Question

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 1

Answer:

The volume is $\frac{16\pi}{e}$

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: $V=\int\limits^b_a {2\pi xf(x)} \, dx$

  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

∴ $V=\int\limits^1_0 {2\pi x(4e^{x}-4e^{-x})} \, dx$

∴ $V=2\pi(4)\int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx$

- To integrate $xe^{x}$ and $xe^{-x}$ we will use

  integration by parts methods $\int\ {uv'=uv-\int{v}\,u' }\,$

∵ 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)

∴ $v=\int\ {e^{x}}\, dx= e^{x}$

∴ $\int\ {xe^{x}} \, dx=xe^{x}-\int\ e^{x}\, dx=xe^{x}-e^{x}$

- 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)

∴ $v=\int\ {e^{-x}} \, dx=-e^{-x}$

∴ $\int\ {x}e^{-x}\, dx=-xe^{-x}+\int\ {e^{-x}} \, dx=-xe^{-x}-e^{-x}$

∴ V = $8\pi \int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx=8\pi[xe^{x}-e^{x}+xe^{-x}+e^{-x}]$ from 0 to 1

- Lets substitute x = 1 minus x = 0

∴ $V=8\pi[(1)(e^{1})-(e^{1})+(1)(e^{-1})+(e^{-1})-(0)(e^{0})+(e^{0})-(0)(e^{0})-(e^{0})]$

∴ $V=8\pi[e^{1}-e^{1}+e^{-1}+e^{-1}-0+1-0-1]=8\pi[2e^{-1}]=16\pi e^{-1}$

∵ $e^{-1}=\frac{1}{e}$

∴ $V=\frac{16\pi}{e}$