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 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.

give an answer in simplest form.

Answer 1

bot, make it more simple

Answer 2

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

**Step 1: Volume of the Hemisphere**
We know that the volume of the hemisphere of ice cream on top of the cone is given as:

\[
V_{\text{hemisphere}} = \frac{2}{3} \pi r^3
\]

We are given that this volume is \(4 \, \text{in}^3\):

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

To find \(r\), we can solve for \(r^3\):

\[
r^3 = \frac{4 \cdot 3}{2\pi} = \frac{12}{2\pi} = \frac{6}{\pi}
\]

Next, we calculate \(r\):

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

**Step 2: Volume of the Ice Cream Cone**
We also know that the ice cream cone has a height \(h\) equal to its diameter, which is \(2r\). Therefore, the height of the cone is:

\[
h = 2r
\]

The volume of a cone is given by the formula:

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

Substituting for \(h\):

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

From step 1, we found that:

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

\[
V_{\text{cone}} = \frac{2}{3} \pi \left(\frac{6}{\pi}\right) = \frac{2}{3} \cdot \frac{12}{\pi} = \frac{24}{3\pi} = \frac{8}{\pi}
\]

**Step 3: Total Volume of Ice Cream**
Finally, we add the volumes of the hemisphere and the cone:

\[
V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = \frac{8}{\pi} + 4
\]

To add these volumes, we can rewrite \(4\) with a common denominator of \(\pi\):

\[
4 = \frac{4\pi}{\pi}
\]

Now we can add the two volumes:

\[
V_{\text{total}} = \frac{8}{\pi} + \frac{4\pi}{\pi} = \frac{8 + 4\pi}{\pi}
\]

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

\[
\boxed{\frac{8 + 4\pi}{\pi}}
\]

This represents the simplest form of the total volume expressed in terms of \(\pi\).

Answer 3

Sure! Let's break it down into simpler steps.

**Step 1: Volume of the Hemisphere**

We know the volume of the hemisphere is given as \(4 \, \text{in}^3\).

\[
\text{Volume of Hemisphere} = \frac{2}{3} \pi r^3 = 4
\]

To find \(r^3\), we multiply both sides by \(\frac{3}{2\pi}\):

\[
r^3 = \frac{4 \times 3}{2\pi} = \frac{12}{2\pi} = \frac{6}{\pi}
\]

**Step 2: Volume of the Cone**

Since the diameter of the cone equals its height, the cone's height is \(2r\) (twice the radius).

The volume of the cone is given by the formula:

\[
\text{Volume of Cone} = \frac{1}{3} \pi r^2 h
\]

Substituting \(h = 2r\):

\[
\text{Volume of Cone} = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3
\]

We already know \(r^3 = \frac{6}{\pi}\), so we substitute that in:

\[
\text{Volume of Cone} = \frac{2}{3} \pi \left(\frac{6}{\pi}\right) = \frac{12}{3} = 4
\]

**Step 3: Total Volume of Ice Cream**

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

\[
\text{Total Volume} = \text{Volume of Cone} + \text{Volume of Hemisphere} = 4 + 4 = 8
\]

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

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

This is the total volume of ice cream in a simpler format!