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

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

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

∵ $e^{-1}=\frac{1}{e}$

∴ $V=\frac{16\pi}{e}$

Answer 2

Volume is the measure of the amount of 3D space occupied by a given figure. The volume of intended solid of revolution is $V = 24e^2\pi \: \rm unit^3$ approx.

How to use cylindrical shell method to find the volume?

Volume of a  cylinder with radius r units, and height h is $\pi r^2h$ cubic units.

When ta region is drawn, and rotated(say about y axis), there is formed a solid due to revolution around the axis of revolution. The volume of this solid can be found by various techniques.

Let we discuss cylindrical shell technique.

Suppose that revolution is being done on y axis. Cut dy amount of height in that solid, and in that short height region, you can find its volume by: Subtracting bigger cylinder's volume from smaller cylinder's volume. Now integrate such volumes over y axis.

For the given case, the diagram of the region which is being rotated is plotted below.

When we rotate it around y axis, there are 2 cylinders on each small partition dy, smaller cylinder is due to the curve $y = 4e^{-x}$, and the bigger cylinder is due to the curve $y = 4e^x$

Their volume is given as:

$\pi (4e^x)^2 \times dy - \pi (4e^{-x}x)^2 \times dy = 16e^2\pi(e^x - e^{-x})dy$

This dy height cut will be integrated from y = 4/e to y = 1 (as the region is bounded by x = 1)

Thus,

the final  volume is:

$V = \int^{4/e}_1 16e^2\pi(e^x - e^{-x})dy = [16e^2\pi(e^x + e^{-x})]^{4/e}_1\\\\V = 16e^2\pi [e^{4/e} + e^{-4/e} -(e+e^{-1} )]\\\\V \approx 16e^2\pi(1.5) = 24e^2\pi \: \rm unit^3$

Thus, The volume of intended solid of revolution is $V = 24e^2\pi \: \rm unit^3$ approx.

Learn more about cylindrical shell integration here:

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