To calculate the mass of a thin rod using its linear density function, we apply the following general formula:

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

where:

• $\rho(x)$ is the linear density function (mass per unit length),
• $a$ and $b$ are the limits of the interval where the rod is located on the axis.

In this case:

• The rod is oriented along the interval $[8, 10]$,
• The linear density function is given as $\rho(x) = 5x^2 + 5$.

Now, let's calculate the mass by integrating the density function over the interval $[8, 10]$:

$\text{Mass} = \int_{8}^{10} (5x^2 + 5) \, dx$

We can break the integral into two parts:

$\text{Mass} = \int_{8}^{10} 5x^2 \, dx + \int_{8}^{10} 5 \, dx$

Let me compute this integral to find the mass.The mass of the thin rod over the interval $[8, 10]$ is $\frac{2470}{3}$, or approximately 823.33 units of mass.

Would you like any further details or clarifications?

Here are 5 related questions for practice:

1. How would the mass change if the density function were linear, such as $\rho(x) = 3x + 2$?
2. What is the physical interpretation of a density function in real-world applications?
3. How would you calculate the center of mass of the rod using the same density function?
4. What would the mass be if the rod were oriented along a different interval, such as $[2, 5]$?
5. How does the shape of the density function affect the total mass distribution along the rod?

Tip: In applications involving mass and density, always verify that the density function units are consistent with the physical quantity you are calculating.