**Instruction:**Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the curves \( y = 3 + 2x - x^2 \) and \( y + x = 3 \) about the \( y \)-axis. Below is a graph of the bounded region.**Graph Explanation:**- The graph displays two curves: \( y = 3 + 2x - x^2 \) (a downward-opening parabola) and \( y = 3 - x \) (a straight line).- The area of interest is shaded in blue, indicating the region between these two curves.- The shaded region starts from where the two curves intersect, forming a closed area.- The x-axis is marked with values from -1 to 1.8, and the y-axis is marked with values up to 7.- This region will be rotated about the y-axis to generate a volume.**Input Field:**Volume = [ ]**Note:**You can click on the graph to enlarge the image.**Options:**- Preview My Answers- Submit Answers

Let a solid be formed by revolving a region R, bounded by x = a and x = b about y- axis . Let r(x) represent the distance from the axis of rotation to x (i.e., the radius of a sample shell) and let h(x) represent the height of the solid at x (i.e., the height of the shell). The volume of the solid is given by V=2π∫abr(x).h(x) dx When the region R is bounded above by y = f(x) and below y = g(x) then h(x) = f(x) - g(x) When the axis of rotation is the y-axis then r(x) = x. In given problem the curves bounded by R are given as y=3+2x-x2 and y+x=3 so we will take hx= 3+2x-x2 - 3-x ⇒ hx = 3x-x2 Since it rotates about y-axis so r(x) = x, where x = 0 to 3 The required volume V=2π∫03(x).(3x-x2) dx ⇒ V=2π∫03(3x2-x3) dx ⇒ V=2π3x33-x4403 ⇒ V=2π33-344 ⇒ V=2π274 ⇒ V=13.5π Answer : V=13.5π