To solve this problem, we'll approach each question individually, using the given graph where density (in g/cm) is plotted against the rod’s length (in cm). The mass of each section of the rod corresponds to the area under the density curve in that section.

1. Find the Mass of the Left Half of the Rod $(0 \leq x \leq 8)$:

   This region includes two parts:

   • From $x = 0$ to $x = 2$, the density is approximately constant at $1 \, \text{g/cm}$.
   • From $x = 2$ to $x = 8$, the density linearly increases from $1 \, \text{g/cm}$ to $10 \, \text{g/cm}$.

   We can calculate the mass as follows:

   • For $0 \leq x \leq 2$: The area of this rectangle is $\text{width} \times \text{height} = 2 \times 1 = 2 \, \text{g}$.
   • For $2 \leq x \leq 8$: The area of this trapezoid is given by: $\frac{1}{2} \times (\text{base}_1 + \text{base}_2) \times \text{height} = \frac{1}{2} \times (1 + 10) \times 6 = \frac{1}{2} \times 11 \times 6 = 33 \, \text{g}.$

   Therefore, the mass of the left half is: $2 + 33 = 35 \, \text{g}.$

2. Find the Mass of the Right Half of the Rod $(8 \leq x \leq 16)$:

   This region shows a constant density of $10 \, \text{g/cm}$ from $x = 8$ to $x = 16$.

   • The area of this rectangle is $\text{width} \times \text{height} = 8 \times 10 = 80 \, \text{g}$.

   Thus, the mass of the right half is $80 \, \text{g}$.

3. Find the Mass of the Entire Rod $(0 \leq x \leq 16)$:

   To find the total mass, we add the masses of the left and right halves: $35 + 80 = 115 \, \text{g}.$

4. Estimate the Center of Mass:

   The center of mass $\bar{x}$ is the point along the length where the rod balances, given by: $\bar{x} = \frac{\int x \cdot \text{density}(x) \, dx}{\text{Total Mass}}.$ However, since we have discrete areas, we can estimate the center of mass by calculating weighted averages for each segment:

   • For $0 \leq x \leq 2$, with mass $2$ g centered around $x = 1$.
   • For $2 \leq x \leq 8$, with mass $33$ g centered around the midpoint $x = 5$.
   • For $8 \leq x \leq 16$, with mass $80$ g centered around $x = 12$.

   Using these values, we approximate $\bar{x}$: $\bar{x} = \frac{(2 \cdot 1) + (33 \cdot 5) + (80 \cdot 12)}{115} = \frac{2 + 165 + 960}{115} = \frac{1127}{115} \approx 9.8 \, \text{cm}.$

Summary of Results

• Mass of left half: $35 \, \text{g}$
• Mass of right half: $80 \, \text{g}$
• Total mass: $115 \, \text{g}$
• Center of mass: approximately $9.8 \, \text{cm}$

Would you like more details or explanations on any part?

Here are related questions to explore further:

1. How would you calculate the center of mass for a continuous density function?
2. How does the density variation affect the balance point of the rod?
3. What are other methods to approximate the mass from a density graph?
4. How could we verify our results with integral calculus?
5. How would the results change if the rod's density distribution were exponential?

Tip: For non-uniform objects, the center of mass shifts toward denser regions.