 steel rod of length L cm has variable cross-section area A(x) cm2, where x is measured in
centimeters from one end of the rod. The rod has uniform mass density ρ g/cm3. (a) Using
Riemann sums, derive an integral for the total mass of the rod M . (hint: divide the rod into
n slices, find the mass of each slice, . . .) (b) Compute the total mass for the case L = 25 cm,
A(x) = (1 + √x ) cm2, ρ = 8 g/cm3. Express M in kilograms (kg).

### (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.

Learn how to derive an integral for the total mass of a steel rod with a variable cross-section area using Riemann sums. Compute the mass for specific parameters: L = 25 cm, A(x) = (1 + √x) cm², ρ = 8 g/cm³. Includes step-by-step calculations and conversions to kilograms.

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