(a) Deriving the Integral for the Total Mass of the Rod

To find the total mass $M$ of the rod, we can use the concept of Riemann sums. The strategy involves dividing the rod into $n$ small slices and summing up the mass of each slice.

1. Divide the rod into $n$ slices:

   • Let the rod be divided into $n$ slices of equal length $\Delta x$. Thus, each slice has a length $\Delta x = \frac{L}{n}$.

2. Mass of each slice:

   • Consider a small slice of the rod at position $x_i$ with length $\Delta x$.
   • The cross-sectional area at position $x_i$ is $A(x_i)$.
   • The volume of this small slice is approximately $A(x_i) \cdot \Delta x$ cm$^3$.
   • Since the mass density is uniform, the mass of this slice is: $\text{mass of slice} \approx \rho \cdot A(x_i) \cdot \Delta x \text{ grams}.$

3. Sum the masses of all slices:

   • The total mass $M$ of the rod is approximately the sum of the masses of all slices: $M \approx \sum_{i=1}^{n} \rho \cdot A(x_i) \cdot \Delta x.$

4. Take the limit as $n$ approaches infinity:

   • As $n$ increases, $\Delta x$ becomes infinitesimally small, and the sum becomes an integral: $M = \lim_{n \to \infty} \sum_{i=1}^{n} \rho \cdot A(x_i) \cdot \Delta x = \rho \int_{0}^{L} A(x) \, dx.$

Thus, the integral that represents the total mass $M$ of the rod is: $M = \rho \int_{0}^{L} A(x) \, dx.$

(b) Computing the Total Mass for Given Parameters

Given:

• Length of the rod $L = 25$ cm,
• Cross-sectional area $A(x) = 1 + \sqrt{x}$ cm$^2$,
• Mass density $\rho = 8$ g/cm$^3$.

Substitute the given $A(x)$ into the integral derived above: $M = 8 \int_{0}^{25} (1 + \sqrt{x}) \, dx.$

Let's compute this integral:

1. Integrate $A(x)$: $\int (1 + \sqrt{x}) \, dx = \int 1 \, dx + \int \sqrt{x} \, dx = x + \frac{2}{3}x^{3/2}.$

2. Evaluate the definite integral: $\int_{0}^{25} (1 + \sqrt{x}) \, dx = \left[ x + \frac{2}{3}x^{3/2} \right]_{0}^{25}.$ At $x = 25$: $25 + \frac{2}{3}(25)^{3/2} = 25 + \frac{2}{3}(125) = 25 + \frac{250}{3} = 25 + 83.\overline{3} = 108.\overline{3}.$ At $x = 0$: $0 + \frac{2}{3}(0)^{3/2} = 0.$ So the integral evaluates to: $\int_{0}^{25} (1 + \sqrt{x}) \, dx = 108.\overline{3} \text{ cm}^3.$

3. Multiply by the density $\rho$: $M = 8 \times 108.\overline{3} = 866.\overline{6} \text{ grams}.$

4. Convert grams to kilograms: $M = \frac{866.\overline{6}}{1000} = 0.866\overline{6} \text{ kg}.$

Thus, the total mass $M$ of the rod is approximately $0.867$ kilograms.

---

Would you like further details or have any questions?

Here are five related questions:

1. How would the total mass change if the density $\rho$ was not uniform?
2. Can you derive the mass if the cross-sectional area varies as a quadratic function of $x$?
3. How would the result differ if the rod length $L$ was increased?
4. What would the total mass be if the cross-sectional area was constant?
5. How would the integral change if the density $\rho$ depended on $x$?

Tip: Always check units at each step of the calculation to avoid mistakes, especially when converting between different units.