https://file.aiquickdraw.com/mathbot/file/1727271824398r9onq2uw.jpg

The problem involves finding the moment of inertia \( I_x \) of a slender metal arch, which lies along the semicircle \( y^2 + z^2 = 1 \) in the \( yz \)-plane, with the condition that \( z \geq 0 \). The density at any point on the arch is given by the function \( \delta(x, y, z) = 6 - z \).

### Steps to find \( I_x \):

1. **Parametrize the curve:**
   The arch lies on the semicircle \( y^2 + z^2 = 1 \), which can be parametrized using the angle \( \theta \) such that:
   \[
   y = \sin(\theta), \quad z = \cos(\theta), \quad \text{where } 0 \leq \theta \leq \pi.
   \]
   This corresponds to the top half of the circle in the \( yz \)-plane.

2. **Arc length element (differential length):**
   The differential length \( ds \) along the arch is given by:
   \[
   ds = d\theta,
   \]
   because the semicircle has radius 1.

3. **Moment of Inertia formula:**
   The moment of inertia about the \( x \)-axis is given by:
   \[
   I_x = \int_{C} \delta(x, y, z) \cdot y^2 \, ds,
   \]
   where \( \delta(x, y, z) = 6 - z = 6 - \cos(\theta) \) and \( y = \sin(\theta) \).

4. **Substitute the parameters:**
   In terms of \( \theta \):
   \[
   I_x = \int_0^\pi (6 - \cos(\theta)) \cdot \sin^2(\theta) \, d\theta.
   \]

5. **Solve the integral:**
   Break the integral into two parts:
   \[
   I_x = 6 \int_0^\pi \sin^2(\theta) \, d\theta - \int_0^\pi \cos(\theta) \sin^2(\theta) \, d\theta.
   \]
   
   - The first integral can be simplified using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
     \[
     \int_0^\pi \sin^2(\theta) \, d\theta = \int_0^\pi \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{\pi}{2}.
     \]
     So the first part is:
     \[
     6 \times \frac{\pi}{2} = 3\pi.
     \]
   
   - The second integral can be solved using integration by parts or recognizing a standard integral:
     \[
     \int_0^\pi \cos(\theta) \sin^2(\theta) \, d\theta = \frac{2}{3}.
     \]

6. **Final result:**
   Thus, the moment of inertia is:
   \[
   I_x = 3\pi - \frac{2}{3} = 2(3\pi - 1).
   \]
   This matches the given answer in the image.

Let me know if you'd like a further breakdown of any steps or additional details!

Here are 5 follow-up questions that could extend your understanding:

1. What would change if the density function was a constant?
2. How would you calculate the moment of inertia about a different axis, such as the \( z \)-axis?
3. Can this problem be solved using Cartesian coordinates instead of parametric?
4. How would the solution change if the arch were part of a full circle instead of a semicircle?
5. What are some applications of calculating the moment of inertia in physics and engineering?

**Tip**: Familiarize yourself with trigonometric identities and integrals, as they frequently appear in problems involving curves and moments of inertia.

A slender metal arch lies along the semicircle y^2 + z^2 = 1, z >= 0, in the yz-plane. The density at a point (x, y, z) on the arch is given by δ(x, y, z) = 6 - z. Find Ix, the moment of inertia about the x-axis.

Solve the moment of inertia Ix for a metal arch along the semicircle y^2 + z^2 = 1 in the yz-plane with a density function δ(x, y, z) = 6 - z. This problem involves parametric equations, integration, and trigonometric identities, suitable for undergraduate-level calculus students.

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