To calculate the mass of the disk using the radial density function $p(x) = 1 + \sin\left(\frac{\pi}{4}x\right)$, we need to use the following formula for the mass of a disk with variable radial density:

$M = \int_0^R 2\pi x \cdot p(x) \, dx$

Where:

• $M$ is the mass of the disk.
• $R$ is the radius of the disk.
• $p(x)$ is the radial density function.
• $2\pi x$ accounts for the circumference of the circular slice at radius $x$.

Given $R = 4$ inches, we can now substitute the radial density $p(x) = 1 + \sin\left(\frac{\pi}{4}x\right)$ and compute the integral:

$M = \int_0^4 2\pi x \left(1 + \sin\left(\frac{\pi}{4}x\right)\right) \, dx$

Now, let's break this into two integrals:

$M = 2\pi \int_0^4 x \, dx + 2\pi \int_0^4 x \sin\left(\frac{\pi}{4}x\right) \, dx$

I'll calculate this step by step.The mass of the disk is given by:

$M = 2\pi \left(\frac{16}{\pi} + 8\right) = 32 + 16\pi$

Thus, the exact mass of the disk is $32 + 16\pi$ units. If you would like a decimal approximation:

$M \approx 32 + 50.27 \approx 82.27 \text{ units}$

Let me know if you'd like more details or clarifications!

Here are some follow-up questions you may explore:

1. How would the mass change if the density function was altered to $p(x) = 1 + \sin(\frac{\pi}{2}x)$?
2. What would the mass be for a disk with radius 6 inches using the same density function?
3. How do you interpret the effect of the sine term in the radial density function on the overall mass distribution?
4. What would happen if the disk had a uniform density $p(x) = 1$?
5. How would you calculate the center of mass for a disk with this variable density?

Tip: When integrating functions involving trigonometric terms, it can be helpful to use known formulas for integrals of products like $x \sin(ax)$.