Use the method of cylindrical shells to find the volume v generated by rotating the region bounded by the curves about the given axis. y = 4ex, y = 4e−x, x = 1; about the y-axis

Answer:The volume is [tex]\frac{16\pi}{e}[/tex]Step-by-step explanation:* Lets talk about the shell method- The shell method is to finding the volume by decomposing  a solid of revolution into cylindrical shells- Consider a region in the plane that is divided into thin vertical    rectangle- If each vertical rectangle is revolved about the y-axis, we  obtain a cylindrical shell, with the top and bottom removed.  - The resulting volume of the cylindrical shell is the surface area    of the cylinder times the thickness of the cylinder- The formula for the volume will be: [tex]V=\int\limits^b_a {2\pi xf(x)} \, dx[/tex]   where 2πx · f(x) is the surface area of the cylinder shell and  dx is its   thickness* Lets solve the problem- To find the volume V generated by rotating the region bounded   by the curves y = 4e^x and y = 4e^-x about the y-axis by use   cylindrical shells- Consider that the height of the cylinder is y = (4e^x - 4e^-x)- Consider that the radius of the cylinder is x- The limits are x = 0 and x = 1- Lets take 2π and 4 as a common factor out the integration∴ [tex]V=\int\limits^1_0 {2\pi x(4e^{x}-4e^{-x})} \, dx[/tex]∴ [tex]V=2\pi(4)\int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx[/tex] - To integrate [tex]xe^{x}[/tex] and [tex]xe^{-x}[/tex] we will use   integration by parts methods [tex]\int\ {uv'=uv-\int{v}\,u' }\,[/tex]∵ u = x∴ u' = du/dx = 1 ⇒ differentiation x with respect to x is 1∵ v' = dv/dx = e^x- The integration e^x is e^x ÷ differentiation of x (1)∴ [tex]v=\int\ {e^{x}}\, dx= e^{x}[/tex] ∴ [tex]\int\ {xe^{x}} \, dx=xe^{x}-\int\ e^{x}\, dx=xe^{x}-e^{x}[/tex]- Similar we will integrate xe^-x∵ u = x∴ u' = du/dx = 1∵ v' = dv/dx = e^-x- The integration e^-x is e^x ÷ differentiation of -x (-1)∴ [tex]v=\int\ {e^{-x}} \, dx=-e^{-x}[/tex]∴ [tex]\int\ {x}e^{-x}\, dx=-xe^{-x}+\int\ {e^{-x}} \, dx=-xe^{-x}-e^{-x}[/tex]∴ V = [tex]8\pi \int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx=8\pi[xe^{x}-e^{x}+xe^{-x}+e^{-x}][/tex] from 0 to 1- Lets substitute x = 1 minus x = 0∴ [tex]V=8\pi[(1)(e^{1})-(e^{1})+(1)(e^{-1})+(e^{-1})-(0)(e^{0})+(e^{0})-(0)(e^{0})-(e^{0})][/tex]∴ [tex]V=8\pi[e^{1}-e^{1}+e^{-1}+e^{-1}-0+1-0-1]=8\pi[2e^{-1}]=16\pi e^{-1}[/tex]∵ [tex]e^{-1}=\frac{1}{e}[/tex]∴ [tex]V=\frac{16\pi}{e}[/tex]