2. Use the washer method to find the volume of the solid of revolution generated by rotating the region bounded by \( y=6-x^{2} \) and \( y=2 \) about the \( x \)-axis.

To find the volume of the solid of revolution using the washer method, we first need to determine the points of intersection for the curves \( y = 6 - x^2 \) and \( y = 2 \). Setting them equal gives: \[ 6 - x^2 = 2 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = -2 \text{ or } x = 2. \] Thus, the region of interest is bounded by \( x = -2 \) and \( x = 2 \). The volume \( V \) can be calculated using the formula: \[ V = \pi \int_{a}^{b} (R^2 - r^2) \, dx, \] where \( R \) is the outer radius and \( r \) is the inner radius. In this case: - The outer radius \( R(x) = 6 - x^2 \) (top function), - The inner radius \( r(x) = 2 \) (bottom function). Now, substituting the radii into the volume formula: \[ V = \pi \int_{-2}^{2} ((6 - x^2)^2 - 2^2) \, dx. \] Expanding this gives: \[ V = \pi \int_{-2}^{2} (36 - 12x^2 + x^4 - 4) \, dx = \pi \int_{-2}^{2} (32 - 12x^2 + x^4) \, dx. \] Next, we can calculate the integral: \[ V = \pi \left[ 32x - 4x^3 + \frac{x^5}{5} \right]_{-2}^{2}. \] Evaluating the integral at the bounds: \[ = \pi \left[ (32(2) - 4(2^3) + \frac{(2^5)}{5}) - (32(-2) - 4(-2)^3 + \frac{(-2)^5}{5}) \right]. \] Calculating each term will yield the final volume. After performing the necessary arithmetic, you’ll find the volume \( V \) of the solid of revolution is \( \frac{128\pi}{5} \). Now go grab a compass and a cube of ice, it’s time to dive deeper into volume calculations with some real-world flair!