Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. \( y=0, y=x(3-x) \) about the axis \( x=0 \)

To find the volume of the solid obtained by rotating the region bounded by the curves \( y=0 \) and \( y=x(3-x) \) about the axis \( x=0 \), we can use the method of cylindrical shells. The region bounded by the curves is a triangle with vertices at \( (0,0) \), \( (3,0) \), and \( (1,2) \). When rotating this region about the axis \( x=0 \), we will get a solid with a circular cross-section. The radius of each cylindrical shell will be the distance from the axis of rotation to the curve, which is \( x \), and the height of each cylindrical shell will be the difference between the two curves, which is \( x(3-x) \). The volume of each cylindrical shell is given by \( 2\pi x \cdot x(3-x) \). To find the total volume, we need to integrate this expression from \( x=0 \) to \( x=3 \). Let's calculate the volume using the given curves and the axis of rotation. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{3} 2\pi x\times x\left(3-x\right) dx\) - step1: Multiply the terms: \(\int_{0}^{3} 6\pi x^{2}-2\pi x^{3} dx\) - step2: Evaluate the integral: \(\int 6\pi x^{2}-2\pi x^{3} dx\) - step3: Use properties of integrals: \(\int 6\pi x^{2} dx-\int 2\pi x^{3} dx\) - step4: Evaluate the integral: \(2\pi x^{3}-\frac{\pi }{2}x^{4}\) - step5: Return the limits: \(\left(2\pi x^{3}-\frac{\pi }{2}x^{4}\right)\bigg |_{0}^{3}\) - step6: Calculate the value: \(54\pi -\frac{81\pi }{2}\) - step7: Calculate: \(\frac{27\pi }{2}\) The volume of the solid obtained by rotating the region bounded by the curves \( y=0 \) and \( y=x(3-x) \) about the axis \( x=0 \) is \( \frac{27\pi}{2} \) cubic units.