Question Consider the region bounded by the functions \( f(x)=x^{2}+1 \) and \( g(x)=1 \) over the interval \( [0,2] \). Find the volume solid of revolution generated by rotating this region around the \( x \)-axis.

To find the volume of the solid of revolution generated by rotating the region bounded by the functions \( f(x) = x^2 + 1 \) and \( g(x) = 1 \) around the \( x \)-axis over the interval \( [0,2] \), we can use the disk method. First, identify the area between the two curves from \( 0 \) to \( 2 \). Notice that \( f(x) \) is above \( g(x) \) since \( f(x) - g(x) = x^2 \), which is always non-negative in the interval \( [0,2] \). To set up the volume integral, we use the formula: \[ V = \pi \int_{a}^{b} (f(x)^2 - g(x)^2) \, dx \] where \( a = 0 \), \( b = 2 \), \( f(x) = x^2 + 1 \), and \( g(x) = 1 \). Now calculate \( f(x)^2 \) and \( g(x)^2 \): - \( f(x)^2 = (x^2 + 1)^2 = x^4 + 2x^2 + 1 \) - \( g(x)^2 = 1^2 = 1 \) Thus, the expression \( f(x)^2 - g(x)^2 \) becomes: \[ f(x)^2 - g(x)^2 = (x^4 + 2x^2 + 1) - 1 = x^4 + 2x^2 \] Now integrate this from \( 0 \) to \( 2 \): \[ V = \pi \int_{0}^{2} (x^4 + 2x^2) \, dx \] Calculating the integral: \[ = \pi \left[\frac{x^5}{5} + \frac{2x^3}{3}\right]_{0}^{2} = \pi \left( \left(\frac{2^5}{5} + \frac{2 \cdot 2^3}{3}\right) - \left(0 + 0\right) \right) \] Calculating the terms: \[ = \pi \left( \frac{32}{5} + \frac{16}{3} \right) \] To combine the fractions, find a common denominator (15): \[ = \pi \left( \frac{32 \cdot 3}{15} + \frac{16 \cdot 5}{15} \right) = \pi \left( \frac{96 + 80}{15} \right) = \pi \left( \frac{176}{15} \right) \] Thus, the volume \( V = \frac{176\pi}{15} \) cubic units. The volume of the solid of revolution generated by rotating the bounded region around the \( x \)-axis is: \[ \boxed{\frac{176\pi}{15}} \]