Question 1182678: The area bounded by y = 3, x = 2, y = -3 and x = 0 is revolved about the y-axis.
The moment of inertia of the solid is
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to calculate the moment of inertia of the solid:
1. **Describe the shape:** The region bounded by y = 3, x = 2, y = -3, and x = 0 is a rectangle. When revolved about the y-axis, this rectangle forms a hollow cylinder (or a cylindrical shell).
2. **Dimensions of the cylinder:**
* Inner radius (r₁): 0
* Outer radius (r₂): 2
* Height (h): 3 - (-3 We are not given a density, so we'll assume a uniform density (ρ). Then, we will find the moment of inertia *in terms of ρ*. If you're given a density, you would multiply by it at the end to get a numerical moment of inertia.
4. **Moment of Inertia of a hollow cylinder:** The moment of inertia (I) of a hollow cylinder about its central axis is given by the formula:
I = (1/2) * m * (r₁² + r₂²)
where m is the mass of the cylinder.
5. **Mass in terms of density and volume:** The mass of the cylinder is equal to the product of its density and volume.
* The volume of the hollow cylinder is given by V = πh(r₂² - r₁²)
* Volume: V = π * 6 * (2² - 0²) = 24π
* Thus, the mass m = ρ * V = 24πρ
6. **Substitute and calculate:**
I = (1/2) * (24πρ) * (0² + 2²)
I = 12πρ * 4
I = 48πρ
Therefore, the moment of inertia of the solid is $\boxed{48\pi\rho}$. If the density (ρ) were given, you would multiply by that value to find the numerical moment of inertia.
Answer by ikleyn(52901)  (Show Source):
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