To calculate the mass of the disk within a radius of 3 cm, we can use the following steps:

1. Mass formula for a disk:
   The mass $M$ of a disk with variable radial density is given by the formula: $M = \int_0^R \rho(r) \cdot 2\pi r \, dr$ where:

   • $\rho(r)$ is the radial density function (in this case, $p(x) = 4r + 5$).
   • $R$ is the radius of the disk (in this case, 3 cm).
   • $2\pi r$ accounts for the circumference at a given radius $r$.

2. Substitute the density function $\rho(r) = 4r + 5$ into the mass formula: $M = \int_0^3 (4r + 5) \cdot 2\pi r \, dr$

3. Solve the integral: First, expand the expression: $M = 2\pi \int_0^3 (4r^2 + 5r) \, dr$ Now, compute the integral: $M = 2\pi \left[ \frac{4r^3}{3} + \frac{5r^2}{2} \right]_0^3$ Substituting $r = 3$ into the expression: $M = 2\pi \left[ \frac{4(3)^3}{3} + \frac{5(3)^2}{2} \right]$ Simplify: $M = 2\pi \left[ \frac{4(27)}{3} + \frac{5(9)}{2} \right]$ $M = 2\pi \left[ 36 + 22.5 \right]$ $M = 2\pi (58.5)$ $M = 117\pi$

4. Approximate the result: $M \approx 117 \times 3.1416 = 367.5667$

Thus, the mass of the disk is approximately 367.5667 grams (rounded to four decimal places).

Would you like further details or have any questions?

Here are five related questions you might find useful:

1. How does the density function affect the mass of the disk?
2. What happens if the radius is increased to 5 cm?
3. Can you explain why $2\pi r$ is used in the formula?
4. What would the mass be if the density was constant (e.g., $\rho(r) = 5$)?
5. How would the mass change if the density function was quadratic, say $p(x) = r^2 + 4r + 5$?

Tip: When dealing with circular objects, always ensure you include $2\pi r$ when integrating over the radius for mass or area-based quantities!