Question
aimee packs ice cream into an ice cream cone. she then puts a perfect hemisphere of ice cream on the top of the cone that has a volume of 4 in to the 3rd. the diameter of the ice cream cone is equal to its height. what is the the total volume of the 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.
Answers
GPT-4o mini
To find the total volume of the ice cream in and on top of the cone, we first need to understand the given dimensions and relationships.
1. **Volume of the Hemisphere**:
The volume \( V \) of a hemisphere is given by the formula:
\[
V = \frac{2}{3} \pi r^3
\]
We know from the problem that the volume of the hemisphere is \( 4 \, \text{in}^3 \). We can set up the equation:
\[
\frac{2}{3} \pi r^3 = 4
\]
To solve for \( r^3 \), first, multiply both sides by \( \frac{3}{2} \):
\[
\pi r^3 = 6
\]
Now, divide both sides by \( \pi \):
\[
r^3 = \frac{6}{\pi}
\]
Next, we can find \( r \):
\[
r = \left(\frac{6}{\pi}\right)^{1/3}
\]
2. **Dimensions of the Cone**:
The problem states that the diameter of the cone is equal to its height. Therefore, if we let \( h \) be the height of the cone, then the diameter \( d \) can be given by:
\[
d = h
\]
Since the radius \( r_c \) of the cone is half the diameter:
\[
r_c = \frac{h}{2}
\]
3. **Volume of the Cone**:
The volume \( V \) of a cone is given by:
\[
V = \frac{1}{3} \pi r_c^2 h
\]
Substituting \( r_c = \frac{h}{2} \):
\[
r_c^2 = \left(\frac{h}{2}\right)^2 = \frac{h^2}{4}
\]
Thus, the volume of the cone becomes:
\[
V = \frac{1}{3} \pi \left(\frac{h^2}{4}\right) h = \frac{1}{3} \pi \frac{h^3}{4} = \frac{\pi h^3}{12}
\]
4. **Connecting the Cone's Height to the Hemisphere**:
From the previous steps, we have established that the radius of the hemisphere (\( r \)) is also related to the cone's height (\( h \)). To find \( h \) in terms of \( r \), we note that the height (\( h \)) must also relate back to the volume we calculated for the hemisphere. Since \( r \) is equal to the radius of the hemisphere, it can be concluded that the radius of the hemisphere at the top corresponds to the radius of the entire cone. Thus, \( r = \frac{h}{2} \):
\[
\frac{h}{2} = \left(\frac{6}{\pi}\right)^{1/3}
\]
To express \( h \):
\[
h = 2 \left(\frac{6}{\pi}\right)^{1/3}
\]
5. **Total Volume Calculation**:
Now substitute this expression for \( h \) back into the cone's volume formula:
\[
V = \frac{\pi}{12} h^3 = \frac{\pi}{12} \left(2 \left(\frac{6}{\pi}\right)^{1/3}\right)^3
\]
\[
h^3 = 2^3 \left(\frac{6}{\pi}\right) = 8 \cdot \frac{6}{\pi}
\]
\[
V = \frac{\pi}{12} \cdot \frac{48}{\pi} = \frac{48}{12} = 4
\]
6. **Total Ice Cream Volume**:
The total volume of the ice cream with the cone and the hemisphere on top:
\[
\text{Total Volume} = V_{\text{cone}} + V_{\text{hemisphere}} = 4 + 4 = 8 \text{ in}^3
\]
Thus, the total volume of the ice cream in and on top of the cone is \( \boxed{8 \text{ in}^3} \).
1. **Volume of the Hemisphere**:
The volume \( V \) of a hemisphere is given by the formula:
\[
V = \frac{2}{3} \pi r^3
\]
We know from the problem that the volume of the hemisphere is \( 4 \, \text{in}^3 \). We can set up the equation:
\[
\frac{2}{3} \pi r^3 = 4
\]
To solve for \( r^3 \), first, multiply both sides by \( \frac{3}{2} \):
\[
\pi r^3 = 6
\]
Now, divide both sides by \( \pi \):
\[
r^3 = \frac{6}{\pi}
\]
Next, we can find \( r \):
\[
r = \left(\frac{6}{\pi}\right)^{1/3}
\]
2. **Dimensions of the Cone**:
The problem states that the diameter of the cone is equal to its height. Therefore, if we let \( h \) be the height of the cone, then the diameter \( d \) can be given by:
\[
d = h
\]
Since the radius \( r_c \) of the cone is half the diameter:
\[
r_c = \frac{h}{2}
\]
3. **Volume of the Cone**:
The volume \( V \) of a cone is given by:
\[
V = \frac{1}{3} \pi r_c^2 h
\]
Substituting \( r_c = \frac{h}{2} \):
\[
r_c^2 = \left(\frac{h}{2}\right)^2 = \frac{h^2}{4}
\]
Thus, the volume of the cone becomes:
\[
V = \frac{1}{3} \pi \left(\frac{h^2}{4}\right) h = \frac{1}{3} \pi \frac{h^3}{4} = \frac{\pi h^3}{12}
\]
4. **Connecting the Cone's Height to the Hemisphere**:
From the previous steps, we have established that the radius of the hemisphere (\( r \)) is also related to the cone's height (\( h \)). To find \( h \) in terms of \( r \), we note that the height (\( h \)) must also relate back to the volume we calculated for the hemisphere. Since \( r \) is equal to the radius of the hemisphere, it can be concluded that the radius of the hemisphere at the top corresponds to the radius of the entire cone. Thus, \( r = \frac{h}{2} \):
\[
\frac{h}{2} = \left(\frac{6}{\pi}\right)^{1/3}
\]
To express \( h \):
\[
h = 2 \left(\frac{6}{\pi}\right)^{1/3}
\]
5. **Total Volume Calculation**:
Now substitute this expression for \( h \) back into the cone's volume formula:
\[
V = \frac{\pi}{12} h^3 = \frac{\pi}{12} \left(2 \left(\frac{6}{\pi}\right)^{1/3}\right)^3
\]
\[
h^3 = 2^3 \left(\frac{6}{\pi}\right) = 8 \cdot \frac{6}{\pi}
\]
\[
V = \frac{\pi}{12} \cdot \frac{48}{\pi} = \frac{48}{12} = 4
\]
6. **Total Ice Cream Volume**:
The total volume of the ice cream with the cone and the hemisphere on top:
\[
\text{Total Volume} = V_{\text{cone}} + V_{\text{hemisphere}} = 4 + 4 = 8 \text{ in}^3
\]
Thus, the total volume of the ice cream in and on top of the cone is \( \boxed{8 \text{ in}^3} \).