in}^3\). The volume \(V\) of a cone can be calculated using the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
where \(r\) is the radius and \(h\) is the height of the cone.
Step 2: Relationship between Diameter and Height
The problem states that the diameter of the cone is equal to its height. Therefore, if we let the height of the cone be \(h\), the radius \(r\) can be expressed as:
\[ r = \frac{h}{2} \]
Step 3: Volume of the Cone
We can substitute the expression for the radius in terms of height into the volume formula.
Substituting \(r\) into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h \]
This simplifies to:
\[ V = \frac{1}{3} \pi \left(\frac{h^2}{4}\right) h = \frac{1}{12} \pi h^3 \]
According to the problem, this volume equals \(4 , \text{in}^3\):
\[ \frac{1}{12} \pi h^3 = 4 \]
Step 4: Solving for Height \(h\)
Now, we solve for \(h\):
\[ h^3 = 4 \cdot 12 / \pi = \frac{48}{\pi} \]
Step 5: Finding the Radius \(r\)
Now that we have \(h\), we can find \(r\):
\[ r = \frac{h}{2} = \frac{1}{2} \left(\frac{48}{\pi}\right)^{1/3} \]
Step 6: Volume of the Hemisphere
Next, we calculate the volume of the hemisphere. The volume \(V_h\) of a hemisphere is given by:
\[ V_h = \frac{2}{3} \pi r^3 \]
Since the radius of the hemisphere is the same as the radius of the cone:
\[ V_h = \frac{2}{3} \pi \left(\frac{h}{2}\right)^3 = \frac{2}{3} \pi \left(\frac{h^3}{8}\right) = \frac{1}{12} \pi h^3 \]
Step 7: Total Volume
Now we sum the volumes of the cone and the hemisphere:
\[ V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = 4 + \frac{1}{12} \pi h^3 \]
Since we know \( \frac{1}{12} \pi h^3 = 4 \):
\[ V_{\text{hemisphere}} = 4 \]
So, the total volume becomes:
\[ V_{\text{total}} = 4 + 4 = 8 , \text{in}^3 \]
Final Answer:
The total volume of ice cream in and on top of the cone is
\[ \boxed{8 , \text{in}^3} \]
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