Question

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
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.

Answer 1

To solve this problem, we need to find the total volume of ice cream in the cone and the volume of the hemisphere on top of the cone.

### Step 1: Understand the given data

1. **Volume of the hemisphere**: Given as \(4 \, \text{in}^3\).
2. **Cone Information**: The diameter of the cone is equal to its height.

### Step 2: Calculate the radius of the hemisphere

The volume \(V\) of a hemisphere with radius \(r\) is given by the formula:

\[
V = \frac{2}{3} \pi r^3
\]

Given that the volume of the hemisphere is \(4 \, \text{in}^3\), we can set up the equation:

\[
\frac{2}{3} \pi r^3 = 4
\]

Now, we can solve for \(r\):

1. Multiply both sides by \(\frac{3}{2}\):

\[
\pi r^3 = 6
\]

2. Divide both sides by \(\pi\):

\[
r^3 = \frac{6}{\pi}
\]

3. Take the cube root to find \(r\):

\[
r = \sqrt[3]{\frac{6}{\pi}}
\]

### Step 3: Relate the cone dimensions to the hemisphere

Since the diameter of the cone is equal to its height, and the radius \(r\) of the hemisphere is equal to the radius of the cone's base, we have:

- Radius of the cone's base \(R = r = \sqrt[3]{\frac{6}{\pi}}\)
- Diameter of the cone (and the height) \(H = 2R = 2\sqrt[3]{\frac{6}{\pi}}\)

### Step 4: Calculate the volume of the cone

The volume \(V\) of a cone with radius \(R\) and height \(H\) is given by:

\[
V = \frac{1}{3} \pi R^2 H
\]

Substituting \(R\) and \(H\) into the formula:

\[
V = \frac{1}{3} \pi \left(\sqrt[3]{\frac{6}{\pi}}\right)^2 \left(2\sqrt[3]{\frac{6}{\pi}}\right)
\]

Calculating \(R^2\) and substituting:

\[
R^2 = \left(\sqrt[3]{\frac{6}{\pi}}\right)^2 = \sqrt[3]{\left(\frac{6}{\pi}\right)^2}
\]

Now substituting this into the volume formula:

\[
V = \frac{1}{3} \pi \cdot \sqrt[3]{\left(\frac{6}{\pi}\right)^2} \cdot 2\sqrt[3]{\frac{6}{\pi}}
\]

Combining the terms inside the volume calculation:

\[
V = \frac{2}{3} \pi \cdot \sqrt[3]{\left(\frac{6^2}{\pi^2}\right) \cdot \frac{6}{\pi}} = \frac{2}{3} \pi \cdot \sqrt[3]{\frac{6^3}{\pi^3}} = \frac{2}{3} \pi \cdot \frac{6}{\pi} = \frac{2 \cdot 6}{3} = 4
\]

### Step 5: Total Volume of Ice Cream

Now we add the volume of the cone and the hemisphere:

\[
\text{Total Volume} = \text{Volume of Cone} + \text{Volume of Hemisphere}
\]

\[
\text{Total Volume} = 4 \, \text{in}^3 + 4 \, \text{in}^3 = 8 \, \text{in}^3
\]

### Conclusion

Thus, the total volume of ice cream in the cone and on top of the cone is

\[
\boxed{8 \, \text{in}^3}
\]