To find the mass of the thin bar, we can use the formula for mass when the density function is given:

$\text{Mass} = \int_{a}^{b} \rho(x) \, dx$

Where:

• $\rho(x)$ is the density function,
• $a$ and $b$ are the bounds of the region where the bar lies.

In this case:

• $\rho(x) = 1 + \sin(x)$,
• $a = \frac{\pi}{6}$,
• $b = \frac{\pi}{2}$.

Thus, the mass of the bar is:

$\text{Mass} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (1 + \sin(x)) \, dx$

Now, let's compute the integral.

Step 1: Separate the integral

$\text{Mass} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 1 \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin(x) \, dx$

Step 2: Evaluate each integral

1. The first integral:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 1 \, dx = x \Big|_{\frac{\pi}{6}}^{\frac{\pi}{2}} = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$

2. The second integral:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin(x) \, dx = -\cos(x) \Big|_{\frac{\pi}{6}}^{\frac{\pi}{2}}$

Evaluating this:

$-\cos\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{6}\right) = 0 + \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

Step 3: Combine the results

Now, sum the results:

$\text{Mass} = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

This is the mass of the thin bar. You can leave it in this exact form or approximate the result numerically.

If you'd like a decimal approximation:

$\frac{\pi}{3} \approx 1.047 \quad \text{and} \quad \frac{\sqrt{3}}{2} \approx 0.866$

So, the mass is approximately:

$1.047 + 0.866 = 1.913 \, \text{units}$

Therefore, the mass of the thin bar is:

$\boxed{ \frac{\pi}{3} + \frac{\sqrt{3}}{2} \text{ units} }$

Or approximately 1.913 units.

Would you like further details on this solution, or do you have any questions?

Related questions:

1. How would the mass change if the density function was different, like $\rho(x) = 1 + 2\sin(x)$?
2. How does the integral formula for mass apply in cases where the density is not a function of $x$, but a constant?
3. How do you compute the center of mass for a thin bar with varying density?
4. What are the implications of changing the bounds from $\frac{\pi}{6}$ to $\pi$ in the density function?
5. How do you handle more complex density functions, such as ones that are piecewise defined?

Tip: When integrating functions involving trigonometric terms, like $\sin(x)$, it often helps to break the integrals into parts and use standard integral formulas for the trigonometric functions.