Find the volume of the solid obtained by rotating the region enclosed by \( y = 3x - x^2 \) and the x-axis for \( 0 \leq x \leq 3 \) around the x-axis using washers.

To find the volume of the solid obtained by rotating the region enclosed by the curve \(y = 3x - x^2\) and the x-axis from \(x = 0\) to \(x = 3\) around the x-axis using the washer method, we need to first determine the outer and inner radii for the washers. The area we are interested in is bounded by the function \(y = 3x - x^2\) (which is the upper curve) and \(y = 0\) (the x-axis). The volume \(V\) of the solid of revolution can be calculated using the formula: \[ V = \pi \int_{a}^{b} (R(x)^2 - r(x)^2) \, dx \] Here, \(R(x)\) is the outer radius and \(r(x)\) is the inner radius. For our case, since the region is above the x-axis and there is no hole (the inner radius is zero), we have \(R(x) = 3x - x^2\) and \(r(x) = 0\). Thus, the volume formula simplifies to: \[ V = \pi \int_{0}^{3} (3x - x^2)^2 \, dx \] First, we simplify the integrand: \((3x - x^2)^2 = 9x^2 - 6x^3 + x^4\) Now we can set up the integral: \[ V = \pi \int_{0}^{3} (9x^2 - 6x^3 + x^4) \, dx \] Next, we compute the integral: \[ \int (9x^2 - 6x^3 + x^4) \, dx = 3x^3 - \frac{6}{4}x^4 + \frac{1}{5}x^5 = 3x^3 - \frac{3}{2}x^4 + \frac{1}{5}x^5 \] Now, we evaluate this from 0 to 3: \[ V = \pi \left[ \left(3(3)^3 - \frac{3}{2}(3)^4 + \frac{1}{5}(3)^5\right) - \left(0\right) \right] \] Calculating each term: - \(3(3)^3 = 3 \times 27 = 81\) - \(-\frac{3}{2}(3)^4 = -\frac{3}{2} \times 81 = -121.5\) - \(\frac{1}{5}(3)^5 = \frac{1}{5} \times 243 = 48.6\) Now combine these results: \[ V = \pi \left( 81 - 121.5 + 48.6 \right) = \pi \left( 81 + 48.6 - 121.5 \right) = \pi \left( 9.1 \right) \] Thus, the final volume of the solid is: \[ V = 9.1\pi \text{ cubic units.} \]