Use differentials to estimate the amount of metal (in cm3) in a closed cylindrical can that is 22 cm high and 8 cm in diameter if the metal in the top and the bottom is 0.4 cm thick and the metal in the sides is 0.05 cm thick. (Round your answer to two decimal places.)

We are tasked with estimating the amount of metal in a closed cylindrical can using differentials. The problem involves a cylinder with a height of 22 cm and a diameter of 8 cm, where the metal in the top and bottom is 0.4 cm thick, and the metal in the sides is 0.05 cm thick.

### Step 1: Basic formula for volume of a cylinder
The volume \( V \) of a cylinder is given by the formula:

\[
V = \pi r^2 h
\]

where \( r \) is the radius and \( h \) is the height.

Given:
- Height of the can \( h = 22 \, \text{cm} \),
- Diameter of the can \( = 8 \, \text{cm} \), so the radius \( r = \frac{8}{2} = 4 \, \text{cm} \).

### Step 2: Calculate the volume of the entire can (outer volume)
The volume of the entire can (assuming it is a solid) can be calculated as:

\[
V_{\text{outer}} = \pi r^2 h = \pi (4)^2 (22) = \pi (16) (22) = 352\pi \, \text{cm}^3
\]

So,

\[
V_{\text{outer}} = 1106.64 \, \text{cm}^3
\]

### Step 3: Estimate the thickness using differentials
We can break this down into two parts: the volume of the metal in the top and bottom, and the volume of the metal in the sides.

#### (a) Volume of metal in the top and bottom
Each top and bottom is a disk with a thickness of \( 0.4 \, \text{cm} \). The volume of one disk is:

\[
V_{\text{disk}} = \pi r^2 t
\]

where \( t \) is the thickness (0.4 cm). So for each disk:

\[
V_{\text{disk}} = \pi (4)^2 (0.4) = \pi (16)(0.4) = 6.4\pi \, \text{cm}^3
\]

For two disks (top and bottom), the total volume of metal is:

\[
V_{\text{top+bottom}} = 2 \times 6.4\pi = 12.8\pi \, \text{cm}^3
\]

So,

\[
V_{\text{top+bottom}} = 40.21 \, \text{cm}^3
\]

#### (b) Volume of metal in the sides
The thickness of the metal in the sides is \( 0.05 \, \text{cm} \). The side of the cylinder is essentially a rectangular sheet of metal, which when wrapped forms the side of the cylinder. The surface area of the side (before considering thickness) is:

\[
A_{\text{side}} = 2\pi r h
\]

Now, we multiply this area by the thickness \( t \) to get the volume of the metal in the sides:

\[
V_{\text{side}} = 2\pi r h \cdot t = 2\pi (4)(22)(0.05) = 8.8\pi \, \text{cm}^3
\]

So,

\[
V_{\text{side}} = 27.64 \, \text{cm}^3
\]

### Step 4: Total volume of metal
Now, we can add up the volumes of the metal in the top, bottom, and sides to get the total amount of metal used in the can:

\[
V_{\text{total}} = V_{\text{top+bottom}} + V_{\text{side}} = 40.21 \, \text{cm}^3 + 27.64 \, \text{cm}^3 = 67.85 \, \text{cm}^3
\]

Thus, the total amount of metal in the can is approximately:

\[
\boxed{67.85 \, \text{cm}^3}
\]

---

Do you want more details or clarifications on any part of this solution?

Here are 5 related questions:
1. How would the answer change if the metal thickness in the sides increased to 0.1 cm?
2. How would you calculate the volume of metal if the can had no bottom?
3. How does the volume of the can change if the height increases by 10%?
4. What is the surface area of the entire can?
5. How would the volume calculation change if the can's top were conical?

**Tip:** Understanding how differentials can estimate changes in volumes or areas is a powerful tool in physics and engineering!

Learn how to use differentials to estimate the volume of metal in a closed cylindrical can. This detailed solution involves applying geometry and volume estimation techniques for a cylinder that is 22 cm high and 8 cm in diameter, with varying thickness in the top, bottom, and sides. Suitable for Grades 10-12.

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