Question What is the volume of the solid of revolution obtained by rotating the region bounded by \( f(x)=x^{2}+1 \) and the \( x \) - over the interval \( [1,3] \) about the \( y \)-axis? Enter your answer in terms of \( \pi \).

To find the volume of the solid of revolution obtained by rotating the region bounded by \( f(x) = x^2 + 1 \) and the \( x \)-axis over the interval \([1, 3]\) about the \( y \)-axis, we can use the **shell method**. Here's a step-by-step solution: ### Shell Method Overview When rotating around the \( y \)-axis, the shell method involves integrating cylindrical shells with: - **Radius**: \( x \) - **Height**: \( f(x) - 0 = x^2 + 1 \) - **Thickness**: \( dx \) ### Volume Formula The volume \( V \) is given by: \[ V = 2\pi \int_{a}^{b} (\text{radius}) \times (\text{height}) \, dx \] Substituting the values: \[ V = 2\pi \int_{1}^{3} x (x^2 + 1) \, dx \] ### Computing the Integral Expand the integrand: \[ x (x^2 + 1) = x^3 + x \] So, \[ V = 2\pi \int_{1}^{3} (x^3 + x) \, dx \] Integrate term by term: \[ \int x^3 \, dx = \frac{x^4}{4}, \quad \int x \, dx = \frac{x^2}{2} \] Therefore, \[ V = 2\pi \left[ \left( \frac{x^4}{4} + \frac{x^2}{2} \right) \bigg|_{1}^{3} \right] \] Evaluate the definite integral: \[ V = 2\pi \left( \left( \frac{3^4}{4} + \frac{3^2}{2} \right) - \left( \frac{1^4}{4} + \frac{1^2}{2} \right) \right) \] \[ V = 2\pi \left( \left( \frac{81}{4} + \frac{9}{2} \right) - \left( \frac{1}{4} + \frac{1}{2} \right) \right) \] \[ V = 2\pi \left( \frac{81 + 18 - 1 - 2}{4} \right) = 2\pi \left( \frac{96}{4} \right) = 2\pi \times 24 = 48\pi \] ### Final Answer \[ 48\pi \]