Find the volume of the eolid of revolution formed by rotating the region bounded by \( y=2 x+1, y=0, x=1 \) and \( s=2 \) ls rotated about the \( x-a x i e \) A \( \pi\left(\frac{11}{3}+7\right) \) unitis B. None of the options C. \( \pi\left(\frac{29}{3}+7\right) \) unitis \( ^{3} \) D. \( \pi\left(\frac{21}{2}-1\right) \) units \( ^{3} \) E. \( -\pi\left(\frac{2 \pi}{3}-7\right) \) units \( ^{3} \)

To find the volume of the solid of revolution formed by rotating the region bounded by the lines \( y = 2x + 1 \), \( y = 0 \), \( x = 1 \), and \( x = 2 \) about the x-axis, we can use the disk method. 1. **Identify the region**: The region is bounded by: - The line \( y = 2x + 1 \) - The x-axis \( y = 0 \) - The vertical lines \( x = 1 \) and \( x = 2 \) 2. **Set up the volume integral**: The volume \( V \) of the solid of revolution is given by the integral: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where \( f(x) = 2x + 1 \), \( a = 1 \), and \( b = 2 \). 3. **Calculate the integral**: \[ V = \pi \int_{1}^{2} (2x + 1)^2 \, dx \] First, expand \( (2x + 1)^2 \): \[ (2x + 1)^2 = 4x^2 + 4x + 1 \] Now, substitute this into the integral: \[ V = \pi \int_{1}^{2} (4x^2 + 4x + 1) \, dx \] Now, compute the integral: \[ V = \pi \left[ \frac{4}{3}x^3 + 2x^2 + x \right]_{1}^{2} \] Evaluate at the bounds: - At \( x = 2 \): \[ \frac{4}{3}(2^3) + 2(2^2) + 2 = \frac{4}{3}(8) + 2(4) + 2 = \frac{32}{3} + 8 + 2 = \frac{32}{3} + \frac{24}{3} + \frac{6}{3} = \frac{62}{3} \] - At \( x = 1 \): \[ \frac{4}{3}(1^3) + 2(1^2) + 1 = \frac{4}{3} + 2 + 1 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \] Now, subtract the two results: \[ V = \pi \left( \frac{62}{3} - \frac{10}{3} \right) = \pi \left( \frac{52}{3} \right) \] 4. **Final volume**: \[ V = \frac{52\pi}{3} \] Now, we need to compare this result with the options provided. None of the options directly match \( \frac{52\pi}{3} \). Thus, the answer is: **B. None of the options**.