Question Define \( R \) to be the region bounded above by the graph of \( f(x)=2 x^{2}+x \) and below by the \( x \)-axis over the interval \( [1,2] \). Use the method of cylindrical shells to find the volume of the solid obtained by rotating \( R \) about the line \( x=-4 \). Enter your answer in terms of \( \pi \).

To find the volume using the method of cylindrical shells, we start with the formula for the volume generated by rotating around a vertical line, given by: \[ V = 2\pi \int_{a}^{b} (r)(h) \, dx \] where \( r \) is the distance from the axis of rotation to the shell, and \( h \) is the height of the shell. In this case, our region \( R \) is bounded above by \( f(x) = 2x^2 + x \) and below by the \( x \)-axis over the interval \([1, 2]\). When rotating around the line \( x = -4 \): 1. The height \( h \) of the shell is \( f(x) = 2x^2 + x \). 2. The radius \( r \) of the shell is given by the distance from the shell to the line \( x = -4 \), which is \( r = x - (-4) = x + 4 \). Now, we substitute \( h \) and \( r \) into the volume formula: \[ V = 2\pi \int_{1}^{2} (x + 4)(2x^2 + x) \, dx \] Next, we expand the integrand: \[ (x + 4)(2x^2 + x) = 2x^3 + x^2 + 8x^2 + 4x = 2x^3 + 9x^2 + 4x \] Now, we can compute the integral: \[ V = 2\pi \int_{1}^{2} (2x^3 + 9x^2 + 4x) \, dx \] Integrating term by term: \[ \int (2x^3) \, dx = \frac{2}{4} x^4 = \frac{1}{2} x^4, \quad \int (9x^2) \, dx = 3x^3, \quad \int (4x) \, dx = 2x^2 \] Thus we get: \[ V = 2\pi \left[ \left( \frac{1}{2} x^4 + 3x^3 + 2x^2 \right) \bigg|_{1}^{2} \right] \] Calculating this at the boundaries: At \( x = 2 \): \[ \frac{1}{2}(2^4) + 3(2^3) + 2(2^2) = \frac{1}{2}(16) + 3(8) + 2(4) = 8 + 24 + 8 = 40 \] At \( x = 1 \): \[ \frac{1}{2}(1^4) + 3(1^3) + 2(1^2) = \frac{1}{2}(1) + 3(1) + 2(1) = \frac{1}{2} + 3 + 2 = \frac{1}{2} + 5 = \frac{11}{2} \] Subtracting these results: \[ 40 - \frac{11}{2} = \frac{80}{2} - \frac{11}{2} = \frac{69}{2} \] Finally, substituting this back into the volume formula: \[ V = 2\pi \left( \frac{69}{2} \right) = 69\pi \] The volume of the solid obtained by rotating the region \( R \) about the line \( x = -4 \) is: \[ \boxed{69\pi} \]