Question
Amy 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 volumes of cones and spears, to help solve this problem.
Answers
GPT-4o mini
To solve this problem, we will first calculate the volume of the ice cream cone based on the information given and then find the total volume of ice cream including both the cone and the hemisphere on top.
We know that the volume of the hemisphere of ice cream is given as 4 in³. The formula for the volume \( V \) of a hemisphere is:
\[
V = \frac{2}{3} \pi r^3
\]
Setting this equal to 4 in³ to find the radius \( r \) of the hemisphere:
\[
\frac{2}{3} \pi r^3 = 4
\]
To isolate \( r^3 \), we first multiply both sides by \( \frac{3}{2} \):
\[
\pi r^3 = 6
\]
Next, we divide both sides by \( \pi \):
\[
r^3 = \frac{6}{\pi}
\]
Now we can take the cube root to find \( r \):
\[
r = \sqrt[3]{\frac{6}{\pi}}
\]
Now, we also know that the diameter of the ice cream cone is equal to its height, which means if the cone has a diameter \( d \), then the height \( h \) is also \( d \). If the radius of the cone's base is \( R = \frac{d}{2} \) and the height \( h = d \), then we can express \( h \) in terms of \( R \):
\[
h = 2R
\]
The formula for the volume \( V \) of a cone is:
\[
V = \frac{1}{3} \pi R^2 h
\]
Substituting \( h = 2R \) into the volume formula gives:
\[
V = \frac{1}{3} \pi R^2 (2R) = \frac{2}{3} \pi R^3
\]
Next, we need to relate the radius \( R \) of the cone's base to the radius \( r \) of the hemisphere. Since the diameter of the cone is equal to its height and the hemisphere sits perfectly on top of the cone, we set the base radius of the hemisphere equal to the radius of the cone:
\[
R = r
\]
Now substituting \( R = r \) into the volume formula of the cone:
\[
V_{cone} = \frac{2}{3} \pi r^3
\]
Since we already have \( r^3 = \frac{6}{\pi} \):
\[
V_{cone} = \frac{2}{3} \pi \left(\frac{6}{\pi}\right) = \frac{2}{3} \cdot 6 = 4 \text{ in}^3
\]
Now, we can find the total volume of ice cream:
\[
\text{Total Volume} = V_{cone} + V_{hemisphere} = 4 \text{ in}^3 + 4 \text{ in}^3 = 8 \text{ in}^3
\]
Therefore, the total volume of ice cream in and on top of the cone is:
\[
\boxed{8} \text{ in}³
\]
We know that the volume of the hemisphere of ice cream is given as 4 in³. The formula for the volume \( V \) of a hemisphere is:
\[
V = \frac{2}{3} \pi r^3
\]
Setting this equal to 4 in³ to find the radius \( r \) of the hemisphere:
\[
\frac{2}{3} \pi r^3 = 4
\]
To isolate \( r^3 \), we first multiply both sides by \( \frac{3}{2} \):
\[
\pi r^3 = 6
\]
Next, we divide both sides by \( \pi \):
\[
r^3 = \frac{6}{\pi}
\]
Now we can take the cube root to find \( r \):
\[
r = \sqrt[3]{\frac{6}{\pi}}
\]
Now, we also know that the diameter of the ice cream cone is equal to its height, which means if the cone has a diameter \( d \), then the height \( h \) is also \( d \). If the radius of the cone's base is \( R = \frac{d}{2} \) and the height \( h = d \), then we can express \( h \) in terms of \( R \):
\[
h = 2R
\]
The formula for the volume \( V \) of a cone is:
\[
V = \frac{1}{3} \pi R^2 h
\]
Substituting \( h = 2R \) into the volume formula gives:
\[
V = \frac{1}{3} \pi R^2 (2R) = \frac{2}{3} \pi R^3
\]
Next, we need to relate the radius \( R \) of the cone's base to the radius \( r \) of the hemisphere. Since the diameter of the cone is equal to its height and the hemisphere sits perfectly on top of the cone, we set the base radius of the hemisphere equal to the radius of the cone:
\[
R = r
\]
Now substituting \( R = r \) into the volume formula of the cone:
\[
V_{cone} = \frac{2}{3} \pi r^3
\]
Since we already have \( r^3 = \frac{6}{\pi} \):
\[
V_{cone} = \frac{2}{3} \pi \left(\frac{6}{\pi}\right) = \frac{2}{3} \cdot 6 = 4 \text{ in}^3
\]
Now, we can find the total volume of ice cream:
\[
\text{Total Volume} = V_{cone} + V_{hemisphere} = 4 \text{ in}^3 + 4 \text{ in}^3 = 8 \text{ in}^3
\]
Therefore, the total volume of ice cream in and on top of the cone is:
\[
\boxed{8} \text{ in}³
\]