Part 1: Setting up the Integral

The given density function for the thin bar is:

$\rho(x) = 1 + \sin(x)$

The mass of a thin bar with a variable density along its length is given by the integral of the density function over the specified interval, multiplied by the differential element of length. In this case, the length of the bar is from $\frac{\pi}{3}$ to $\pi$, and the density function is $\rho(x) = 1 + \sin(x)$.

So, the mass $M$ is given by the integral:

$M = \int_{\frac{\pi}{3}}^{\pi} (1 + \sin(x)) \, dx$

Part 2: Solving the Integral

Now, we will evaluate the integral:

$M = \int_{\frac{\pi}{3}}^{\pi} (1 + \sin(x)) \, dx$

This can be split into two integrals:

$M = \int_{\frac{\pi}{3}}^{\pi} 1 \, dx + \int_{\frac{\pi}{3}}^{\pi} \sin(x) \, dx$

1. The first integral is:

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

2. The second integral is:

$\int_{\frac{\pi}{3}}^{\pi} \sin(x) \, dx = -\cos(x) \Big|_{\frac{\pi}{3}}^{\pi}$

Evaluating this:

$-\cos(\pi) + \cos\left(\frac{\pi}{3}\right) = -(-1) + \left(\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2}$

Thus, the total mass is:

$M = \frac{2\pi}{3} + \frac{3}{2}$

Now, we combine these two terms by finding a common denominator:

$M = \frac{4\pi}{6} + \frac{9}{6} = \frac{4\pi + 9}{6}$

Final Answer:

The mass of the thin bar is:

$M = \frac{4\pi + 9}{6} \text{ units.}$

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Would you like a detailed explanation of any part of the solution? Or would you like me to assist you with other questions?

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Here are 5 related questions for further practice:

1. How do you interpret the physical meaning of the density function $\rho(x)$?
2. What is the mass of a thin bar with a constant density $\rho = 1$ over the same interval $\frac{\pi}{3} \leq x \leq \pi$?
3. How would the mass change if the density function was $\rho(x) = 1 + \cos(x)$ instead of $1 + \sin(x)$?
4. What would happen to the mass if the limits of integration were changed to $0 \leq x \leq \pi$?
5. How do you compute the mass of a thin bar with varying density in three dimensions?

Tip: For integrals involving trigonometric functions like $\sin(x)$ or $\cos(x)$, always recall the standard antiderivatives: $\int \sin(x) \, dx = -\cos(x)$ and $\int \cos(x) \, dx = \sin(x)$.