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. (4 points)

1 answer

Given that the volume of the hemisphere of ice cream is \( 4 , \text{in}^3 \), we can use the formula for the volume of a hemisphere with radius \( r \):

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

Setting this equal to the volume given:

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

To solve for \( r^3 \), we 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 take the cube root to find \( r \):

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

Next, we need to find the total volume of ice cream in and on top of the cone. The cone's height \( h \) is equal to its diameter, which means:

\[ h = 2r \]

Since the diameter of the cone is equal to its height, its radius \( R \) is half of that:

\[ R = \frac{h}{2} = \frac{2r}{2} = r \]

Now we can calculate the volume of the cone using its volume formula \( V = \frac{1}{3}\pi R^2 h \):

We substitute \( R \) and \( h \):

\[ V_{\text{cone}} = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3 \]

Since we already have \( r^3 = \frac{6}{\pi} \), we can substitute that into the equation:

\[ V_{\text{cone}} = \frac{2}{3}\pi \left(\frac{6}{\pi}\right) = \frac{2}{3} \times 6 = 4 , \text{in}^3 \]

Now we can find the total volume of ice cream in and on top of the cone by adding the volume of the cone and the volume of the hemisphere:

\[ V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = 4 , \text{in}^3 + 4 , \text{in}^3 = 8 , \text{in}^3 \]

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

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