Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the specified axis. y=4x-x^2 y=3 about x=1

Answer:[tex]\frac{16\pi}{3}[/tex].Step-by-step explanation:I graphed the region in the image below. The blue line is y=3, the purple line is x=1 and the green curve is y = [tex]4x-x^{2}[/tex]. The shaded region in blue is the region we are going to rotate.Now, to find the volume  v= [tex]2\pi \int\limits^a_b {p(x)h(x)} \, dx[/tex] where a=1, b=3 (left and right points of the region), p(x) is the distance from the rotation axis to the diferential Δx, we say p(x)=x and h(x) is the height of the region, in this case is h(x)= [tex]4x-x^{2}-3[/tex]. Then,v =  [tex]2\pi \int\limits^1_3 {x(4x-x^{2}-3)} \, dx[/tex]=  [tex]2\pi \int\limits^1_3 {4x^{2}-x^{3}-3x} \, dx[/tex]= [tex]2\pi (\frac{4x^{3}}{3}-\frac{x^{4}}{4}-\frac{3x^{2}}{2})^{3}_1[/tex]= [tex]2\pi (36-\frac{81}{4}-\frac{27}{2}-\frac{4}{3}+\frac{1}{4}+\frac{3}{2})[/tex]= [tex]2\pi*\frac{8}{3}[/tex]=  [tex]\frac{16\pi}{3}[/tex].