Question What is the volume of the solid of revolution obtained by rotating the region bounded by \( f(x)=x^{2}+3 \) and the \( x \)-axis over the interval \( [2,4] \) about the \( y \)-axis? Enter your answer in terms of \( \pi \).

To find the volume of the solid obtained by rotating the region bounded by \( f(x) = x^2 + 3 \) and the \( x \)-axis over the interval \([2, 4]\) about the \( y \)-axis, we can use the **method of cylindrical shells**. **Step 1: Set Up the Integral** Using the cylindrical shell formula, the volume \( V \) is given by: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] Here, \( f(x) = x^2 + 3 \), \( a = 2 \), and \( b = 4 \). Plugging these into the formula: \[ V = 2\pi \int_{2}^{4} x (x^2 + 3) \, dx \] **Step 2: Simplify the Integrand** Expand the integrand: \[ x (x^2 + 3) = x^3 + 3x \] So the integral becomes: \[ V = 2\pi \int_{2}^{4} (x^3 + 3x) \, dx \] **Step 3: Compute the Integral** Find the antiderivative: \[ \int (x^3 + 3x) \, dx = \frac{1}{4}x^4 + \frac{3}{2}x^2 \] Evaluate from 2 to 4: \[ \left[\frac{1}{4}(4)^4 + \frac{3}{2}(4)^2\right] - \left[\frac{1}{4}(2)^4 + \frac{3}{2}(2)^2\right] = (64 + 24) - (4 + 6) = 88 - 10 = 78 \] **Step 4: Multiply by \( 2\pi \)** \[ V = 2\pi \times 78 = 156\pi \] **Final Answer:** \( \boxed{156\,\pi} \)